How To Prove The Nvarance Of The Hankel Transform Of A Sequence

Transcription

1 Journal of Integer Sequences, Vol. 9 (2006, Artcle The k-bnomal Transforms and the Hankel Transform Mchael Z. Spvey Department of Mathematcs and Computer Scence Unversty of Puget Sound Tacoma, Washngton USA [email protected] Laura L. Stel Department of Mathematcs and Computer Scence Samford Unversty Brmngham, Alabama USA [email protected] Abstract We gve a new proof of the nvarance of the Hankel transform under the bnomal transform of a sequence. Our method of proof leads to three varatons of the bnomal transform; we call these the k-bnomal transforms. We gve a smple means of constructng these transforms va a trangle of numbers. We show how the exponental generatng functon of a sequence changes after our transforms are appled, and we use ths to prove that several sequences n the On-Lne Encyclopeda of Integer Sequences are related va our transforms. In the process, we prove three conectures n the OEIS. Addressng a queston of Layman, we then show that the Hankel transform of a sequence s nvarant under one of our transforms, and we show how the Hankel transform changes after the other two transforms are appled. Fnally, we use these results to determne the Hankel transforms of several nteger sequences. 1

2 1 Introducton Gven a sequence A = {a 0, a 1,...}, defne the bnomal transform B of a sequence A to be the sequence B(A = {b n }, where b n s gven by b n = =0 ( n a. Defne the Hankel matrx of order n of A to be the (n + 1 (n + 1 upper left submatrx of a 0 a 1 a 2 a 3 a 1 a 2 a 3 a 4 a 2 a 3 a 4 a 5. a 3 a 4 a 5 a Let h n denote the determnant of the Hankel matrx of order n. Then defne the Hankel transform H of A to be the sequence H(A = {h 0, h 1, h 2,..., }. For example, the Hankel matrx of order 3 of the derangement numbers, {D n } = {1, 0, 1, 2, 9, 44, 265,...}, s The determnant of ths matrx s 144, whch s (0! 2 (1! 2 (2! 2 (3! 2. Flaolet [4] and Radoux [9] have shown that the Hankel transform of the derangement numbers s { n =0 (n!2 }. Although the determnants of Hankel matrces had been studed before, the term Hankel transform was ntroduced n 2001 by Layman [7]. Other recent papers nvolvng Hankel determnants nclude those by Ehrenborg [3], Peart [8], and Woan and Peart [15]. Layman [7] proves that the Hankel transform s nvarant under the bnomal transform, n the sense that H(B(A = H(A. He also proves that the Hankel transform s nvarant under the nvert transform (see the On-Lne Encyclopeda of Integer Sequences [11] for the defnton, and he asks f there are other transforms for whch the Hankel transform s nvarant. Ths artcle partly addresses Layman s queston. We provde a new proof of the nvarance of the Hankel transform under the bnomal transform. Our method of proof generalzes to three varatons of the bnomal transform that we call the k-bnomal transform, the rsng k-bnomal transform, and the fallng k-bnomal transform. Collectvely, we refer to these as the k-bnomal transforms. (There should be no confuson n context. We gve a smple means of constructng these transforms usng a trangle of numbers, and we provde combnatoral nterpretatons of the transforms as well. We show how the exponental generatng functon of a sequence changes after applyng our transforms, and we use these results to prove that several sequences n the On-Lne Encyclopeda of Integer Sequences (OEIS [11] are related by the transforms. In the process, we prove three conectures lsted n the OEIS concernng the bnomal mean transform. Then, gvng an answer to Layman s queston, we. 2

3 show that the Hankel transform s nvarant under the fallng k-bnomal transform. The Hankel transform s not nvarant under the k-bnomal transform and the rsng k-bnomal transform, but we gve a formula showng how the Hankel transform changes under these two transforms. These results, together wth our proofs of relatonshps between sequences n the OEIS, determne the Hankel transforms of several sequences n the OEIS. (Unfortunately, ths s some dscrepancy n the lterature n the defntons of the Hankel determnant and the bnomal transform of an nteger sequence. The defntons of the Hankel determnant n Layman [7] and n Ehrenborg [3] are slghtly dfferent from each other, and ours s slghtly dfferent from both of these. As many sequences of nterest begn ndexng wth 0, we defne the Hankel determnant and the Hankel transform so that the frst elements n the sequences A and H(A are ndexed by 0. Layman s defnton begns ndexng A and H(A by 1, whereas Ehrenborg s defnton results n ndexng A begnnng wth 0 and H(A begnnng wth 1. Our defnton of the bnomal transform s that used by Layman [7] and the OEIS [11]. Somewhat dfferent defntons are gven n Knuth [6] (p. 136 and by MathWorld [13]. 2 Invarance of the Hankel transform under the bnomal transform Layman [7] proves that H(B(A = H(A for any sequence A. He does ths by showng that the Hankel matrx of order n of B(A can be obtaned by multplyng the Hankel matrx of order n of A by certan upper and lower trangular matrces, each of whch have determnant 1. We present a new proof of ths result. Our proof technque suggests generalzatons of the bnomal transform, whch we dscuss n subsequent sectons. We requre the followng lemma. Lemma 2.1. Gven a sequence A = {a 0, a 1, a 2,...}, create a trangle of numbers T usng the followng rule: 1. The left dagonal of the trangle conssts of the elements of A. 2. Any number off the left dagonal s the sum of the number to ts left and the number dagonally above t to the left. Then the sequence on the rght dagonal s the bnomal transform of A. For example, the bnomal transform of the derangement numbers s the factoral numbers [7]. Fgure 1 llustrates how the factoral numbers can be generated from the derangement numbers usng the trangle descrbed n Lemma 2.1. Although they do not use the term bnomal transform, Lemma 2.1 s essentally proven by Graham, Knuth, and Patashnk [5] (pp We present a dfferent proof, one that allows us to prove smlar results for the k-bnomal transforms we dscuss n subsequent sectons. 3

