A054377 -id:A054377 - cp's OEIS frontend

A000058 Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(0) = 2.

2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443, 12864938683278671740537145998360961546653259485195807

Offset: 0

N. J. A. Sloane

Also called Euclid numbers, because a(n) = a(0)*a(1)*...*a(n-1) + 1 for n>0, with a(0)=2. - Jonathan Sondow, Jan 26 2014
Another version of this sequence is given by A129871, which starts with 1, 2, 3, 7, 43, 1807, ... .
The greedy Egyptian representation of 1 is 1 = 1/2 + 1/3 + 1/7 + 1/43 + 1/1807 + ... .
Take a square. Divide it into 2 equal rectangles by drawing a horizontal line. Divide the upper rectangle into 2 squares. Now you can divide the lower one into another 2 squares, but instead of doing so draw a horizontal line below the first one so you obtain a (2+1 = 3) X 1 rectangle which can be divided in 3 squares. Now you have a 6 X 1 rectangle at the bottom. Instead of dividing it into 6 squares, draw another horizontal line so you obtain a (6+1 = 7) X 1 rectangle and a 42 X 1 rectangle left, etc. - Néstor Romeral Andrés, Oct 29 2001
More generally one may define f(1) = x_1, f(2) = x_2, ..., f(k) = x_k, f(n) = f(1)*...*f(n-1)+1 for n > k and natural numbers x_i (i = 1, ..., k) which satisfy gcd(x_i, x_j) = 1 for i <> j. By definition of the sequence we have that for each pair of numbers x, y from the sequence gcd(x, y) = 1. An interesting property of a(n) is that for n >= 2, 1/a(0) + 1/a(1) + 1/a(2) + ... + 1/a(n-1) = (a(n)-2)/(a(n)-1). Thus we can also write a(n) = (1/a(0) + 1/a(1) + 1/a(2) + ... + 1/a(n-1) - 2 )/( 1/a(0) + 1/a(1) + 1/a(2) + ... + 1/a(n-1) - 1). - Frederick Magata (frederick.magata(AT)uni-muenster.de), May 10 2001; [corrected by Michel Marcus, Mar 27 2019]
A greedy sequence: a(n+1) is the smallest integer > a(n) such that 1/a(0) + 1/a(1) + 1/a(2) + ... + 1/a(n+1) doesn't exceed 1. The sequence gives infinitely many ways of writing 1 as the sum of Egyptian fractions: Cut the sequence anywhere and decrement the last element. 1 = 1/2 + 1/3 + 1/6 = 1/2 + 1/3 + 1/7 + 1/42 = 1/2 + 1/3 + 1/7 + 1/43 + 1/1806 = ... . - Ulrich Schimke, Nov 17 2002; [corrected by Michel Marcus, Mar 27 2019]
Consider the mapping f(a/b) = (a^3 + b)/(a + b^3). Starting with a = 1, b = 2 and carrying out this mapping repeatedly on each new (reduced) rational number gives 1/2, 1/3, 4/28 = 1/7, 8/344 = 1/43, ..., i.e., 1/2, 1/3, 1/7, 1/43, 1/1807, ... . Sequence contains the denominators. Also the sum of the series converges to 1. - Amarnath Murthy, Mar 22 2003
a(1) = 2, then the smallest number == 1 (mod all previous terms). a(2n+6) == 443 (mod 1000) and a(2n+7) == 807 (mod 1000). - Amarnath Murthy, Sep 24 2003
An infinite coprime sequence defined by recursion.
Apart from the initial 2, a subsequence of A002061. It follows that no term is a square.
It appears that a(k)^2 + 1 divides a(k+1)^2 + 1. - David W. Wilson, May 30 2004. This is true since a(k+1)^2 + 1 = (a(k)^2 - a(k) + 1)^2 +1 = (a(k)^2-2*a(k)+2)*(a(k)^2 + 1) (a(k+1)=a(k)^2-a(k)+1 by definition). - Pab Ter (pabrlos(AT)yahoo.com), May 31 2004
In general, for any m > 0 coprime to a(0), the sequence a(n+1) = a(n)^2 - m*a(n) + m is infinite coprime (Mohanty). This sequence has (m,a(0))=(1,2); (2,3) is A000215; (1,4) is A082732; (3,4) is A000289; (4,5) is A000324.
Any prime factor of a(n) has -3 as its quadratic residue (Granville, exercise 1.2.3c in Pollack).
Note that values need not be prime, the first composites being 1807 = 13 * 139 and 10650056950807 = 547 * 19569939581. - Jonathan Vos Post, Aug 03 2008
If one takes any subset of the sequence comprising the reciprocals of the first n terms, with the condition that the first term is negated, then this subset has the property that the sum of its elements equals the product of its elements. Thus -1/2 = -1/2, -1/2 + 1/3 = -1/2 * 1/3, -1/2 + 1/3 + 1/7 = -1/2 * 1/3 * 1/7, -1/2 + 1/3 + 1/7 + 1/43 = -1/2 * 1/3 * 1/7 * 1/43, and so on. - Nick McClendon, May 14 2009
(a(n) + a(n+1)) divides a(n)*a(n+1)-1 because a(n)*a(n+1) - 1 = a(n)*(a(n)^2 - a(n) + 1) - 1 = a(n)^3 - a(n)^2 + a(n) - 1 = (a(n)^2 + 1)*(a(n) - 1) = (a(n) + a(n)^2 - a(n) + 1)*(a(n) - 1) = (a(n) + a(n+1))*(a(n) - 1). - Mohamed Bouhamida, Aug 29 2009
This sequence is also related to the short side (or hypotenuse) of adjacent right triangles, (3, 4, 5), (5, 12, 13), (13, 84, 85), ... by A053630(n) = 2*a(n) - 1. - Yuksel Yildirim, Jan 01 2013, edited by M. F. Hasler, May 19 2017
For n >= 4, a(n) mod 3000 alternates between 1807 and 2443. - Robert Israel, Jan 18 2015
The set of prime factors of a(n)'s is thin in the set of primes. Indeed, Odoni showed that the number of primes below x dividing some a(n) is O(x/(log x log log log x)). - Tomohiro Yamada, Jun 25 2018
Sylvester numbers when reduced modulo 864 form the 24-term arithmetic progression 7, 43, 79, 115, 151, 187, 223, 259, 295, 331, ..., 763, 799, 835 which repeats itself until infinity. This was first noticed in March 2018 and follows from the work of Sondow and MacMillan (2017) regarding primary pseudoperfect numbers which similarly form an arithmetic progression when reduced modulo 288. Giuga numbers also form a sequence resembling an arithmetic progression when reduced modulo 288. - Mehran Derakhshandeh, Apr 26 2019
Named after the English mathematician James Joseph Sylvester (1814-1897). - Amiram Eldar, Mar 09 2024
Guy askes if it is an irrationality sequence (see Guy, 1981). - Stefano Spezia, Oct 13 2024

