A014566 -id:A014566 - cp's OEIS frontend

A121999 Primes p such that p^2 divides Sierpinski number A014566((p-1)/2).

Original entry on oeis.org

29, 37, 3373

Offset: 1

Alexander Adamchuk, Sep 11 2006

Subsequence of A003628.

No other terms below 10^11. - Max Alekseyev, Sep 18 2010

Eric Weisstein's World of Mathematics, Sierpinski Number of the First Kind.

Cf. A014566, A003628.

Do[p=Prime[n];f=((p-1)/2)^((p-1)/2)+1;If[IntegerQ[f/p^2],Print[p]],{n,1,3373}]

{ forprime(p=3, 10^11, if(Mod((p-1)/2, p^2)^((p-1)/2)==-1, print(p); )) } \\ Max Alekseyev, Sep 18 2010

Elements of A125854 that are congruent to 5 or 7 modulo 8, i.e., primes p such that p == 5 or 7 (mod 8) and 2^(p-1) == 1+p (mod p^2). - Max Alekseyev, Sep 18 2010

A124899 Sierpinski quotient ((2n-1)^(2n-1) + 1)/(2n) = A014566(2n-1)/(2n).

Original entry on oeis.org

1, 7, 521, 102943, 38742049, 23775972551, 21633936185161, 27368368148803711, 45957792327018709121, 98920982783015679456199, 265572137199362841880960201, 870019499993663001431459704607, 3416070845000481662841943594125601

Offset: 1

Alexander Adamchuk, Nov 12 2006

nonn

2n divides Sierpinski number A014566(2n-1).
2^n divides A014566(2^n-1); A014566(2^n - 1) / 2^n = A081216(2^n - 1) = A122000(n) = {1, 7, 102943, 27368368148803711, 533411691585101123706582594658103586126397951, ...}.
p+1 divides A014566(p) for prime p; A014566(p)/(p+1) = A056852(n) = {7, 521, 102943, 23775972551, 21633936185161, ...}.
Primes in this sequence are {7, 521, 45957792327018709121}.

Seiichi Manyama, Table of n, a(n) for n = 1..194
Eric Weisstein, World of Mathematics. Sierpinski Numbers of the First Kind.

Cf. A014566 (Sierpinski numbers of the first kind: n^n + 1).

Cf. A056826, A056852, A081216, A122000.

List([1..15],n->((2*n-1)^(2*n-1)+1)/(2*n)); # Muniru A Asiru, Apr 08 2018

seq(((2*n-1)^(2*n-1)+1)/(2*n),n=1..20); # Muniru A Asiru, Apr 08 2018

Mathematica
```
Table[((2n-1)^(2n-1)+1)/(2n),{n,1,20}]
```

a(n) = ((2*n-1)^(2*n-1) + 1)/(2*n); \\ Michel Marcus, Apr 08 2018

a(n) = ((2n-1)^(2n-1) + 1)/(2n) = A014566(2n-1)/(2n).

(2n-1)^(a(n)-1) == 1 (mod a(n)). - Thomas Ordowski, Mar 16 2021

A001008 a(n) = numerator of harmonic number H(n) = Sum_{i=1..n} 1/i.

Original entry on oeis.org

1, 3, 11, 25, 137, 49, 363, 761, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 14274301, 275295799, 55835135, 18858053, 19093197, 444316699, 1347822955, 34052522467, 34395742267, 312536252003, 315404588903, 9227046511387

