A333849 a(n) = gcd(A333848(n), 2*(2*n+1)), for n >= 0.
2, 1, 2, 1, 1, 1, 2, 2, 2, 1, 6, 1, 2, 1, 2, 1, 2, 2, 2, 6, 2, 1, 2, 1, 1, 2, 2, 10, 6, 1, 2, 6, 2, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 6, 2, 2, 2, 2, 1, 6, 1, 2, 6, 2, 2, 6, 2, 1, 2, 2, 1, 6, 1, 2, 2, 2, 1, 2, 2, 2, 6, 2, 1, 2, 10, 2, 2, 2, 1, 10, 1, 2, 18, 2, 2, 2, 1, 2
Offset: 0
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..10000
- Wolfdieter Lang, On the Equivalence of Three Complete Cyclic Systems of Integers, arXiv:2008.04300 [math.NT], 2020.
Programs
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Mathematica
{2}~Join~Table[GCD[Total@ Select[Range[1, m, 2], GCD[#, m] == 1 &], 2 m], {m, Array[2 # + 1 &, 85]}] (* Michael De Vlieger, Oct 15 2020 *)
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PARI
f(n) = if (n==0, 0, my(m=2*n+1); vecsum(select(x->((gcd(m, x)==1) && (x%2)), [1..m]))); \\ A333848 a(n) = gcd(f(n), 2*(2*n+1)); \\ Michel Marcus, May 05 2020
Formula
a(n) = gcd(A333848(n), 2*(2*n+1)), for n >= 0.
Comments