4 Fgure 1: Derangement trangle Proof. Let t n be the n th element on the rght dagonal of the trangle. By constructon of the trangle, we can see from Fgure 2 that the number of tmes element a contrbutes to the value of t n s the number of paths from a to t n. To move from a to t n requres n path segments, of whch move drectly to the rght. Thus there are ways to choose whch of the n ordered segments are the rghtward-movng segments, and the down segments are completely determned by ths choce. Therefore the contrbuton of a to t n s a, and the value of t n, n terms of the elements on the left dagonal, s n =0 a. But ths s the defnton of the bnomal transform, makng the rght dagonal of the trangle the bnomal transform of A. a 0 t 0 a 1 t 1 a 2 t 2 a 3 t 3 a 4 t 4 Fgure 2: Drected graph underlyng the bnomal transform The bnomal transform has the followng combnatoral nterpretaton: If a n represents the number of arrangements of n labeled obects wth some property P, then b n represents the number of ways of dvdng n labeled obects nto two groups such that the frst group has property P. In terms of the derangement and factoral numbers, then, as D n s the number of permutatons of n ordered obects n whch no obect remans n ts orgnal poston, n! s the number of ways that one can dvde n labeled obects nto two groups, order the obects n the frst group, and then permute the frst group obects so that none remans n ts orgnal poston. The numbers n the trangle descrbed n Lemma 2.1, not ust the rght and left dagonals, can also have combnatoral nterpretatons. For nstance, the trangle of numbers n Fgure 1 4

5 s dscussed as a combnatoral entty n ts own rght n another artcle by the frst author [12]. The number n row, poston, n the trangle s the number of permutatons of ordered obects such that every obect after does not reman n ts orgnal poston. We now gve our proof of Layman s result. Theorem 2.1. (Layman The Hankel transform s nvarant under the bnomal transform. Proof. We defne a procedure for transformng the Hankel matrx of order n of a sequence A to the Hankel matrx of order n of B(A usng only matrx row and column addton. Whle perhaps more complcated than Layman s proof, ours has the vrtue of beng easly modfed to gve proofs for the Hankel transforms of the k-bnomal transforms that we dscuss subsequently. The procedure s as follows: 1. Gven a sequence A = {a 0, a 1,...}, create the trangle of numbers descrbed n Lemma 2.1, where T, s the (, th entry n the trangle. 2. Let T n be the followng matrx consstng of numbers from the left dagonal of T : T 0,0 T 1,0 T 2,0 T n,0 T 1,0 T 2,0 T 3,0 T n+1,0 T 2,0 T 3,0 T 4,0 T n+2, T n,0 T n+1,0 T n+2,0 T 2n,0 Snce a = T,0, T n s the Hankel matrx of order n of A. 3. Then apply the followng transformatons to T n, where rows and columns of the matrx are ndexed begnnng wth 0. (a Let range from 1 to n. Durng stage, for each row, add row 1 to row and replace row wth the result. (b Then let agan range from 1 to n. Durng stage, for each column, add column 1 to column and replace column wth the result. Clam 1: After stage n 3(a, row m of the matrx s of the followng form: [ Tm,m T m+1,m T n+m,m ], f m ; [ Tm, T m+1, T n+m, ], f m >. The clam s clearly true ntally, when = 0. Now, assume the clam s true for all values of from 0 to k 1. Then, n stage = k, the only rows that change are rows k, k + 1,..., n. Row m, for m k, s the sum of rows m and m 1 from the prevous teraton: [ Tm,k 1 + T m 1,k 1 T m+1,k 1 + T m,k 1 T n+m,k 1 + T n+m 1,k 1 ]. But, by the defnton of T, T, + T 1, = T,+1. Thus ths row s equal to [ Tm,k T m+1,k T n+m,k ], 5

