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A074944
Number of k with 1 <= k <= n such that gcd(n,k) = tau(k), where tau is A000005, number of divisors function.
1
1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 4, 1, 2, 2, 2, 1, 3, 1, 3, 2, 2, 1, 4, 1, 2, 1, 3, 1, 5, 1, 3, 2, 2, 1, 3, 1, 2, 2, 3, 1, 5, 1, 3, 2, 2, 1, 5, 1, 2, 2, 3, 1, 3, 1, 4, 2, 2, 1, 5, 1, 2, 2, 5, 1, 5, 1, 3, 2, 2, 1, 5, 1, 2, 2, 3, 1, 5, 1, 4, 2, 2, 1, 7, 1, 2, 2, 5, 1, 4, 1, 3, 2, 2, 1, 9, 1, 2, 2, 3, 1, 5, 1, 6, 2
OFFSET
1,2
FORMULA
Sum_{i=1..n} a(i) seems to be asymptotic to c*n*log(n) with 0.5 < c < 0.6.
The above conjecture seems to be wrong. See the graph in the Links section. - Amiram Eldar, Mar 10 2026
MATHEMATICA
a[n_] := Sum[If[GCD[n, k] == DivisorSigma[0, k], 1, 0], {k, 1, n}]; Array[a, 105] (* Amiram Eldar, Mar 10 2026 *)
PROG
(PARI) a(n)=sum(k=1, n, if(gcd(n, k)-numdiv(k), 0, 1))
CROSSREFS
Cf. A000005.
Sequence in context: A161288 A185217 A131456 * A245041 A161315 A161249
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, Oct 05 2002
STATUS
approved