A130335 Smallest k > 0 such that gcd(n*(n+1)/2, (n+k)*(n+k+1)/2) = 1.
1, 2, 7, 2, 2, 4, 2, 2, 4, 2, 2, 10, 2, 2, 7, 2, 2, 4, 2, 2, 4, 2, 2, 13, 2, 2, 10, 2, 2, 7, 2, 2, 4, 2, 2, 10, 2, 2, 7, 2, 2, 4, 2, 2, 7, 2, 2, 10, 2, 2, 7, 2, 2, 4, 2, 2, 4, 2, 2, 13, 2, 2, 10, 2, 2, 4, 2, 2, 4, 2, 2, 10, 2, 2, 7, 2, 2, 4, 2, 2, 4, 2, 2, 22, 2, 2, 7, 2, 2, 16, 2, 2, 4, 2, 2, 10, 2, 2, 7, 2
Offset: 1
Keywords
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
f[n_] := Block[{k = If[ n == 1 || Mod[n, 3] == 0, 1, 2]}, While[ GCD[n(n + 1)/2, (n + k)(n + k + 1)/2] != 1, k += 3 ]; k]; Array[f, 100] (* Robert G. Wilson v, Jun 03 2007 *)
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PARI
a(n) = my(k=1); while (gcd(n*(n+1)/2, (n+k)*(n+k+1)/2) != 1, k++); k;
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Python
from math import gcd def A130335(n): k, Tn, Tm = 1, n*(n+1)//2, (n+1)*(n+2)//2 while gcd(Tn,Tm) != 1: k += 1 Tm += k+n return k # Chai Wah Wu, Sep 16 2021
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