A065925 Smallest k such that sopf(n+k) = sopf(k), where sopf = A008472.
5, 2, 7, 4, 114, 2, 5, 8, 13, 10, 25, 4, 5, 2, 19, 16, 85, 6, 5, 5, 209, 22, 25, 3, 493, 26, 31, 4, 20, 2, 5, 32, 7, 34, 516, 12, 33, 38, 10, 10, 99, 6, 5, 44, 57, 46, 25, 6, 5, 50, 49, 52, 52, 18, 855, 8, 61, 58, 295, 4, 261, 2, 91, 64, 602, 6, 5, 68, 21, 10, 25, 9, 7, 74, 13, 76
Offset: 1
Keywords
Examples
a(6) = 2 because A008472(2) = A008472(6+2) = 2, but A008472(1) = 0 doesn't equal A008472(6+1) = 7.
References
- J. Earls, Mathematical Bliss, Pleroma Publications, 2009, pages 99-100. ASIN: B002ACVZ6O [From Jason Earls, Nov 26 2009]
Links
- Harry J. Smith, Table of n, a(n) for n=1..1000
- Carlos Rivera, Conjecture 25. sopf(n) = sopf(n+k), The Prime Puzzles and Problems Connection.
Programs
-
Mathematica
Table[k = 1; While[Total[FactorInteger[n + k][[All, 1]]] != Total[FactorInteger[k][[All, 1]]], k++]; k, {n, 76}] (* Michael De Vlieger, Jan 11 2017 *) -
PARI
sopf(n) = local(fac, i); fac=factor(n); sum(i=1,matsize(fac)[1],fac[i,1]) A065925(m)={local(k,n); for(k=1,m,n=1; while(sopf(n)!=sopf(n+k), n++); print1(n,","))} \\ Klaus Brockhaus
-
Python
from sympy import primefactors from itertools import count, dropwhile def sopf(n): return sum(p for p in primefactors(n)) def a(n): k = 1 while sopf(n+k) != sopf(k): k += 1 return k print([a(n) for n in range(1, 77)]) # Michael S. Branicky, May 02 2021