			a(0)=2, a(1) = 2+1 = 3, a(2) = 2*3 + 1 = 7, a(3) = 2*3*7 + 1 = 43.

Graham Everest, Alf van der Poorten, Igor Shparlinski and Thomas Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 6.7.
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994.
Richard K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E24.
Richard K. Guy and Richard Nowakowski, Discovering primes with Euclid. Delta, Vol. 5 (1975), pp. 49-63.
Amarnath Murthy, Smarandache Reciprocal partition of unity sets and sequences, Smarandache Notions Journal, Vol. 11, 1-2-3, Spring 2000.
Amarnath Murthy and Charles Ashbacher, Generalized Partitions and Some New Ideas on Number Theory and Smarandache Sequences, Hexis, Phoenix; USA 2005. See Section 1.1.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 0..12
David Adjiashvili, Sandro Bosio and Robert Weismantel, Dynamic Combinatorial Optimization: a complexity and approximability study, 2012.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.
Gennady Bachman and Troy Kessler, On divisibility properties of certain multinomial coefficients—II, Journal of Number Theory, Volume 106, Issue 1, May 2004, Pages 1-12.
Andreas Bäuerle, Sharp volume and multiplicity bounds for Fano simplices, arXiv:2308.12719 [math.CO], 2023.
Kevin S. Brown, Odd, Greedy and Stubborn (Unit Fractions).
Eunice Y. S. Chan and Robert M. Corless, Minimal Height Companion Matrices for Euclid Polynomials, Mathematics in Computer Science, Vol. 13, No. 1-2 (2019), pp. 41-56, arXiv preprint, arXiv:1712.04405 [math.NA], 2017.
Hung Viet Chu, A Threshold for the Best Two-term Underapproximation by the Greedy Algorithm, arXiv:2306.12564 [math.NT], 2023.
Matthew Brendan Crawford, On the Number of Representations of One as the Sum of Unit Fractions, Master's Thesis, Virginia Polytechnic Institute and State University (2019).
D. R. Curtiss, On Kellogg's Diophantine problem, Amer. Math. Monthly, Vol. 29, No. 10 (1922), pp. 380-387.
Mehran Derakhshandeh, Why do Sylvester numbers, when reduced modulo 864, form an arithmetic progression 7,43,79,115,151,187,223,...?
Paul Erdős and E. G. Straus, On the Irrationality of Certain Ahmes Series, J. Indian Math. Soc. (N.S.), 27(1964), pp. 129-133.
Steven Finch, Exercises in Iterational Asymptotics, arXiv:2411.16062 [math.NT], 2024. See p. 10.
Solomon W. Golomb, On the sum of the reciprocals of the Fermat numbers and related irrationalities, Canad. J. Math., 15 (1963), 475-478.
Solomon W. Golomb, On certain nonlinear recurring sequences, Amer. Math. Monthly 70 (1963), 403-405.
Richard K. Guy and Richard Nowakowski, Discovering primes with Euclid, Research Paper No. 260 (Nov 1974), The University of Calgary Department of Mathematics, Statistics and Computing Science.
János Kollár, Which powers of holomorphic functions are integrable?, arXiv:0805.0756 [math.AG], 2008.
E. Lemoine, Sur la décomposition d'un nombre en ses carrés maxima, Assoc. Française pour L'Avancement des Sciences (1896), 73-77.
Zheng Li and Quanyu Tang, On a conjecture of Erdős and Graham about the Sylvester's sequence, arXiv:2503.12277 [math.NT], 2025. See p. 2.
Nick Lord, A uniform construction of some infinite coprime sequences, The Mathematical Gazette, vol. 92, no. 523, March 2008, pp.66-70.
Romeo Meštrović, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012. - _N. J. A. Sloane_, Jun 13 2012