6 provng the clam. After the transformatons n 3(a are appled, then, we have the matrx T 0,0 T 1,0 T 2,0 T n,0 T 1,1 T 2,1 T 3,1 T n+1,1 T 2,2 T 3,2 T 4,2 T n+2, T n,n T n+1,n T n+2,n T 2n,n Clam 2: After stage n 3(b, column m of the matrx s of the followng form: [ ] T Tm,m T m+1,m+1 T n+m,n+m, f m ; [ Tm, T m+1,+1 T n+m,n+ ] T, f m >. The proof s almost the same as that for Clam 1. The clam s clearly true for = 0. Assume the clam s true for all values of from 0 to k 1. In stage = k, the only columns that change are columns k, k + 1,..., n. Column m, for m k, s the sum of columns m and m 1 from the prevous teraton: [ Tm,k 1 + T m 1,k 1 T m+1,k + T m,k T n+m,n+k 1 + T n+m 1,n+k 1 ] T. Agan, by the defnton of T, T, + T 1, = T,+1. Thus ths column s equal to [ Tm,k T m+1,k+1 T n+m,n+k ] T, whch proves the clam. After applyng the transformatons n 3(b, then, we have the matrx T 0,0 T 1,1 T 2,2 T n,n T 1,1 T 2,2 T 3,3 T n+1,n+1 T 2,2 T 3,3 T 4,4 T n+2,n T n,n T n+1,n+1 T n+2,n+2 T 2n,2n But ths s the Hankel matrx of order n of B(A, as B(A s the rght dagonal of trangle T. Snce the only matrx manpulatons we used were addng a row to another row and addng a column to another column, and the determnant of a matrx s nvarant under these operatons [2] (p. 262, the determnant of the Hankel matrx of order n of A s equal to the determnant of the Hankel matrx of order n of B(A. The operatons descrbed n the proof of Theorem 2.1, when done n the order prescrbed by the procedure, have a smple nterpretaton n terms of the trangle. Addng a row to another row shfts a left dagonal n the trangle one place to the rght, and addng a column to another column shfts a rght dagonal one place to the rght. We can see ths n the case 6

7 of the Hankel matrx of order 2 of the derangement numbers by a comparson of Fgure 1 and the sequence of matrces arsng from the procedure descrbed n the proof of Theorem The k-bnomal transforms We now consder three varatons of the bnomal transform. All three transforms take two parameters: the nput sequence A and a scalar k. The k-bnomal transform W of a sequence A s the sequence W (A, k = {w n }, where w n s gven by { n w n = =0 k n a = k n n =0 a, f k 0 or n 0; a 0, f k = 0, n = 0. The rsng k-bnomal transform R of a sequence A s the sequence R(A, k = {r n }, where r n s gven by { n r n = =0 k a, f k 0; a 0, f k = 0. The fallng k-bnomal transform F of a sequence A s the sequence F (A, k = {f n }, where f n s gven by { n f n = =0 k n a, f k 0; a n, f k = 0. The case k = 0 must be dealt wth separately because 0 0 would occur n the formulas otherwse. Our defntons effectvely take 0 0 to be 1. These turn out to be good defntons, n the sense that all the results dscussed subsequently hold under our defntons for the k = 0 case. When k = 0, the k-bnomal transform of A s the sequence {a 0, 0, 0, 0,...}, the rsng k-bnomal transform of A s {a 0, a 0, a 0,...}, and the fallng k-bnomal transform s the dentty transform. The k-bnomal transform when k = 1/2 s of specal nterest; ths s the bnomal mean transform, defned n the OEIS [11], sequence A When k s a postve nteger, these varatons of the bnomal transform all have combnatoral nterpretatons smlar to that of the bnomal transform, although, unlke the bnomal transform, they have a two-dmensonal component. If a n represents the number of arrangements of n labeled obects wth some property P, then w n represents the number of ways of dvdng n obects such that In one dmenson, the n obects are dvded nto two groups so that the frst group has property P. In a second dmenson, the n obects are dvded nto k labeled groups. The nterpretaton of the second dmenson could be somethng as smple as a colorng of each obect from a choce of k colors, ndependent of the dvson of the obects n the 7

8 frst dmenson. For example, f the nput sequence s the derangement numbers, w n s the number of ways of dvdng n labeled obects nto two groups such that the obects n the frst group are deranged and each of the n obects has been colored one of k colors, ndependently of the ntal dvson nto two groups. Wth ths nterpretaton of w n n mnd, r n represents the number of ways of dvdng n labeled obects nto two groups such that the frst group has property P and each obect n the frst group s further placed nto one of k labeled groups (e.g., colored usng one of k colors. Smlarly, f n represents the number of ways of dvdng n labeled obects nto two groups such that the frst group has property P and the obects n the second group are further placed nto k labeled groups (e.g., colored usng one of k colors. Each of the k-bnomal transforms can also be replcated va a trangle lke that descrbed n Lemma 2.1. Theorem 3.1. Gven a sequence A = {a 0, a 1, a 2,...}, 1. Create a trangle of numbers so that the left dagonal conssts of the elements of A, and any number off the left dagonal s k tmes the sum of the numbers to ts left and dagonally above t to the left. Then the rght dagonal s the k-bnomal transform W (A, k of A. 2. Create a trangle of numbers so that the left dagonal conssts of the elements of A, and any number off the left dagonal s the sum of the number dagonally above t to the left and k tmes the number to ts left. Then the rght dagonal s the rsng k-bnomal transform R(A, k of A. 3. Create a trangle of numbers so that the left dagonal conssts of the elements of A, and any number off the left dagonal s the sum of the number to ts left and k tmes the number dagonally above t to the left. Then the rght dagonal s the fallng k-bnomal transform F (A, k of A. For example, applyng the procedure n Part 1 wth k = 2 to the derangement numbers (sequence A yelds the trangle of numbers n Fgure ,200 46,080 Fgure 3: Derangement trangle wth 2-bnomal transform As we shall see, the sequence on the rght dagonal s the double factoral numbers (sequence A000165, the 2-bnomal transform of the derangement numbers. We now prove Theorem

