OFFSET
1,2
COMMENTS
Conjecture: a(n) is divisible by prime(n) for n > 1.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 1..1000
EXAMPLE
G.f.: A(x) = x - 3*x^2 + 10*x^3 - 35*x^4 + 121*x^5 - 390*x^6 + 1037*x^7 - 1083*x^8 - 14030*x^9 + 137837*x^10 - 382106*x^11 - 8791718*x^12 + 199408912*x^13 - 2701500413*x^14 + 28888970650*x^15 + ...
where A(x) = x*F'(x)/F(x) with F(x) being the g.f. of A381353.
It appears that the coefficient of x^n in A(x) is divisible by prime(n) for n > 1;
a(2)/prime(2) = -3/3 = -1;
a(3)/prime(3) = 10/5 = 2;
a(4)/prime(4) = -35/7 = -5;
a(5)/prime(5) = 121/11 = 11;
a(6)/prime(6) = -390/13 = -30;
a(7)/prime(7) = 1037/17 = 61;
a(8)/prime(8) = -1083/19 = -57;
a(9)/prime(9) = -14030/23 = -610;
a(10)/prime(10) = 137837/29 = 4753;
a(11)/prime(11) = -382106/31 = -12326;
a(12)/prime(12) = -8791718/37 = -237614; ...
to form the (conjectural) sequence of integers starting with
[-1, 2, -5, 11, -30, 61, -57, -610, 4753, -12326, -237614, 4863632, -62825591, 614658950, -4949571887, 35267329190, -227575820550, 780036553533, 9046120663235, -300534607725610, 5137779629093346, -72047366128046665, ...].
As a logarithmic series, exponentiation yields F(x), the g.f. of A381353
F(x) = exp(x - 3*x^2/2 + 10*x^3/3 - 35*x^4/4 + 121*x^5/5 - 390*x^6/6 + 1037*x^7/7 + ...) = 1 + x - x^2 + 2*x^3 - 5*x^4 + 13*x^5 - 31*x^6 + 48*x^7 + 129*x^8 - 2035*x^9 + 12963*x^10 + ... + A381353(n)*x^n + ...
RELATED SEQUENCE A381353.
Let F(x) be the g.f. of A381353, then F(x) satisfies
[x^n] F(x)^prime(n) = 0 for n > 1,
as illustrated by the following table.
The table of coefficients of x^k in F(x)^n begins
n\k: 0 1 2 3 4 5 6
1: [1, 1, -1, 2, -5, 13, -31, ...];
2: [1, 2, -1, 2, -5, 12, -22, ...];
3: [1, 3, (0), 1, -3, 6, -1, ...];
4: [1, 4, 2, 0, -1, 0, 18, ...];
5: [1, 5, 5, (0), 0, -4, 30, ...];
6: [1, 6, 9, 2, 0, -6, 35, ...];
7: [1, 7, 14, 7, (0), -7, 35, ...];
8: [1, 8, 20, 16, 2, -8, 32, ...];
9: [1, 9, 27, 30, 9, -9, 27, ...];
10: [1, 10, 35, 50, 25, -8, 20, ...];
11: [1, 11, 44, 77, 55, (0), 11, ...];
12: [1, 12, 54, 112, 105, 24, 2, ...];
13: [1, 13, 65, 156, 182, 78, (0), ...];
...
in which the coefficient of x^n in F(x)^prime(n) equals 0 for n > 1.
PROG
(PARI) {a(n) = my(F=[1, 1], A); for(m=2, n+1, F=concat(F, 0);
F[#F] = -polcoef(Ser(F)^prime(m)/prime(m), m) ); A=Vec(Ser(F)'/Ser(F)); A[n]}
for(n=1, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
sign
AUTHOR
Paul D. Hanna, Mar 11 2025
STATUS
approved