Melvyn B. Nathanson, Underapproximation by Egyptian fractions, arXiv:2202.00191 [math.NT], 2022.
Benjamin Nill, Volume and lattice points of reflexive simplices, Discrete & Computational Geometry, Vol. 37, No. 2 (2007), pp. 301-320, arXiv preprint, arXiv:math/0412480 [math.AG], 2004-2007.
R. W. K. Odoni, On the prime divisors of the sequence w_{n+1}=1+w_1 ... w_n, J. London Math. Soc. 32 (1985), 1-11.
Michael Penn, An intriguing integer sequence — Sylvester’s Sequence, YouTube video (2022).
Simon Plouffe, A set of formulas for primes, arXiv:1901.01849 [math.NT], 2019.
Paul Pollack, Analytic and Combinatorial Number Theory Course Notes, p. 5. [?Broken link]
Paul Pollack, Analytic and Combinatorial Number Theory Course Notes, p. 5.
Filip Saidak, A New Proof of Euclid's Theorem, Amer. Math. Monthly, Vol. 113, No. 10 (Dec., 2006), pp. 937-938.
N. J. A. Sloane, A Nasty Surprise in a Sequence and Other OEIS Stories, Experimental Mathematics Seminar, Rutgers University, Oct 10 2024, Youtube video; Slides [Mentions this sequence]
Jonathan Sondow and Kieren MacMillan, Primary pseudoperfect numbers, arithmetic progressions, and the Erdős-Moser equation, Amer. Math. Monthly, Vol. 124, No. 3 (2017), pp. 232-240, arXiv preprint, arXiv:math/1812.06566 [math.NT], 2018.
J. J. Sylvester, On a Point in the Theory of Vulgar Fractions, Amer. J. Math., Vol. 3, No. 4 (1880), pp. 332-335.
Burt Totaro, The ACC conjecture for log canonical thresholds, Séminaire Bourbaki no. 1025 (juin 2010).
Burt Totaro and Chengxi Wang, Varieties of general type with small volume, arXiv:2104.12200 [math.AG], 2021.
Akiyoshi Tsuchiya, The delta-vectors of reflexive polytopes and of the dual polytopes, Discrete Mathematics, Vol. 339, No. 10 (2016), pp. 2450-2456, arXiv preprint, arXiv:1411.2122 [math.CO], 2014-2016.
Stephan Wagner and Volker Ziegler, Irrationality of growth constants associated with polynomial recursions, arXiv:2004.09353 [math.NT], 2020.
Eric Weisstein's World of Mathematics, Quadratic Recurrence Equation.
Eric Weisstein's World of Mathematics, Sylvester's Sequence.
Wikipedia, Sylvester's sequence.
Bowen Yao, A note on fraction decompositions of integers, The American Mathematical Monthly, 127(10), 928-932, Dec 2020.
Index entries for sequences of form a(n+1)=a(n)^2 + ....
Index entries for "core" sequences.

Cf. A005267, A000945, A000946, A005265, A005266, A075442, A007018, A014117, A054377, A002061, A118227, A126263, A007996 (primes dividing some term), A323605 (smallest prime divisors), A091335 (number of prime divisors), A129871 (a variant starting with 1).

a000058 0 = 2
a000058 n = a000058 m ^ 2 - a000058 m + 1 where m = n - 1
-- James Spahlinger, Oct 09 2012

a000058_list = iterate a002061 2  -- Reinhard Zumkeller, Dec 18 2013

a(n) = n == 0 ? BigInt(2) : a(n - 1)*(a(n - 1) - 1) + 1
[a(n) for n in 0:8] |> println # Peter Luschny, Dec 15 2020

A[0]:= 2:
for n from 1 to 12 do
A[n]:= A[n-1]^2 - A[n-1]+1
od:
seq(A[i],i=0..12); # Robert Israel, Jan 18 2015

a[0] = 2; a[n_] := a[n - 1]^2 - a[n - 1] + 1; Table[ a[ n ], {n, 0, 9} ]
NestList[#^2-#+1&,2,10] (* Harvey P. Dale, May 05 2013 *)
RecurrenceTable[{a[n + 1] == a[n]^2 - a[n] + 1, a[0] == 2}, a, {n, 0, 10}] (* Emanuele Munarini, Mar 30 2017 *)

a(n) := if n = 0 then 2 else a(n-1)^2-a(n-1)+1 $
makelist(a(n),n,0,8); /* Emanuele Munarini, Mar 23 2017 */

a(n)=if(n<1,2*(n>=0),1+a(n-1)*(a(n-1)-1))

