cp's OEIS Frontend

A360344 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n))^(n+1) for n >= 0.

%I A360344 #7 Feb 06 2023 04:14:43
%S A360344 1,1,4,45,820,19820,582007,19812744,760177656,32275309743,
%T A360344 1497313010037,75208566398988,4062020902196139,234638046113989856,
%U A360344 14432573619909530980,941883830760366274935,65013065172020161949992,4733236746727327140204578,362575149419405494321544263
%N A360344 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n))^(n+1) for n >= 0.
%H A360344 Paul D. Hanna, <a href="/A360344/b360344.txt">Table of n, a(n) for n = 0..300</a>
%F A360344 G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
%F A360344 (1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n))^(n+1) for n>=0.
%F A360344 (2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(2*n))^(n+1) / (n+1).
%F A360344 a(n) ~ c * d^n * n! * n^alpha, where d = 3.93464558322824528799..., alpha = 1.1169011279372..., c = 0.052820142023857... - _Vaclav Kotesovec_, Feb 06 2023
%e A360344 G.f.: A(x) = 1 + x + 4*x^2 + 45*x^3 + 820*x^4 + 19820*x^5 + 582007*x^6 + 19812744*x^7 + 760177656*x^8 + 32275309743*x^9 + ...
%e A360344 RELATED SERIES.
%e A360344 G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
%e A360344 B(x) = 1 + x + 5*x^2 + 58*x^3 + 1057*x^4 + 25471*x^5 + 746143*x^6 + 25364298*x^7 + 972602305*x^8 + ... + b(n)*x^n + ...
%e A360344 such that b(n) = [x^n] (1 + x*A(x)^(2*n))^(n+1) / (n+1),
%e A360344 as well as b(n) = [x^n] A(x)^(n+1) / (n+1),
%e A360344 so that b(n) begin:
%e A360344 [1/1, 2/2, 15/3, 232/4, 5285/5, 152826/6, 5223001/7, 202914384/8, ...].
%e A360344 ILLUSTRATION OF DEFINITION.
%e A360344 The table of coefficients of x^k in A(x)^(n+1) begins:
%e A360344 n=0: [1, 1,  4,  45,   820,  19820,  582007,  19812744, ...];
%e A360344 n=1: [1, 2,  9,  98,  1746,  41640, 1212239,  41021862, ...];
%e A360344 n=2: [1, 3, 15, 160,  2790,  65643, 1894300,  63714729, ...];
%e A360344 n=3: [1, 4, 22, 232,  3965,  92028, 2632070,  87984416, ...];
%e A360344 n=4: [1, 5, 30, 315,  5285, 121011, 3429725, 113930075, ...];
%e A360344 n=5: [1, 6, 39, 410,  6765, 152826, 4291758, 141657348, ...];
%e A360344 n=6: [1, 7, 49, 518,  8421, 187726, 5223001, 171278801, ...];
%e A360344 n=7: [1, 8, 60, 640, 10270, 225984, 6228648, 202914384, ...]; ...
%e A360344 Compare to the table of coefficients in (1 + x*A(x)^(2*n))^(n+1):
%e A360344 n=0: [1, 1,   0,    0,     0,      0,       0,         0, ...];
%e A360344 n=1: [1, 2,   5,   22,   218,   3724,   87245,   2516506, ...];
%e A360344 n=2: [1, 3,  15,   91,   888,  13929,  308182,   8594133, ...];
%e A360344 n=3: [1, 4,  30,  232,  2397,  35712,  742902,  19860536, ...];
%e A360344 n=4: [1, 5,  50,  470,  5285,  77631, 1530000,  38965400, ...];
%e A360344 n=5: [1, 6,  75,  830, 10245, 152826, 2900808,  70300080, ...];
%e A360344 n=6: [1, 7, 105, 1337, 18123, 280140, 5223001, 121085308, ...];
%e A360344 n=7: [1, 8, 140, 2016, 29918, 485240, 9053576, 202914384, ...]; ...
%e A360344 to see that the main diagonals of the tables are the same.
%o A360344 (PARI) {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(2*m))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]}
%o A360344 for(n=0, 20, print1(a(n), ", "))
%Y A360344 Cf. A360342, A360343, A360345, A360346, A360347.
%Y A360344 Cf. A302702, A302703.
%K A360344 nonn
%O A360344 0,3
%A A360344 _Paul D. Hanna_, Feb 05 2023