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A118193
Column 0 of the matrix inverse of triangle A118190(n,k) = 5^(k*(n-k)).
3
1, -1, 4, -76, 7124, -3326876, 7760553124, -90490361296876, 5275336666748203124, -1537656615631182860546876, 2240970675863910673065189453124, -16329855533286908545970966339091796876, 594974481262862479448134839533519744970703124
OFFSET
0,3
COMMENTS
The entire matrix inverse of triangle A118190 is determined by column 0 (this sequence): [A118190^-1](n,k) = a(n-k)*5^(k*(n-k)) for n>=k>=0. Any g.f. of the form: Sum_{k>=0} b(k)*x^k may be expressed as: Sum_{n>=0} c(n)*x^n/(1-5^n*x) by applying the inverse transformation: c(n) = Sum_{k=0..n} a(n-k)*b(k)*5^(k*(n-k)).
LINKS
FORMULA
G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1-5^n*x).
0^n = Sum_{k=0..n} a(k)*5^(k*(n-k)) for n>=0.
a(n) = - Sum_{j=0..n-1} 5^(j*(n-j))*a(j) with a(0) = 1 and a(1) = -1. - G. C. Greubel, Jun 29 2021
EXAMPLE
Recurrence at n=4: 0 = a(0)*(5^0)^4 +a(1)*(5^1)^3 +a(2)*(5^2)^2 +a(3)*(5^3)^1 +a(4)*(5^4)^0 = 1*(5^0) - 1*(5^3) + 4*(5^4) - 76*(5^3) + 7124*(5^0).
The g.f. is illustrated by: 1 = 1/(1-x) - 1*x/(1-5*x) + 4*x^2/(1-25*x) - 76*x^3/(1-125*x) + 7124*x^4/(1-625*x) - 3326876*x^5/(1-3125*x) + 7760553124*x^6/(1-15625*x) +...
MATHEMATICA
a[n_]:= a[n]= If[n<2, (-1)^n, -Sum[5^(j*(n-j))*a[j], {j, 0, n-1}]];
Table[a[n], {n, 0, 30}] (* G. C. Greubel, Jun 29 2021 *)
PROG
(PARI) {a(n)=local(T=matrix(n+1, n+1, r, c, if(r>=c, (5^(c-1))^(r-c)))); return((T^-1)[n+1, 1])}
(Sage)
@CachedFunction
def a(n): return (-1)^n if (n<2) else -sum(5^(j*(n-j))*a(j) for j in (0..n-1))
[a(n) for n in (0..30)] # G. C. Greubel, Jun 29 2021
CROSSREFS
Cf. A118190.
Sequence in context: A012155 A350489 A325214 * A052271 A184272 A080989
KEYWORD
sign
AUTHOR
Paul D. Hanna, Apr 15 2006
STATUS
approved