A000058(n,p=2)={for(k=1,n,p=(p-1)*p+1);p} \\ give Mod(2,m) as 2nd arg to calculate a(n) mod m. - M. F. Hasler, Apr 25 2014

a=vector(20); a[1]=3; for(n=2, #a, a[n]=a[n-1]^2-a[n-1]+1); concat(2, a) \\ Altug Alkan, Apr 04 2018

A000058 = [2]
for n in range(1, 10):
    A000058.append(A000058[n-1]*(A000058[n-1]-1)+1)
# Chai Wah Wu, Aug 20 2014

a(n) = 1 + a(0)*a(1)*...*a(n-1).
a(n) = a(n-1)*(a(n-1)-1) + 1; Sum_{i>=0} 1/a(i) = 1. - Néstor Romeral Andrés, Oct 29 2001
Vardi showed that a(n) = floor(c^(2^(n+1)) + 1/2) where c = A076393 = 1.2640847353053011130795995... - Benoit Cloitre, Nov 06 2002 (But see the Aho-Sloane paper!)
a(n) = A007018(n+1)+1 = A007018(n+1)/A007018(n) [A007018 is a(n) = a(n-1)^2 + a(n-1), a(0)=1]. - Gerald McGarvey, Oct 11 2004
a(n) = sqrt(A174864(n+1)/A174864(n)). - Giovanni Teofilatto, Apr 02 2010
a(n) = A014117(n+1)+1 for n = 0,1,2,3,4; a(n) = A054377(n)+1 for n = 1,2,3,4. - Jonathan Sondow, Dec 07 2013
a(n) = f(1/(1-(1/a(0) + 1/a(1) + ... + 1/a(n-1)))) where f(x) is the smallest integer > x (see greedy algorithm above). - Robert FERREOL, Feb 22 2019
From Amiram Eldar, Oct 29 2020: (Start)
Sum_{n>=0} (-1)^n/(a(n)-1) = A118227.
Sum_{n>=0} (-1)^n/a(n) = 2 * A118227 - 1. (End)

A031971 a(n) = Sum_{k=1..n} k^n.

Original entry on oeis.org

1, 5, 36, 354, 4425, 67171, 1200304, 24684612, 574304985, 14914341925, 427675990236, 13421957361110, 457593884876401, 16841089312342855, 665478473553144000, 28101527071305611528, 1262899292504270591313, 60182438244917445266889, 3031284048960901518840700

Offset: 1

Chris du Feu (chris(AT)beckingham0.demon.co.uk)

From Alexander Adamchuk, Jul 21 2006: (Start)
p^(3k - 1) divides a(p^k) for prime p > 2 and k = 1, 2, 3, 4, ... or p^2 divides a(p) for prime p > 2. p^5 divides a(p^2) for prime p > 2. p^8 divides a(p^3) for prime p > 2. p^11 divides a(p^4) for prime p > 2.
p^2 divides a(2p) for prime p > 3. p^3 divides a(3p) for prime p > 2. p^2 divides a(4p) for prime p > 5. p^3 divides a(5p) for prime p > 3. p^2 divides a(6p) for prime p > 7.
p divides a(2p - 1) for all prime p. p^3 divides a(2p^2 - 1) for all prime p. p^5 divides a(2p^3 - 1) for all prime p.
p divides a((p - 1)/2) for p = 5, 13, 17, 29, 37, 41, 53, 61, ... = A002144 Pythagorean primes: primes of form 4n + 1.
(End)
If p prime then a(p-1) == -1 (mod p) [see De Koninck & Mercier reference]. Example: for p = 7, a(6) = 67171 = 7 * 9596 - 1. - Bernard Schott, Mar 06 2020

J.-M. De Koninck et A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 327 pp. 48-200, Ellipses, Paris (2004).
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 21.

T. D. Noe, Table of n, a(n) for n = 1..100
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 [math.NT], 2013-2014.

A diagonal of array A103438.
Cf. A002144, A014117, A089072, A229303, A256016, A261490.
For a(n) mod n see A182398.

a031971 = sum . a089072_row  -- Reinhard Zumkeller, Mar 18 2013

[&+[(k)^n: k in [0..n]]: n in [1..30]]; // Vincenzo Librandi, Apr 18 2011

a := n->sum('i'^n,'i'=1..n);
# alternative
A031971 := proc(n)
    (bernoulli(n+1,n+1)-bernoulli(n+1))/(n+1) ;
end proc: # R. J. Mathar, May 10 2013

