A377096 - OEIS

%I #11 Nov 04 2024 21:11:51

%S 1,1,-3,21,-143,1149,-9770,86972,-800114,7550103,-72668622,710843094,

%T -7046801413,70648735836,-715146994085,7299753288947,-75056915001279,

%U 776741439506769,-8084658799971145,84585069659108863,-889118226617804356,9385960967322224518,-99471211651693732205,1057994867315360083603

%N G.f. A(x) satisfies [x^(2*n)] A(x)^(5*n) = -5 and [x^(2*n+1)] A(x)^(5*n+2) = [x^(2*n+1)] A(x)^(5*n+3) for n >= 1.

%C Let D(x) = 1 + x*D(x)^3 be the g.f. of A001764, then

%C (1) [x^(2*n)] 1/D(-x)^(5*n) = 0 for n >= 1, and

%C (2) [x^(2*n+1)] 1/D(-x)^(5*n+2) = [x^(2*n+1)] 1/D(-x)^(5*n+3) = 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)^(5*n+2) = (5*n+2) * Product_{k=1..n} (5*k + 3) for n >= 1, a conjecture which has been verified for the initial 1001 terms. Compare with A377097, the dual of this sequence.

%H Paul D. Hanna, <a href="/A377096/b377096.txt">Table of n, a(n) for n = 0..600</a>

%F G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

%F (1) [x^(2*n)] A(x)^(5*n) = -5 for n >= 1.

%F (2) [x^(2*n+1)] A(x)^(5*n+2) = [x^(2*n+1)] A(x)^(5*n+3) for n >= 1.

%F (3) [x^(2*n+1)] A(x)^(5*n+2) = (5*n+2) * Product_{k=1..n} (5*k + 3) for n >= 1 (conjecture).

%e G.f.: A(x) = 1 + x - 3*x^2 + 21*x^3 - 143*x^4 + 1149*x^5 - 9770*x^6 + 86972*x^7 - 800114*x^8 + 7550103*x^9 - 72668622*x^10 + 710843094*x^11 - 7046801413*x^12 + ...

%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

%e  1: [1,  1,  -3,  21, -143, 1149,  -9770,  86972, ...];

%e  2: [1,  2,  -5,  36, -235, 1886, -15943, 141504, ...];

%e  3: [1,  3,  -6,  46, -285, 2301, -19353, 171507, ...];

%e  4: [1,  4,  -6,  52, -301, 2472, -20696, 183456, ...];

%e  5: [1,  5,  -5,  55, -290, 2466, -20545, 182585, ...];

%e  6: [1,  6,  -3,  56, -258, 2340, -19364, 173064, ...];

%e  7: [1,  7,   0,  56, -210, 2142, -17521, 158159, ...];

%e  8: [1,  8,   4,  56, -150, 1912, -15300, 140376, ...];

%e  9: [1,  9,   9,  57,  -81, 1683, -12912, 121590, ...];

%e 10: [1, 10,  15,  60,   -5, 1482, -10505, 103160, ...];

%e 11: [1, 11,  22,  66,   77, 1331,  -8173,  86031, ...];

%e 12: [1, 12,  30,  76,  165, 1248,  -5964,  70824, ...];

%e 13: [1, 13,  39,  91,  260, 1248,  -3887,  57915, ...];

%e 14: [1, 14,  49, 112,  364, 1344,  -1918,  47504, ...];

%e 15: [1, 15,  60, 140,  480, 1548,     -5,  39675, ...];

%e 16: [1, 16,  72, 176,  612, 1872,   1928,  34448, ...];

%e 17: [1, 17,  85, 221,  765, 2329,   3978,  31824, ...];

%e 18: [1, 18,  99, 276,  945, 2934,   6261,  31824, ...];

%e 19: [1, 19, 114, 342, 1159, 3705,   8911,  34523, ...];

%e 20: [1, 20, 130, 420, 1415, 4664,  12080,  40080, -5, ...];

%e 21: [1, 21, 147, 511, 1722, 5838,  15939,  48765, 51240, 1023701, ...];

%e 22: [1, 22, 165, 616, 2090, 7260,  20680,  60984, 105787, 947232, ...];

%e 23: [1, 23, 184, 736, 2530, 8970,  26519,  77303, 167325, 947232, ...];

%e ...

%e in which we see that [x^(2*n)] A(x)^(5*n) = -5 for n >= 1.

%e Also, [x^(2*n+1)] A(x)^(5*n+2) = [x^(2*n+1)] A(x)^(5*n+3) for n >= 1;

%e these coefficients begin

%e [56, 1248, 31824, 947232, 32550336, 1273079808, ...],

%e and appear to equal (5*n + 2) * Product_{k=1..n} (5*k + 3) 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^(5*m), 2*m) + 5)/(5*m), polcoef(A^(5*m+2), 2*m+1) - polcoef(A^(5*m+3), 2*m+1) ) ); V[n+1]}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A377097, A376225, A377095.

%K sign

%O 0,3

%A _Paul D. Hanna_, Nov 04 2024