A328099 a(n) = min(A003415(n), A276086(n)).

0, 1, 1, 4, 1, 5, 1, 12, 6, 7, 1, 16, 1, 9, 8, 32, 1, 21, 1, 24, 10, 13, 1, 44, 10, 15, 27, 32, 1, 7, 1, 21, 14, 19, 12, 35, 1, 21, 16, 68, 1, 41, 1, 48, 39, 25, 1, 112, 14, 45, 20, 56, 1, 81, 16, 92, 22, 31, 1, 49, 1, 33, 51, 192, 18, 61, 1, 72, 26, 59, 1, 156, 1, 39, 55, 80, 18, 71, 1, 176, 108, 43, 1, 124, 22, 45, 32, 140, 1, 123, 20, 96, 34, 49, 24

Offset: 1

Author

Antti Karttunen, Oct 07 2019

Keywords

• nonn

Comments

Always choosing the lesser of A003415(n) and A276086(n) is often a good heuristic when trying to find the shortest path to zero. However, this doesn't always guarantee the optimal result. E.g., if we define b(0) = 0; and for n > 0, b(n) = 1+(a(n)), then we have b(8) = 8 > A327969(8) = 6, b(12) = 7 > A327969(12) = 5, and b(15) = 9 > A327969(15) = 6.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A003415, A048103, A129251, A276085, A276086, A327858, A327859, A327928, A327929, A327963, A327965, A327969, A328097, A328098, A328112.

Programs

PARI

A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A276086(n) = { my(i=0,m=1,pr=1,nextpr); while((n>0),i=i+1; nextpr = prime(i)*pr; if((n%nextpr),m*=(prime(i)^((n%nextpr)/pr));n-=(n%nextpr));pr=nextpr); m; };
A328099(n) = min(A003415(n),A276086(n));