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Search: id:A118193
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| A118193 |
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Column 0 of the matrix inverse of triangle A118190(n,k) = (5^k)^(n-k). |
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+0
3
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| 1, -1, 4, -76, 7124, -3326876, 7760553124, -90490361296876, 5275336666748203124, -1537656615631182860546876, 2240970675863910673065189453124, -16329855533286908545970966339091796876, 594974481262862479448134839533519744970703124
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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The entire matrix inverse of triangle A118193 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).
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FORMULA
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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.
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EXAMPLE
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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) +...
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PROGRAM
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(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])}
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CROSSREFS
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Cf. A118190.
Sequence in context: A012047 A012010 A012155 this_sequence A052271 A080989 A006267
Adjacent sequences: A118190 A118191 A118192 this_sequence A118194 A118195 A118196
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KEYWORD
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sign
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Apr 15 2006
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