A363331 a(n) is the sum of divisors of n that are both coreful and infinitary.
1, 2, 3, 4, 5, 6, 7, 14, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 42, 25, 26, 39, 28, 29, 30, 31, 50, 33, 34, 35, 36, 37, 38, 39, 70, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 78, 55, 98, 57, 58, 59, 60, 61, 62, 63, 84, 65, 66, 67, 68
Offset: 1
Examples
a(8) = 14 since 8 has 3 divisors that are both infinitary and coreful, 2, 4 and 8, and 2 + 4 + 8 = 14.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
f[p_, e_] := Times @@ (1 + Flatten[p^(2^(-1 + Position[Reverse@ IntegerDigits[e, 2], ?(# == 1 &)]))]) - 1; a[1] = 1; a[n] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
-
PARI
a(n) = {my(f = factor(n), b); prod(i = 1, #f~, b = binary(f[i, 2]); prod(k = 1, #b, if(b[k], f[i, 1]^(2^(#b - k)) + 1, 1)) - 1);}
Formula
Multiplicative with a(p^e) = (Product_{k>=0} (p^(2^k*(b(k)+1)) - 1)/(p^(2^k) - 1)) - 1, where e = Sum_{k >= 0} b(k) * 2^k is the binary representation of e.
a(n) >= n, with equality if and only if n is in A138302.
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} ((1 - 1/p) * Sum_{k>=1} a(p^k)/p^(2*k)) = 0.53906337497505398777... .
Comments