9 Proof. Suppose k 0. (Part 1. The proof s smlar to that of Lemma 2.1. Let t n be the n th element on the rght dagonal of the trangle. There are stll n path segments from a to t n ; however, traversng a path segment now ncurs a multplcatve factor of k rather than the understood factor of 1 ncurred wth the bnomal transform. Thus each path from a to t n contrbutes k n a to the value of t n. Wth moves to the rght, there are stll dfferent paths from a to t n ; thus the total contrbuton from a to the value of t n s ( n k n a. Therefore, t n = n =0 k n a, the n th element of W (A, k. (Part 2. The proof s smlar to that of Part 1. The dfference s that the multplcatve factor of k only appears on moves to the rght, not moves down. Wth rghtward-movng path segments, the contrbuton from a along one path to t n s thus k a. The total contrbuton from a over all paths s therefore ( n k a, and we have t n = n =0 k a, the n th element of R(A, k. (Part 3. The proof s dentcal to that of Part 2, except that the multplcatve factor of k occurs on down path segments rather than on rght-movng segments. As there are n down path segments from a to t n, we have t n = n =0 k n a, the n th element of F (A, k. Now, suppose k = 0. For the trangle descrbed n Part 1, every path from the left dagonal to t n s a zero path, except for the empty path from a 0 to t 0. Thus t 0 = a 0, and t n = 0 for n 0. For the trangle n Part 2, the only nonzero path to t n from the left dagonal s the one straght down from a 0 ; thus t n = a 0 for all n. For Part 3, the only nonzero path to t n from the left dagonal s the one straght rght from a n ; therefore, t n = a n for each n. There are also some nce relatonshps between our transforms, the bnomal transform, and the nverse bnomal transform. Gven a sequence A = {a 0, a 1,...}, the nverse bnomal transform of A s defned to be the sequence B 1 (A = {c n }, where c n s gven by c n = =0 ( n ( 1 n a. It s easy to show that B 1 (B(A = B(B 1 (A = A. For k Z +, let B k (A denote k successve applcatons of the bnomal transform;.e., B k (A = B(B( B(A, where B occurs k tmes. For k Z, let B k (A denote k successve applcatons of the nverse bnomal transform. Fnally, let B 0 (A denote the dentty transform. Then we have Theorem 3.2. If k Z, B k (A = F (A, k. In other words, k successve applcatons of the bnomal transform (or the nverse bnomal transform, f k Z s equvalent to the fallng k-bnomal transform. Proof. The theorem s clearly true when k = 1. Assume the theorem s true for values of k from 1 to m. Let b k n denote the n th element n the sequence B k (A, and let f k n denote the 9

10 n th element n the sequence F (A, k. For k = m + 1, we have b m+1 n = = = =0 ( n b m = = =0 ( n f m = =0 m a = a (m + 1 n = f m+1 n. ( n a ( m a = = = ( n m = ( m a n a l=0 The second equalty holds by the nducton hypothess, the ffth by trnomal revson [10] (p. 104, and the second-to-last by the Bnomal Theorem. Ths proves the relatonshp for postve values of k. The proof for k Z s an almost dentcal nducton wth k = 1 as the base case and movng down. When k = 0 we have the dentty transform n both cases. We also have the followng relatonshp between the W, R, and F transforms. Theorem 3.3. W (A, k = F (R(A, k, k 1. In other words, the k-bnomal transform s equvalent to the composton of the fallng (k 1-bnomal transform wth the rsng k-bnomal transform. Proof. For k = 1, we have W (A, 1 = B(A, and F (R(A, 1, 0 = F (B(A, 0 = B(A. For k = 0, R(A, 0 = {a 0, a 0, a 0,...}. Then the n th term of F (R(A, 0, 1 s gven by =0 ( n ( 1 n a 0 = a 0 =0 ( n ( 1 n. But the summaton on the rght-hand sde s 1 f n = 0 and 0 otherwse [10] (p Thus F (R(A, 0, 1 = {a 0, 0, 0,...}, whch s the sequence W (A, 0. Assumng k 0 and k 1, the n th term of F (R(A, k, k 1 s gven by =0 = = ( n (k 1 n = n k a l=0 ( k a = (k 1 n k a = The last term s the n th term of W (A, k. l ( = (k 1 n l = (k 1 n k a k a = ( n k a k n = ( n (k 1 n ( n k n a. l m l 10

