|
%I #11 Aug 31 2020 05:16:23
%S 1,1,2,11,130,2450,63012,2040779,79377914,3594766694,185457776252,
%T 10725423627006,686721189003668,48200778475446916,3679104677398632520,
%U 303348177377608050219,26865664102518601306154,2543352040870175109554654,256296085507636954980717708,27390678829206902911266889386
%N O.g.f. A(x) satisfies: [x^n] exp( n*(n+1) * x ) / A(x)^(n+1) = 0 for n > 0.
%C It is remarkable that this sequence consists entirely of integers.
%H Paul D. Hanna, <a href="/A337458/b337458.txt">Table of n, a(n) for n = 0..300</a>
%F Given o.g.f. A(x), define B(x) = A(x/B(x)), then B(x) is the o.g.f. of A337457 and satisfies [x^n] exp( n*(n-1)*x/B(x) ) = 0 for n>0.
%F a(n) ~ sqrt(1-c) * 2^(2*n - 3/2) * n^(n - 3/2) / (sqrt(Pi) * c^n * (2-c)^(n-1) * exp(n)), where c = -A226775 = -LambertW(-2*exp(-2)). - _Vaclav Kotesovec_, Aug 31 2020
%e O.g.f.: A(x) = 1 + x + 2*x^2 + 11*x^3 + 130*x^4 + 2450*x^5 + 63012*x^6 + 2040779*x^7 + 79377914*x^8 + 3594766694*x^9 + 185457776252*x^10 + ...
%e ILLUSTRATION OF DEFINITION.
%e The table of coefficients of x^k/k! in exp(n*(n+1)*x) / A(x)^(n+1) begins:
%e n=0: [1, -1, -2, -48, -2616, -262080, -41718240, -9630270720, ...];
%e n=1: [1, 0, -6, -112, -5592, -547968, -86345120, -19809990912, ...];
%e n=2: [1, 3, 0, -222, -10728, -958824, -144971712, -32519314080, ...];
%e n=3: [1, 8, 52, 0, -18648, -1693248, -236690784, -50727983616, ...];
%e n=4: [1, 15, 210, 2420, 0, -2739720, -399251600, -80125144800, ...];
%e n=5: [1, 24, 558, 12192, 221184, 0, -616918320, -131299591680, ...];
%e n=6: [1, 35, 1204, 40278, 1272768, 33597312, 0, -196436730672, ...];
%e n=7: [1, 48, 2280, 106688, 4869552, 210771456, 7654459648, 0, ...]; ...
%e in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n*(n+1)*x ) / A(x)^(n+1) = 0 for n>0.
%e RELATED SERIES.
%e Define B(x) = A(x/B(x)), which begins
%e B(x) = 1 + x + x^2 + 7*x^3 + 93*x^4 + 1859*x^5 + 49357*x^6 + 1629227*x^7 + 64149805*x^8 + 2929386667*x^9 + ... + A337457(n)*x^n + ...
%e then the table of coefficients of x^k/k! in exp( n*(n-1)*x/B(x) ) begins:
%e n=0: [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...];
%e n=1: [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...];
%e n=2: [1, 2, 0, -16, -320, -21888, -2648576, -494325760, ...];
%e n=3: [1, 6, 24, 0, -1728, -88704, -9621504, -1715198976, ...];
%e n=4: [1, 12, 120, 864, 0, -281088, -26873856, -4328017920, ...];
%e n=5: [1, 20, 360, 5600, 65920, 0, -66944000, -10207436800, ...];
%e n=6: [1, 30, 840, 21600, 492480, 8784000, 0, -22098355200, ...];
%e n=7: [1, 42, 1680, 63504, 2237760, 71229312, 1814690304, 0, ...]; ...
%e in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n*(n-1)*x/B(x) ) = 0 for n>0.
%e Also note that B(x) = x/Series_Reversion( x*A(x) ) and A(x) = B(x*A(x)).
%o (PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(m*(m-1)*x +x*O(x^m)) / Ser(A)^m )[m]/m);A[n+1]}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A337457, A304319.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Aug 28 2020
|
