1,2

The polynomial p(n,x) is defined as the numerator when the sum 1 + 1/(n*x + 1) + 1/((n*x + 1)(n*x + 2)) + ... + 1/((n*x + 1)(n*x + 2)...(n*x + n - 1)) is written as a fraction with denominator (n*x + 1)(n*x + 2)...(n*x + n - 1). For more, see A248664.

Since p(n,x) is a sum of products of terms (n*x + i), the only coefficient which is not necessarily divisible by n is the coefficient of x^0 = A000522(n-1). On the other hand, the coefficient of x^(n-1) is n^n. Therefore n is in this sequence iff gcd(n, A000522(n-1)) = 1. - Peter J. Taylor, Apr 08 2022

Conjecture: any composite term can be represented as a product a(i)*a(j) (i > 1, j > 1) in at least one way. - Mikhail Kurkov, Apr 09 2022

Michel Marcus, Table of n, a(n) for n = 1..2000

The first six polynomials with GCD(coefficients) shown just to the right of "=":

p(1,x) = 1,

p(2,x) = 2*(x + 1),

p(3,x) = 1*(9*x^2 + 12*x + 5),

p(4,x) = 4*(16*x^3 + 28*x^2 + 17*x + 4),

p(5,x) = 5*(125*x^4 + 275*x^3 + 225*x^2 + 84*x + 13),

p(6,x) = 2*(3888*x^5 + 10368*x^4 + 10800*x^3 + 5562*x^2 + 1455*x + 163),

so that a(1) = 1 and a(2) = 3.

t[x_, n_, k_] := t[x, n, k] = Product[n*x + n - i, {i, 1, k}];

p[x_, n_] := Sum[t[x, n, k], {k, 0, n - 1}];

TableForm[Table[Factor[p[x, n]], {n, 1, 6}]]

c[n_] := c[n] = CoefficientList[p[x, n], x];

TableForm[Table[c[n], {n, 1, 10}]] (* A248664 array *)

u = Table[Apply[GCD, c[n]], {n, 1, 60}] (* A248666 *)

Flatten[Position[u, 1]] (* this sequence *)

Table[Apply[Plus, c[n]], {n, 1, 60}] (* A248668 *)

(PARI) isok(k) = gcd(k, sum(i=0, k-1, binomial(k-1, i)*i!)) == 1; \\ Michel Marcus, Mar 02 2026

Cf. A000522, A248664, A248665, A248666, A248668, A248669.

Sequence in context: A287914 A291348 A186890 * A075607 A256465 A275386

Adjacent sequences: A248664 A248665 A248666 * A248668 A248669 A248670

nonn,easy

Clark Kimberling, Oct 11 2014

approved