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Amiram Eldar, <a href="/A051712/b051712.txt">Table of n, a(n) for n = 1..10000</a>
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]
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]
bc(n) = a(n+1) is multiplicative with bc(2^e) = 1 2^(e-3) if p = e > 2, b and 1 otherwise, c(3^e) = p3^(e-1), 3, and c(p^e) = p^e if p >= 5. [corrected by _Amiram Eldar_, Nov 20 2022]
M. Masanobu Kaneko, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL3/KANEKO/AT-kaneko.html">The Akiyama-Tanigawa algorithm for Bernoulli numbers</a>, J. Integer Sequences, 3 (2000), #Article 00.2.9.
Sum_{k=1..n} a(k) ~ (301/1152) * n^2. - Amiram Eldar, Nov 20 2022
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 *)
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M. Kaneko, <a href="httphttps://www.cs.uwaterloo.ca/journals/JIS/indexVOL3/KANEKO
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