11 4 Integer sequences related by the k-bnomal transforms The k-bnomal transforms relate many sequences lsted n the On-Lne Encyclopeda of Integer Sequences [11]. Several of these relatonshps are gven n Layman [7]. He lsts a number of tables of nteger sequences related by repeated applcatons of the bnomal transform and thus (va Theorem 3.2 by the fallng k-bnomal transform. The sequences n Tables 1, 2, and 3 are related or appear to be related by the k-bnomal transform, the rsng k-bnomal transform, and the fallng k-bnomal transform, respectvely. (Table 3 s short because we do not duplcate Layman s lsts [7]. The tables were mostly obtaned by an nvestgaton of several of the entres n the OEIS; undoubtedly there are more sequences related by these transforms. Under the status column, K ( known means that we were able to fnd a reference for the transform relatonshp, P ( proved means that we could not fnd a reference for the transform relatonshp but that t can be proved va the technques n the followng secton, and C ( conecture means that we were not able to fnd a reference for the transform relatonshp and were not able to prove t. In addton, the expresson (E n front of a sequence means that the sequence has been shfted; ts frst term has been deleted. Status A Name (f common k W (A, k Name (f common P A Factoral Numbers 2 A P A Factoral Numbers 3 A P A Factoral Numbers 4 A P A Factoral Numbers 5 A P A Derangement Numbers 2 A Double Factoral Numbers P A Derangement Numbers 3 A Trple Factoral Numbers P A Derangement Numbers 4 A Quadruple Factoral Numbers P A Derangement Numbers 5 A Quntuple Factoral Numbers K A /2 A Factoral Numbers P (EA /2 A P A A Quadruple Factoral Numbers K A Euler or Secant Numbers 1/2 A C A /2 A P A Motzkn Numbers 2 A C A /2 A P A /2 A Double Factoral Numbers P A /2 A NSW Numbers P A Central Trnomal Coeffcents 2 A C A /2 A K A /2 A Table 1: k-bnomal transforms 11

12 Status A Name (f common k R(A, k Name (f common K A Lucas Numbers 2 A Even Lucas Numbers, or L 3n K A Fbonacc Numbers 2 A Even Fbonacc Numbers, or F 3n C (EA Catalan Numbers 2 A P A Factoral Numbers 2 A P A Factoral Numbers 3 A P A Factoral Numbers 4 A P A Factoral Numbers 5 A P A Central Bnomal Coeffcents 2 A Table 2: Rsng k-bnomal transforms Status A Name (f common k F (A, k Name (f common P A A Table 3: Fallng k-bnomal transforms A close examnaton of Table 1 ndcates that a couple of new bnomal transform relatonshps should exst as well; these are gven n Table 4. Status A Name (f common B(A Name (f common P A A Double Factoral Numbers P A A Quadruple Factoral Numbers Table 4: New bnomal transforms The relatonshps n the tables were found prmarly by determnng that the frst several terms of a sequence correspond to a transform of another sequence. Ths, of course, does not consttute proof, and so the relatonshps that we were unable to prove and for whch we could not fnd references reman conectures. The relatonshps nvolvng the Lucas numbers and the Fbonacc numbers are gven n Benamn and Qunn [1] (p. 135, and the other known relatonshps are mentoned n ther respectve entres n the OEIS [11]. The relatonshps ndcated by a P n the status column we were able to prove va the generatng functon method we dscuss n the next secton. In partcular, the entres n Table 1 nvolvng the bnomal mean transform W (A, 1/2 and the nput sequences (EA000354, A001907, and A are lsted as conectures n the OEIS, and we prove these conectures n the next secton. Fnally, we note that the Hankel transforms of many of these sequences, such as the derangement numbers, the factoral numbers, and the Catalan numbers, are known. Thus Tables 1, 2, 3, and 4, together wth Theorems 6.1, 6.2, and 6.3 (proved n Secton 6, contan several results and conectures concernng the Hankel transforms of varous nteger sequences. 12

13 5 Exponental generatng functons of the k-bnomal transforms The method we use to prove the relatonshps ndcated wth a P n Tables 1, 2, 3, and 4 s that of generatng functons. (See Wlf s text [14] for an extensve dscusson of generatng functons and ther uses. The exponental generatng functon (egf f of a sequence A = {a 0, a 1,...} s the formal power seres f(x = a 0 + a 1 x + a 2 x 2 /2! + = =0 a x /!. We have the followng: Theorem 5.1. If g(x s the exponental generatng functon of a sequence A = {a 0, a 1,...}, then: 1. The exponental generatng functon of the sequence F (A, k s e kx g(x. 2. The exponental generatng functon of the sequence W (A, k s e kx g(kx. 3. The exponental generatng functon of the sequence R(A, k s e x g(kx. Proof. For the specal case k = 0, the theorem states that the egf of F (A, k s g(x, whch s the egf of A. The theorem gves the egf of W (A, k to be g(0, whch generates the sequence {a 0, 0, 0, 0,...}. The theorem gves the egf of R(A, k to be e x g(0, whch generates the sequence {a 0, a 0, a 0,...} [14] (p. 52. These are all consstent wth the defntons of the k-bnomal transforms when k = 0. Now, suppose k 0. (Part 1. It s known [14] (p. 42 that f f s the egf of some sequence {a n } and h s the egf of some sequence {b n }, then fh s the egf of the sequence { =0 ( } b a b n. (1 (Ths s known as the bnomal convoluton of {a n } and {b n }. F (A, k s, agan, the sequence { =0 ( } b k n a. Wth (1 n mnd, then, b n = k n for each n. The egf of the sequence {1, k, k 2,...} s e kx [5] (p Thus the egf of F (A, k s e kx g(x. (Part 2. By defnton, W (A, k = {k n b n }, where {b n } = B(A. From the proof of Part 1, we know that the egf of B(A s e x g(x. By defnton, t s also =0 b x /!. The egf of W (A, k s thus =0 k b x /! = =0 b (kx /! = e kx g(kx. (Part 3. By Theorem 3.3, W (A, k = F (R(A, k, k 1. Thus the egf of the sequence on the left must equal that of the sequence on the rght. Lettng G(x denote the egf of R(A, k, we have, by Parts 1 and 2, e kx g(kx = e (k 1x G(x. Thus G(x = e x g(kx. We now use Theorem 5.1 to prove three relatonshps lsted n Table 1 that are gven as conectures n the OEIS. They all nvolve the bnomal mean transform W (A, 1/2. The 13

