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Triangle T, read by rows, where matrix inverse T^-1 has -2^n in the secondary diagonal: [T^-1](n+1,n) = -2^n, with all 1's in the main diagonal and zeros elsewhere.
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%I #14 Dec 31 2024 15:37:07

%S 1,1,1,2,2,1,8,8,4,1,64,64,32,8,1,1024,1024,512,128,16,1,32768,32768,

%T 16384,4096,512,32,1,2097152,2097152,1048576,262144,32768,2048,64,1,

%U 268435456,268435456,134217728,33554432,4194304,262144,8192,128,1

%N Triangle T, read by rows, where matrix inverse T^-1 has -2^n in the secondary diagonal: [T^-1](n+1,n) = -2^n, with all 1's in the main diagonal and zeros elsewhere.

%C More generally, if a lower triangular matrix T to the power p is given by: [T^p](n,k) = C(r,n-k)*p^(n-k)*q^(n*(n-1)/2-k*(k-1)/2) then, for all m, [T^m](n,k) = [prod_{j=0..n-k-1}(m*r-p*j)]/(n-k)!*q^(n*(n-1)/2-k*(k-1)/2) for n>k>=0, with T(n,n) = 1. This triangle results when m=1, p=-1, q=2, r=1.

%C T(n,k) is the number of simple labeled graphs G on [n] such that the subgraph of G induced by the vertices labeled 1,2,...,k is a clique of size k. Cf A277219. - _Geoffrey Critzer_, May 05 2024

%F T(n,k) = 2^(n*(n-1)/2 - k*(k-1)/2).

%e Triangle T begins:

%e 1;

%e 1,1;

%e 2,2,1;

%e 8,8,4,1;

%e 64,64,32,8,1;

%e 1024,1024,512,128,16,1;

%e 32768,32768,16384,4096,512,32,1;

%e 2097152,2097152,1048576,262144,32768,2048,64,1;

%e 268435456,268435456,134217728,33554432,4194304,262144,8192,128,1;

%e Matrix inverse T^-1 has -2^n in the 2nd diagonal:

%e 1,

%e -1,1,

%e 0,-2,1,

%e 0,0,-4,1,

%e 0,0,0,-8,1,

%e 0,0,0,0,-16,1,

%e 0,0,0,0,0,-32,1,

%e ...

%p T := (n, k) -> 2^(((n + k - 1)*(n - k))/2):

%p seq(seq(T(n, k), k = 0..n), n = 0..8); # _Peter Luschny_, Dec 31 2024

%t Flatten[Table[2^((n(n-1))/2-(k(k-1))/2),{n,0,10},{k,0,n}]] (* _Harvey P. Dale_, Sep 19 2013 *)

%o (PARI) {T(n,k)=local(m=1,p=-1,q=2,r=1);prod(j=0,n-k-1,m*r-p*j)/(n-k)!*q^((n-k)*(n+k-1)/2)}

%Y Cf. A006125 (column 0); variants: A117250 (p=q=2), A117252 (p=q=3), A117254 (p=q=4), A117256 (p=q=5), A117258 (p=2, q=4), A117262 (p=-1, q=3), A117265 (p=-2, q=2).

%K nonn,tabl

%O 0,4

%A _Paul D. Hanna_, Mar 14 2006