A376109 a(n) is the length of the longest arithmetic progression ending at n consisting of numbers with the same number of prime factors as n, counted with multiplicity.
1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 2, 3, 2, 1, 3, 2, 3, 2, 3, 3, 4, 2, 2, 3, 2, 3, 5, 2, 3, 1, 3, 3, 3, 2, 2, 3, 2, 2, 4, 3, 4, 3, 2, 4, 3, 2, 3, 2, 3, 3, 5, 2, 4, 3, 3, 5, 4, 2, 3, 3, 3, 1, 3, 3, 3, 3, 4, 3, 3, 2, 3, 3, 2, 3, 3, 3, 4, 2, 2, 4, 4, 3, 3, 4, 5, 3, 3, 2, 4, 4, 4, 3, 3, 2, 4, 3, 3
Offset: 1
Keywords
Examples
a(7) = 3 because 7 is prime and there is an arithmetic progression of 3 primes, namely 3, 5, 7, ending with 7 but no such arithmetic progression of 4 primes.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
M:= Array(1..10): for n from 2 to 100 do v:= numtheory:-bigomega(n); if M[v] = 0 then M[v]:= n else M[v]:= M[v],n fi; od: for i from 1 to 10 do M[i]:= [M[i]] od: f:= proc(s) local n,i,m,d,v,j; m:= 1; v:= numtheory:-bigomega(s); member(s,M[v],n); for i from n-1 to 1 by -1 do d:= s - M[v][i]; if s - m*d < M[v][1] then return m fi; for j from 2 while ListTools:-BinarySearch(M[v],s-j*d) <> 0 do od: m:= max(m,j); od; m; end proc: f(1):= 1: map(f, [$1..100]);
Comments