

A214057


Least m>0 such that 2^n1+m and nm have a common divisor > 1.


3



1, 2, 1, 4, 1, 3, 1, 8, 1, 10, 1, 3, 1, 14, 1, 16, 1, 3, 1, 5, 1, 2, 1, 3, 1, 26, 1, 28, 1, 3, 1, 4, 1, 6, 1, 2, 1, 15, 1, 5, 1, 2, 1, 5, 1, 17, 1, 3, 1, 50, 1, 8, 1, 2, 1, 56, 1, 58, 1, 3, 1, 2, 1, 6, 1, 3, 1, 31, 1, 70, 1, 3, 1, 4, 1, 6, 1, 3, 1, 5, 1, 2, 1
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OFFSET

1,2


LINKS

Clark Kimberling, Table of n, a(n) for n = 1..1000


EXAMPLE

gcd(2^611,61) = gcd(62,5) = 1
gcd(2^612,62) = gcd(61,4) = 1
gcd(2^613,63) = gcd(60,3) = 3,
so that a(6) = 3.


MATHEMATICA

b[n_] := 2^n1; c[n_] := n;
Table[m = 1; While[GCD[b[n] + m, c[n]  m] == 1, m++]; m, {n, 1, 100}]


CROSSREFS

Cf. A214056.
Sequence in context: A127461 A329900 A101261 * A067614 A341308 A113901
Adjacent sequences: A214054 A214055 A214056 * A214058 A214059 A214060


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Jul 22 2012


STATUS

approved



