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%I #22 Nov 05 2024 00:28:04
%S 1,1,-2,10,-43,245,-1415,8699,-55146,359502,-2391307,16187568,
%T -111112275,772031484,-5419579222,38392912974,-274157203525,
%U 1971830165065,-14274069360446,103944171679331,-761061075929465,5600738429715754,-41413279720692033,307606387883900380,-2294690968394992619,17189393071158368580
%N G.f. A(x) satisfies [x^(2*n)] A(x)^(3*n) = -3 and [x^(2*n+1)] A(x)^(3*n+1) = [x^(2*n+1)] A(x)^(3*n+2) for n >= 1.
%C Let C(x) = 1 + x*C(x)^2 be the g.f. of the Catalan numbers (A000108), then
%C (1) [x^(2*n)] 1/C(-x)^(3*n) = 0 for n >= 1, and
%C (2) [x^(2*n+1)] 1/C(-x)^(3*n+1) = [x^(2*n+1)] 1/C(-x)^(3*n+2) = 0 for n >= 1.
%C The g.f. A(x) of this sequence satisfies similar conditions, with the surprisingly simple formula [x^(2*n+1)] A(x)^(3*n+1) = (3*n+1) * Product_{k=1..n} (3*k + 2) for n >= 1, a conjecture which has been verified for the initial 1001 terms. Compare with A377095, the dual of this sequence.
%H Paul D. Hanna, <a href="/A376225/b376225.txt">Table of n, a(n) for n = 0..501</a>
%F G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
%F (1) [x^(2*n)] A(x)^(3*n) = -3 for n >= 1.
%F (2) [x^(2*n+1)] A(x)^(3*n+1) = [x^(2*n+1)] A(x)^(3*n+2) for n >= 1.
%F (3) [x^(2*n+1)] A(x)^(3*n+1) = (3*n+1) * Product_{k=1..n} (3*k + 2) for n >= 1 (conjecture).
%e G.f.: A(x) = 1 + x - 2*x^2 + 10*x^3 - 43*x^4 + 245*x^5 - 1415*x^6 + 8699*x^7 - 55146*x^8 + 359502*x^9 - 2391307*x^10 + ...
%e RELATED TABLES.
%e The table of coefficients of x^k in A(x)^n begins
%e n\k 0 1 2 3 4 5 6 7 8
%e 1: [1, 1, -2, 10, -43, 245, -1415, 8699, -55146, ...];
%e 2: [1, 2, -3, 16, -62, 364, -2068, 12728, -80485, ...];
%e 3: [1, 3, -3, 19, -63, 399, -2216, 13758, -86898, ...];
%e 4: [1, 4, -2, 20, -51, 384, -2052, 13016, -82158, ...];
%e 5: [1, 5, 0, 20, -30, 346, -1715, 11375, -71625, ...];
%e 6: [1, 6, 3, 20, -3, 306, -1299, 9432, -58821, ...];
%e 7: [1, 7, 7, 21, 28, 280, -861, 7575, -45899, ...];
%e 8: [1, 8, 12, 24, 62, 280, -428, 6040, -34019, ...];
%e 9: [1, 9, 18, 30, 99, 315, -3, 4959, -23643, ...];
%e 10: [1, 10, 25, 40, 140, 392, 430, 4400, -14760, ...];
%e 11: [1, 11, 33, 55, 187, 517, 902, 4400, -7051, ...];
%e 12: [1, 12, 42, 76, 243, 696, 1456, 4992, -3, ...];
%e 13: [1, 13, 52, 104, 312, 936, 2145, 6227, 7020, 80080, ...];
%e 14: [1, 14, 63, 140, 399, 1246, 3031, 8192, 14742, 80080, ...];
%e 15: [1, 15, 75, 185, 510, 1638, 4185, 11025, 23970, 90320, -3, ...]; ...
%e in which we see that [x^(2*n)] A(x)^(3*n) = -3 for n >= 1.
%e Also, [x^(2*n+1)] A(x)^(3*n+1) = [x^(2*n+1)] A(x)^(3*n+2) for n >= 1;
%e these coefficients begin
%e [20, 280, 4400, 80080, 1675520, 39793600, 1059766400, 31311280000, ...],
%e and appear to equal (3*n+1) * Product_{k=1..n} (3*k + 2) for n >= 1.
%o (PARI) {a(n) = my(V=[1,1,0,0],A); for(i=0,n, V=concat(V,0); A = Ser(V); m=(#V-1)\2-1;
%o V[#V-2] = if(#V%2 == 1, -(polcoef(A^(3*m), 2*m) + 3)/(3*m), polcoef(A^(3*m+1), 2*m+1) - polcoef(A^(3*m+2), 2*m+1) ) );V[n+1]}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A377095, A377096, A377097.
%K sign
%O 0,3
%A _Paul D. Hanna_, Oct 26 2024