A101508 Product of binomial matrix and the Mobius matrix A051731.

1, 2, 1, 4, 2, 1, 8, 4, 3, 1, 16, 8, 6, 4, 1, 32, 16, 11, 10, 5, 1, 64, 32, 21, 20, 15, 6, 1, 128, 64, 42, 36, 35, 21, 7, 1, 256, 128, 85, 64, 70, 56, 28, 8, 1, 512, 256, 171, 120, 127, 126, 84, 36, 9, 1, 1024, 512, 342, 240, 220, 252, 210, 120, 45, 10, 1, 2048, 1024, 683, 496, 385, 463, 462, 330, 165, 55, 11, 1

Offset: 0

Author

Paul Barry, Dec 05 2004

Keywords

• easy
• nonn
• tabl

Comments

Row sums are A101509. Diagonal sums are A101510.

The matrix inverse appears to be A128313. - R. J. Mathar, Mar 22 2013

Read as upper triangular matrix, this can be seen as "recurrences in A135356 applied to A023531" [Paul Curtz, Mar 03 2017]. - The columns are: A000079, A131577, A024495, A000749, A139761, ... Column n differs after the (n+1)-th nonzero term on from the binomial coefficients C(k,n). - M. F. Hasler, Mar 05 2017

Examples

			Rows begin
  1;
  2,1;
  4,2,1;
  8,4,3,1;
  16,8,6,4,1;
  ...
		

Links

• Alois P. Heinz, Rows n = 0..140, flattened

Programs

Maple

A101508 := proc(n,k)
    a := 0 ;
    for i from 0 to n do
        if modp(i+1,k+1) = 0 then
            a := a+binomial(n,i) ;
        end if;
    end do:
    return a;
end proc: # R. J. Mathar, Mar 22 2013

Mathematica

t[n_, k_] := Sum[If[Mod[i + 1, k + 1] == 0, Binomial[n, i], 0], {i, 0, n}]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 24 2014 *)

PARI

T(n,k)=sum(i=0,n, if((i+1)%(k+1)==0, binomial(n, i))) \\ M. F. Hasler, Mar 05 2017

Formula

T(n, k) = Sum_{i=0..n} if(mod(i+1, k+1)=0, binomial(n, i), 0).

Rows have g.f. x^k/((1-x)^(k+1)-x^(k+1)).