14 proofs of the other relatonshps ndcated wth a P n Tables 1, 2, 3, and 4 are smlar n flavor to these, and so we do not prove them explctly here. In fact, most can be proved smply by applyng Theorem 5.1 to a drect comparson of the egf s gven n the OEIS [11]. Corollary 5.1. The bnomal mean transform of sequence (EA s sequence A Proof. The egf of sequence A s gven by the OEIS as e x /(1 2x. The egf of sequence (EA s thus d/dx[e x /(1 2x] [14] (p. 40, whch s e x (1 + 2x (1 2x 2. By Theorem 5.1, the egf of the bnomal mean transform of (EA s therefore ( e e x/2 x/2 (1 + x = 1 + x (1 x 2 (1 x, 2 whch s the egf of sequence A as lsted n the OEIS. Corollary 5.2. The bnomal mean transform of sequence A s sequence A (the double factoral numbers. Proof. The egf of sequence A s gven by the OEIS as e x /(1 4x. Thus the egf of the bnomal mean transform of A s ( e e x/2 x/2 = 1 1 2x 1 2x, whch, accordng to the OEIS, s the egf of sequence A Corollary 5.3. The bnomal mean transform of sequence A s sequence A Proof. The OEIS does not lst egf s for these sequences. It does, however, lst two-term recurrence relatonshps for the sequences, and we can use these relatonshps to set up dfferental equatons for the egf s. Sequence A s generated by the recursve relatonshp a n = 6a n 1 a n 2, for n 2, wth ntal values a 0 = 1, a 1 = 7. Ths means that the egf f of sequence A satsfes the ntal-value problem f 6f + f = 0, f(0 = 1, f (0 = 7 [14] (p. 40. The characterstc equaton for the dfferental equaton s r 2 6r +1 = 0, whch has soluton r = 3 ± 2 2. Thus the soluton to the dfferental equaton, and therefore the egf of A002315, s f(x = c 1 e (3+2 2x +c 2 e (3 2 2x. Usng the ntal condtons, the constants are found to be c 1 = 1/2 + 1/ 2, c 2 = 1/2 1/ 2. The recurrence relaton for sequence A s gven by a n = 4a n 1 2a n 2, for n 2, wth a 0 = 1, a 1 = 4. Usng the same method as before, ts egf s found to be c 1 e (2+ 2x + c 2 e (2 2x, for the same values of c 1 and c 2 found prevously. Ths s also the egf of the bnomal mean transform of sequence A002315: e x/2[ c 1 e (3/2+ 2x + c 2 e (3/2 2x ] = c 1 e (2+ 2x + c 2 e (2 2x. 14

15 Theorem 5.1 can also be used to prove many addtonal relatonshps between sequences n the OEIS nvolvng compostons of the k-bnomal transforms. For example, let D(a, b denote the sequence of central coeffcents of (1 + ax + bx 2 n, and let C denote the sequence of central bnomal coeffcents (A Then we have Corollary 5.4. D(a, b = F (R(C, b, a 2 b 1. Proof. The central bnomal coeffcents have e 2x BesselI(0, 2x as ther exponental generatng functon. Thus the egf of F (R(C, b, a 2 b 1 s e (a 2 b 1x e x e 2 bx BesselI(0, 2 bx, whch s e ax BesselI(0, 2 bx. Ths last expresson, accordng to the comments under sequence A n the OEIS, s the egf of the central coeffcents of (1 + ax + bx 2 n. Another example nvolves the Motzkn numbers (sequence A and the super- Catalan numbers wth the frst element deleted (sequence (EA Denotng the sequence of Motzkn numbers by M and the sequence of super-catalan numbers by S, we have Corollary 5.5. (ES = F (R(M, 2, 2 2. Proof. The Motzkn numbers have egf e x BesselI(1, 2x/x. The egf of F (R(M, 2, 2 2 s therefore e (2 2x e x e 2x BesselI(1, 2 2x/( 2x. Smplfed, ths s e 3x BesselI(1, 2 2x/( 2x, whch s the egf of the super-catalan numbers wth the frst element deleted. 6 Hankel transforms of the k-bnomal transforms At the end of hs artcle [7], Layman asks f there are transforms besdes the bnomal and nvert under whch the Hankel transform s nvarant. We now address ths queston for the three k-bnomal transforms. Theorem 6.1. The Hankel transform s nvarant under the fallng k-bnomal transform. In other words, for a gven sequence A = {a 0, a 1,...}, H(F (A, k = H(A. For k Z, ths follows from Theorems 2.1 and 3.2. However, the theorem s true for k Z as well. Our proof of ths entals a slght modfcaton of the proof of Theorem 2.1. Proof. Create a trangle of numbers T by lettng the left dagonal consst of A and every number off of the left dagonal be the sum of the number to ts left and k tmes the number dagonally above t to the left. Then, va Theorem 3.1, the rght dagonal conssts of the sequence F (A, k. Construct the matrx T n as n the proof of Theorem 2.1. Change the transformaton rules n part 3 of the procedure descrbed n that proof so that one adds k tmes row (column 1 to row (column and replaces row (column wth the result. By thus mmckng the rule for the creaton of the numbers off the left dagonal of T, Clams 1 and 2 n the proof of Theorem 2.1 stll hold. Therefore, the fnal matrx resultng from the transformaton procedure s the Hankel matrx of order n of F (A, k. Snce the transformaton rules only nvolve addng a multple of a row to another row and addng a multple of a column to another column, and the determnant s nvarant under these 15

