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

Author

Paul Barry, Apr 15 2010

Keywords

• easy
• nonn

Comments

Hankel transform is A176333.

Let A(n) denote the n X n array such that the i-th row of this array is the sequence obtained by applying the partial sum operator i-1 times to the tuple ((sqrt(-1))^m, 1 <= m <= n). Then the negative of the real part of the (n, n)-entry of A(n) equals a(n-2) for all n > 2. - John M. Campbell, Jan 20 2019

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A000108, A176331, A176333.

Cf. A005317, A024718, A360185, A360211.

Programs

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

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

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

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

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