A107594 -id:A107594 - cp's OEIS frontend

A107595 G.f. satisfies: A(x) = Sum_{n>=0} x^n * A(x)^(n^2).

1, 1, 2, 7, 31, 158, 884, 5292, 33385, 219797, 1500449, 10573815, 76688602, 571232869, 4363912280, 34161879247, 273906591562, 2248935278231, 18909284838057, 162842178607893, 1436660527685476, 12988076148036405, 120345643023918566, 1143054910071718088, 11129160383826078389

Offset: 0

Paul D. Hanna, May 17 2005

			G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 31*x^4 + 158*x^5 + 884*x^6 + 5292*x^7 +...
Let A = g.f. A(x) then
A = 1 + x*A^1 + x^2*A^4 + x^3*A^9 + x^4*A^16 + x^5*A^25 ...
= 1 + x*(1 + x + 2*x^2 + 7*x^3 + 31*x^4 + 158*x^5 + 884*x^6 +...)
+ x^2*(1 + 4*x + 14*x^2 + 56*x^3 + 257*x^4 + 1312*x^5 +...)
+ x^3*(1 + 9*x + 54*x^2 + 291*x^3 + 1557*x^4 + 8568*x^5 +..)
+ x^4*(1 + 16*x + 152*x^2 + 1152*x^3 + 7836*x^4 +...)
+ x^5*(1 + 25*x + 350*x^2 + 3675*x^3 + 32625*x^4 +...)
+ x^6*(1 + 36*x + 702*x^2 + 9912*x^3 + 114201*x^4 +...) +...
= 1 + x + 2*x^2 + 7*x^3 + 31*x^4 + 158*x^5 + 884*x^6 +...

Paul D. Hanna, Table of n, a(n) for n = 0..200

Cf. A176719, A107590, A107594, A107596.

m = 25; A[_] = 0;
Do[A[x_] = 1 + Sum[x^k A[x]^(k^2) + O[x]^j, {k, 1, j}], {j, m}];
CoefficientList[A[x], x] (* Jean-François Alcover, Nov 05 2019 *)

{a(n)=local(A=1+x+x*O(x^n)); for(k=1,n,A=1+sum(j=1,n,x^j*A^(j^2)+x*O(x^n)));polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

G.f. A(x) = (1/x)*Series_Reversion(x/F(x)) and thus A(x) = F(x*A(x)) where F(x) is the g.f. of A107594.
G.f. A(x) = x/Series_Reversion(x*G(x)) and thus A(x) = G(x/A(x)) where G(x) is the g.f. of A107596.
From Paul D. Hanna, Apr 23 2010: (Start)
Let A = g.f. A(x), then A satisfies the continued fraction:
A = 1/(1 - A*x/(1 - (A^3-A)*x/(1 - A^5*x/(1 - (A^7-A^3)*x/(1 - A^9*x/(1- (A^11-A^5)*x/(1 - A^13*x/(1 - (A^15-A^7)*x/(1 - ...)))))))))
due to an identity of a partial elliptic theta function. (End)
From Paul D. Hanna, May 04 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^(4k-3)) / (1 - x*A^(4k-1))
due to a q-series identity. (End)

A107596 G.f. satisfies: A(x) = Sum_{n>=0} x^n * A(x)^(n^2+n).

Original entry on oeis.org

1, 1, 3, 14, 80, 514, 3567, 26153, 199900, 1579107, 12816020, 106421359, 901430144, 7771535382, 68085001080, 605420138920, 5459655601753, 49904765136264, 462228258349278, 4337787743946224, 41249375376404380, 397572319756235577

Offset: 0

Paul D. Hanna, May 17 2005

			A = 1 + x*A^2 + x^2*A^6 + x^3*A^12 + x^4*A^20 + x^5*A^30 ...
= 1 + (x + 2*x^2 + 7*x^3 + 34*x^4 + 197*x^5 + 1272*x^6 +...)
+ (x^2 + 6*x^3 + 33*x^4 + 194*x^5 + 1230*x^6 +...)
+ (x^3 + 12*x^4 + 102*x^5 + 784*x^6 +...)
+ (x^4 + 20*x^5 + 250*x^6 +...) +...
= 1 + x + 3*x^2 + 14*x^3 + 80*x^4 + 514*x^5 + 3567*x^6 +...

Cf. A107592, A107594, A107595.

{a(n)=local(A=1+x+x*O(x^n)); for(k=1,n,A=1+sum(j=1,n,x^j*A^(j^2+j)+x*O(x^n)));polcoeff(A,n)}

G.f. A(x)^2 = (1/x)*series-reversion(x/G107594(x)^2) and thus A(x) = G107594(x*A(x)^2) where G107594(x) is the g.f. of A107594. G.f. A(x) = (1/x)*series-reversion(x/G107595(x)) and thus A(x) = G107595(x*A(x)) where G107595(x) is the g.f. of A107595.
Contribution from Paul D. Hanna, Apr 25 2010: (Start)
Let A = g.f. A(x), then A satisfies the continued fraction:
A = 1/(1- A^2*x/(1- (A^4-A^2)*x/(1- A^6*x/(1- (A^8-A^4)*x/(1- A^10*x/(1- (A^12-A^6)*x/(1- A^14*x/(1- (A^16-A^8)*x/(1- A^18*x)))))))))
due to an identity of a partial elliptic theta function.
(End)

cp's OEIS Frontend

A107595 G.f. satisfies: A(x) = Sum_{n>=0} x^n * A(x)^(n^2).

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