Table[Zeta[-n] - Zeta[-n, n + 1], {n, 25}] (* Alexander Adamchuk, Jul 21 2006 *)
Table[Total[Range[n]^n], {n,25}] (* T. D. Noe, Apr 19 2011 *)
Table[HarmonicNumber[n, -n], {n, 1, 25}] (* Jean-François Alcover, Apr 09 2015 *)

a(n)=sum(k=1,n,k^n) \\ Charles R Greathouse IV, Jun 05 2015

from sympy import harmonic
def A031971(n):
    return harmonic(n,-n) # Chai Wah Wu, Feb 15 2020

a(n) is asymptotic to (e/(e - 1))*n^n. - Benoit Cloitre, Dec 17 2003
a(n) = zeta(-n) - zeta(-n, n + 1), where zeta(s) is the Riemann zeta function and zeta(s, a) is the Hurwitz zeta function, a generalization of the Riemann zeta function. - Alexander Adamchuk, Jul 21 2006
a(n) == 1 (mod n) <==> n is in A014117 = 1, 2, 6, 42, 1806 (see the link "On the congruence ..."). - Jonathan Sondow, Oct 18 2013
a(A054377(n)) = A233045(n). - Jonathan Sondow, Dec 11 2013
a(n) = n! * [x^n] exp(x)*(exp(n*x) - 1)/(exp(x) - 1). - Ilya Gutkovskiy, Apr 07 2018
a(n) ~ ((e*n+1)/((e-1)*(n+1))) * n^n. - N. J. A. Sloane, Oct 13 2018, based on email from Claude F. Leibovici who claims this is slightly better than Cloitre's version when n is small.

A069359 a(n) = n * Sum_{p|n} 1/p where p are primes dividing n.

Original entry on oeis.org

0, 1, 1, 2, 1, 5, 1, 4, 3, 7, 1, 10, 1, 9, 8, 8, 1, 15, 1, 14, 10, 13, 1, 20, 5, 15, 9, 18, 1, 31, 1, 16, 14, 19, 12, 30, 1, 21, 16, 28, 1, 41, 1, 26, 24, 25, 1, 40, 7, 35, 20, 30, 1, 45, 16, 36, 22, 31, 1, 62, 1, 33, 30, 32, 18, 61, 1, 38, 26, 59, 1, 60, 1, 39, 40, 42, 18, 71, 1, 56

Offset: 1

Benoit Cloitre, Apr 15 2002

nonn

Coincides with arithmetic derivative on squarefree numbers: a(A005117(n)) = A068328(n) = A003415(A005117(n)). - Reinhard Zumkeller, Jul 20 2003, Clarified by Antti Karttunen, Nov 15 2019
a(n) = n-1 iff n = 1 or n is a primary pseudoperfect number A054377. - Jonathan Sondow, Apr 16 2014
a(1) = 0 by the standard convention for empty sums.
“Seva” on the MathOverflow link asks if the iterates of this sequence are all eventually 0. - Charles R Greathouse IV, Feb 15 2019

			a(12) = 10 because the prime divisors of 12 are 2 and 3 so we have: 12/2 + 12/3 = 6 + 4 = 10. - _Geoffrey Critzer_, Mar 17 2015

Antti Karttunen, Table of n, a(n) for n = 1..16384 (first 10000 terms from Franklin T. Adams-Watters)
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000
MathOverflow, A recursion with a number-theoretic function (2019)
Joshua Zelinsky, The sum of the reciprocals of the prime divisors of an odd perfect or odd primitive non-deficient number, arXiv:2402.14234 [math.NT], 2024. See also Integers (2025) Vol. 25, Art. No. A59, page 2.

Cf. A003415, A005117, A068328, A010051, A000027, A054377, A180253, A230593, A292786, A306369, A326690, A329029, A329350, A329352.
Cf. A322068 (partial sums), A323599 (Inverse Möbius transform).
Sequences of the form n^k * Sum_{p|n, p prime} 1/p^k for k = 0..10: A001221 (k=0), this sequence (k=1), A322078 (k=2), A351242 (k=3), A351244 (k=4), A351245 (k=5), A351246 (k=6), A351247 (k=7), A351248 (k=8), A351249 (k=9), A351262 (k=10).

[0] cat [n*&+[1/p: p in PrimeDivisors(n)]:n in [2..80]]; // Marius A. Burtea, Jan 21 2020

A069359 := n -> add(n/d, d = select(isprime, numtheory[divisors](n))):
seq(A069359(i), i = 1..20); # Peter Luschny, Jan 31 2012
# second Maple program:
a:= n-> n*add(1/i[1], i=ifactors(n)[2]):
seq(a(n), n=1..100);  # Alois P. Heinz, Oct 23 2019

f[list_, i_] := list[[i]]; nn = 100; a = Table[n, {n, 1, nn}]; b =
Table[If[PrimeQ[n], 1, 0], {n, 1, nn}]; Table[DirichletConvolve[f[a, n], f[b, n], n, m], {m, 1, nn}] (* Geoffrey Critzer, Mar 17 2015 *)

a(n) = n*sumdiv(n, d, isprime(d)/d); \\ Michel Marcus, Mar 18 2015

a(n) = my(ps=factor(n)[,1]~);sum(k=1,#ps,n\ps[k]) \\ Franklin T. Adams-Watters, Apr 09 2015

from sympy import primefactors
def A069359(n): return sum(n//p for p in primefactors(n)) # Chai Wah Wu, Feb 05 2022

def A069359(n) :
    D = filter(is_prime, divisors(n))
    return add(n/d for d in D)
print([A069359(i) for i in (1..20)]) # Peter Luschny, Jan 31 2012

