A352412 E.g.f.: 2*x / LambertW( 2*x/(1-x) ).
1, 1, -4, 20, -224, 3392, -67232, 1629728, -46799104, 1552143104, -58386807296, 2455954797056, -114222622662656, 5819845970653184, -322384671892123648, 19290013218140254208, -1239886482366130946048, 85200320552417960394752
Offset: 0
Keywords
Examples
E.g.f.: A(x) = 1 + x - 4*x^2/2! + 20*x^3/3! - 224*x^4/4! + 3392*x^5/5! - 67232*x^6/6! + 1629728*x^7/7! - 46799104*x^8/8! + ... such that A(x) = (1-x) * exp(2*x/A(x)), where exp(2*x/A(x)) = 1 + 2*x + 20*x^3/3! - 144*x^4/4! + 2672*x^5/5! - 51200*x^6/6! + 1271328*x^7/7! - 36628480*x^8/8! + ... Related series. The e.g.f. A(x) satisfies A( x/(exp(-2*x) + x) ) = 1/(exp(-2*x) + x), where 1/(exp(-2*x) + x) = 1 + x - 2*x^2/2! - 10*x^3/3! + 24*x^4/4! + 312*x^5/5! - 560*x^6/6! + ... + A336958(n)*(-x)^n/n! + ... Related table. Another defining property of the e.g.f. A(x) is illustrated here. The table of coefficients of x^k/k! in A(x)^n begins: n=1: [1, 1, -4, 20, -224, 3392, -67232, 1629728, ...]; n=2: [1, 2, -6, 16, -192, 2944, -58880, 1434752, ...]; n=3: [1, 3, -6, -6, -48, 1296, -29664, 776544, ...]; n=4: [1, 4, -4, -40, 88, 128, -7424, 263936, ...]; n=5: [1, 5, 0, -80, 120, 280, -320, 38720, ...]; n=6: [1, 6, 6, -120, -24, 1872, -3312, 768, ...]; n=7: [1, 7, 14, -154, -392, 4424, -3920, -22288, ...]; ... from which we can illustrate that the partial sum of coefficients of x^k, k=0..n, in A(x)^n equals zero, for n > 1, as follows: n=1: 2 = 1 + 1; n=2: 0 = 1 + 2 + -6/2!; n=3: 0 = 1 + 3 + -6/2! + -6/3!; n=4: 0 = 1 + 4 + -4/2! + -40/3! + 88/4!; n=5: 0 = 1 + 5 + 0/2! + -80/3! + 120/4! + 280/5!; n=6: 0 = 1 + 6 + 6/2! + -120/3! + -24/4! + 1872/5! + -3312/6!; n=7: 0 = 1 + 7 + 14/2! + -154/3! + -392/4! + 4424/5! + -3920/6! + -22288/7!; ...
Programs
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PARI
{a(n) = n!*polcoeff( x/serreverse( x/(exp(-2*x +x^2*O(x^n)) + x) ),n)} for(n=0,30,print1(a(n),", ")) -
PARI
my(x='x+O('x^30)); Vec(serlaplace(2*x/lambertw(2*x/(1-x)))) \\ Michel Marcus, Mar 17 2022
Formula
E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies:
(1) A(x) = 2*x / LambertW( 2*x/(1-x) ).
(2) A(x) = (1-x) * exp( 2*x/A(x) ).
(3) A(x) = 2*x / log( A(x)/(1-x) ).
(4) A( x/(exp(-2*x) + x) ) = 1/(exp(-2*x) + x).
(5) A(x) = x / Series_Reversion( x/(exp(-2*x) + x) ).
(6) Sum_{k=0..n} [x^k] A(x)^n = 0, for n > 1.
(7) [x^(n+1)/(n+1)!] A(x)^n = -(-2)^(n+1) * n for n >= (-1).
a(n) ~ (-1)^(n+1) * exp(-1) * sqrt(2) * (2 - exp(-1))^(n - 1/2) * n^(n-1). - Vaclav Kotesovec, Mar 15 2022
Comments