A333340 -id:A333340 - cp's OEIS frontend

A333335 a(n) is the smallest positive number k such that n divides 4^k + k.

1, 2, 2, 4, 1, 2, 5, 8, 2, 4, 10, 8, 4, 6, 11, 16, 13, 2, 12, 4, 5, 10, 22, 8, 21, 4, 11, 48, 28, 14, 30, 32, 17, 16, 31, 20, 7, 12, 29, 24, 40, 26, 42, 68, 11, 22, 44, 32, 5, 44, 86, 4, 52, 38, 51, 48, 59, 28, 50, 44, 60, 30, 47, 64, 4, 68, 3, 16, 158, 94, 70

Offset: 1

Jinyuan Wang, Apr 14 2020

Brazil National Olympiad, 2005, Problem 6

Cf. A247248, A333334, A333336, A333340.

a(n) = for(k=1, oo, if(Mod(4, n)^k==-k, return(k)));

a(4^m) = 4^m for m >= 0.

A333341 a(n) is the smallest positive number k such that n divides 5^k - k.

Original entry on oeis.org

1, 1, 4, 1, 5, 5, 16, 5, 4, 5, 9, 5, 5, 17, 5, 5, 11, 11, 16, 5, 16, 9, 2, 5, 25, 5, 4, 17, 74, 5, 56, 21, 16, 11, 100, 29, 13, 101, 5, 5, 43, 17, 27, 9, 40, 61, 8, 5, 32, 25, 11, 5, 28, 29, 45, 61, 16, 149, 21, 5, 3, 63, 58, 53, 5, 47, 75, 133, 4, 145, 76, 29

Offset: 1

Jinyuan Wang, Apr 14 2020

nonn

For any positive integer n, if k = a(n) + n*m*A007736(n) and m >= 0 then 5^k - k is divisible by n.

Cf. A007736, A072872, A333336, A333339, A333340.

a(n) = for(k=1, oo, if(Mod(5, n)^k==k, return(k)));

a(5^m) = 5^m for m >= 0.

cp's OEIS Frontend

A333335 a(n) is the smallest positive number k such that n divides 4^k + k.

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