16 operatons, the determnant of the Hankel matrx of order n of F (A, k s equal to the determnant of the Hankel matrx of order n of A. Ths proof technque works for determnng how the Hankel transforms of A, W (A, k, and R(A, k relate as well. However, the relatonshps are more complcated, as the Hankel transform s not nvarant under W (A, k and R(A, k. We begn wth the transform for R(A, k. Theorem 6.2. Gven a sequence A = {a 0, a 1,...}, let H(A = {h n }. Then H(R(A, 0 = {a 0, 0, 0,...}. If k 0, H(R(A, k = {k n(n+1 h n }. Proof. If k = 0, then H n s the (n + 1 (n + 1 matrx whose entres are all a 0. Thus det(h 0 = a 0, and for n > 0, det(h n = 0. Now, assume k 0. Create a trangle of numbers T by lettng the left dagonal consst of A and lettng each number off the left dagonal be the sum of the number dagonally above t to the left and k tmes the number to ts left. By Theorem 3.1, the rght dagonal of T s R(A, k. Construct the matrx T n as n the proofs of Theorems 2.1 and 6.1, but change the transformaton rule so that one adds row (column 1 to k tmes row (column and replaces row (column wth that result. Agan, Clams 1 and 2 n the proof of Theorem 2.1 hold, and so the fnal matrx resultng from the transformaton procedure s the Hankel matrx of order n of R(A, k. The transformaton rule can be broken nto two parts: Frst multply a row (column by k. Then replace that row (column wth the sum of tself and the prevous row (column. Multplyng a row of a matrx by a factor of k changes the determnant by a factor of k, whle replacng a row by the sum of tself and another row does not affect the determnant. The same s true for columns [2] (p In order to determne how the Hankel transform changes under the rsng k-bnomal transform, then, we need to determne the number of tmes a row or column s multpled by a factor of k. Accordng to the transformaton procedure, row s multpled by a factor of k n stages 1, 2,...,. Thus row s multpled by a factor of k a total of tmes. Therefore the total number of tmes a row s multpled by a factor of k s n =0 = n(n + 1/2. Smlarly, the number of tmes a column s multpled by a factor of k s n(n + 1/2. Thus the determnant of the Hankel matrx of order n of R(A, k s k n(n+1 tmes that of the Hankel matrx of order n of A. Fnally, we have Theorem 6.3. Gven a sequence A = {a 0, a 1,...}, let H(A = {h n }. Then H(W (A, 0 = {a 0, 0, 0,...}. If k 0, H(W (A, k = {k n(n+1 h n }. Proof. By Theorems 3.3 and 6.1, H(W (A, k = H(F (R(A, k, k 1 = H(R(A, k. Alternatvely, one can prove ths result wth a modfcaton of the proofs of Theorems 6.1 and 6.2. The transformaton rule would be addng row (column 1 to row (column and replacng row (column wth k tmes that sum. Ths s equvalent to frst multplyng row (column by a factor of k and then addng k tmes row (column 1 to that and replacng row (column wth ths sum. As we ust noted, multplyng a row or column by a factor of 16

