A357879 Number of divisors of n with the same sum of prime indices as their quotient. Central column of A321144, taking gaps as 0's.
1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2
Offset: 1
Keywords
Examples
The a(3600) = 5 divisors, their prime indices, and the prime indices of their quotients: 45: {2,2,3} * {1,1,1,1,3} 50: {1,3,3} * {1,1,1,2,2} 60: {1,1,2,3} * {1,1,2,3} 72: {1,1,1,2,2} * {1,3,3} 80: {1,1,1,1,3} * {2,2,3}
Links
Crossrefs
Programs
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Mathematica
sumprix[n_]:=Total[Cases[FactorInteger[n],{p_,k_}:>k*PrimePi[p]]]; Table[Length[Select[Divisors[n],sumprix[#]==sumprix[n]/2&]],{n,100}]
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PARI
A056239(n) = if(1==n, 0, my(f=factor(n)); sum(i=1, #f~, f[i, 2] * primepi(f[i, 1]))); A357879(n) = sumdiv(n,d, A056239(d)==A056239(n/d)); \\ Antti Karttunen, Jan 20 2025
Formula
a(n) = Sum_{d|n} [A056239(d) = A056239(n/d)], where [ ] is the Iverson bracket. - Antti Karttunen, Jan 20 2025
Extensions
Data section extended to a(108) by Antti Karttunen, Jan 20 2025
Comments