A055032 -id:A055032 - cp's OEIS frontend

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

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

A055030 (Sum(m^(p-1),m=1..p-1)+1)/p as p runs through the primes.

Original entry on oeis.org

1, 2, 71, 9596, 1355849266, 1032458258547, 1653031004194447737, 3167496749732497119310, 22841077183004879532481321652, 1768861419039838982256898243427529138091, 10293527624511391856267274608237685758691696

Offset: 1

N. J. A. Sloane, Jun 11 2000

nonn

It is conjectured that (Sum(m^(n-1),m=1..n-1)+1)/n is an integer iff n is 1 or a prime.
Always an integer from little Fermat theorem. Converse is conjectured to be true: if p | (1+1^(p-1)+2^(p-1)+3^(p-1)+...+(p-1)^(p-1)) and p > 1, then p is prime. That was checked by Giuga up to p <= 10^1000. [Benoit Cloitre, Jun 09 2002]
For Sum(m^p, m=1..p-1)/p as p runs through the odd primes, see A219550. - Jonathan Sondow, Oct 31 2017

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

Seiichi Manyama, Table of n, a(n) for n = 1..76
K. MacMillan and J. Sondow, Proofs of power sum and binomial coefficient congruences via Pascal's identity, Amer. Math. Monthly, 118 (2011), 549-551.

Cf. A055031, A055032, A055023, A201560, A204187, A219550, A294507.

A055030 := proc(n)
    p := ithprime(n) ;
    add(m^(p-1),m=1..p-1) ;
    (1+%)/p ;
end proc:
seq(A055030(n),n=1..5) ; # R. J. Mathar, Jan 09 2017

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

for(n=1,20,print1((1+sum(i=1, prime(n)-1,i^(prime(n)-1)))/prime(n), ",")) /* Benoit Cloitre, Jun 09 2002*/

a(n) = (1+A225578(n))/A000040(n). - R. J. Mathar, Jan 09 2017

Comments corrected by Jonathan Sondow, Jan 11 2012

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

Original entry on oeis.org

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

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).

A323072 a(n) = n/A323071(n) = n/gcd(n, 1+A060681(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, 11, 56, 19, 29, 1, 60, 1, 31, 63, 64, 65, 33, 1, 68, 69, 35, 1, 72, 1, 37, 25, 76, 77, 39, 1, 80, 81, 41, 1, 84, 85, 43, 87, 88, 1, 45, 91, 92, 31, 47

Offset: 1

Antti Karttunen, Jan 04 2019

nonn

Differs from A055032 at n = 55, 105, 155, ..., (A323070).

Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000

Cf. A055032, A060681, A323070, A323071.

A060681(n) = (n-if(1==n,n,n/vecmin(factor(n)[,1])));
A323071(n) = gcd(n,1+A060681(n));
A323072(n) = (n/A323071(n));

a(n) = n/A323071(n) = n/gcd(n, 1+A060681(n)).

A340079 a(n) = n / gcd(n, 1+A018804(n)), where A018804(n) = Sum_{k=1..n} gcd(k, 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, 4, 1, 15, 1, 32, 33, 17, 35, 36, 1, 19, 13, 40, 1, 3, 1, 44, 9, 23, 1, 48, 49, 25, 51, 52, 1, 27, 11, 56, 19, 29, 1, 60, 1, 31, 63, 64, 65, 33, 1, 68, 69, 35, 1, 72, 1, 37, 75, 76, 77, 39, 1, 80, 81, 41, 1, 84, 85, 43, 87, 88

Offset: 1

Antti Karttunen, Dec 30 2020

nonn

It is conjectured that this is 1 iff n is 1 or a prime. See Thomas Ordowski's Oct 22 2014 comment in A018804.

Antti Karttunen, Table of n, a(n) for n = 1..8191
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Cf. A018804, A340078, A340080.

Cf. also A055032, A323072 (similar but different sequences).

A018804(n) = sumdiv(n, d, n*eulerphi(d)/d); \\ From A018804
A340079(n) = (n/gcd(n,1+A018804(n)));

a(n) = n / A340078(n) = n / gcd(n, 1+A018804(n)).

A323070 Numbers k such that A055023(k) != A323071(k), where A323071(k) = gcd(k, 1+A060681(k)).

Original entry on oeis.org

55, 105, 155, 203, 253, 355, 405, 455, 497, 595, 655, 689, 705, 737, 755, 791, 955, 979, 1005, 1027, 1055, 1081, 1221, 1255, 1305, 1355, 1379, 1555, 1605, 1655, 1673, 1703, 1711, 1751, 1855, 1905, 1955, 1967, 2065, 2155, 2189, 2205, 2255, 2261, 2329, 2455, 2505, 2555, 2755, 2805, 2849, 2855, 3055

Offset: 1

Antti Karttunen, Jan 04 2019

nonn

Equivalently, numbers k for which A055032(k) != A323072(k).

Neither primes nor prime powers present?

Cf. A055023, A055032, A060681, A323071, A323072.

A055023(n) = (n/denominator((sum(m=1, n - 1, m^(n - 1)) + 1)/n)); \\ From A055023.
A060681(n) = (n-if(1==n,n,n/vecmin(factor(n)[,1])));
A323071(n) = gcd(n,1+A060681(n));
is_A323070(n) = (A055023(n)!=A323071(n));

cp's OEIS Frontend

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

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A055030 (Sum(m^(p-1),m=1..p-1)+1)/p as p runs through the primes.

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A204187 a(n) = Sum_{m=1..n-1} m^(n-1) modulo 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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A323072 a(n) = n/A323071(n) = n/gcd(n, 1+A060681(n)).

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A340079 a(n) = n / gcd(n, 1+A018804(n)), where A018804(n) = Sum_{k=1..n} gcd(k, n).

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