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A107591
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G.f. satisfies: A(x) = Sum_{n>=0} x^n * A(x)^(n*(n+1)/2).
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8
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1, 1, 2, 6, 22, 91, 408, 1939, 9635, 49614, 263140, 1431301, 7959568, 45152340, 260847526, 1532825675, 9154581802, 55537885743, 342147577227, 2140251570508, 13594688301758, 87702596534110, 574815620158265, 3829029514213952
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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FORMULA
| G.f. A(x) = (1/x)*series-reversion(x/G107590(x)) and thus A(x) = G107590(x*A(x)) where G107591(x) is the g.f. of A107590. G.f. A(x) = x/series-reversion(x*G107592(x)) and thus A(x) = G107592(x/A(x)) where G107592(x) is the g.f. of A107592.
Contribution from Paul D. Hanna (pauldhanna(AT)juno.com), Apr 24 2010: (Start)
Let A = g.f. A(x), then A satisfies the continued fraction:
A = 1/(1- A*x/(1- (A^2-A)*x/(1- A^3*x/(1- (A^4-A^2)*x/(1- A^5*x/(1- (A^6-A^3)*x/(1- A^7*x/(1- (A^8-A^4)*x/(1- ...)))))))))
due to an identity of a partial elliptic theta function.
(End)
Contribution from Paul D. Hanna (pauldhanna(AT)juno.com), May 5 2010: (Start)
Let A = g.f. A(x), then A satisfies:
A = Sum_{n>=0} x^n*A^n*Product_{k=1..n} (1-x*A^(2k-1))/(1-x*A^(2k))
due to a q-series identity.
(End)
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EXAMPLE
| A = 1 + x*A^1 + x^2*A^3 + x^3*A^6 + x^4*A^10 + x^5*A^15 ...
= 1 + (x + x^2 + 2*x^3 + 6*x^4 + 22*x^5 + 91*x^6 +...)
+ (x^2 + 3*x^3 + 9*x^4 + 31*x^5 + 120*x^6 +...)
+ (x^3 + 6*x^4 + 27*x^5 + 116*x^6 +...)
+ (x^4 + 10*x^5 + 65*x^6 +...) +...
= 1 + x + 2*x^2 + 6*x^3 + 22*x^4 + 91*x^5 + 408*x^6 +...
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PROG
| (PARI) {a(n)=local(A=1+x+x*O(x^n)); for(k=1, n, A=1+sum(j=1, n, x^j*A^(j*(j+1)/2)+x*O(x^n))); polcoeff(A, n)}
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CROSSREFS
| Cf. A107590, A107592.
Sequence in context: A150271 A150272 A124293 * A155866 A150273 A001181
Adjacent sequences: A107588 A107589 A107590 * A107592 A107593 A107594
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KEYWORD
| eigen,nonn
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AUTHOR
| Paul D. Hanna (pauldhanna(AT)juno.com), May 17 2005, May 05 2010
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