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%I #10 Nov 16 2023 12:54:59
%S 1,1,5,61,1161,28857,864141,29861749,1160382737,49854838897,
%T 2340623599381,119051103325613,6516915195123097,381912592990453545,
%U 23856225840952434333,1582482450123627473637,111113139625779846025761,8234335766045466358238433
%N G.f. A(x) satisfies: A(x) = 1 + x*A(x)^2 / (A(x) - 4*x*A'(x)).
%H Vaclav Kotesovec, <a href="/A352238/b352238.txt">Table of n, a(n) for n = 0..363</a>
%F G.f. A(x) satisfies:
%F (1) [x^n] A(x)^(4*n+1) = [x^(n-1)] (4*n+1) * A(x)^(4*n+1) for n >= 1.
%F (2) A(x) = 1 + x*A(x)^2/(A(x) - 4*x*A'(x)).
%F (3) A'(x) = A(x) * (1 + x*A(x)/(1 - A(x))) / (4*x).
%F (4) A(x) = exp( Integral (1 + x*A(x)/(1 - A(x)))/(4*x) dx ).
%F a(n) ~ c * 4^n * n! * n^(5/4), where c = 0.0440035900116077498469559... - _Vaclav Kotesovec_, Nov 16 2023
%e G.f.: A(x) = 1 + x + 5*x^2 + 61*x^3 + 1161*x^4 + 28857*x^5 + 864141*x^6 + 29861749*x^7 + 1160382737*x^8 + ...
%e such that A(x) = 1 + x*A(x)^2/(A(x) - 4*x*A'(x)).
%e Related table.
%e The table of coefficients of x^k in A(x)^(4*n+1) begins:
%e n=0: [1, 1, 5, 61, 1161, 28857, 864141, ...];
%e n=1: [1, 5, 35, 415, 7430, 176286, 5107530, ...];
%e n=2: [1, 9, 81, 993, 17127, 389583, 10916559, ...];
%e n=3: [1, 13, 143, 1859, 31564, 693212, 18802212, ...];
%e n=4: [1, 17, 221, 3077, 52309, 1118549, 29427153, ...];
%e n=5: [1, 21, 315, 4711, 81186, 1704906, 43640030, ...];
%e n=6: [1, 25, 425, 6825, 120275, 2500555, 62513875, ...]; ...
%e in which the following pattern holds:
%e [x^n] A(x)^(4*n+1) = [x^(n-1)] (4*n+1) * A(x)^(4*n+1), n >= 1,
%e as illustrated by
%e [x^1] A(x)^5 = 5 = [x^0] 5*A(x)^5 = 5*1;
%e [x^2] A(x)^9 = 81 = [x^1] 9*A(x)^9 = 9*9;
%e [x^3] A(x)^13 = 1859 = [x^2] 13*A(x)^13 = 13*143;
%e [x^4] A(x)^17 = 52309 = [x^3] 17*A(x)^17 = 17*3077;
%e [x^5] A(x)^21 = 1704906 = [x^4] 21*A(x)^21 = 21*81186;
%e [x^6] A(x)^25 = 62513875 = [x^5] 25*A(x)^25 = 25*2500555; ...
%e Also, compare the above terms along the diagonal to the series
%e B(x) = A(x*B(x)^4) = 1 + x + 9*x^2 + 143*x^3 + 3077*x^4 + 81186*x^5 + 2500555*x^6 + 87388600*x^7 + ...
%e where B(x)^4 = (1/x) * Series_Reversion( x/A(x)^4 ).
%o (PARI) /* Using A(x) = 1 + x*A(x)^2/(A(x) - 3*x*A'(x)) */
%o {a(n) = my(A=1); for(i=1,n, A = 1 + x*A^2/(A - 4*x*A' + x*O(x^n)) );
%o polcoeff(A,n)}
%o for(n=0,20, print1(a(n),", "))
%o (PARI) /* Using [x^n] A(x)^(4*n+1) = [x^(n-1)] (4*n+1)*A(x)^(4*n+1) */
%o {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
%o A[#A] = polcoeff((x*Ser(A)^(4*(#A)-3) - Ser(A)^(4*(#A)-3)/(4*(#A)-3)),#A-1));A[n+1]}
%o for(n=0,20, print1(a(n),", "))
%Y Cf. A088715, A286797, A317352, A352235, A352236, A352237.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Mar 08 2022
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