G.f.: Sum(x^p(j)/(1-x^p(j))^2,j>=1), where p(j) is the j-th prime. - Vladeta Jovovic, Mar 29 2006
a(n) = A230593(n) - n. a(n) = A010051(n) (*) A000027(n), where operation (*) denotes Dirichlet convolution, that is, convolution of type: a(n) = Sum_(d|n) b(d) * c(n/d) = Sum_{d|n} A010051(d) * A000027(n/d). - Jaroslav Krizek, Nov 07 2013
a(A054377(n)) = A054377(n) - 1. - Jonathan Sondow, Apr 16 2014
Dirichlet g.f.: zeta(s - 1)*primezeta(s). - Geoffrey Critzer, Mar 17 2015
Sum_{k=1..n} a(k) ~ A085548 * n^2 / 2. - Vaclav Kotesovec, Feb 04 2019
From Antti Karttunen, Nov 15 2019: (Start)
a(n) = Sum_{d|n} A008683(n/d)*A323599(d).
a(n) = A003415(n) - A329039(n) = A230593(n) - n = A306369(n) - A000010(n).
a(n) = A276085(A329350(n)) = A048675(A329352(n)).
a(A276086(n)) = A329029(n), a(A328571(n)) = A329031(n).
(End)
a(n) = Sum_{d|n} A000010(d) * A001221(n/d). - Torlach Rush, Jan 21 2020
a(n) = Sum_{k=1..n} omega(gcd(n, k)). - Ilya Gutkovskiy, Feb 21 2020
a(p^k) = p^(k-1) for p prime and k>=1. - Wesley Ivan Hurt, Jul 15 2025

A007850 Giuga numbers: composite numbers n such that p divides n/p - 1 for every prime divisor p of n.

Original entry on oeis.org

30, 858, 1722, 66198, 2214408306, 24423128562, 432749205173838, 14737133470010574, 550843391309130318, 244197000982499715087866346, 554079914617070801288578559178, 1910667181420507984555759916338506

Offset: 1

D. Borwein, J. M. Borwein, P. B. Borwein and R. Girgensohn

There are no other Giuga numbers with 8 or fewer prime factors. I did an exhaustive search using a PARI script which implemented Borwein and Girgensohn's method for finding n factor solutions given n - 2 factors. - Fred Schneider, Jul 04 2006
One further Giuga number is known with 10 prime factors, namely:
420001794970774706203871150967065663240419575375163060922876441614\
2557211582098432545190323474818 =
2 * 3 * 11 * 23 * 31 * 47059 * 2217342227 * 1729101023519 * 8491659218261819498490029296021 * 58254480569119734123541298976556403 but this may not be the next term. (See the Butske et al. paper.)
Conjecture: Giuga numbers are the solution of the differential equation n' = n + 1, where n' is the arithmetic derivative of n. - Paolo P. Lava, Nov 16 2009
n is a Giuga number if and only if n' = a*n + 1 for some integer a > 0 (see our preprint in arXiv:1103.2298). - José María Grau Ribas, Mar 19 2011
A composite number n is a Giuga number if and only if Sum_{i = 1..n-1} i^phi(n) == -1 (mod n), where phi(n) = A000010(n). - Jonathan Sondow, Jan 03 2014
A composite number n is a Giuga number if and only if Sum_{prime p|n} 1/p = 1/n + an integer. (In fact, all known Giuga numbers n satisfy Sum_{prime p|n} 1/p = 1/n + 1.) - Jonathan Sondow, Jan 08 2014
The prime factors of a(n) are listed as n-th row of A236434. - M. F. Hasler, Jul 13 2015
Conjecture: let k = a(n) and k be the product of x(n) distinct prime factors where x(n) <= x(n+1). Then, for any even n, n/2 + 2 <= x(n) <= n/2 + 3 and, for any odd n, (n+1)/2 + 2 <= x(n) <= (n+1)/2 + 3. For any n > 1, there are y "old" distinct prime factors o(1)...o(y) such that o(1) = 2, o(2) = 3, and z "new" distinct prime factors n(1)...n(z) such that none of them - unlike the "old" ones - can be a divisor of a(q) while q < n; n(1) > o(y), y = x(n) - z >= 2, 2 <= z <= b where b is either 4, or 1/2*n. - Sergey Pavlov, Feb 24 2017
Conjecture: a composite n is a Giuga number if and only if Sum_{k=1..n-1} k^lambda(n) == -1 (mod n), where lambda(n) = A002322(n). - Thomas Ordowski and Giovanni Resta, Jul 25 2018
A composite number n is a Giuga number if and only if A326690(n) = 1. - Jonathan Sondow, Jul 19 2019
A composite n is a Giuga number if and only if n * A027641(phi(n)) == - A027642(phi(n)) (mod n^2). Note: Euler's phi function A000010 can be replaced by the Carmichael lambda function A002322. - Thomas Ordowski, Jun 07 2020
By von Staudt and Clausen theorem, a composite n is a Giuga number if and only if n * A027759(phi(n)) == A027760(phi(n)) (mod n^2). Note: Euler's phi function can be replaced by the Carmichael lambda function. - Thomas Ordowski, Aug 01 2020

