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G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n))^(n+1) for n >= 0.
7

%I #7 Feb 06 2023 04:14:43

%S 1,1,4,45,820,19820,582007,19812744,760177656,32275309743,

%T 1497313010037,75208566398988,4062020902196139,234638046113989856,

%U 14432573619909530980,941883830760366274935,65013065172020161949992,4733236746727327140204578,362575149419405494321544263

%N 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 Paul D. Hanna, <a href="/A360344/b360344.txt">Table of n, a(n) for n = 0..300</a>

%F G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:

%F (1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n))^(n+1) for n>=0.

%F (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 a(n) ~ c * d^n * n! * n^alpha, where d = 3.93464558322824528799..., alpha = 1.1169011279372..., c = 0.052820142023857... - _Vaclav Kotesovec_, Feb 06 2023

%e 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 RELATED SERIES.

%e G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:

%e 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 such that b(n) = [x^n] (1 + x*A(x)^(2*n))^(n+1) / (n+1),

%e as well as b(n) = [x^n] A(x)^(n+1) / (n+1),

%e so that b(n) begin:

%e [1/1, 2/2, 15/3, 232/4, 5285/5, 152826/6, 5223001/7, 202914384/8, ...].

%e ILLUSTRATION OF DEFINITION.

%e The table of coefficients of x^k in A(x)^(n+1) begins:

%e n=0: [1, 1, 4, 45, 820, 19820, 582007, 19812744, ...];

%e n=1: [1, 2, 9, 98, 1746, 41640, 1212239, 41021862, ...];

%e n=2: [1, 3, 15, 160, 2790, 65643, 1894300, 63714729, ...];

%e n=3: [1, 4, 22, 232, 3965, 92028, 2632070, 87984416, ...];

%e n=4: [1, 5, 30, 315, 5285, 121011, 3429725, 113930075, ...];

%e n=5: [1, 6, 39, 410, 6765, 152826, 4291758, 141657348, ...];

%e n=6: [1, 7, 49, 518, 8421, 187726, 5223001, 171278801, ...];

%e n=7: [1, 8, 60, 640, 10270, 225984, 6228648, 202914384, ...]; ...

%e Compare to the table of coefficients in (1 + x*A(x)^(2*n))^(n+1):

%e n=0: [1, 1, 0, 0, 0, 0, 0, 0, ...];

%e n=1: [1, 2, 5, 22, 218, 3724, 87245, 2516506, ...];

%e n=2: [1, 3, 15, 91, 888, 13929, 308182, 8594133, ...];

%e n=3: [1, 4, 30, 232, 2397, 35712, 742902, 19860536, ...];

%e n=4: [1, 5, 50, 470, 5285, 77631, 1530000, 38965400, ...];

%e n=5: [1, 6, 75, 830, 10245, 152826, 2900808, 70300080, ...];

%e n=6: [1, 7, 105, 1337, 18123, 280140, 5223001, 121085308, ...];

%e n=7: [1, 8, 140, 2016, 29918, 485240, 9053576, 202914384, ...]; ...

%e to see that the main diagonals of the tables are the same.

%o (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 for(n=0, 20, print1(a(n), ", "))

%Y Cf. A360342, A360343, A360345, A360346, A360347.

%Y Cf. A302702, A302703.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Feb 05 2023