%I #23 Feb 05 2023 03:20:14

%S 1,1,7,87,1495,31865,793769,22290228,689397657,23116772771,

%T 831159921411,31787496335409,1285410740283302,54708408148614317,

%U 2441969507507612684,113988651908380638224,5551479742274622439616,281540748098045175486249,14843765603832700589293465

%N G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n+5))^(n+1) for n >= 0.

%C Sequences with g.f. A(x,k) such that [x^n] A(x,k)^(n+1) = [x^n] (1 + x*A(x,k)^(n+k))^(n+1) have a rate of growth: a(n) ~ c(k) * d^n * n! * n^alfa(k), where d = A360279 = 2.1246065836242897918278825746989... (independent on k) and alfa(k) = 1.256334309718765863868089027485828533429844901971596190707510781... + k*0.94499537543174739679595101598799077876961098786349034... - _Vaclav Kotesovec_, Feb 05 2023

%H Paul D. Hanna, <a href="/A360237/b360237.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)^(n+5))^(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)^(n+5))^(n+1) / (n+1).

%F a(n) ~ c * d^n * n! * n^alfa, where d = A360279 = 2.12460658362428979..., alfa = 5.981311186877502847847844107425..., c = 0.000055660090340764345672306890127... - _Vaclav Kotesovec_, Jan 31 2023

%e G.f.: A(x) = 1 + x + 7*x^2 + 87*x^3 + 1495*x^4 + 31865*x^5 + 793769*x^6 + 22290228*x^7 + 689397657*x^8 + 23116772771*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 + 8*x^2 + 109*x^3 + 1984*x^4 + 43816*x^5 + 1116182*x^6 + 31810516*x^7 + 994086874*x^8 + ... + b(n)*x^n + ...

%e such that b(n) = [x^n] (1 + x*A(x)^(n+5))^(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, 24/3, 436/4, 9920/5, 262896/6, 7813274/7, 254484128/8, ...].

%e ILLUSTRATION OF DEFINITION.

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

%e n=0: [1, 1, 7, 87, 1495, 31865, 793769, 22290228, ...];

%e n=1: [1, 2, 15, 188, 3213, 67938, 1679767, 46874234, ...];

%e n=2: [1, 3, 24, 304, 5175, 108627, 2666476, 73945899, ...];

%e n=3: [1, 4, 34, 436, 7403, 154368, 3763020, 103713764, ...];

%e n=4: [1, 5, 45, 585, 9920, 205626, 4979200, 136401955, ...];

%e n=5: [1, 6, 57, 752, 12750, 262896, 6325530, 172251150, ...];

%e n=6: [1, 7, 70, 938, 15918, 326704, 7813274, 211519589, ...];

%e n=7: [1, 8, 84, 1144, 19450, 397608, 9454484, 254484128, ...]; ...

%e Compare to the table of coefficients in (1 + x*A(x)^(n+5))^(n+1):

%e n=0: [1, 1, 5, 45, 585, 9920, 205626, 4979200, ...];

%e n=1: [1, 2, 13, 126, 1654, 27688, 563565, 13415580, ...];

%e n=2: [1, 3, 24, 253, 3402, 56679, 1135813, 26574702, ...];

%e n=3: [1, 4, 38, 436, 6065, 101400, 2008616, 46226504, ...];

%e n=4: [1, 5, 55, 685, 9920, 167686, 3299580, 74828790, ...];

%e n=5: [1, 6, 75, 1010, 15285, 262896, 5165838, 115758780, ...];

%e n=6: [1, 7, 98, 1421, 22519, 396109, 7813274, 173599042, ...];

%e n=7: [1, 8, 124, 1928, 32022, 578320, 11506804, 254484128, ...]; ...

%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)^(m+5))^(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. A360231, A302702, A302703, A360234, A360235, A360236.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jan 30 2023