Offset: 1

N. J. A. Sloane

H(n)/2 is the maximal distance that a stack of n cards can project beyond the edge of a table without toppling.
By Wolstenholme's theorem, p^2 divides a(p-1) for all primes p > 3.
From Alexander Adamchuk, Dec 11 2006: (Start)
p divides a(p^2-1) for all primes p > 3.
p divides a((p-1)/2) for primes p in A001220.
p divides a((p+1)/2) or a((p-3)/2) for primes p in A125854.
a(n) is prime for n in A056903. Corresponding primes are given by A067657. (End)
a(n+1) is the numerator of the polynomial A[1, n](1) where the polynomial A[genus 1, level n](m) is defined to be Sum_{d = 1..n - 1} m^(n - d)/d. (See the Mathematica procedure generating A[1, n](m) below.) - Artur Jasinski, Oct 16 2008
Better solutions to the card stacking problem have been found by M. Paterson and U. Zwick (see link). - Hugo Pfoertner, Jan 01 2012
a(n) = A213999(n, n-1). - Reinhard Zumkeller, Jul 03 2012
a(n) coincides with A175441(n) if and only if n is not from the sequence A256102. The quotient a(n) / A175441(n) for n in A256102 is given as corresponding entry of A256103. - Wolfdieter Lang, Apr 23 2015
For a very short proof that the Harmonic series diverges, see the Goldmakher link. - N. J. A. Sloane, Nov 09 2015
All terms are odd while corresponding denominators (A002805) are all even for n > 1 (proof in Pólya and Szegő). - Bernard Schott, Dec 24 2021

			H(n) = [ 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, ... ].
Coincidences with A175441: the first 19 entries coincide because 20 is the first entry of A256102. Indeed, a(20)/A175441(20) = 55835135 / 11167027 = 5 = A256103(1). - _Wolfdieter Lang_, Apr 23 2015

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 258-261.
H. W. Gould, Combinatorial Identities, Morgantown Printing and Binding Co., 1972, # 1.45, page 6, #3.122, page 36.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 259.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, page 347.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 615.
G. Pólya and G. Szegő, Problems and Theorems in Analysis, volume II, Springer, reprint of the 1976 edition, 1998, problem 251, p. 154.
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).

Kenny Lau, Table of n, a(n) for n = 1..2295 (first 200 terms provided by T. D. Noe)
David H. Bailey, Jonathan M. Borwein, and Roland Girgensohn, Experimental evaluation of Euler sums, Exper. Math. 3(1) (1994), 17-30; they evaluate the constants Sum_{k>=1} H_k^m/(k+1)^n.
Hongwei Chen, Evaluations of Some Variant Euler Sums, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.3.
Hongwei Chen, Interesting Series Associated with Central Binomial Coefficients, Catalan Numbers and Harmonic Numbers, J. Int. Seq. 19 (2016), #16.1.5.
R. M. Dickau, Harmonic numbers and the book-stacking problem.
Leo Goldmakher, A short(er) proof of the divergence of the harmonic series.
Antal Iványi, Leader election in synchronous networks, Acta Univ. Sapientiae, Mathematica, 5, 2 (2013), 54-82.
Fredrik Johansson, How (not) to compute harmonic numbers. Feb 21 2009.
Masanobu Kaneko, The Akiyama-Tanigawa algorithm for Bernoulli numbers, J. Integer Sequences, 3 (2000), #00.2.9.
Romeo Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv preprint arXiv:1111.3057 [math.NT], 2011.
Jerry Metzger and Thomas Richards, A Prisoner Problem Variation, Journal of Integer Sequences, Vol. 18 (2015), Article 15.2.7.
Hisanori Mishima, Factorizations of Wolstenholme numbers, n=1..100, n=101..200, n=201..300.
Mike Paterson et al., Maximum Overhang.
Maxie D. Schmidt, Generalized j-Factorial Functions, Polynomials, and Applications, J. Int. Seq. 13 (2010), 10.6.7, Section 4.3.
Peter Shiu, The denominators of harmonic numbers, arXiv:1607.02863 [math.NT], 2016.
N. J. A. Sloane, Illustration of initial terms.
Jonathan Sondow and Eric W. Weisstein, MathWorld: Harmonic Number.
Roberto Tauraso, Some Congruences for Central Binomial Sums Involving Fibonacci and Lucas Numbers, JIS 19 (2016) # 16.5.4.
Eric Weisstein's World of Mathematics, Book Stacking Problem, Wolstenholme's Theorem, Harmonic Mean, Digamma Function.
Wikipedia, Harmonic number.

