OFFSET
1,4
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
FORMULA
a(n) = A003959(n) - n.
a(n) = A348508(n) + n.
From Antti Karttunen, Nov 06 2021: (Start)
a(n) = Sum_{d|n} A348971(d).
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)
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 *)
PROG
(PARI)
A003959(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1]++); factorback(f); };
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Oct 30 2021
STATUS
editing