			From _M. F. Hasler_, Jul 13 2015: (Start)
The prime divisors of 30 are {2, 3, 5}, and 2 divides 30/2-1 = 14, 3 divides 30/3-1 = 9, and 5 divides 30/5-1 = 5.
The prime divisors of 858 are {2, 3, 11, 13} and 858/2-1 = 428 is even, 858/3-1 = 285 is divisible by 3, 858/11-1 = 77 is a multiple of 11, and 858/13-1 = 65 = 13*5.
(End)

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 30, pp 11, Ellipses, Paris 2008.

M. A. Alekseyev, J. M. Grau, A. M. Oller-Marcen. Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n). Discrete Applied Mathematics, 2018. doi:10.1016/j.dam.2018.05.022 arXiv:1602.02407 [math.NT], 2016.
D. Borwein, J. M. Borwein, P. B. Borwein and R. Girgensohn, Giuga's Conjecture on Primality, Amer. Math. Monthly 103, No. 1, 40-50 (1996).
J. M. Borwein and E. Wong, A Survey of Results Relating to Giuga's Conjecture on Primality, Vinet, Luc (ed.): Advances in Mathematical Sciences: CRM's 25 Years. Providence, RI: American Mathematical Society. CRM Proc. Lect. Notes. 11, 13-27 (1997).
William Butske, Lynda M. Jaje, and Daniel R. Mayernik, On the equation Sum_{p | N} 1/p + (1/N)=1, pseudoperfect numbers and perfectly weighted graphs, Math. Comp. 69 (2000), no. 229, 407-420.
José María Grau and Antonio M. Oller-Marcén, Giuga Numbers and the arithmetic derivative., arXiv:1103.2298 [math.NT], 2011; J. Int. Seq. 15 (2012) 12.4.1
José María Grau and Antonio M. Oller-Marcén, Generalizing Giuga's conjecture, arXiv:1103.3483 [math.NT], 2011.
J. M. Grau and A. M. Oller-Marcén, On the congruence sum_{j=1}^{n-1} j^{k(n-1)} == -1 (mod n); k-strong Giuga and k-Carmichael numbers, arXiv preprint arXiv:1311.3522 [math.NT], 2013.
J. M. Grau, A. M. Oller-Marcén, and D. Sadornil, On µ-Sondow Numbers, arXiv:2111.14211 [math.NT], 2021.
John Machacek, Egyptian Fractions and Prime Power Divisors, arXiv:1706.01008 [math.NT], 2017.
Mersenne Forum, Giuga numbers
Romeo Meštrović, Generalizations of Carmichael numbers I, arXiv:1305.1867 [math.NT], May 04 2013.
R. Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.
J. Sondow and K. MacMillan, Primary pseudoperfect numbers, arithmetic progressions, and the Erdős-Moser equation, Amer. Math. Monthly, 124 (2017) 232-240; arXiv:math/1812.06566 [math.NT], 2018.
J. 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.
Eric Weisstein's World of Mathematics, Giuga Number.
Wikipedia, Agoh-Giuga conjecture
Wikipedia, Giuga number

Cf. A054377, A216823, A216824, A235137, A235138, A235140, A235363, A236434, A326690.

fQ[n_] := AllTrue[First /@ FactorInteger@ n, Divisible[n/# - 1, #] &]; Select[Range@ 100000, CompositeQ@ # && fQ@ # &] (* Michael De Vlieger, Oct 05 2015 *)

is(n)=if(isprime(n), return(0)); my(f=factor(n)[,1]); for(i=1,#f, if((n/f[i])%f[i]!=1, return(0))); n>1 \\ Charles R Greathouse IV, Apr 28 2015

from itertools import count, islice
from sympy import isprime, primefactors
def A007850_gen(startvalue=2): # generator of terms >= startvalue
    return filter(lambda x: not isprime(x) and all((x//p-1) % p == 0 for p in primefactors(x)), count(max(startvalue,2)))
A007850_list = list(islice(A007850_gen(),4)) # Chai Wah Wu, Feb 19 2022

Sum_{i = 1..a(n)-1} i^phi(a(n)) == -1 (mod a(n)). - Jonathan Sondow, Jan 03 2014

a(12) from Fred Schneider, Jul 04 2006
Further references from Fred Schneider, Aug 19 2006
Definition corrected by Jonathan Sondow, Sep 16 2012

A014117 Numbers n such that m^(n+1) == m (mod n) holds for all m.