Cf. A002805 (denominators), A007406, A007408, A007410, A075135, A001220, A125854, A121999, A014566, A056903, A067657, A177427, A177690.
Cf. A145609-A145640. - Artur Jasinski, Oct 16 2008
Cf. A003506. - Paul Curtz, Nov 30 2013
The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358.
Cf. A195505.

List([1..30],n->NumeratorRat(Sum([1..n],i->1/i))); # Muniru A Asiru, Dec 20 2018

import Data.Ratio ((%), numerator)
a001008 = numerator . sum . map (1 %) . enumFromTo 1
a001008_list = map numerator $ scanl1 (+) $ map (1 %) [1..]
-- Reinhard Zumkeller, Jul 03 2012

[Numerator(HarmonicNumber(n)): n in [1..30]]; // Bruno Berselli, Feb 17 2016

A001008 := proc(n)
    add(1/k,k=1..n) ;
    numer(%) ;
end proc:
seq( A001008(n),n=1..40) ; # Zerinvary Lajos, Mar 28 2007; R. J. Mathar, Dec 02 2016

Table[Numerator[HarmonicNumber[n]], {n, 30}]
(* Procedure generating A[1,n](m) (see Comments section) *) m =1; aa = {}; Do[k = 0; Do[k = k + m^(r - d)/d, {d, 1, r - 1}]; AppendTo[aa, k], {r, 1, 20}]; aa (* Artur Jasinski, Oct 16 2008 *)
Numerator[Accumulate[1/Range[25]]] (* Alonso del Arte, Nov 21 2018 *)
Numerator[Table[((n - 1)/2)*HypergeometricPFQ[{1, 1, 2 - n}, {2, 3}, 1] + 1, {n, 1, 29}]] (* Artur Jasinski, Jan 08 2021 *)

A001008(n) = numerator(sum(i=1,n,1/i)) \\ Michael B. Porter, Dec 08 2009

H1008=List(1); A001008(n)={for(k=#H1008,n-1,listput(H1008,H1008[k]+1/(k+1))); numerator(H1008[n])} \\ about 100x faster for n=1..1500. - M. F. Hasler, Jul 03 2019

from sympy import Integer
[sum(1/Integer(i) for i in range(1, n + 1)).numerator() for n in range(1, 31)]  # Indranil Ghosh, Mar 23 2017

def harmonic(a, b):  # See the F. Johansson link.
    if b - a == 1:
        return 1, a
    m = (a+b)//2
    p, q = harmonic(a,m)
    r, s = harmonic(m,b)
    return p*s+q*r, q*s
def A001008(n): H = harmonic(1,n+1); return numerator(H[0]/H[1])
[A001008(n) for n in (1..29)] # Peter Luschny, Sep 01 2012