17 k changes the determnant by a factor of k, whle addng a multple of a row to another row or a multple of a column to another column does not change the determnant. The analyss of the number of tmes a row or column s multpled by a factor of k s exactly the same as that n the proof of Theorem 6.2. Theorems 6.1, 6.2, and 6.3, together the relatonshps gven n Tables 1, 2, 3, and 4, allow us to determne the Hankel transforms of several nteger sequences. Gven that the factoral numbers (A and the derangement numbers (A have Hankel transform { n =0 (!2 } (sequence A055209, the Motzkn numbers (A have Hankel transform {1, 1, 1,...} (sequence A000012, and the central bnomal coeffcents (A and trnomal coeffcents (A have Hankel transform {2 n } (sequence A [7], we have the Hankel transforms of several nteger sequences gven n Table 5. Hankel Transform Some sequences wth ths Hankel transform {2 n(n+1 } A {2 n(n+2 } A059304, A {2 n(n+1 n =0 (!2 } A000165, A000354, A010844, A {3 n(n+1 n =0 (!2 } A010845, A032031, A {4 n(n+1 n =0 (!2 } A001907, A047053, A056545, A {5 n(n+1 n =0 (!2 } A052562, A056546, A Table 5: Hankel transforms of some nteger sequences In addton, Theorems 6.1, 6.2, and 6.3 can be used to relate the Hankel transforms of sequences that are themselves related by compostons of the k-bnomal transforms. For example, Theorems 6.1 and 6.2 appled to Corollary 5.4 tell us that the Hankel transform of the central coeffcents of (1 + ax + bx 2 n s {2 n b n(n+1/2 }. Appled to Corollary 5.5, they tell us that the Hankel transform of the super-catalan numbers wth the frst element deleted s {2 n(n+1/2 }. 7 Conclusons We have ntroduced three generalzatons of the bnomal transform: the k-bnomal transform, the rsng k-bnomal transform, and the fallng k-bnomal transform. We have gven a smple method for constructng these transforms, and we have gven combnatoral nterpretatons of each of them. We have also shown how the generatng functon of a sequence changes after applyng one of the transforms. Ths allows us to prove that several sequences n the On-Lne Encyclopeda of Integer Sequences are related by one of these transforms, as well as prove three specfc conectures n the OEIS. In addton, we have shown how the Hankel transform of a sequence changes after applyng one of the transforms. These results determne the Hankel transforms of several sequences lsted n the OEIS. We see several areas of further study. One nvolves contnung to answer Layman s queston [7]. We have proved that the Hankel transform s nvarant under the fallng k-bnomal 17

18 transform, and he proves that the Hankel transform s nvarant under the bnomal and nvert transforms. Are there other nterestng transforms under whch the Hankel transform s nvarant? Second, are there varatons of our trangle method that would prove for other transforms T how H(A and H(T (A relate? Thrd, we conecture four relatonshps n Tables 1 and 2: Sequences A002078, A007052, and A are the bnomal mean transforms of A000609, A001653, and A007696, respectvely, and sequence A s the rsng 2-bnomal transform of (EA (The frst three are also conectured n the OEIS. Can these conectures be proved? Fnally, what sequences, other than those lsted n Layman [7] and n Tables 1, 2, 3, and 4, are related by the k-bnomal transforms? References [1] Arthur T. Benamn and Jennfer J. Qunn, Proofs That Really Count, MAA, Washngton, D. C., [2] Otto M. Bretscher, Lnear Algebra wth Applcatons, Pearson Prentce Hall, Upper Saddle Rver, NJ, thrd ed., [3] Rchard Ehrenborg, The Hankel determnant of exponental polynomals, Amer. Math. Monthly, 107 (2000, [4] Phlppe Flaolet, On congruences and contnued fractons for some classcal combnatoral quanttes, Dscrete Math., 41 (1982, [5] R. L. Graham, D. E. Knuth, and O. Patashnk, Concrete Mathematcs, Addson-Wesley, Readng, MA, second ed., [6] Donald E. Knuth, The Art of Computer Programmng, Vol. III: Sortng and Searchng, Addson-Wesley, Readng, MA, second ed., [7] John W. Layman, The Hankel transform and some of ts propertes, J. Integer Seq., 4 (2001, Artcle [8] Paul Peart, Hankel determnants va Steltes matrces, Congr. Numer., 144 (2000, [9] Chrstan Radoux, Détermnant de Hankel construt sur des polynômes lés aux nombres de dérangements, European J. Combn., 12 ( [10] Kenneth H. Rosen, ed., Handbook of Dscrete and Combnatoral Mathematcs, CRC Press, Boca Raton, FL, [11] Nel J. A. Sloane, The On-Lne Encyclopeda of Integer Sequences, 2005, publshed electroncally at nas/sequences/. 18

19 [12] Mchael Z. Spvey, Exams wth deranged questons, n preparaton. [13] Erc W. Wessten, Bnomal transform, from MathWorld A Wolfram Web Resource, [14] Herbert S. Wlf, Generatngfunctonology, Academc Press, Boston, second ed., [15] Wn-Jn Woan and Paul Peart, Determnants of Hankel matrces of aerated sequences, Congressus Numerantum, 157 (2002, Mathematcs Subect Classfcaton: Prmary 11B65; Secondary 11B75. Keywords: bnomal transform, Hankel transform. (Concerned wth sequences A000012, A000032, A000045, A000079, A000108, A000142, A000165, A000166, A000354, A000364, A000609, A000984, A001003, A001006, A001653, A001907, A002078, A002315, A002426, A002801, A003645, A005799, A007052, A007070, A007680, A007696, A010844, A010845, A014445, A014448, A032031, A047053, A052562, A055209, A056545, A056546, A059231, A059304, A075271, A075272, A082032, A084770, A084771, A097814, A097815, and A Receved June ; revsed verson receved November Publshed n Journal of Integer Sequences, November Return to Journal of Integer Sequences home page. 19