A337457 - OEIS

%I #14 Sep 02 2020 03:55:53

%S 1,1,1,7,93,1859,49357,1629227,64149805,2929386667,152027131261,

%T 8830653890299,567303319553421,39924294419453931,3053895154472856285,

%U 252244319795920299419,22373037117819632459821,2120745476831765696381387,213946972632171665440620925,22887117259538879173402222075

%N O.g.f. A(x) satisfies: [x^n] exp( n*(n-1) * x/A(x) ) = 0 for n > 0.

%C It is remarkable that this sequence consists entirely of integers.

%H Paul D. Hanna, <a href="/A337457/b337457.txt">Table of n, a(n) for n = 0..300</a>

%F Given o.g.f. A(x), define B(x) = A(x*B(x)), then B(x) is the o.g.f. of A337458 and satisfies [x^n] exp( n*(n+1)*x ) / B(x)^(n+1) = 0 for n>0.

%F a(n) ~ c * d^n * n! / n^2, where d = -4 / (LambertW(-2*exp(-2)) * (2 + LambertW(-2*exp(-2)))) = 6.17655460948348035823168... and c = 0.083103344220475784... - _Vaclav Kotesovec_, Aug 31 2020

%e O.g.f.: A(x) = 1 + x + x^2 + 7*x^3 + 93*x^4 + 1859*x^5 + 49357*x^6 + 1629227*x^7 + 64149805*x^8 + 2929386667*x^9 + 152027131261*x^10 + ...

%e ILLUSTRATION OF DEFINITION.

%e The table of coefficients of x^k/k! in exp( n*(n-1)*x/A(x) ) begins:

%e n=0: [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...];

%e n=1: [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...];

%e n=2: [1, 2, 0, -16, -320, -21888, -2648576, -494325760, ...];

%e n=3: [1, 6, 24, 0, -1728, -88704, -9621504, -1715198976, ...];

%e n=4: [1, 12, 120, 864, 0, -281088, -26873856, -4328017920, ...];

%e n=5: [1, 20, 360, 5600, 65920, 0, -66944000, -10207436800, ...];

%e n=6: [1, 30, 840, 21600, 492480, 8784000, 0, -22098355200, ...];

%e n=7: [1, 42, 1680, 63504, 2237760, 71229312, 1814690304, 0, ...]; ...

%e in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n*(n-1)*x/A(x) ) = 0 for n>0.

%e RELATED SERIES.

%e Define B(x) = A(x*B(x)), which begins

%e B(x) = 1 + x + 2*x^2 + 11*x^3 + 130*x^4 + 2450*x^5 + 63012*x^6 + 2040779*x^7 + 79377914*x^8 + 3594766694*x^9 + ... + A337458(n)*x^n + ...

%e then the table of coefficients of x^k/k! in exp(n*(n+1)*x) / B(x)^(n+1) begins:

%e n=0: [1, -1, -2, -48, -2616, -262080, -41718240, -9630270720, ...];

%e n=1: [1, 0, -6, -112, -5592, -547968, -86345120, -19809990912, ...];

%e n=2: [1, 3, 0, -222, -10728, -958824, -144971712, -32519314080, ...];

%e n=3: [1, 8, 52, 0, -18648, -1693248, -236690784, -50727983616, ...];

%e n=4: [1, 15, 210, 2420, 0, -2739720, -399251600, -80125144800, ...];

%e n=5: [1, 24, 558, 12192, 221184, 0, -616918320, -131299591680, ...];

%e n=6: [1, 35, 1204, 40278, 1272768, 33597312, 0, -196436730672, ...];

%e n=7: [1, 48, 2280, 106688, 4869552, 210771456, 7654459648, 0, ...]; ...

%e in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n*(n+1)*x ) / B(x)^(n+1) = 0 for n>0.

%e Also note that B(x) = (1/x)*Series_Reversion( x/A(x) ) and A(x) = B(x/A(x)).

%o (PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(m*(m-1)*x/Ser(A) ))[m+1]/(m*(m-1)) );A[n+1]}

%o for(n=0, 20, print1(a(n), ", "))

%Y Cf. A337458.

%K nonn

%O 0,4

%A _Paul D. Hanna_, Aug 28 2020