H(n) ~ log n + gamma + O(1/n). [See Hardy and Wright, Th. 422.]
log n + gamma - 1/n < H(n) < log n + gamma + 1/n [follows easily from Hardy and Wright, Th. 422]. - David Applegate and N. J. A. Sloane, Oct 14 2008
G.f. for H(n): log(1-x)/(x-1). - Benoit Cloitre, Jun 15 2003
H(n) = sqrt(Sum_{i = 1..n} Sum_{j = 1..n} 1/(i*j)). - Alexander Adamchuk, Oct 24 2004
a(n) is the numerator of Gamma/n + Psi(1 + n)/n = Gamma + Psi(n), where Psi is the digamma function. - Artur Jasinski, Nov 02 2008
H(n) = 3/2 + 2*Sum_{k = 0..n-3} binomial(k+2, 2)/((n-2-k)*(n-1)*n), n > 1. - Gary Detlefs, Aug 02 2011
H(n) = (-1)^(n-1)*(n+1)*n*Sum_{k = 0..n-1} k!*Stirling2(n-1, k) * Stirling1(n+k+1,n+1)/(n+k+1)!. - Vladimir Kruchinin, Feb 05 2013
H(n) = n*Sum_{k = 0..n-1} (-1)^k*binomial(n-1,k)/(k+1)^2. (Wenchang Chu) - Gary Detlefs, Apr 13 2013
H(n) = (1/2)*Sum_{k = 1..n} (-1)^(k-1)*binomial(n,k)*binomial(n+k, k)/k. (H. W. Gould) - Gary Detlefs, Apr 13 2013
E.g.f. for H(n) = a(n)/A002805(n): (gamma + log(x) - Ei(-x)) * exp(x), where gamma is the Euler-Mascheroni constant, and Ei(x) is the exponential integral. - Vladimir Reshetnikov, Apr 24 2013
H(n) = residue((psi(-s)+gamma)^2/2, {s, n}) where psi is the digamma function and gamma is the Euler-Mascheroni constant. - Jean-François Alcover, Feb 19 2014
H(n) = Sum_{m >= 1} n/(m^2 + n*m) = gamma + digamma(1+n), numerators and denominators. (see Mathworld link on Digamma). - Richard R. Forberg, Jan 18 2015
H(n) = (1/2) Sum_{j >= 1} Sum_{k = 1..n} ((1 - 2*k + 2*n)/((-1 + k + j*n)*(k + j*n))) + log(n) + 1/(2*n). - Dimitri Papadopoulos, Jan 13 2016
H(n) = (n!)^2*Sum_{k = 1..n} 1/(k*(n-k)!*(n+k)!). - Vladimir Kruchinin, Mar 31 2016
a(n) = Stirling1(n+1, 2) / gcd(Stirling1(n+1, 2), n!) = A000254(n) / gcd(A000254(n), n!). - Max Alekseyev, Mar 01 2018
From Peter Bala, Jan 31 2019: (Start)
H(n) = 1 + (1 + 1/2)*(n-1)/(n+1) + (1/2 + 1/3)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/3 + 1/4)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
H(n)/n  = 1 + (1/2^2 - 1)*(n-1)/(n+1) + (1/3^2 - 1/2^2)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/4^2 - 1/3^2)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
For odd n >= 3, (1/2)*H((n-1)/2) = (n-1)/(n+1) + (1/2)*(n-1)*(n-3)/((n+1)*(n+3)) + 1/3*(n-1)*(n-3)*(n-5)/((n+1)*(n+3)*(n+5)) + ... . Cf. A195505. See the Bala link in A036970. (End)
H(n) = ((n-1)/2) * hypergeom([1,1,2-n], [2,3], 1) + 1. - Artur Jasinski, Jan 08 2021
Conjecture: for nonzero m, H(n) = (1/m)*Sum_{k = 1..n} ((-1)^(k+1)/k) * binomial(m*k,k)*binomial(n+(m-1)*k,n-k). The case m = 1 is well-known; the case m = 2 is given above by Detlefs (dated Apr 13 2013). - Peter Bala, Mar 04 2022
a(n) = the (reduced) numerator of the continued fraction 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n-1)^2/(2*n-1))))). - Peter Bala, Feb 18 2024
H(n) = Sum_{k=1..n} (-1)^(k-1)*binomial(n,k)/k (H. W. Gould). - Gary Detlefs, May 28 2024

Edited by Max Alekseyev, Oct 21 2011

Changed title, deleting the incorrect name "Wolstenholme numbers" which conflicted with the definition of the latter in both Weisstein's World of Mathematics and in Wikipedia, as well as with OEIS A007406. - Stanislav Sykora, Mar 25 2016

A007571 a(n) = largest prime factor of n^n + 1.

Original entry on oeis.org

2, 5, 7, 257, 521, 97, 911, 673, 530713, 27961, 58367, 2227777, 79301, 176597, 142111, 67280421310721, 45957792327018709121, 33388093, 870542161121, 4406613081041681, 22864311556633, 73194743542229, 1522029233, 27250359649

Offset: 1

N. J. A. Sloane, Robert G. Wilson v

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Tyler Busby, Table of n, a(n) for n = 1..148 (terms 1..123 from Hugo Pfoertner, terms 124..135 from Yurii Ivanov)
factordb, Status of 148^148+1 ... 167^167+1.

Cf. A006530, A006486, A014566, A056790.

