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%I #29 Aug 31 2020 06:48:46
%S 1,1,5,54,935,22417,685592,25431764,1106630687,55174867339,
%T 3097872254493,193283918695494,13260815963831108,991928912663646012,
%U 80325879518096889760,7000127337189146831092,653156403671376068448047,64963788042207845593775999,6861040250464949653809027311,766815367797924824316405828466,90417908118862070187113849296815
%N O.g.f. A(x) satisfies: [x^n] exp( n^2 * x ) / A(x) = 0 for n>0.
%C It is conjectured that the coefficients of o.g.f. A(x) consist entirely of integers.
%C Equals row 2 of table A304320.
%C O.g.f. A(x) = 1/(1 - x*B(x)), where B(x) is the o.g.f. of A107668.
%C Logarithmic derivative of o.g.f. A(x), A'(x)/A(x), equals o.g.f. of A304312.
%C Conjecture: given o.g.f. A(x), the coefficient of x^n in A'(x)/A(x) is the number of connected n-state finite automata with 2 inputs (A006691).
%H Paul D. Hanna, <a href="/A304322/b304322.txt">Table of n, a(n) for n = 0..400</a>
%F a(n) ~ sqrt(1-c) * 2^(2*n - 1/2) * n^(n - 1/2) / (sqrt(Pi) * c^n * (2-c)^n * exp(n)), where c = -A226775 = -LambertW(-2*exp(-2)). - _Vaclav Kotesovec_, Aug 31 2020
%e O.g.f.: A(x) = 1 + x + 5*x^2 + 54*x^3 + 935*x^4 + 22417*x^5 + 685592*x^6 + 25431764*x^7 + 1106630687*x^8 + 55174867339*x^9 + 3097872254493*x^10 + ...
%e ILLUSTRATION OF DEFINITION.
%e The table of coefficients of x^k/k! in exp(n^2*x) / A(x) begins:
%e n=0: [1, -1, -8, -270, -19584, -2427000, -455544000, -120136161600, ...];
%e n=1: [1, 0, -9, -296, -20715, -2527704, -470405285, -123376631664, ...];
%e n=2: [1, 3, 0, -350, -24672, -2867256, -518870528, -133753337280, ...];
%e n=3: [1, 8, 55, 0, -29547, -3559056, -614943333, -153534305160, ...];
%e n=4: [1, 15, 216, 2674, 0, -4291704, -783235520, -187656684864, ...];
%e n=5: [1, 24, 567, 12880, 251541, 0, -948897125, -243358236600, ...];
%e n=6: [1, 35, 1216, 41634, 1372320, 38884296, 0, -295870371264, ...];
%e n=7: [1, 48, 2295, 109000, 5106453, 230531544, 8944955227, 0, ...];
%e n=8: [1, 63, 3960, 248050, 15443328, 949131144, 56257429312, 2865412167360, 0, ...]; ...
%e in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n^2*x ) / A(x) = 0 for n>=0.
%e LOGARITHMIC DERIVATIVE.
%e The logarithmic derivative of A(x) yields the o.g.f. of A304312:
%e A'(x)/A(x) = 1 + 9*x + 148*x^2 + 3493*x^3 + 106431*x^4 + 3950832*x^5 + 172325014*x^6 + 8617033285*x^7 + 485267003023*x^8 + 30363691715629*x^9 + ... + A304312(n)*x^n +...
%e INVERT TRANSFORM.
%e 1/A(x) = 1 - x*B(x), where B(x) is the o.g.f. of A107668:
%e B(x) = 1 + 4*x + 45*x^2 + 816*x^3 + 20225*x^4 + 632700*x^5 + 23836540*x^6 + 1048592640*x^7 + 52696514169*x^8 + ... + A107668(n)*x^n + ...
%o (PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)^2 +x*O(x^m)) / Ser(A) )[m] ); A[n+1]}
%o for(n=0,25, print1( a(n),", "))
%Y Cf. A304320, A304312, A304321, A304323, A304324, A304325.
%Y Cf. A006691, A107668.
%K nonn
%O 0,3
%A _Paul D. Hanna_, May 11 2018
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