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A248667 Numbers k for which coefficients of the polynomial p(k,x) defined in Comments are relatively prime.

%I A248667 #40 Apr 22 2024 21:43:51
%S A248667 1,3,7,9,11,17,19,21,23,27,29,31,33,41,43,47,49,51,53,57,59,61,63,67,
%T A248667 69,71,73,77,79,81,83,87,89,93,97,99,101,103,107,109,113,119,121,123,
%U A248667 127,129,131,133,137,139,141,147,149,151,153,157,159,161,163,167
%N A248667 Numbers k for which coefficients of the polynomial p(k,x) defined in Comments are relatively prime.
%C A248667 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.
%C A248667 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
%C A248667 From _Mikhail Kurkov_, Apr 09 2022: (Start)
%C A248667 False conjecture (which still gives many correct values): {b(n)} is a subsequence of {a(n)} where {b(n)} are the numbers m for which Sum(abs(Moebius(p_j+1))) = 0 with m = Product(p_j^k_j). This conjecture was disproved by _Peter J. Taylor_. The first counterexample, i.e., the smallest m which belongs to {b(n)} and does not belong to {a(n)}, is m = 463. All other counterexamples computed up to 2.5*10^4 have the form 463*b(n). Are there any other numbers q such that q and q*b(n) are counterexamples for any n > 0? [verification needed]
%C A248667 Conjecture: any composite a(n) can be represented as a product a(i)*a(j) (i > 1, j > 1) in at least one way. (End)
%e A248667 The first six polynomials with GCD(coefficients) shown just to the right of "=":
%e A248667 p(1,x) = 1
%e A248667 p(2,x) = 2*(x + 1)
%e A248667 p(3,x) = 1*(9x^2 + 12 x +  5)
%e A248667 p(4,x) = 4*(16 x^3 + 28 x^2 + 17 x + 4)
%e A248667 p(5,x) = 5*(125 x^4 + 275 x^3 + 225 x^2 + 84 x + 13)
%e A248667 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 A248667 t[x_, n_, k_] := t[x, n, k] = Product[n*x + n - i, {i, 1, k}];
%t A248667 p[x_, n_] := Sum[t[x, n, k], {k, 0, n - 1}];
%t A248667 TableForm[Table[Factor[p[x, n]], {n, 1, 6}]]
%t A248667 c[n_] := c[n] = CoefficientList[p[x, n], x];
%t A248667 TableForm[Table[c[n], {n, 1, 10}]] (* A248664 array *)
%t A248667 u = Table[Apply[GCD, c[n]], {n, 1, 60}]  (* A248666 *)
%t A248667 Flatten[Position[u, 1]]  (* this sequence *)
%t A248667 Table[Apply[Plus, c[n]], {n, 1, 60}] (* A248668 *)
%Y A248667 Cf. A000522, A248664, A248665, A248666, A248668, A248669.
%K A248667 nonn,easy
%O A248667 1,2
%A A248667 _Clark Kimberling_, Oct 11 2014