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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A051159 Triangle read by rows: T(n, k) = binomial(n mod 2, k mod 2) * binomial(n div 2, k div 2), where 'div' denotes integer division.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 1, 1, 2, 2, 1, 1, 1, 0, 3, 0, 3, 0, 1, 1, 1, 3, 3, 3, 3, 1, 1, 1, 0, 4, 0, 6, 0, 4, 0, 1, 1, 1, 4, 4, 6, 6, 4, 4, 1, 1, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1, 1, 1, 5, 5, 10, 10, 10, 10, 5, 5, 1, 1, 1, 0, 6, 0, 15, 0, 20, 0, 15, 0, 6, 0, 1, 1, 1, 6, 6, 15, 15, 20, 20, 15, 15, 6, 6, 1, 1
Offset: 0

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Author

Michael Somos, Oct 14 1999

Keywords

Comments

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

Examples

			Triangle starts:
{1},
{1,  1},
{1,  0,  1},
{1,  1,  1,  1},
{1,  0,  2,  0,  1},
{1,  1,  2,  2,  1,  1},
{1,  0,  3,  0,  3,  0,  1},
{1,  1,  3,  3,  3,  3,  1,  1},
{1,  0,  4,  0,  6,  0,  4,  0,  1},
{1,  1,  4,  4,  6,  6,  4,  4,  1,  1},
{1,  0,  5,  0, 10,  0, 10,  0,  5,  0,  1},
{1,  1,  5,  5, 10, 10, 10, 10,  5,  5,  1,  1}
... - _Roger L. Bagula_ and _Gary W. Adamson_, Dec 04 2009

Links

Crossrefs

Cf. A007318, A051160, A016116, A034851, A169623, A036355.

Programs

  • Haskell
    a051159 n k = a051159_tabl !! n !! k
a051159_row n = a051159_tabl !! n
a051159_tabl = [1] : f [1] [1,1] where
   f us vs = vs : f vs (zipWith (+) ([0,0] ++ us) (us ++ [0,0]))
-- Reinhard Zumkeller, Apr 25 2013
  • Maple
    T:= proc(n, k) option remember; `if`(n=0 and k=0, 1,
      `if`(n<0 or k<0, 0, `if`(irem(n, 2)=1 or
       irem(k, 2)=0, T(n-1, k-1) + T(n-1, k), 0)))
    end:
seq(seq(T(n, k), k=0..n), n=0..14);  # Alois P. Heinz, Jul 12 2014
  • Mathematica
    T[ n_, k_] := QBinomial[n, k, -1]; (* Michael Somos, Jun 14 2011; since V7 *)
Clear[p, n, x, a]
w = 0;
p[x, 1] := 1;
p[x_, n_] := p[x, n] = If[Mod[n, 2] == 0, (x + 1)*p[x, n - 1], (x^2 + w*x + 1)^Floor[n/2]]
a = Table[CoefficientList[p[x, n], x], {n, 1, 12}]
Flatten[a] (* Roger L. Bagula and Gary W. Adamson, Dec 04 2009 *)
  • PARI
    {T(n, k) = binomial(n%2, k%2) * binomial(n\2, k\2)};
  • Python
    from math import comb as binomial
def T(n, k): return binomial(n%2, k%2) * binomial(n//2, k//2)
print([T(n, k) for n in range(14) for k in range(n+1)])  # Peter Luschny, Oct 17 2024
  • SageMath
    @cached_function
def T(n, k):
    if k == 0 or k == n: return 1
    return T(n-1, k-1) + (-1)^k*T(n-1, k)
for n in (0..12): print([T(n, k) for k in (0..n)]) # Peter Luschny, Jul 06 2021

Formula

T(n, k) = T(n-1, k-1) + T(n-1, k) if n odd or k even, else 0. T(0, 0) = 1.
T(n, k) = T(n-2, k-2) + T(n-2, k). T(0, 0) = T(1, 0) = T(1, 1) = 1.
Square array made by setting first row/column to 1's (A(i, 0) = A(0, j) = 1); A(1, 1) = 0; A(1, j) = A(1, j-2); A(i, 1) = A(i-2, 1); other entries A(i, j) = A(i-2, j) + A(i, j-2). - Gerald McGarvey, Aug 21 2004
Sum_{k=0..n} k * T(n,k) = A093968(n); A093968 = S-D transform of A001477. - Philippe Deléham, Aug 02 2006
Equals 2*A034851 - A007318. - Gary W. Adamson, Dec 31 2007. [Corrected by Yosu Yurramendi, Aug 07 2008]
A051160(n, k) = (-1)^floor(k/2) * T(n, k).
Sum_{k = 0..n} T(n,k)*x^k = A000012(n), A016116(n+1), A056487(n), A136859(n+2) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Mar 11 2014
G.f.: (1+x+x*y)/(1-x^2-y^2*x^2). - Philippe Deléham, Mar 11 2014
For n,k >= 1, T(n, k) = 0 when n odd and k even; otherwise, T(n, k) = binomial(floor((n-1)/2), floor((k-1)/2)). - Christian Barrientos, Mar 14 2020
From Werner Schulte, Jun 25 2021: (Start)
T(n,k) = T(n-1,k-1) + (-1)^k * T(n-1,k) for 0 < k < n with initial values T(n,0) = T(n,n) = 1 for n >= 0.
Matrix inverse is T^-1(n,k) = (-1)^((n-k)*(n+k+1)/2) * T(n,k) for 0 <= k <= n. (End)
From Peter Bala, Aug 08 2021: (Start)
Double Riordan array ( 1/(1 - x); x/(1 + x), x/(1 - x) ) in the notation of Davenport et al.
G.f. for column 2*n: (1 + x)*x^(2*n)/(1 - x^2)^(n+1); G.f. for column 2*n+1: x^(2*n+1)/(1 - x^2)^(n+1)
Row polynomials: R(2*n,x) = (1 + x^2)^n; R(2*n+1,x) = (1 + x)*(1 + x^2)^n.
The infinitesimal generator of this triangle has the sequence [1, 0, 1, 0, 1, 0, ...] on the main subdiagonal, the sequence [1, 1, 2, 2, 3, 3, 4, 4, ...] on the diagonal immediately below and zeros elsewhere.
Let T denote this lower triangular array. Then T^a, for a in C, is the double Riordan array ( (1 + a*x)/(1 - a*x^2); x/(1 + a*x), (1 + a*x)/(1 - a*x^2) ) with o.g.f. (1 + x*(a + y))/(1 - x^2*(a + y^2)) = 1 + (a + y)*x + (a + y^2)*x^2 + (a^2 + a*y + a*y^2 + y^3)*x^3 + (a^2 + 2*a*y^2 + y^4)*x^4 + ....
The (2*n)-th row polynomial of T^a is (a + y^2)^n; The (2*n+1)-th row polynomial of T^a is (a + y)*(a + y^2)^n. (End)

Extensions

New name using a formula of the author by Peter Luschny, Oct 17 2024

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