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A235138
a(n) = Sum_{k=1..n} k^phi(n) (mod n) where phi(n) = A000010(n).
4
0, 1, 2, 2, 4, 1, 6, 4, 6, 3, 10, 2, 12, 5, 7, 8, 16, 3, 18, 6, 11, 9, 22, 4, 20, 11, 18, 10, 28, 29, 30, 16, 19, 15, 23, 6, 36, 17, 23, 12, 40, 1, 42, 18, 21, 21, 46, 8, 42, 15, 31, 22, 52, 9, 39, 20, 35, 27, 58, 58, 60, 29, 33, 32, 47, 5, 66, 30, 43, 11, 70, 12, 72, 35, 35, 34, 59, 7, 78, 24, 54, 39, 82, 2, 63, 41, 55, 36, 88, 87, 71, 42, 59, 45, 71, 16, 96, 35, 57, 30
OFFSET
1,3
COMMENTS
a(n) = n-1 if and only if n is prime or is a Giuga number A007850.
a(n) = 1 if (and probably only if) n is a primary pseudoperfect number A054377.
a(2^k*p) = 2^(k-1)*p-2^k if p is an odd prime. - Robert Israel, Apr 25 2017
LINKS
Jonathan Sondow and K. MacMillan, Reducing the Erdos-Moser equation 1^n + 2^n + . . . + k^n = (k+1)^n modulo k and k^2, Integers 11 (2011), #A34.
Jonathan Sondow and E. Tsukerman, The p-adic order of power sums, the Erdos-Moser equation, and Bernoulli numbers, arXiv:1401.0322 [math.NT], 2014; see section 4.
Wikipedia, Giuga number
FORMULA
a(n) = A235137(n) (mod n).
Conjecture : a(n) = Sum_{d|n} phi(n/d)*d^phi(n) (mod n). - Ridouane Oudra, Feb 17 2024
EXAMPLE
a(4) = 30 (mod 4) = 2 since 1^(phi(4)) + 2^(phi(4)) + 3^(phi(4)) + 4^(phi(4))= 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30.
MAPLE
f:= proc(n) local q; q:= numtheory:-phi(n);
add(k&^q, k=1..n) mod n
end proc:
map(f, [$1..100]); # Robert Israel, Apr 25 2017
MATHEMATICA
a[n_] := Mod[Sum[PowerMod[i, EulerPhi@n, n], {i, n}], n]
PROG
(PARI) a(n)=my(p=eulerphi(n)); sum(k=1, n, k^p) \\ Charles R Greathouse IV, Jan 04 2014
CROSSREFS
KEYWORD
nonn,look
AUTHOR
Jonathan Sondow and Emmanuel Tsukerman, Jan 03 2014
STATUS
approved