A342195 a(n) = Sum_{k=0..floor(n/2)} A323833(n,k) if A323833 is read as a triangle.

0, 1, 1, -5, -8, 61, 130, -1385, -3680, 50521, 160816, -2702765, -10026368, 199360981, 844583440, -19391512145, -92369507840, 2404879675441, 12722897618176, -370371188237525, -2154662195222528, 69348874393137901, 440001333689382400, -15514534163557086905, -106615331831035289600, 4087072509293123892361

Offset: 0

Author

Petros Hadjicostas, Mar 04 2021

Keywords

• sign

Comments

Because A323833(n,n/2) = 0 for n even (if A323833 is read as a triangle), we also have a(n) = Sum_{k=0..ceiling((n-1)/2)} A323833(n,k).

Examples

			a(3) = -2 - 3 = -5.
a(4) = -5 - 3 = -8.
a(5) = 16 + 21 + 24 = 61.
a(6) = 61 + 45 + 24 = 130.
a(7) = -272 - 333 - 378 - 402 = -1385.
		

Crossrefs

Cf. A000111, A028296, A163747, A323833.

Programs

PARI

{b(n) = local(v=[1], t); if( n<0, 0, for(k=2, n+2, t=0; v = vector(k, i, if( i>1, t+= v[k+1-i]))); v[2])}; \\ Michael Somos's PARI program for A000111
c(n) = if(n==0, 0, (-1)^(n+floor(n/2))*b(n))
a(n) = sum(k=0, floor(n/2), sum(i=0, n-k, binomial(n-k,i)*(-1)^(k+i)*c(k+i)))

Formula

a(n) = Sum_{k=0..floor(n/2)} A323833(n-k,k) if A323833 is read as a square array (by upwards antidiagonals).

a(2*n+1) = -A028296(n+1).

a(n) = Sum_{k=0..floor(n/2)} Sum_{i=0..n-k} binomial(n-k,i) * (-1)^(k+i) * A163747(k+i).