A182398 a(n) = (Sum_{k=1..2n} k^2n) mod 2n.

1, 2, 1, 4, 5, 2, 7, 8, 3, 6, 11, 4, 13, 14, 5, 16, 17, 6, 19, 12, 1, 22, 23, 8, 25, 26, 9, 28, 29, 58, 31, 32, 11, 34, 35, 12, 37, 38, 13, 24, 41, 2, 43, 44, 15, 46, 47, 16, 49, 30, 17, 52, 53, 18, 45, 56, 19, 58, 59, 116, 61, 62, 3, 64, 65, 22, 67, 68, 23

Offset: 1

Author

Michel Lagneau, Apr 27 2012

Keywords

• nonn

Comments

Sum_{k=1..n} k^n (mod n) = 0 if n odd.

Properties of this sequence:

a(n) = 1 for n = 1, 3, 21, 903, ...

a(n) = n if n not divisible by 3;

a(3*n) = n except for n = 7, 10, 14, 20, 21, 26, 28, 30, 35, ...

a(21*n) = n, except for n = 10, 20, 26, 30, 40, 43, 50, 52, ...

a(903*n) = n, except for n = 10, ....

It appears that a(A007018(n)/2) = 1 and conjecturally a(m*A007018(n)/2) = m for a majority of value m.

No, a(A007018(n)/2) <> 1 for n > 4. (For example, a(A007018(5)/2) = a(1631721) = 1807.) - Jonathan Sondow, Oct 18 2013

0 < a(n) < 10 for n: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 18, 21, 24, 27, 42, 63, 84, 105, 126, 147, 168, 189, 903, 1806, 2709, 3612, 4515, 5418, 6321, 7224, 8127, .... Search limit was 25000. - Robert G. Wilson v, Jun 18 2015

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000 (terms 1..2499 from Michel Lagneau)
• J. M. Grau, A. M. Oller-Marcen, and J. Sondow, On the congruence 1^m + 2^m + ... + m^m == n (mod m) with n|m, arXiv 1309.7941, 2013.
• Mathematics Stack Exchange, 1^n + 2^n + ... + (p-1)^n mod p = ?

Crossrefs

Cf. A007018, A014117, A031971, A229307, A229311.

Programs

Maple

for n from 1 to 100 do: s:=sum('k^(2*n)', 'k'=1..2*n)
: x:=irem(s,2*n): printf(`%d, `,x):od:
# second Maple program:
a:= n-> add(k&^(2*n) mod (2*n), k=1..2*n) mod (2*n):
seq(a(n), n=1..100);

Mathematica

Table[Mod[Total[PowerMod[Range[2*n], 2*n, 2*n]], 2*n], {n, 100}] (* T. D. Noe, Apr 28 2012 *)

Formula

a(n) = A031971(2n) mod 2n. - Jonathan Sondow, Oct 18 2013