[Max(PrimeDivisors(n^n+1)):n in [1..24]]; // Marius A. Burtea, Aug 24 2019

Table[ FactorInteger[ n^n + 1, FactorComplete -> True ] [ [ -1, 1 ] ], {n, 1, 25} ]

for(k=1, 24, my(x=factor(k^k+1), f=x[#x[, 1], 1]); print1(f,", ")) \\ Hugo Pfoertner, Aug 23 2019

a(n) = A006530(A014566(n)). - Michel Marcus, Aug 24 2019

A110932 Numbers k such that 2*k^k + 1 is prime.

Original entry on oeis.org

0, 1, 12, 18, 251, 82992

Offset: 0

Ray G. Opao, Sep 25 2005

As a "list of numbers such that ...", the sequence should have offset 1, but to preserve the validity of formulas referring to this sequence, the offset was set to 0 when the initial value a(0)=0 was added. - M. F. Hasler, Sep 02 2012

Cf. A110931, A121270 (= primes in A014566), A088790, A160360, A160600.

The primes 2n^n+1, for k<4, n=a(k)<251, are listed at A216148(k) = A216147(a(k)). - M. F. Hasler, Sep 02 2012

Join[{0}, Select[Range[1000], PrimeQ[2*#^# + 1] &]] (* Robert Price, Mar 27 2019 *)

is_A110932(n)=ispseudoprime(n^n*2+1) \\ M. F. Hasler, Sep 02 2012

a(5) from Serge Batalov, Apr 08 2018

A121270 Prime Sierpinski numbers of the first kind: primes of the form k^k+1.

Original entry on oeis.org

2, 5, 257

Offset: 1

Alexander Adamchuk, Aug 23 2006

Sierpinski proved that k>1 must be of the form 2^(2^j) for k^k+1 to be a prime. All a(n) > 2 must be the Fermat numbers F(m) with m = j+2^j = A006127(j). [Edited by Jeppe Stig Nielsen, Jul 09 2023]

See e.g. pp. 156-157 in M. Krizek, F. Luca & L. Somer, 17 Lectures on Fermat Numbers, Springer-Verlag NY 2001. - Walter Nissen, Mar 20 2010

Eric Weisstein's World of Mathematics, Sierpinski Number of the First Kind

Primes of form b*k^k + 1: this sequence (b=1), A216148 (b=2), A301644 (b=3), A301641 (b=4), A301642 (b=16).

Cf. A014566, A048861, A006127, A000215.

Do[f=n^n+1;If[PrimeQ[f],Print[{n,f}]],{n,1,1000}]

for(n=1,9,if(ispseudoprime(t=n^n+1),print1(t", "))) \\ Charles R Greathouse IV, Feb 01 2013

Definition rewritten by Walter Nissen, Mar 20 2010

A081216 a(n) = (n^n-(-1)^n)/(n+1).

Original entry on oeis.org

0, 1, 1, 7, 51, 521, 6665, 102943, 1864135, 38742049, 909090909, 23775972551, 685853880635, 21633936185161, 740800455037201, 27368368148803711, 1085102592571150095, 45957792327018709121, 2070863582910344082917, 98920982783015679456199

Offset: 0

Vladeta Jovovic, Apr 17 2003

a(n) is prime for n = {3, 5, 17, 157} = A056826(n) Primes p such that (p^p + 1)/(p + 1) is a prime. Prime a(n) are {7, 521, 45957792327018709121, ...}. Bisection of a(n) is Sierpinski quotient a(2n-1) = A124899(n) = ((2n-1)^(2n-1) + 1)/(2n) = A014566(2n-1)/(2n). - Alexander Adamchuk, Nov 12 2006

This is related to the dimension of the primitive middle cohomology of Dwork hypersurfaces x1**n+x2**n+...+xn**n=n*psi*x1*x2*...*xn. [F. Chapoton, Dec 11 2009]

Seiichi Manyama, Table of n, a(n) for n = 0..387
Philippe Goutet, Isotypic Decomposition of the Cohomology and Factorization of the Zeta Functions of Dwork Hypersurfaces, arXiv:0912.2075 [math.NT], 2009.

