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A302060 G.f. A(x) satisfies: [x^n] A(x)^(n*(n+1)-1) / (x*A(x)^n)' = 0 for n>1. 5

%I #20 Oct 20 2020 03:22:50

%S 1,1,3,107,11627,2513589,949355653,575357369483,525974349806337,

%T 691365121056215549,1257552573597625318887,3067926576692255188333527,

%U 9781672352885807666285800891,39881788154276616499389883709989,204117604287379008572673888063188001,1290628051526744629398741843471306433463

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

%C Compare to: [x^n] (x*F(x)^n)' / F(x)^(n*(n+1)) = 0 for n>1 holds when F(0) = 1.

%C Note that [x^n] G(x,k)^(k*(n+1)-1) / (x*G(x,k)^k)' = 0 is satisfied by an integer series G(x,k) when k is a fixed positive integer; the g.f. of this sequence explores the case when k varies with n.

%H Paul D. Hanna, <a href="/A302060/b302060.txt">Table of n, a(n) for n = 0..300</a>

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

%F a(n) ~ c * 2^n * n!^3 / n, where c = 0.2754104367997800017735528000516901713205624917... - _Vaclav Kotesovec_, Oct 20 2020

%e G.f.: A(x) = 1 + x + 3*x^2 + 107*x^3 + 11627*x^4 + 2513589*x^5 + 949355653*x^6 + 575357369483*x^7 + 525974349806337*x^8 + 691365121056215549*x^9 + ...

%e such that [x^n] A(x)^(n*(n+1)-1) / (x*A(x)^n)' = 0 for n>1.

%e ILLUSTRATION OF DEFINITION.

%e The table of coefficients in A(x)^(n*(n+1)-1) / (x*A(x)^n)' begins:

%e n=0: [1, -1, -2, -102, -11412, -2490030, -944283630, -573448825894, ...];

%e n=1: [1, -1, -4, -304, -45436, -12414490, -5655451828, -4009336016960, ...];

%e n=2: [1, 1, 0, -296, -56621, -17380683, -8487839136, -6303946190960, ...];

%e n=3: [1, 5, 22, 0, -43410, -17309652, -9440759988, -7462899694108, ...];

%e n=4: [1, 11, 86, 874, 0, -11796810, -8449485806, -7468455619310, ...];

%e n=5: [1, 19, 228, 3068, 88298, 0, -5377376960, -6278167743244, ...];

%e n=6: [1, 29, 496, 8136, 256299, 19641657, 0, -3822351028528, ...];

%e n=7: [1, 41, 950, 18924, 581824, 50072326, 8025251308, 0, ...]; ...

%e in which the main diagonal consists of all zeros after the initial terms, illustrating that [x^n] A(x)^(n*(n+1)-1) / (x*A(x)^n)' = 0 for n>1.

%e RELATED SERIES.

%e log(A(x)) = x + 5*x^2/2 + 313*x^3/3 + 46073*x^4/4 + 12508771*x^5/5 + 5680881713*x^6/6 + 4020812685695*x^7/7 + 4203174178089489*x^8/8 + 6217540835502410521*x^9/9 + ...

%o (PARI) {a(n) = my(A=[1, 1]); for(i=1, n, A = concat(A, 0); A[#A] = Vec( Ser(A)^(#A*(#A-1)-1)/(x*Ser(A)^(#A-1))' )[#A] ); A[n+1]}

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

%Y Cf. A300995, A300994, A300627, A302059.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Mar 31 2018

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Last modified March 29 09:59 EDT 2024. Contains 371268 sequences. (Running on oeis4.)