A381733 Number of divisors d of n such that 2^omega(n + d) = tau(n + d), where omega = A001221 and tau = A000005.
1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 3, 2, 2, 1, 1, 1, 2, 1, 3, 2, 2, 1, 2, 2, 1, 1, 4, 2, 3, 1, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 1, 1, 3, 1, 2, 1, 1, 0, 2, 1, 3, 1, 2, 2, 3, 2, 2, 1, 5, 2, 1, 2, 2, 4, 3, 1, 4, 2, 3, 1, 3, 2, 1, 1, 4, 2, 2, 1, 2, 1, 2, 1, 5, 3, 2, 1, 2, 1, 4, 1, 4, 2, 2, 2, 2, 1, 1, 2, 4
Offset: 1
Keywords
Programs
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Magma
[#[d: d in Divisors(n) | 2^#PrimeDivisors(n+d) eq #Divisors(n+d)]: n in [1..100]];
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Mathematica
a[n_]:=Length[Select[Divisors[n], DivisorSigma[0, #+n]==2^PrimeNu[#+n]&]]; Array[a,100] (* Stefano Spezia, Mar 07 2025 *)