Main diagonal of A062160.

Cf. A056826, A124899, A014566 (Sierpinski numbers of the first kind: n^n + 1).

a:= n-> (n^n-(-1)^n)/(n+1):
seq(a(n), n=0..20);  # Alois P. Heinz, May 11 2023

a(n) = (n^n-(-1)^n)/(n+1); \\ Michel Marcus, Jul 29 2017

[((n - 1)**(n - 1) + (-1)**n) // n for n in range(1, 16)]

Edited by F. Chapoton, Feb 03 2011

A110567 a(n) = n^(n+1) + 1.

Original entry on oeis.org

1, 2, 9, 82, 1025, 15626, 279937, 5764802, 134217729, 3486784402, 100000000001, 3138428376722, 106993205379073, 3937376385699290, 155568095557812225, 6568408355712890626, 295147905179352825857, 14063084452067724991010

Offset: 0

Jonathan Vos Post, Sep 12 2005

For n >= 2, a(n) = the n-th positive integer such that a(n) (base n) has a block of exactly n consecutive zeros.
Comments from Alexander Adamchuk, Nov 12 2006 (Start)
(2n+1)^2 divides a(2n). a(2n)/(2n+1)^2 = {1,1,41,5713,1657009,826446281,633095889817,691413758034721,...} = A081215(2n).
p divides a(p-1) for prime p. a(p-1)/p = {1,3,205,39991,9090909091,8230246567621,...} = A081209(p-1) = A076951(p-1).
p^2 divides a(p-1) for an odd prime p. a(p-1)/p^2 = {1,41,5713,826446281,633095889817,1021273028302258913,1961870762757168078553, 14199269001914612973017444081,...} = A081215(p-1).
Prime p divides a((p-3)/2) for p = {13,17,19,23,37,41,43,47,61,67,71,89, 109,113,137,139,157,163,167,181,191,...}.
Prime p divides a((p-5)/4) for p = {29,41,61,89,229,241,281,349,421,509,601,641,661,701,709,769,809,821,881,...} = A107218(n) Primes of the form 4x^2+25y^2.
Prime p divides a((p-7)/6) for p = {79,109,127,151,313,421,541,601,613,751,757,787,...}.
Prime p divides a((p-9)/8) for p = {41,337,401,521,569,577,601,857,929,937,953,977,...} A subset of A007519(n) Primes of form 8n+1.
Prime p divides a((p-11)/10) for p = {41,181,331,601,761,1021,1151,1231,1801,...}.
Prime p divides a((p-13)/12) for p = {313,337,433,1621,1873,1993,2161,2677,2833,...}. (End)

			Examples illustrating the Comment:
a(2) = 9 because the first positive integer (base 2) with a block of 2 consecutive zeros is 100 (base 2) = 4, and the 2nd is 1001 (base 2) = 9 = 1 + 2^3.
a(3) = 82 because the first positive integer (base 3) with a block of 3 consecutive zeros is 1000 (base 3) = 81, the 2nd is 2000 (base 3) = 54 and the 3rd is 10001 (base 3) = 82 = 1 + 3^4.
a(4) = 1025 because the first positive integer (base 4) with a block of 4 consecutive zeros is 10000 (base 4) = 256, the 2nd is 20000 (base 4) = 512, the 3rd is 30000 (base 4) = 768 and the 4th 100001 (base 4) = 1025 = 1 + 4^5. and the 2nd is 1001 (base 2) = 9 = 1 + 2^3.

G. C. Greubel, Table of n, a(n) for n = 0..385

Cf. A007778: n^(n+1); A000312: n^n; A014566: Sierpinski numbers of the first kind: n^n + 1.

Cf also A081209, A076951, A081215. A110567.

[n^(n+1) + 1: n in [0..25]]; // G. C. Greubel, Oct 16 2017

Table[n^(n+1)+1,{n,0,30}] (* Harvey P. Dale, Oct 30 2015 *)

for(n=0,25, print1(1 + n^(n+1), ", ")) \\ G. C. Greubel, Aug 31 2017

a(n) = A007778(n) + 1.

a(n) = A110567(n) for n > 1. - Georg Fischer, Oct 20 2018

Entry revised by N. J. A. Sloane, Oct 20 2018 at the suggestion of Georg Fischer.

