A325973 Arithmetic mean of {sum of unitary divisors} and {sum of squarefree divisors}: a(n) = (1/2) * (A034448(n) + A048250(n)).

1, 3, 4, 4, 6, 12, 8, 6, 7, 18, 12, 16, 14, 24, 24, 10, 18, 21, 20, 24, 32, 36, 24, 24, 16, 42, 16, 32, 30, 72, 32, 18, 48, 54, 48, 31, 38, 60, 56, 36, 42, 96, 44, 48, 42, 72, 48, 40, 29, 48, 72, 56, 54, 48, 72, 48, 80, 90, 60, 96, 62, 96, 56, 34, 84, 144, 68, 72, 96, 144, 72, 51, 74, 114, 64, 80, 96, 168, 80, 60, 43, 126

Offset: 1

Author

Antti Karttunen, Jun 02 2019

Keywords

• nonn

Comments

This is not multiplicative: a(4) = 4, a(9) = 7, but a(36) = 31, not 28. However, the function acts multiplicatively on certain subsequences of natural numbers, like for example when restricted to A048107, where this sequence coincides with A326043.

Examples

			For n = 36, its divisors are 1, 2, 3, 4, 6, 9, 12, 18, 36. Of these, unitary divisors are 1, 4, 9 and 36, so A034448(36) = 1+4+9+36 = 50, while the squarefree divisors are 1, 2, 3 and 6, so A048250(36) = 1+2+3+6 = 12, thus a(36) = (50+12)/2 = 31.
For n = 495, its divisors are 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495. Of these, unitary are 1, 5, 9, 11, 45, 55, 99, 495, whose sum is A034448(495) = 720, while the squarefree divisors are 1, 3, 5, 11, 15, 33, 55, 165, and their sum is A048250(495) = 288. Thus a(495) = (720+288)/2 = 504. Also for 495, whose prime factorization is 3^2 * 5^1 * 11^1 this can be computed faster as the average of ((3^2)+1)*(5+1)*(11+1) and (3+1)*(5+1)*(11+1), thus (1/2)*(3+(3^2)+2)*(5+1)*(11+1) = 504.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A000203, A034448, A048107, A048250, A052548, A289521, A306633, A325974, A325975, A325977, A325978, A325981, A326043.

Programs

Mathematica

Array[(1/2) If[# == 1, 2, Times @@ (1 + Power @@@ #) + Times @@ (1 + #[[;; , 1]]) &@ FactorInteger[#]] &, 90] (* Michael De Vlieger, Jun 06 2019, after Giovanni Resta at A034448 and Amiram Eldar at A048250. *)

PARI

A034448(n) = { my(f=factorint(n)); prod(k=1, #f~, 1+(f[k, 1]^f[k, 2])); }; \\ After code in A034448
A048250(n) = factorback(apply(p -> p+1,factor(n)[,1]));
A325973(n) = ((A034448(n)+A048250(n))/2);

PARI

A325973(n) = (1/2)*sumdiv(n, d, d*(issquarefree(d) + (1==gcd(d, n/d))));

Formula

a(n) = (1/2) * (A034448(n) + A048250(n)).

a(n) = A000203(n) - A325974(n).

a(n) = n + A325977(n).

a(A048107(n)) = A326043(A048107(n)).

For n >= 1, a(2^n) = A052548(n-1) = 2^(n-1) + 2.

For n >= 1, a(3^n) = A289521(n) = (3^n + 5)/2.

Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(2)/zeta(3) + 1)/4 = 0.5921081944... . - Amiram Eldar, Feb 22 2024