cp's OEIS Frontend

Previous name: Triangular array made of three copies of Pascal's triangle.

Computing each term modulo 2 also gives A047999, i.e., a(n) mod 2 = A007318(n) mod 2 for all n. (The triangle is paritywise isomorphic to Pascal's Triangle.) - Antti Karttunen

5th row/column gives entries of A000217 (triangular numbers C(n+1,2)) repeated twice and every other entry in 6th row/column form A000217. 7th row/column gives entries of A000292 (Tetrahedral (or pyramidal) nos: C(n+3,3)) repeated twice and every other entry in 8th row/column form A000292. 9th row/column gives entries of A000332 (binomial coefficients binomial(n,4)) repeated twice and every other entry in 10th row/column form A000332. 11th row/column gives entries of A000389 (binomial coefficients C(n,5)) repeated twice and every other entry in 12th row/column form A000389. - Gerald McGarvey, Aug 21 2004

If Sum_{k=0..n} A(k)*T(n,k) = B(n), the sequence B is the S-D transform of the sequence A. - Philippe Deléham, Aug 02 2006

Number of n-bead black-white reversible strings with k black beads; also binary grids; string is palindromic. - Yosu Yurramendi, Aug 07 2008

Row sums give A016116(n+1). - Yosu Yurramendi, Aug 07 2008 [corrected by Petros Hadjicostas, Nov 04 2017]

Coefficients in expansion of (x + y)^n where x and y anticommute (y x = -x y), that is, q-binomial coefficients when q = -1. - Michael Somos, Feb 16 2009

The sequence of coefficients of a general polynomial recursion that links at w=2 to the Pascal triangle is here w=0. Row sums are {1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, ...}. - Roger L. Bagula and Gary W. Adamson, Dec 04 2009

T(n,k) is the number of palindromic compositions of n+1 with exactly k+1 parts. T(6,4) = 3 because we have the following compositions of n+1=7 with length k+1=5: 1+1+3+1+1, 2+1+1+1+2, 1+2+1+2+1. - Geoffrey Critzer, Mar 15 2014 [corrected by Petros Hadjicostas, Nov 03 2017]

Let P(n,k) be the number of palindromic compositions of n with exactly k parts. MacMahon (1893) was the first to prove that P(n,k) = T(n-1,k-1), where T(n,k) are the numbers in this sequence (see the comment above by G. Critzer). He actually proved that, for 1 <= s <= m, we have P(2*m,2*s) = P(2*m,2*s-1) = P(2*m-1, 2*s-1) = bin(m-1, s-1), but P(2*m-1, 2*s) = 0. For the current sequence, this can be translated into T(2*m-1, 2*s-1) = T(2*m-1,2*s-2) = T(2*m-2, 2*s-2) = bin(m-1,s-1), but T(2m-2, 2*s-1) = 0 (valid again for 1 <= s <= m). - Petros Hadjicostas, Nov 03 2017

T is the infinite lower triangular matrix for this sequence; define two others, U and V; let U(n,k)=e_k(-1,2,-3,...,(-1)^n n), where e_k is the k-th elementary symmetric polynomial, and let V be the diagonal matrix A057077 (periodic sequence 1,1,-1,-1). Clearly V^-1 = V. Conjecture: U = U^-1, T = U . V, T^-1 = V . U, and |T| = |U|. - George Beck, Dec 16 2017

Let T*(n,k)=T(n,k) except when n is odd and k=(n+1)/2, where T*(n,k) = T(n,k)+2^((n-1)/2). Thus, T*(n,k) is the number of non-isomorphic symmetric stairs with n cells and k steps, i.e., k-1 changes of direction. See A016116. - Christian Barrientos and Sarah Minion, Jul 29 2018

Binomial transform of A093968.

Number of permutations of the n-th row of Pascal's triangle.

Can be seen as the multiplicative equivalent to the generalized pentagonal numbers. - Peter Luschny, Oct 13 2012

a(n) is the number of permutations of [n] in which all ascents start at an even position. For example, a(3) = 3 counts 213, 312, 321. - David Callan, Nov 25 2021

This sequence can be calculated by a recursive algorithm:

Let B1 be an array of finite length, the "1" denotes that it is the first generation. Let B1' be the reversed version of B1. Let C be the element-wise product C = B1 * B1'. Then B2 is a concatenation of taking each element of B1 and add all divisors of the corresponding element in C. If we start with B1 = {1} then we get this sequence of arrays: B2 = {2}, B3 = {3, 4, 6}, ... . a(n) is the length of the array Bn. In short the length of Bn+1 and so a(n+1) is the sum over A000005(Bn * Bn').

The transform used in the definition of this sequence is its own inverse, so if c = S(b) then b = S(c). The eigensequence is 2^n = S(2^n).

There exist some transformation pairs of infinite sequences in the database:

A000027 <--> A000142; A000079 <--> A000079; A000045 <--> A001654;

A026549 <--> A038754; A100071 <--> A001405; A058295 <--> A------;

A111286 <--> A098011; A093968 <--> A205825; A166447 <--> A------;

A098558 <--> A004277; A010551 <--> A087811; A100538 <--> A329227;

A079352 <--> A------; A082458 <--> A------; A008233 <--> A264635;

A138278 <--> A------; A006501 <--> A264557; A336496 <--> A------;

A019464 <--> A------; A062112 <--> A------; A171647 <--> A359039;

A279312 <--> A------; A031923 <--> A------.

These transformation pairs are conjectured:

A137326 <--> A------; A066332 <--> A300902; A208147 <--> A308546;

A057895 <--> A------; A349080 <--> A------; A019442 <--> A------;

A349079 <--> A------.

("A------" means not yet in the database.)

Some sequences in the lists above may need offset adjustment to force a beginning with 1,2,... in the transformation.

If we allowed signed rational numbers, further interesting transformation pairs could be observed. For example, 1/n will transform into factorials with alternating sign. 2^(-n) transforms into ones with alternating sign and 1/A000045(n) into A000045 with alternating sign.