A055030 -id:A055030 - cp's OEIS frontend

A071871 Duplicate of A055030.

1, 2, 71, 9596, 1355849266, 1032458258547, 1653031004194447737

Offset: 1

dead

A204187 a(n) = Sum_{m=1..n-1} m^(n-1) modulo n.

0, 1, 2, 0, 4, 3, 6, 0, 6, 5, 10, 0, 12, 7, 10, 0, 16, 9, 18, 0, 14, 11, 22, 0, 20, 13, 18, 0, 28, 15, 30, 0, 22, 17, 0, 0, 36, 19, 26, 0, 40, 21, 42, 0, 21, 23, 46, 0, 42, 25, 34, 0, 52, 27, 0, 0, 38, 29, 58, 0, 60, 31, 42, 0, 52, 33, 66, 0, 46, 35, 70, 0

Offset: 1

Jonathan Sondow, Jan 12 2012

a(n) = n - 1 if n is 1 or a prime, by Fermat's little theorem. It is conjectured that the converse is also true; see A055032 and A201560 and note that a(n) = n-1 <==> A055032(n) = 1 <==> A201560(n) = 0.

As of 1991, Giuga and Bedocchi had verified no composite n < 10^1700 satisfies a(n) = n - 1 (Ribemboim, 1991). - Alonso del Arte, May 10 2013

			Sum(m^3, m = 1 .. 3) = 1^3 + 2^3 + 3^3 = 36 == 0 (mod 4), so a(4) = 0.

Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 6 of Hong Kong Mathematical Olympiad 2007 (find a(7)), page 134.
Richard K. Guy, Unsolved Problems in Number Theory, A17.
Paulo Ribemboim, The Little Book of Big Primes. New York: Springer-Verlag (1991): 17.

Ivan Neretin, Table of n, a(n) for n = 1..10000
John Clark, On a conjecture involving Fermat's Little Theorem, Thesis, 2008, University of South Florida.
Hong Kong Mathematics Olympiad (2007-2008), Final Event 2 (Group), problem 2, p. 437.
Bernard Schott, Proof that a(n)=n/2 if n is of the form 4k+2.
Index to sequences related to Olympiads.

Cf. A055023, A055030, A055031, A055032, A201560.

Cf. A191677 (zeros).

Table[Mod[Sum[i^(n - 1), {i, n - 1}], n], {n, 75}] (* Alonso del Arte, May 10 2013 *)

a(n) = lift(sum(i=1, n, Mod(i, n)^(n-1))); \\ Michel Marcus, Feb 23 2020

def a(n): return sum(pow(m, n-1, n) for m in range(1, n))%n
print([a(n) for n in range(1, 73)]) # Michael S. Branicky, Jan 02 2022

a(p) = p - 1 if p is prime, and a(4n) = 0.
a(n) + 1 == A201560(n) (mod n).
a(n) = n/2 iff n is of the form 4k+2 (conjectured). - Ivan Neretin, Sep 23 2016
a(4*k+2) = 2*k+1; for a proof see corresponding link. - Bernard Schott, Dec 29 2021

A055023 a(n) = n/A055032(n).

Original entry on oeis.org

1, 2, 3, 1, 5, 2, 7, 1, 1, 2, 11, 1, 13, 2, 1, 1, 17, 2, 19, 1, 3, 2, 23, 1, 1, 2, 1, 1, 29, 2, 31, 1, 1, 2, 1, 1, 37, 2, 3, 1, 41, 2, 43, 1, 1, 2, 47, 1, 1, 2, 1, 1, 53, 2, 1, 1, 3, 2, 59, 1, 61, 2, 1, 1, 1, 2, 67, 1, 1, 2, 71, 1, 73, 2, 3, 1, 1, 2, 79, 1, 1, 2, 83, 1, 1

Offset: 1

N. J. A. Sloane, Jun 11 2000

nonn

It is conjectured that this is n iff n is 1 or a prime.

R. K. Guy, Unsolved Problems Number Theory, A17.

Indranil Ghosh, Table of n, a(n) for n = 1..1000

Cf. A055030, A055031, A055032, A201560, A204187.

Table[n/Denominator[(Sum[m^(n - 1), {m, n - 1}] + 1)/n], {n, 10}] (* Indranil Ghosh, May 17 2017 *)

a(n) = n/denominator((sum(m=1, n - 1, m^(n - 1)) + 1)/n); \\ Indranil Ghosh, May 17 2017

from sympy import Integer
def a(n): return Integer(n)/((sum(m**(n - 1) for m in range(1, n)) + 1)/Integer(n)).denominator # Indranil Ghosh, May 17 2017

A055032 Denominator of (Sum(m^(n-1),m=1..n-1)+1)/n.

