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Search: id:A118194
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| A118194 |
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Column 0 of the matrix log of triangle A118190, after term in row n is multiplied by n: a(n) = n*[LOG(A118190)](n,0), where A118190(n,k) = (5^k)^(n-k). |
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2
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| 0, 1, -3, 53, -4871, 2262505, -5269940619, 61424345593757, -3580474937256484367, 1043606492389898678125009, -1520932783784930699920673828115, 11082945991224258678496051788222656261, -403804307486446123171767495567989349951171863
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OFFSET
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0,3
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COMMENT
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The entire matrix log of triangle A118190 is determined by column 0 (this sequence): [LOG(A118190)](n,k) = a(n-k)/(n-k)*(5^k)^(n-k) for n>k>=0.
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FORMULA
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G.f.: x/(1-x)^2 = Sum_{n>=0} a(n)*x^n/(1-5^n*x). By using the inverse transformation: a(n) = Sum_{k=0..n} k*A118193(n-k)*(5^k)^(n-k) for n>=0.
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EXAMPLE
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Column 0 of LOG(A118190) = [0, 1, -3/2, 53/3, -4871/4, ...].
The g.f. is illustrated by:
x/(1-x)^2 = x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 6*x^6 + 7*x^7 +...
= x/(1-5*x) -3*x^2/(1-25*x) +53*x^3/(1-125*x) -4871*x^4/(1-625*x) + 2262505*x^5/(1-3125*x) - 5269940619*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))), L=sum(m=1, #T, -(T^0-T)^m/m)); return(n*L[n+1, 1])}
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CROSSREFS
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Cf. A118190.
Sequence in context: A012823 A098932 A100444 this_sequence A093164 A092448 A045481
Adjacent sequences: A118191 A118192 A118193 this_sequence A118195 A118196 A118197
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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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