A381613 If n = Product (p_j^k_j) then a(n) = Product (min(p_j, k_j)), with a(1) = 1.
1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 3, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 2, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1
Offset: 1
Examples
a(18) = 2 because 18 = 2^1*3^2, min(2,1) = 1, min(3,2) = 2 and 1*2 = 2. a(300) = 4 because 300 = 2^2*3^1*5^2, min(2,2) = 2, min(3,1) = 1, min(5,2) = 2 and 2*1*2 = 4.
Links
- Paolo Xausa, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
A381613[n_] := Times @@ Min @@@ FactorInteger[n]; Array[A381613, 100]
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PARI
a(n) = my(f=factor(n)); prod(i=1, #f~, min(f[i,1], f[i,2])); \\ Michel Marcus, Mar 02 2025
Formula
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + (1/p - 1/p^p)/(p-1)) = 1.59383299054679951264... . - Amiram Eldar, Mar 07 2025
Comments