A209287 Minimal m>=0 such that prime(n)+2*m-1 has form 2^k*p, where k>=0 and p is prime.
1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 2, 1, 1, 0, 0, 0, 2, 1, 0, 2, 0, 1, 2, 0, 2, 0, 1, 1, 1, 0, 0, 0, 2, 1, 0, 3, 2, 1, 1, 0, 2, 0, 3, 2, 2, 0, 0, 0, 1, 1, 2, 1, 0, 4, 2, 1, 0, 1, 4, 0, 2, 0, 0, 1, 2, 2, 0, 0, 1, 2, 2, 2, 1, 3, 2, 4, 2, 0
Offset: 1
Keywords
Examples
Let n=7. Then prime(7)=17 and, for m=0, 17+2m-1=16=2^3*p, where p=2. Thus a(7)=0.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A074781.
Programs
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Maple
f:= proc(n) local v,m,p; p:= ithprime(n)-3; for m from 0 do p:= p+2; v:= p/2^padic:-ordp(p,2); if v=1 or isprime(v) then return m fi od; end proc: f(1):= 1: map(f, [$1..100]); # Robert Israel, Mar 18 2019
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Mathematica
good[n_] := Module[{k = n/2^IntegerExponent[n, 2]}, n > 1 && (k == 1 || PrimeQ[k])]; Table[p = Prime[n]; m = 0; While[! good[p + 2*m - 1], m++]; m, {n, 87}] (* T. D. Noe, Feb 26 2013 *)
Extensions
More terms from T. D. Noe, Feb 26 2013
Comments