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Product of binomial matrix and the Mobius matrix A051731.
5

%I #23 Aug 31 2024 08:33:42

%S 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,

%T 128,64,42,36,35,21,7,1,256,128,85,64,70,56,28,8,1,512,256,171,120,

%U 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

%N Product of binomial matrix and the Mobius matrix A051731.

%C Row sums are A101509. Diagonal sums are A101510.

%C The matrix inverse appears to be A128313. - _R. J. Mathar_, Mar 22 2013

%C 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

%H Alois P. Heinz, <a href="/A101508/b101508.txt">Rows n = 0..140, flattened</a>

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

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

%e Rows begin

%e 1;

%e 2,1;

%e 4,2,1;

%e 8,4,3,1;

%e 16,8,6,4,1;

%e ...

%p A101508 := proc(n,k)

%p a := 0 ;

%p for i from 0 to n do

%p if modp(i+1,k+1) = 0 then

%p a := a+binomial(n,i) ;

%p end if;

%p end do:

%p return a;

%p end proc: # _R. J. Mathar_, Mar 22 2013

%t 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 *)

%o (PARI) T(n,k)=sum(i=0,n, if((i+1)%(k+1)==0, binomial(n, i))) \\ _M. F. Hasler_, Mar 05 2017

%K easy,nonn,tabl

%O 0,2

%A _Paul Barry_, Dec 05 2004