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A319940
O.g.f. A(x) satisfies [x^n] exp( -n*A(x) ) / (1 - n*x)^n = 0, for n > 0.
4
1, 3, 30, 586, 17430, 696744, 34892228, 2095250576, 146470011822, 11669877667640, 1043022527852272, 103294254944725680, 11223660850862809960, 1327297414140637610776, 169690627501555713200460, 23320015259500560303564736, 3428111061331035575475494598, 536769111685159965192282250632, 89187403511916331132476542213808
OFFSET
1,2
COMMENTS
It is remarkable that this sequence should consist entirely of integers.
a(n) is even for n > 2, with a(2^k + 1) = 2 (mod 4) for k >= 1 (conjecture).
LINKS
FORMULA
a(n) ~ c * d^n * n! / n^2, where d = 9.669628447... and c = 0.0559981... - Vaclav Kotesovec, Oct 24 2020
From Seiichi Manyama, Apr 10 2026: (Start)
a(n) = n^(n-1) + (1/n) * Sum_{k=1..n-1} k * c_n(k) * e_n(n-k),
where c_n(k) = n^k/k - a(k) for 1 <= k <= n-1,
and e_n(0) = 1, e_n(k) = (n/k) * Sum_{j=1..k} j * c_n(j) * e_n(k-j) for 1 <= k <= n-1. (End)
EXAMPLE
O.g.f.: A(x) = x + 3*x^2 + 30*x^3 + 586*x^4 + 17430*x^5 + 696744*x^6 + 34892228*x^7 + 2095250576*x^8 + 146470011822*x^9 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp(-n*A(x))/(1 - n*x)^n begins:
n=1: [1, 0, -5, -178, -13983, -2082676, -500286245, ...];
n=2: [1, 2, 0, -344, -30592, -4460832, -1052294144, ...];
n=3: [1, 6, 45, 0, -46323, -7614918, -1758528063, ...];
n=4: [1, 12, 184, 2960, 0, -10429504, -2724259328, ...];
n=5: [1, 20, 495, 14050, 391505, 0, -3527335025, ...];
n=6: [1, 30, 1080, 44712, 2022912, 86720544, 0, ...];
n=7: [1, 42, 2065, 115556, 7166733, 472602158, 28883187781, 0, ...]; ...
in which the coefficient of x^n in row n forms a diagonal of zeros.
RELATED SERIES.
exp(A(x)) = 1 + x + 7*x^2/2! + 199*x^3/3! + 14929*x^4/4! + 2175121*x^5/5! + 516079351*x^6/6! + 179777047927*x^7/7! + ...
exp(-A(x)) = 1 - x - 5*x^2/2! - 163*x^3/3! - 13271*x^4/4! - 2012761*x^5/5! - 487790189*x^6/6! - 172048095115*x^7/7! + ...
PROG
(PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(-m*x*Ser(A))/(1-m*x +x^2*O(x^m))^m)[m+1]/m ); A[n]}
for(n=1, 30, print1(a(n), ", "))
(Ruby)
def A319940(n)
a = [0]
(1..n).each{|i|
c = [0] + (1..i - 1).map{|k| i ** k / k.to_r - a[k]}
e = [1]
(1..i - 1).each{|k| e << i / k.to_r * (1..k).inject(0){|s, j| s + j * c[j] * e[k - j]}}
a << i ** (i - 1) + (1..i - 1).inject(0){|s, k| s + k * c[k] * e[i - k]}.to_i / i
}
a[1..-1]
end
p A319940(20) # Seiichi Manyama, Apr 10 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 11 2018
STATUS
approved