A343932 a(n) = (Sum_{k=1..n} k^k) mod n.
0, 1, 2, 0, 3, 5, 5, 4, 1, 7, 3, 4, 11, 13, 3, 4, 0, 15, 0, 4, 14, 13, 10, 20, 22, 11, 25, 20, 21, 1, 18, 4, 6, 17, 27, 12, 31, 27, 20, 28, 6, 41, 34, 32, 31, 45, 45, 4, 11, 25, 39, 48, 21, 45, 46, 12, 53, 47, 41, 32, 9, 5, 55, 4, 25, 7, 47, 8, 45, 19, 12, 60, 50, 43, 20, 60, 54, 29, 72, 36, 70, 31, 74, 40, 69, 7, 18, 20, 63, 3, 24, 32
Offset: 1
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
a[n_] := Mod[Sum[PowerMod[k, k, n], {k, 1, n}], n]; Array[a, 100] (* Amiram Eldar, May 04 2021 *)
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PARI
a(n) = sum(k=1, n, k^k)%n;
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Python
def A343932(n): return sum(pow(k,k,n) for k in range(1,n+1)) % n # Chai Wah Wu, Jun 18 2022
Formula
a(n) = A001923(n) mod n.