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.
1, 1, 4, 40, 712, 18784, 663424, 29480896, 1581976960, 99585422848, 7198258087936, 587699970912256, 53497834761985024, 5372784803063664640, 590164397145095421952, 70386834555048578596864, 9058611906733586004803584, 1251310862246447324484468736, 184665445630564847038730076160
Offset: 0
Keywords
Examples
E.g.f.: A(x) = 1 + x + 4*x^2/2! + 40*x^3/3! + 712*x^4/4! + 18784*x^5/5! +... such that A(x) = exp(4*x*G(x)^3) / G(x)^3 where G(x) = 1 + x*G(x)^4 is the g.f. of A002293: G(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + 7084*x^6 +... Note that A'(x) = exp(4*x*G(x)^3) = 1 + 4*x + 40*x^2/2! + 712*x^3/3! + 18784*x^4/4! +... LOGARITHMIC DERIVATIVE. The logarithm of the e.g.f. begins: log(A(x)) = x + 3*x^2/2 + 15*x^3/3 + 91*x^4/4 + 612*x^5/5 +... and so A'(x)/A(x) = G(x)^3. TABLE OF POWERS OF E.G.F. Form a table of coefficients of x^k/k! in A(x)^n as follows. n=1: [1, 1, 4, 40, 712, 18784, 663424, 29480896, ...]; n=2: [1, 2, 10, 104, 1840, 47888, 1669696, 73399040, ...]; n=3: [1, 3, 18, 198, 3528, 91152, 3146256, 136990656, ...]; n=4: [1, 4, 28, 328, 5944, 153376, 5257024, 227057728, ...]; n=5: [1, 5, 40, 500, 9280, 240440, 8209600, 352337600, ...]; n=6: [1, 6, 54, 720, 13752, 359424, 12263184, 523933056, ...]; n=7: [1, 7, 70, 994, 19600, 518728, 17737216, 755807920, ...]; n=8: [1, 8, 88, 1328, 27088, 728192, 25020736, 1065353216, ...]; ... in which the main diagonal begins (see A251584): [1, 2, 18, 328, 9280, 359424, 17737216, 1065353216, ...] and is given by the formula: [x^n/n!] A(x)^(n+1) = 4^(n-2) * (n+1)^(n-3) * (3*n^2 + 13*n + 16) for n>=0.
Crossrefs
Programs
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Mathematica
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 *)
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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)} for(n=0, 20, print1(a(n), ", "))
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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) ))} for(n=0, 20, print1(a(n), ", "))
Formula
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:
(1) A'(x)/A(x) = G(x)^3.
(2) A'(x) = exp(4*x*G(x)^3).
(3) A(x) = exp( Integral G(x)^3 dx ).
(5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251584.
(6) A(x) = Sum_{n>=0} A251584(n)*(x/A(x))^n/n! and
(7) [x^n/n!] A(x)^(n+1) = (n+1)*A251584(n),
where A251584(n) = 4^(n-2) * (n+1)^(n-4) * (3*n^2 + 13*n + 16).
a(n) = Sum_{k=0..n} 4^k * n!/k! * binomial(4*n-k-4, n-k) * (k-1)/(n-1) for n>1.
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
a(n) ~ 2^(8*n-9) * n^(n-2) / (3^(3*n-7/2) * exp(n-1)). - Vaclav Kotesovec, Dec 07 2014