A097702 a(n) = (A063880(n) - 108)/216.

0, 2, 3, 5, 6, 8, 9, 11, 14, 15, 17, 18, 20, 21, 23, 26, 27, 29, 30, 32, 33, 35, 36, 38, 39, 41, 42, 44, 45, 47, 48, 50, 51, 53, 54, 56, 57, 59, 63, 65, 66, 68, 69, 71, 72, 74, 75, 77, 78, 80, 81, 83, 86, 89, 90, 92, 93, 95, 96, 98, 99, 101, 102, 104, 105, 107, 108, 110

Offset

1,2

Comments

Conjecture: k is a term iff 6*k+3 is squarefree. - Vladeta Jovovic, Aug 27 2004

It is only a conjecture that all terms are integers (confirmed up to 10^6 by Robert G. Wilson v).

From Amiram Eldar, Aug 31 2024: (Start)

The first conjecture is true. If m = 216*k + 108 = 108 * (2*k + 1) is a term of A063880, then 2*k+1 is a squarefree number coprime to 6. This is because sigma(n)/usigma(n) is multiplicative, equals 1 if and only if n is squarefree and larger than 1 otherwise, sigma(108)/usigma(108) = 2 and sigma(3^k)/usigma(3^k) increases with k. 6*k+3 = 3*(2*k+1) is squarefree because 2*k+1 is a squarefree coprime to 6.

Assuming that (A063880(n) - 108)/216 is an integer for all n, we have a(n) = (A276378(n) - 1)/2. (End)

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Mathematica

usigma[n_] := Block[{d = Divisors[n]}, Plus @@ Select[d, GCD[ #, n/# ] == 1 &]]; (Select[ Range[ 24500], DivisorSigma[1, # ] == 2usigma[ # ] &] - 108)/216 (* Robert G. Wilson v, Aug 28 2004 *)

Prog

(PARI) {u(n)=sumdiv(n, d, if(gcd(d, n/d)==1, d))}
n=2; while(n<50000, n++; if(sigma(n)==2*u(n), print1((n-108)/216", ")))

Crossrefs

Cf. A000203, A005117, A034448, A063880, A097703, A276378.

Sequence in context: A094820 A309793 A093689 * A347912 A082583 A274332

Adjacent sequences: A097699 A097700 A097701 * A097703 A097704 A097705

Keyword

nonn

Author

Ralf Stephan, Aug 26 2004

Status

approved