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%I #6 Nov 03 2024 17:08:27
%S 1,1,1,1,3,8,26,89,324,1225,4786,19170,78408,326275,1377772,5891401,
%T 25467509,111144579,489145720,2168854885,9681072845,43473716527,
%U 196286934526,890640262188,4059500301390,18579693200838,85360357637580,393548515741979,1820335724153452,8445294476235727,39291407672079211
%N G.f. A(x) satisfies: coefficient of x^n in A(x)^(2*n) equals A000172(n) = Sum_{k=0..n} C(n,k)^3, the n-th Franel number.
%F G.f.: sqrt( x / Series_Reversion( x*exp( Sum_{n>=1} A000172(n)*x^n/n ) ) ), where A000172(n) is the n-th Franel number.
%F [x^n] A(x)^(2*n+2) = (n+1)*A166990(n).
%F Convolution square-root of A088220.
%e G.f.: A(x) = 1 + x + x^2 + x^3 + 3*x^4 + 8*x^5 + 26*x^6 + 89*x^7 + 324*x^8 +...
%e Form a table of coefficients in A(x)^(2*n) as follows:
%e [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...];
%e [1, 2, 3, 4, 9, 24, 75, 252, 903, 3376, ...];
%e [1, 4, 10, 20, 43, 108, 316, 1020, 3537, 12908, ...];
%e [1, 6, 21, 56, 138, 354, 1002, 3120, 10485, 37318, ...];
%e [1, 8, 36, 120, 346, 960, 2756, 8448, 27723, 96440, ...];
%e [1, 10, 55, 220, 735, 2252, 6785, 21020, 68340, 233870, ...];
%e [1, 12, 78, 364, 1389, 4716, 15184, 48588, 159186, 541424, ...];
%e [1, 14, 105, 560, 2408, 9030, 31304, 104960, 351792, 1203244, ...];
%e [1, 16, 136, 816, 3908, 16096, 60184, 213152, 739162, 2570464, ...];
%e [1, 18, 171, 1140, 6021, 27072, 109047, 409500, 1480293, 5280932, ...]; ...
%e then the main diagonal forms the Franel numbers:
%e [1, 2, 10, 56, 346, 2252, 15184, 104960, 739162, 5280932, ...].
%o (PARI) {a(n)=polcoeff(sqrt(x/serreverse(x*exp(sum(m=1, n+1, sum(k=0, m, binomial(m, k)^3)*x^m/m +x^2*O(x^n))))),n)}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A166990, A088220, A000172.
%K nonn
%O 0,5
%A _Paul D. Hanna_, May 25 2014