Original entry on oeis.org

1, 1, 1, 4, 1, 3, 1, 8, 9, 5, 1, 12, 1, 7, 15, 16, 1, 9, 1, 20, 7, 11, 1, 24, 25, 13, 27, 28, 1, 15, 1, 32, 33, 17, 35, 36, 1, 19, 13, 40, 1, 21, 1, 44, 45, 23, 1, 48, 49, 25, 51, 52, 1, 27, 55, 56, 19, 29, 1, 60, 1, 31, 63, 64, 65, 33, 1, 68, 69, 35, 1, 72, 1

Offset: 1

N. J. A. Sloane, Jun 11 2000

It is conjectured that this is 1 iff n is 1 or a prime.

R. K. Guy, Unsolved Problems in Number Theory, A17.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A055031, A055030, A055023, A201560, A204187.

a:= proc(n) local S,m;
    S:= 1;
    for m from 1 to n-1 do
      S:= S + m &^(n-1) mod n;
    od:
    denom(S/n);
end proc;
seq(a(n),n=1..1000); # Robert Israel, May 30 2014

Table[Denominator[(Sum[m^(n - 1), {m, 1, n - 1}] + 1)/n], {n, 1, 10}] (* G. C. Greubel, Jun 06 2016 *)

a(n) = denominator((sum(m=1, n - 1, m^(n - 1)) + 1)/n); \\ Indranil Ghosh, May 17 2017

from sympy import Integer
def a(n): return ((sum(m**(n - 1) for m in range(1, n)) + 1)/Integer(n)).denominator # Indranil Ghosh, May 17 2017

A055031 Numerator of (Sum(m^(n-1),m=1..n-1)+1)/n.

Original entry on oeis.org

1, 1, 2, 37, 71, 2213, 9596, 1200305, 24684613, 287152493, 1355849266, 427675990237, 1032458258547, 228796942438201, 16841089312342856, 665478473553144001, 1653031004194447737, 631449646252135295657, 3167496749732497119310

Offset: 1

N. J. A. Sloane, Jun 11 2000

R. K. Guy, Unsolved Problems Number Theory, A17.

Indranil Ghosh, Table of n, a(n) for n = 1..200

Cf. A055032, A055030, A055023, A201560, A204187.

Table[Numerator[(Sum[m^(n - 1), {m, n - 1}] + 1)/n], {n, 50}] (* Indranil Ghosh, May 17 2017 *)

a(n) = numerator((sum(m=1, n - 1, m^(n - 1)) + 1)/n); \\ Indranil Ghosh, May 17 2017

from sympy import Integer
def a(n): return ((sum(m**(n - 1) for m in range(1, n)) + 1)/Integer(n)).numerator # Indranil Ghosh, May 17 2017

A201560 a(n) = (Sum(m^(n-1), m=1..n-1) + 1) modulo n.

Original entry on oeis.org

0, 0, 0, 1, 0, 4, 0, 1, 7, 6, 0, 1, 0, 8, 11, 1, 0, 10, 0, 1, 15, 12, 0, 1, 21, 14, 19, 1, 0, 16, 0, 1, 23, 18, 1, 1, 0, 20, 27, 1, 0, 22, 0, 1, 22, 24, 0, 1, 43, 26, 35, 1, 0, 28, 1, 1, 39, 30, 0, 1, 0, 32, 43, 1, 53, 34, 0, 1, 47, 36, 0, 1, 0, 38, 51, 1, 1

Offset: 1

Jonathan Sondow, Jan 11 2012

nonn

Equals 0 if n is 1 or a prime, by Fermat's little theorem. It is conjectured that the converse is also true; see A055030, A055032, A204187 and note that a(n) = 0 <==> A055032(n) = 1 <==> A204187(n) = n-1.

			Sum(m^3, m=1..3) + 1 = 1^3 + 2^3 + 3^3 + 1 = 37 == 1 (mod 4), so a(4) = 1.

R. K. Guy, Unsolved Problems in Number Theory, A17.

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A055023, A055030, A055031, A055032, A204187.

Table[Mod[Plus @@ PowerMod[Range[n - 1], n - 1, n] + 1, n], {n, 77}] (* Ivan Neretin, Sep 23 2016 *)

a(prime) = 0 and a(4n) = 1.

a(n) == A204187(n) + 1 (mod n).

A294507 Sum(m^p, m=1..p-1) / p^2 as p runs through the odd primes.

Original entry on oeis.org

1, 52, 7689, 1176564625, 915495729492, 1507470694179701824, 2916521865098581522761, 21333370304597346190818638521, 1675481131512375613482932303229309861556, 9784120259254858957467327917016090730358625, 4284997268972399392421947270075253022799901265537333204

Offset: 1

Jonathan Sondow, Nov 01 2017

nonn

			a(1) = (1^3 + 2^3)/3^2 = (1 + 8)/9 = 1.

