A054143 Triangular array T given by T(n,k) = Sum_{0 <= j <= i-n+k, n-k <= i <= n} C(i,j) for n >= 0 and 0 <= k <= n.
1, 1, 3, 1, 4, 7, 1, 5, 11, 15, 1, 6, 16, 26, 31, 1, 7, 22, 42, 57, 63, 1, 8, 29, 64, 99, 120, 127, 1, 9, 37, 93, 163, 219, 247, 255, 1, 10, 46, 130, 256, 382, 466, 502, 511, 1, 11, 56, 176, 386, 638, 848, 968, 1013, 1023, 1, 12, 67, 232, 562, 1024, 1486, 1816, 1981, 2036, 2047
Offset: 0
Examples
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins: 1; 1, 3; 1, 4, 7; 1, 5, 11, 15; 1, 6, 16, 26, 31; 1, 7, 22, 42, 57, 63;
Links
- G. C. Greubel, Rows n = 0..100 of triangle, flattened
Crossrefs
Diagonal sums give A005672. - Paul Barry, Feb 07 2003
Programs
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GAP
Flat(List([0..12], n-> List([0..n], k-> Sum([n-k..n], i-> Sum([0..i-n+k], j-> Binomial(i,j) ))))); # G. C. Greubel, Aug 01 2019
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Magma
T:= func< n,k | (&+[ (&+[ Binomial(i,j): j in [0..i-n+k]]): i in [n-k..n]]) >; [T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019
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Maple
A054143_row := proc(n) add(add(binomial(n,n-i)*x^(k+1),i=0..k),k=0..n-1); coeffs(sort(%)) end; seq(print(A054143_row(n)),n=1..6); # Peter Luschny, Sep 29 2011
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Mathematica
(* First program *) z=10; p[n_,x_]:=(x+1)^n; q[0,x_]:=1;q[n_,x_]:=x*q[n-1,x]+1; p1[n_,k_]:=Coefficient[p[n,x],x^k];p1[n_,0]:=p[n,x]/.x->0; d[n_,x_]:=Sum[p1[n,k]*q[n-1-k,x],{k,0,n-1}] h[n_]:=CoefficientList[d[n,x],{x}] TableForm[Table[Reverse[h[n]],{n,0,z}]] Flatten[Table[Reverse[h[n]],{n,-1,z}]] (* A054143 *) TableForm[Table[h[n],{n,0,z}]] Flatten[Table[h[n],{n,-1,z}]] (* A104709 *) (* Second program *) Table[Sum[Binomial[i, j], {i, n-k, n}, {j,0,i-n+k}], {n,0,12}, {k,0,n}]// Flatten (* G. C. Greubel, Aug 01 2019 *)
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PARI
T(n,k) = sum(i=n-k,n, sum(j=0,i-n+k, binomial(i,j))); for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Aug 01 2019
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Sage
def T(n, k): return sum(sum( binomial(i,j) for j in (0..i-n+k)) for i in (n-k..n)) [[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019
Formula
T(n,k) = Sum_{0 <= j <= i-n+k, n-k <= i <= n} binomial(i,j).
T(n,k) = T(n-1,k) + 3*T(n-1,k-1) - 2*T(n-2,k-1) - 2*T(n-2,k-2), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 30 2013
From Petros Hadjicostas, Jun 05 2020: (Start)
Bivariate o.g.f.: Sum_{n,k >= 0} T(n,k)*x^n*y^k = 1/(1 - x - 3*x*y + 2*x^2*y + 2*x^2*y^2) = 1/((1 - 2*x*y)*(1 - x*(y+1))).
n-th row o.g.f.: ((1 + y)^(n+1) - (2*y)^(n+1))/(1 - y). (End)
Extensions
Name edited by Petros Hadjicostas, Jun 04 2020
Comments