cp's OEIS Frontend

A385200 The sum of the exponents e for the integers k from 1 to n such that the greatest divisor of k that is a unitary divisor of n is a prime power p^e.

%I A385200 #8 Jun 22 2025 16:33:21
%S A385200 0,1,1,2,1,3,1,3,2,5,1,7,1,7,6,4,1,10,1,11,8,11,1,13,2,13,3,15,1,14,1,
%T A385200 5,12,17,10,22,1,19,14,19,1,20,1,23,16,23,1,23,2,26,18,27,1,29,14,25,
%U A385200 20,29,1,34,1,31,20,6,16,32,1,35,24,34,1,38,1,37,28
%N A385200 The sum of the exponents e for the integers k from 1 to n such that the greatest divisor of k that is a unitary divisor of n is a prime power p^e.
%H A385200 Amiram Eldar, <a href="/A385200/b385200.txt">Table of n, a(n) for n = 1..10000</a>
%F A385200 a(n) = uphi(n) * Sum_{p^e || n} (e/(p^e-1)), where uphi = A047994, and p^e || n denotes that the prime power p^e unitarily divides n (i.e., p^e divides n but p^(e+1) does not divide n).
%F A385200 Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = c1 * c2 = 0.31889766457764592387..., c1 = Product_{p prime}(1 - 1/(p*(p+1))) = A065463, and c2 = Sum_{p prime}(p^2/((p^2-1)*(p^2+p-1))) = 0.45269528731431531046... .
%e A385200 For n = 12 and the integers k from 1 to 12, the greatest divisor of k that is a unitary divisor of 12 are 1, 1, 3, 4, 1, 3, 1, 4, 3, 1, 1 and 12, respectively. The prime powers are 3 = 3^1, 4 = 2^2, 3 = 3^1, 4 = 2^2 and 3 = 3^1, and the sum of the exponents is a(12) = 1 + 2 + 1 + 2 + 1 = 7.
%t A385200 a[n_] := Module[{f = FactorInteger[n], p, e}, p = f[[;; , 1]]; e = f[[;; , 2]]; Times @@ (p^e-1) * Total[e/(p^e-1)]]; a[1] = 0; Array[a, 100]
%o A385200 (PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^f[i,2]-1) * sum(i = 1, #f~, f[i,2]/(f[i,1]^f[i,2] - 1));}
%Y A385200 The unitary analog of A122410.
%Y A385200 Cf. A047994, A065463, A077610, A246655, A384047, A385198.
%K A385200 nonn,easy
%O A385200 1,4
%A A385200 _Amiram Eldar_, Jun 21 2025