A176332 Row sums of triangle A176331.
1, 2, 5, 16, 56, 202, 741, 2752, 10318, 38972, 148070, 565280, 2166646, 8332378, 32136205, 124249856, 481433286, 1868972828, 7267804550, 28304698336, 110383060776, 431000853028, 1684754608210, 6592277745536, 25818887839956
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
Programs
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GAP
List([0..30], n -> Sum([0..n], k -> Sum([0..n], j -> (-1)^(n-j)* Binomial(j, n-k)*Binomial(j, k) ))) # G. C. Greubel, Feb 22 2019
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Magma
[(&+[ (&+[(-1)^(n-j)*Binomial(j,n-k)*Binomial(j,k): j in [0..n]]): k in [0..n]]): n in [0..30]]; // G. C. Greubel, Feb 21 2019
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Maple
a:=n->add(add(binomial(j,n-k)*binomial(j,k)*(-1)^(n-j),j=0..n),k=0..n): seq(a(n),n=0..30); # Muniru A Asiru, Jan 23 2019
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Mathematica
f[n_]:= (-1)^n*Sum[Binomial[n+k, k] Cos[Pi(n+k)/2], {k, 0, n}]; Array[f, 24, 0] (* Robert G. Wilson v, Apr 02 2012 *) -
PARI
{a(n) = sum(k=0,n, sum(j=0,n, (-1)^(n-j)*binomial(j,n-k)* binomial(j,k))) };vector(30, n, n--; a(n)) \\ G. C. Greubel, Feb 21 2019 -
PARI
a(n) = {my(v = vector(n, k, I^k)); for (k=1, n-1, v = vector(n, i, sum(j=1, i, v[j]));); -real(v[n]);} \\ Michel Marcus, Feb 25 2019 -
PARI
a(n) = sum(k=0, n\2, (-1)^k*binomial(2*n-2*k, n)); \\ Seiichi Manyama, Jan 29 2023
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PARI
my(N=30, x='x+O('x^N)); Vec(1/(sqrt(1-4*x)*(1+x^2*(2/(1+sqrt(1-4*x)))^2))) \\ Seiichi Manyama, Jan 29 2023 -
Sage
[sum(sum((-1)^(n-j)*binomial(j,n-k)*binomial(j,k) for j in (0..n)) for k in (0..n)) for n in (0..30)] # G. C. Greubel, Feb 21 2019
Formula
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(j,n-k)*binomial(j,k)*(-1)^(n-j).
Logarithm g.f.: arctan(x*C(x)) = Sum_{n>=1} a(n)/n*x^n, where C(x) = (1-sqrt(1-4*x))/(2*x) (A000108). - Vladimir Kruchinin, Aug 10 2010
Conjecture: 6*n*a(n) + 2*(-17*n+10)*a(n-1) + (47*n-60)*a(n-2) + 10*(-3*n+5)*a(n-3) + 4*(2*n-5)*a(n-4) = 0. - R. J. Mathar, Nov 24 2012
Recurrence: 2*n*(5*n-8)*a(n) = 2*(25*n^2 - 50*n + 18)*a(n-1) - (45*n^2 - 92*n + 36)*a(n-2) + 2*(2*n-3)*(5*n-3)*a(n-3). - Vaclav Kotesovec, Feb 12 2014
a(n) ~ 4^(n+1) / (5*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 12 2014
From Seiichi Manyama, Jan 29 2023: (Start)
a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(2*n-2*k,n).
G.f.: 1 / ( sqrt(1-4*x) * (1 + x^2 * c(x)^2) ), where c(x) is the g.f. of A000108. (End)
a(n) = [x^n] 1/((1+x^2) * (1-x)^(n+1)). - Seiichi Manyama, Apr 08 2024
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