Cf. A005097, A055030, A065091, A219550.

Table[ p = Prime[n]; Sum[ m^p, {m, 1, p - 1}] / p^2, {n, 2, 12}]

a(n) = A219550(n) / prime(n+1) = A219550(n) / A065091(n).

a(n) (mod prime(n+1)) = (prime(n+1) - 1)/2 = A005097(n).

A219550 Sum(m^p, m=1..p-1)/p as p runs through the odd primes.

Original entry on oeis.org

3, 260, 53823, 12942210875, 11901444483396, 25627001801054931008, 55413915436873048932459, 490667517005738962388828685983, 48588952813858892791005036793649985985124, 303307728036900627681487165427498812641117375, 158544898951978777519612048992784361843596346824881328548

Offset: 1

Jonathan Sondow, Dec 04 2012

nonn

Always an integer: for an elementary proof that Sum(m^k,m=1..p-1)/p is an integer if p is prime and p-1 does not divide k (and a discussion of other proofs), see MacMillan and Sondow 2011. Applications are in Sondow and MacMillan 2011.
For (Sum(m^(p-1), m=1..p-1)+1)/p as p runs through the primes, see A055030.
For Sum(m^p, m=1..p-1) / p^2 as p runs through the odd primes, see A294507.

			a(1) = (1^3 + 2^3)/3 = (1 + 8)/3 = 3.

K. MacMillan and J. Sondow, Proofs of power sum and binomial coefficient congruences via Pascal's identity, Amer. Math. Monthly, 118 (2011), 549-551.
J. 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.

Cf. A055030, A294507.

Array[Sum[m^#, {m, # - 1}]/# &@ Prime@ # &, 11, 2] (* Michael De Vlieger, Nov 04 2017 *)

A225578 Sum of first (prime(n) - 1) (prime(n) - 1)th powers.

Original entry on oeis.org

1, 5, 354, 67171, 14914341925, 13421957361110, 28101527071305611528, 60182438244917445266889, 525344775209112229247070397995, 51296981152155330485450049059398345004638, 319099356359853147544285512855368258519442575

Offset: 1

Alonso del Arte, May 10 2013

It follows from Fermat's little theorem that a(n) is congruent to -1 mod the n-th prime.

			a(2) = 5 because, since 3 is the second prime, we have 1^2 + 2^2 = 1 + 4 = 5.
a(3) = 354 because, since 5 is the third prime, we have 1^4 + 2^4 + 3^4 + 4^4 = 1 + 4 + 81 + 256 = 354.

R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section A17.
Paulo Ribemboim, The Little Book of Big Primes, New York, Springer-Verlag (1991): 17.

Seiichi Manyama, Table of n, a(n) for n = 1..76

Cf. A055030, A031971, A204187.

Table[Sum[i^(Prime[n] - 1), {i, Prime[n] - 1}], {n, 15}]

a(n) = Sum_{i=1..prime(n)-1} i^(prime(n) - 1).

A338472 (1 + Sum_{k(even)=2..p-1} 2*k^(p-1))/p as p runs through the odd primes.

Original entry on oeis.org

3, 109, 14519, 2024592291, 1536463613637, 2449395996564189425, 4686662617019462175259, 33724155827962966577589860263, 2606282943971359343146382147809434583605, 15159042500551578738018590862773479717960671, 6576976543997974825092367662248938303820921894460988333

Offset: 1

Davide Rotondo, Oct 29 2020

nonn

Conjecture: (1 + Sum_{k(even)=2..p-1} 2*k^(p-1))/p is an integer iff p is an odd prime.

Cf. A055030.

a[n_] := Module[{p = Prime[n + 1]}, (1 + 2 * Sum[k^(p - 1), {k, 2, p - 1, 2}])/p]; Array[a, 11] (* Amiram Eldar, Oct 29 2020 *)

a(n) = my(p=prime(n+1)); (1 + sum(k=1, (p-1)\2, 2*(2*k)^(p-1)))/p; \\ Michel Marcus, Oct 29 2020

cp's OEIS Frontend

A071871 Duplicate of A055030.

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A204187 a(n) = Sum_{m=1..n-1} m^(n-1) modulo n.

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A055023 a(n) = n/A055032(n).

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A055032 Denominator of (Sum(m^(n-1),m=1..n-1)+1)/n.

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A055031 Numerator of (Sum(m^(n-1),m=1..n-1)+1)/n.

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A201560 a(n) = (Sum(m^(n-1), m=1..n-1) + 1) modulo n.

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A294507 Sum(m^p, m=1..p-1) / p^2 as p runs through the odd primes.

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A219550 Sum(m^p, m=1..p-1)/p as p runs through the odd primes.

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