A125073 a(n) = sum of the exponents in the prime factorization of n which are triangular numbers.
0, 1, 1, 0, 1, 2, 1, 3, 0, 2, 1, 1, 1, 2, 2, 0, 1, 1, 1, 1, 2, 2, 1, 4, 0, 2, 3, 1, 1, 3, 1, 0, 2, 2, 2, 0, 1, 2, 2, 4, 1, 3, 1, 1, 1, 2, 1, 1, 0, 1, 2, 1, 1, 4, 2, 4, 2, 2, 1, 2, 1, 2, 1, 6, 2, 3, 1, 1, 2, 3, 1, 3, 1, 2, 1, 1, 2, 3, 1, 1, 0, 2, 1, 2, 2, 2, 2, 4, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 0, 1, 3, 1, 4, 3
Offset: 1
Examples
The prime factorization of 360 is 2^3 *3^2 *5^1. There are two exponents in this factorization which are triangular numbers, 1 and 3. So a(360) = 1 + 3 = 4.
Links
Programs
-
Mathematica
f[n_] := Plus @@ Select[Last /@ FactorInteger[n], IntegerQ[Sqrt[8# + 1]] &];Table[f[n], {n, 110}] (* Ray Chandler, Nov 19 2006 *)
-
PARI
A010054(n) = issquare(8*n + 1); \\ This function from Michael Somos, Apr 27 2000. A125073(n) = vecsum(apply(e -> (A010054(e)*e), factorint(n)[, 2])); \\ Antti Karttunen, Jul 08 2017
Formula
Additive with a(p^e) = A010054(e)*e. - Antti Karttunen, Jul 08 2017
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = -P(2) + Sum_{k>=2} (k*(k+1)/2) * (P(k*(k+1)/2) - P(k*(k+1)/2 + 1)) = -0.10099019472003733178..., where P(s) is the prime zeta function. - Amiram Eldar, Sep 28 2023
Extensions
Extended by Ray Chandler, Nov 19 2006