A051712 Numerator of b(n)-b(n+1), where b(n) = n/((n+1)(n+2)) = A026741/A045896.

0, 1, 1, 1, 1, 5, 1, 7, 1, 3, 5, 11, 1, 13, 7, 5, 2, 17, 3, 19, 5, 7, 11, 23, 1, 25, 13, 9, 7, 29, 5, 31, 4, 11, 17, 35, 3, 37, 19, 13, 5, 41, 7, 43, 11, 15, 23, 47, 2, 49, 25, 17, 13, 53, 9, 55, 7, 19, 29, 59, 5, 61, 31, 21, 8, 65, 11, 67, 17, 23, 35, 71, 3, 73

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• frac
• easy
• mult

Examples

			0, 1/60, 1/60, 1/70, 1/84, 5/504, 1/120, 7/990, 1/165, 3/572,...
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Masanobu Kaneko, The Akiyama-Tanigawa algorithm for Bernoulli numbers, J. Integer Sequences, 3 (2000), Article 00.2.9.

Crossrefs

Cf. A026741, A045896, A051713.

Row 3 of table in A051714/A051715.

Programs

Mathematica

b[n_] := n/((n + 1) (n + 2)); Numerator[-Differences[Array[b, 100]]]
(* or *)
f[p_, e_] := p^e; f[2, e_] := If[e < 3, 1, 2^(e - 3)]; f[3, e_] := 3^(e - 1); a[1] = 0; a[n_] := Times @@ f @@@ FactorInteger[n - 1]; Array[a, 100] (* Amiram Eldar, Nov 20 2022 *)

Formula

c(n) = a(n+1) is multiplicative with c(2^e) = 2^(e-3) if e > 2 and 1 otherwise, c(3^e) = 3^(e-1), and c(p^e) = p^e if p >= 5. [corrected by Amiram Eldar, Nov 20 2022]

Sum_{k=1..n} a(k) ~ (301/1152) * n^2. - Amiram Eldar, Nov 20 2022