Original entry on oeis.org

1, 2, 6, 42, 1806

Offset: 1

David Broadhurst

"Somebody incorrectly remembered Fermat's little theorem as saying that the congruence a^{n+1} = a (mod n) holds for all a if n is prime" (Zagier). The sequence gives the set of integers n for which this property is in fact true.
If i == j (mod n), then m^i == m^j (mod n) for all m. The latter congruence generally holds for any (m, n)=1 with i == j (mod k), k being the order of m modulo n, i.e., the least power k for which m^k == 1 (mod n). - Lekraj Beedassy, Jul 04 2002
Also, numbers n such that n divides denominator of the n-th Bernoulli number B(n) (cf. A106741). Also, numbers n such that 1^n + 2^n + 3^n + ... + n^n == 1 (mod n). Equivalently, numbers n such that B(n)*n == 1 (mod n). Equivalently, Sum_{prime p, (p-1) divides n} n/p == -1 (mod n). It is easy to see that for n > 1, n must be an even squarefree number. Moreover, the set P of prime divisors of all such n satisfies the property: if p is in P, then p-1 is the product of distinct elements of P. This set is P = {2, 3, 7, 43}, implying that the sequence is finite and complete. - Max Alekseyev, Aug 25 2013
In 2005, B. C. Kellner proved E. W. Weisstein's conjecture that denom(B_n) = n only if n = 1806. - Jonathan Sondow, Oct 14 2013
Squarefree numbers n such that b^n == 1 (mod n^2) for every b coprime to n. Squarefree terms of A341858. - Thomas Ordowski, Aug 05 2024
Conjecture: Numbers n such that gcd(d+1, n) > 1 for every proper divisor d of n. Verified up to 10^696. - David Radcliffe, May 29 2025

M. A. Alekseyev, J. M. Grau, and A. M. Oller-Marcen. Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n). Discrete Applied Mathematics, 2018. doi:10.1016/j.dam.2018.05.022 arXiv:1602.02407 [math.NT], 2016-2018.
John H. Castillo and Jhony Fernando Caranguay Mainguez, The set of k-units modulo n, arXiv:1708.06812 [math.NT], 2017.
Yongyi Chen and Tae Kyu Kim, On Generalized Carmichael Numbers, arXiv:2103.04883 [math.NT], 2021.
J. Dyer-Bennet, A Theorem in Partitions of the Set of Positive Integers, Amer. Math. Monthly, 47(1940) pp. 152-4.
L. Halbeisen and N. Hungerbühler, On generalised Carmichael numbers, Hardy-Ramanujan Society, 1999, 22 (2), pp.8-22. (hal-01109575)
J. M. Grau, A. M. Oller-Marcen, and Jonathan Sondow, On the congruence 1^m + 2^m + ... + m^m == n (mod m) with n|m, arXiv 1309.7941 [math.NT], 2013; Monatsh. Math., 177 (2015), 421-436.
B. C. Kellner, The equation denom(B_n) = n has only one solution, preprint 2005.
Don Reble, A014117 and related OEIS sequences (shows there are no other terms)
Jonathan Sondow and K. MacMillan, Reducing the Erdos-Moser equation 1^n + 2^n + … + k^n = (k+1)^n modulo k and k^2, arXiv:1011.2154 [math.NT], 2010; see Prop. 2; Integers, 11 (2011), article A34.
Eric Weisstein's World of Mathematics, Bernoulli Number
D. Zagier, Problems posed at the St Andrews Colloquium, 1996

Squarefree terms of A124240. - Robert Israel and Thomas Ordowski, Jun 23 2017

Cf. A007018, A031971, A054377, A100016, A242927, A341858.

r[n_] := Reduce[ Mod[m^(n+1) - m, n] == 0, m, Integers]; ok[n_] := Range[n]-1 === Simplify[ Mod[ Flatten[ m /. {ToRules[ r[n][[2]] ]}], n], Element[C[1], Integers]]; ok[1] = True; A014117 = {}; Do[ If[ok[n], Print[n]; AppendTo[ A014117, n] ], {n, 1, 2000}] (* Jean-François Alcover, Dec 21 2011 *)
Select[Range@ 2000, Function[n, Times @@ Boole@ Map[Function[m, PowerMod[m, n + 1, n] == Mod[m, n]], Range@ n] > 0]] (* Michael De Vlieger, Dec 30 2016 *)

[n for n in range(1, 2000) if all(pow(m, n+1, n) == m for m in range(n))] # David Radcliffe, May 29 2025

For n <= 5, a(n) = a(n-1)^2 + a(n-1) with a(0) = 1. - Raphie Frank, Nov 12 2012

a(n+1) = A007018(n) = A054377(n) = A100016(n) for n = 1, 2, 3, 4. - Jonathan Sondow, Oct 01 2013

cp's OEIS Frontend

A236433 List of primes generated by factoring successive primary pseudoperfect numbers (A054377).

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A216825 Prime factors of the first 8 primary pseudoperfect numbers A054377.

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A216826 Largest prime factor of the n-th primary pseudoperfect number A054377(n).

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A191975 Least common multiple of all p-1, where prime p divides the n-th primary pseudoperfect number A054377(n).

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A283311 a(n) = (A054377(n) - 6)/36.

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A000058 Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(0) = 2.

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A031971 a(n) = Sum_{k=1..n} k^n.

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