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Search: id:A100235
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| A100235 |
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Triangle, read by rows, of the coefficients of [x^k] in G100234(x)^n such that the row sums are 6^n-1 for n>0, where G100234(x) is the g.f. of A100234. |
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+0 4
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| 1, 1, 4, 1, 8, 26, 1, 12, 63, 139, 1, 16, 116, 436, 726, 1, 20, 185, 965, 2830, 3774, 1, 24, 270, 1790, 7335, 17634, 19601, 1, 28, 371, 2975, 15505, 52444, 106827, 101784, 1, 32, 488, 4584, 28860, 124424, 358748, 633952, 528526, 1, 36, 621, 6681, 49176, 256194
(list; table; graph; listen)
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OFFSET
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0,3
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COMMENT
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The main diagonal forms A100236. Secondary diagonal is: T(n+1,n) = (n+1)*A100237(n). More generally, if g.f. F(x) satisfies: m^n-b^n = Sum_{k=0..n} [x^k]F(x)^n, then F(x) also satisfies: (m+z)^n - (b+z)^n + z^n = Sum_{k=0..n} [x^k](F(x)+z*x)^n for all z and F(x)=(1+(m-1)*x+sqrt(1+2*(m-2*b-1)*x+(m^2-2*m+4*b+1)*x^2))/2; the triangle formed from powers of F(x) will have the g.f.: G(x,y)=(1-2*x*y+m*x^2*y^2)/((1-x*y)*(1-(m-1)*x*y-x^2*y^2-x*(1-x*y))).
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LINKS
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Tanya Khovanova, Recursive Sequences
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FORMULA
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G.f.: A(x, y)=(1-2*x*y+6*x^2*y^2)/((1-x*y)*(1-5*x*y-x^2*y^2-x*(1-x*y))).
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EXAMPLE
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Rows begin:
[1],
[1,4],
[1,8,26],
[1,12,63,139],
[1,16,116,436,726],
[1,20,185,965,2830,3774],
[1,24,270,1790,7335,17634,19601],
[1,28,371,2975,15505,52444,106827,101784],
[1,32,488,4584,28860,124424,358748,633952,528526],...
where row sums form 6^n-1 for n>0:
6^1-1 = 1+4 = 5
6^2-1 = 1+8+26 = 35
6^3-1 = 1+12+63+139 = 215
6^4-1 = 1+16+116+436+726 = 1295
6^5-1 = 1+20+185+965+2830+3774 = 7775.
The main diagonal forms A100236 = [1,4,26,139,726,3774,...],
where Sum_{n>=1} A100236(n)/n*x^n = log((1-x)/(1-5*x-x^2)).
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PROGRAM
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(PARI) {T(n, k, m=6)=if(n<k|k<0, 0, if(k==0, 1, polcoeff(((1+(m-1)*x+sqrt(1+2*(m-3)*x+(m^2-2*m+5)*x^2+x*O(x^k)))/2)^n, k)))}
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CROSSREFS
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Cf. A100234, A100236, A100237, A100232.
Sequence in context: A125129 A013611 A077910 this_sequence A089072 A036177 A141680
Adjacent sequences: A100232 A100233 A100234 this_sequence A100236 A100237 A100238
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KEYWORD
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nonn,tabl
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Nov 29 2004
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