A125250 Square array, read by antidiagonals, where A(1,1) = A(2,2) = 1, A(1,2) = A(2,1) = 0, A(n,k) = 0 if n < 1 or k < 1, otherwise A(n,k) = A(n-2,k-2) + A(n-1,k-2) + A(n-2,k-1) + A(n-1,k-1).
1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 2, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 1, 5, 1, 0, 0, 0, 0, 0, 5, 5, 0, 0, 0, 0, 0, 0, 3, 11, 3, 0, 0, 0, 0, 0, 0, 1, 13, 13, 1, 0, 0, 0, 0, 0, 0, 0, 9, 26, 9, 0, 0, 0, 0, 0, 0, 0, 0, 4, 32, 32, 4, 0, 0, 0, 0, 0, 0, 0, 0, 1, 26, 63, 26, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14, 80, 80, 14, 0, 0, 0, 0, 0
Offset: 1
Examples
Array starts as: 1 0 0 0 0 0 0 ... 0 1 1 0 0 0 0 ... 0 1 2 2 1 0 0 ... 0 0 2 5 5 3 1 0 ... 0 0 1 5 11 13 9 4 1 0... 0 0 0 3 13 26 32 26 14 5 1 0 ... 0 0 0 1 9 32 63 80 71 45 20 6 1 0 ... 0 0 0 0 4 26 80 153 201 191 135 71 27 7 1 0 ... ...
Programs
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Mathematica
T[n_, k_] := Sum[Binomial[i, n-i] Binomial[i, k-i], {i, Floor[(n+1)/2], k}]; Table[T[n-k, k], {n, 0, 13}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 12 2019 *) -
PARI
A=matrix(22,22);A[1,1]=1;A[2,2]=1;A[2,1]=0;A[1,2]=0;A[3,2]=1;A[2,3]=1; for(n=3,22,for(k=3,22,A[n,k]=A[n-2,k-2]+A[n-1,k-2]+A[n-2,k-1]+A[n-1,k-1])); for(n=1,22,for(i=1,n,print1(A[n-i+1,i],", ")))
Formula
A(1,1) = A(2,2) = 1, A(1,2) = A(2,1) = 0, A(n,k) = 0 if n < 1 or k < 1, otherwise A(n,k) = A(n-2,k-2) + A(n-1,k-2) + A(n-2,k-1) + A(n-1,k-1).
From Peter Bala, Nov 07 2017: (Start)
T(n,k) = Sum_{i = floor((n+1)/2)..k} binomial(i,n-i)* binomial(i,k-i).
Square array = A026729 * transpose(A026729), where A026729 is viewed as a lower unit triangular array. Omitting the first row and column of square array = A030528 * transpose(A030528).
O.g.f. 1/(1 - t*(1 + t)*x - t*(1 + t)*x^2) = 1 + (t + t^2)*x + (t + 2*t^2 + 2*t^3 + t^4)*x^2 + .... Cf. A109466 with o.g.f. 1/(1 - t*x - t*x^2).
The n-th row polynomial R(n,t) satisfies R(n,t) = R(n,-1 - t).
R(n,t) = (-1)^n*sqrt(-t*(1 + t))^n*U(n, 1/2*sqrt(-t*(1 + t))), where U(n,x) denotes the n-th Chebyshev polynomial of the second kind.
The sequence of row polynomials R(n,t) is a divisibility sequence of polynomials, that is, if m divides n then R(m,t) divides R(n,t) in the polynomial ring Z[t].
Comments