A125854 Primes p with the property that p divides the Wolstenholme number A001008((p+1)/2).

Original entry on oeis.org

3, 29, 37, 3373, 2001907169

Offset: 1

Alexander Adamchuk, Dec 11 2006

Note that if prime p>3 divides A001008((p+1)/2) then it also divides A001008((p-3)/2).
Note that for a prime p, H([p/2]) == 2*(2^(-p(p-1)) - 1)/p^2 (mod p). Therefore a prime p divides the Wolstenholme number A001008((p+1)/2) if and only if 2^(-p(p-1)) == 1 - p^2 (mod p^3) or, equivalently, 2^(p-1) == 1 + p (mod p^2).
Disjunctive union of the sequences A154998 and A121999 that contain primes congruent respectively to 1,3 and 5,7 modulo 8. (Alekseyev)
a(6) > 5.5*10^12. - Giovanni Resta, Apr 13 2017
Primes p that are base-((p-1)/2) Wieferich primes, that is, primes p such that ((p-1)/2)^(p-1) == 1 (mod p^2). - Jianing Song, Jan 27 2019

			a(1) = 3 because prime 3 divides A001008(2) = 3 and there is no p < 3 that divides A001008((p+1)/2).
a(2) = 29 because 29 divides A001008(15) = 1195757 and there is no prime p (3 < p < 29) that divides A001008((p+1)/2).

Cf. A001008, A121999, A014566, A154998.

Select[Prime[Range[1, 5000]],
Divisible[Numerator[HarmonicNumber[(# + 1)/2]], #] &] (* Robert Price, May 10 2019 *)

Entry revised and a(5) = 2001907169 provided by Max Alekseyev, Jan 18 2009

Edited by Max Alekseyev, Oct 13 2009

A055385 Smallest prime factor of n^n + 1.

Original entry on oeis.org

2, 5, 2, 257, 2, 13, 2, 97, 2, 101, 2, 89, 2, 29, 2, 274177, 2, 5, 2, 148721, 2, 5, 2, 17, 2, 53, 2, 449, 2, 17, 2, 641, 2, 13, 2, 17, 2, 5, 2, 17, 2, 5, 2, 41, 2, 29, 2, 769, 2, 41, 2, 89, 2, 13, 2, 17, 2, 5, 2, 17, 2, 5, 2, 59649589127497217, 2, 37, 2, 41, 2, 13, 2, 97, 2, 149

Offset: 1

Walter Nissen, Jun 24 2000

nonn

If we use the commonly accepted convention that 0^0 = 1, then a(0) = 2. - Chai Wah Wu, Jul 22 2019

			4^4 + 1 = 257 prime, so a(4) = 257;
6^6 + 1 = 13 * 37 * 97, so a(6) = 13.

C. Stanley Ogilvy and John T. Anderson, Excursions in Number Theory. Dover. New York: 1988. Page 82.

Seiichi Manyama, Table of n, a(n) for n = 1..500

Cf. A014566, A055386.

Table[FactorInteger[n^n + 1][[1, 1]], {n, 74}] (* Vincenzo Librandi, Jul 23 2013 *)

a(n) = {if (n % 2, return (2)); return (factor(n^n + 1)[1, 1]);} \\ Michel Marcus, Jul 23 2013

cp's OEIS Frontend

A121999 Primes p such that p^2 divides Sierpinski number A014566((p-1)/2).

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A124899 Sierpinski quotient ((2n-1)^(2n-1) + 1)/(2n) = A014566(2n-1)/(2n).

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A001008 a(n) = numerator of harmonic number H(n) = Sum_{i=1..n} 1/i.

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A007571 a(n) = largest prime factor of n^n + 1.

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A110932 Numbers k such that 2*k^k + 1 is prime.

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A121270 Prime Sierpinski numbers of the first kind: primes of the form k^k+1.

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