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A251574 E.g.f.: exp(4*x*G(x)^3) / G(x)^3 where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.

%I A251574 #13 Jun 04 2025 12:21:43
%S A251574 1,1,4,40,712,18784,663424,29480896,1581976960,99585422848,
%T A251574 7198258087936,587699970912256,53497834761985024,5372784803063664640,
%U A251574 590164397145095421952,70386834555048578596864,9058611906733586004803584,1251310862246447324484468736,184665445630564847038730076160
%N A251574 E.g.f.: exp(4*x*G(x)^3) / G(x)^3 where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.
%F A251574 Let G(x) = 1 + x*G(x)^4 be the g.f. of A002293, then the e.g.f. A(x) of this sequence satisfies:
%F A251574 (1) A'(x)/A(x) = G(x)^3.
%F A251574 (2) A'(x) = exp(4*x*G(x)^3).
%F A251574 (3) A(x) = exp( Integral G(x)^3 dx ).
%F A251574 (4) A(x) = exp( Sum_{n>=1} A006632(n)*x^n/n ), where A006632(n) = binomial(4*n-2,n)/(3*n-1).
%F A251574 (5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251584.
%F A251574 (6) A(x) = Sum_{n>=0} A251584(n)*(x/A(x))^n/n! and
%F A251574 (7) [x^n/n!] A(x)^(n+1) = (n+1)*A251584(n),
%F A251574 where A251584(n) = 4^(n-2) * (n+1)^(n-4) * (3*n^2 + 13*n + 16).
%F A251574 a(n) = Sum_{k=0..n} 4^k * n!/k! * binomial(4*n-k-4, n-k) * (k-1)/(n-1) for n>1.
%F A251574 Recurrence: 3*(3*n-5)*(3*n-4)*(4*n^2 - 23*n + 34)*a(n) = 8*(128*n^5 - 1440*n^4 + 6520*n^3 - 14906*n^2 + 17289*n - 8190)*a(n-1) + 256*(4*n^2 - 15*n + 15)*a(n-2). - _Vaclav Kotesovec_, Dec 07 2014
%F A251574 a(n) ~ 2^(8*n-9) * n^(n-2) / (3^(3*n-7/2) * exp(n-1)). - _Vaclav Kotesovec_, Dec 07 2014
%e A251574 E.g.f.: A(x) = 1 + x + 4*x^2/2! + 40*x^3/3! + 712*x^4/4! + 18784*x^5/5! +...
%e A251574 such that A(x) = exp(4*x*G(x)^3) / G(x)^3
%e A251574 where G(x) = 1 + x*G(x)^4 is the g.f. of A002293:
%e A251574 G(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + 7084*x^6 +...
%e A251574 Note that
%e A251574 A'(x) = exp(4*x*G(x)^3) = 1 + 4*x + 40*x^2/2! + 712*x^3/3! + 18784*x^4/4! +...
%e A251574 LOGARITHMIC DERIVATIVE.
%e A251574 The logarithm of the e.g.f. begins:
%e A251574 log(A(x)) = x + 3*x^2/2 + 15*x^3/3 + 91*x^4/4 + 612*x^5/5 +...
%e A251574 and so A'(x)/A(x) = G(x)^3.
%e A251574 TABLE OF POWERS OF E.G.F.
%e A251574 Form a table of coefficients of x^k/k! in A(x)^n as follows.
%e A251574 n=1: [1, 1,  4,   40,   712,  18784,   663424,   29480896, ...];
%e A251574 n=2: [1, 2, 10,  104,  1840,  47888,  1669696,   73399040, ...];
%e A251574 n=3: [1, 3, 18,  198,  3528,  91152,  3146256,  136990656, ...];
%e A251574 n=4: [1, 4, 28,  328,  5944, 153376,  5257024,  227057728, ...];
%e A251574 n=5: [1, 5, 40,  500,  9280, 240440,  8209600,  352337600, ...];
%e A251574 n=6: [1, 6, 54,  720, 13752, 359424, 12263184,  523933056, ...];
%e A251574 n=7: [1, 7, 70,  994, 19600, 518728, 17737216,  755807920, ...];
%e A251574 n=8: [1, 8, 88, 1328, 27088, 728192, 25020736, 1065353216, ...]; ...
%e A251574 in which the main diagonal begins (see A251584):
%e A251574 [1, 2, 18, 328, 9280, 359424, 17737216, 1065353216, ...]
%e A251574 and is given by the formula:
%e A251574 [x^n/n!] A(x)^(n+1) = 4^(n-2) * (n+1)^(n-3) * (3*n^2 + 13*n + 16) for n>=0.
%t A251574 Flatten[{1,1,Table[Sum[4^k * n!/k! * Binomial[4*n-k-4, n-k] * (k-1)/(n-1),{k,0,n}],{n,2,20}]}] (* _Vaclav Kotesovec_, Dec 07 2014 *)
%o A251574 (PARI) {a(n) = local(G=1);for(i=1,n,G=1+x*G^4 +x*O(x^n)); n!*polcoeff(exp(4*x*G^3)/G^3, n)}
%o A251574 for(n=0, 20, print1(a(n), ", "))
%o A251574 (PARI) {a(n) = if(n==0||n==1, 1, sum(k=0, n, 4^k * n!/k! * binomial(4*n-k-4,n-k) * (k-1)/(n-1) ))}
%o A251574 for(n=0, 20, print1(a(n), ", "))
%Y A251574 Cf. A251584, A251664, A002293, A006632.
%Y A251574 Cf. Variants: A243953, A251573, A251575, A251576, A251577, A251578, A251579, A251580.
%K A251574 nonn
%O A251574 0,3
%A A251574 _Paul D. Hanna_, Dec 06 2014