A348507 a(n) = A003959(n) - n, where A003959 is multiplicative with a(p^e) = (p+1)^e.

0, 1, 1, 5, 1, 6, 1, 19, 7, 8, 1, 24, 1, 10, 9, 65, 1, 30, 1, 34, 11, 14, 1, 84, 11, 16, 37, 44, 1, 42, 1, 211, 15, 20, 13, 108, 1, 22, 17, 122, 1, 54, 1, 64, 51, 26, 1, 276, 15, 58, 21, 74, 1, 138, 17, 160, 23, 32, 1, 156, 1, 34, 65, 665, 19, 78, 1, 94, 27, 74, 1, 360, 1, 40, 69, 104, 19, 90, 1, 406, 175, 44, 1, 204

Offset: 1

Author

Antti Karttunen, Oct 30 2021

Keywords

• nonn

Comments

a(p*(n/p)) - (n/p) = (p+1)*a(n/p) holds for all prime divisors p of n, which can be seen by expanding the left hand side as (A003959(p*(n/p)) - (p*(n/p))) - (n/p) = (p+1)*A003959(n/p)-((p+1)*(n/p)) = (p+1)*(A003959(n/p)-(n/p)) = (p+1)*a(n/p). This implies that a(n) >= A003415(n) for all n. (See also comments in A348970). - Antti Karttunen, Nov 06 2021

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Crossrefs

Cf. A001065, A003415, A003959, A020639, A032742, A065488, A348029, A348508, A348929 [= gcd(n,a(n))], A348950, A348970.

Cf. A348971 (Möbius transform) and A349139, A349140, A349141, A349142, A349143 (other Dirichlet convolutions).

Cf. also A168065 (the arithmetic mean of this and A322582), A168066.

Programs

Mathematica

f[p_, e_] := (p + 1)^e; a[1] = 0; a[n_] := Times @@ f @@@ FactorInteger[n] - n; Array[a, 100] (* Amiram Eldar, Oct 30 2021 *)

PARI

A003959(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1]++); factorback(f); };
A348507(n) = (A003959(n) - n);

PARI

A020639(n) = if(1==n,n,(factor(n)[1, 1]));
A348507(n) = { my(s=0, m=1, spf); while(n>1, spf = A020639(n); n /= spf; s += m*n; m *= (1+spf)); (s); }; \\ (Compare this with similar programs given in A003415 and in A322582) - Antti Karttunen, Nov 06 2021

Formula

a(n) = A003959(n) - n.

a(n) = A348508(n) + n.

a(n) = A001065(n) + A348029(n).

From Antti Karttunen, Nov 06 2021: (Start)

a(n) = Sum_{d|n} A348971(d).

a(n) = A003415(n) + A348970(n).

For all n >= 1, A322582(n) <= A003415(n) <= a(n).

For n > 1, a(n) = a(A032742(n))*(1+A020639(n)) + A032742(n). [See the comments above, and compare this with Reinhard Zumkeller's May 09 2011 recursive formula for A003415] (End)

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = A065488 - 1. - Amiram Eldar, Jun 01 2025