cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

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

Original entry on oeis.org

2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443, 12864938683278671740537145998360961546653259485195807
Offset: 0

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Author

Keywords

Comments

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

Examples

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

References

  • 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).

Crossrefs

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).

Programs

  • Haskell
    a000058 0 = 2
    a000058 n = a000058 m ^ 2 - a000058 m + 1 where m = n - 1
    -- James Spahlinger, Oct 09 2012
    
  • Haskell
    a000058_list = iterate a002061 2  -- Reinhard Zumkeller, Dec 18 2013
    
  • Julia
    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
  • Maple
    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
  • Mathematica
    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 *)
  • Maxima
    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 */
    
  • PARI
    a(n)=if(n<1,2*(n>=0),1+a(n-1)*(a(n-1)-1))
    
  • PARI
    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
    
  • PARI
    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
    
  • Python
    A000058 = [2]
    for n in range(1, 10):
        A000058.append(A000058[n-1]*(A000058[n-1]-1)+1)
    # Chai Wah Wu, Aug 20 2014
    

Formula

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)

A180871 Index of term in Sylvester's sequence A000058 divisible by prime A007996(n).

Original entry on oeis.org

0, 1, 2, 4, 3, 11, 4, 9, 6, 6, 6, 29, 64, 42, 9, 59, 10, 80, 39, 103, 140, 41, 137, 53, 69, 146, 104, 14, 92, 15, 117, 199, 75, 98, 316, 233, 28, 92, 281, 44, 136, 26, 258, 7, 38, 6, 176, 126, 74, 59, 89, 61, 45, 79, 13, 448, 119, 180, 290, 184, 348, 502, 508, 161, 7, 265, 229
Offset: 1

Views

Author

T. D. Noe, Sep 25 2010

Keywords

Comments

Because all terms of Sylvester's sequence are coprime to each other, each prime in A007996 divides only one term of A000058. The Mathematica program computes both the primes in A007996 and the terms in this sequence. Using modular arithmetic, it is easy to see that if prime p divides A000058(k) for some k, then we must have k < p. In practice, k < 5*sqrt(p).
An open problem is to prove that all terms of Sylvester's sequence are squarefree or to find a counterexample. Using the p from A007996 and k found here, it is simple to determine whether A000058(k) = 0 (mod p^2). No p < 10^10 was found to have this property.

Examples

			A000058(4) = 1807 = 43 * 181 = A007996(4) * A007996(7), so a(4) = a(7) = 4. - _Jonathan Sondow_, Jan 26 2014
		

Crossrefs

Programs

  • Mathematica
    t={}; p=1; While[Length[t]<100, p=NextPrime[p]; s=Mod[2,p]; k=0; modSet={}; While[s>0 && !MemberQ[modSet,s], AppendTo[modSet,s]; k++; s=Mod[s^2-s+1,p]]; If[s==0, AppendTo[t,{p,k}]]]; Transpose[t][[2]]

Formula

A000058(a(n)) == 0 (mod A007996(n)) implies a(n) < A007996(n). - Jonathan Sondow, Jan 26 2014

Extensions

Definition clarified by Jonathan Sondow, Jan 26 2014

A236434 Table whose n-th row lists the prime factors of the n-th Giuga number A007850(n).

Original entry on oeis.org

2, 3, 5, 2, 3, 11, 13, 2, 3, 7, 41, 2, 3, 11, 17, 59, 2, 3, 11, 23, 31, 47057, 2, 3, 7, 43, 3041, 4447, 2, 3, 7, 59, 163, 1381, 775807, 2, 3, 7, 71, 103, 67213, 713863, 2, 3, 7, 71, 103, 61559, 29133437, 2, 3, 11, 23, 31, 47137, 28282147, 3892535183, 2, 3, 11, 23, 31, 47059, 2259696349, 110725121051, 2, 3, 7, 43, 1831, 138683, 2861051, 1456230512169437
Offset: 1

Views

Author

Jonathan Sondow, Jan 25 2014

Keywords

Comments

It is unknown whether there are infinitely many Giuga numbers or any odd ones.
See A007850 for other comments, references, links, etc.
For prime factors of primary pseudoperfect numbers, see A236433; of terms in Sylvester's sequence, see A126263.

Examples

			30 = 2 * 3 * 5.
858 = 2 * 3 * 11 * 13.
1722 = 2 * 3 * 7 * 41.
66198 = 2 * 3 * 11 * 17 * 59.
2214408306 = 2 * 3 * 11 * 23 * 31 * 47057.
24423128562 = 2 * 3 * 7 * 43 * 3041 * 4447.
432749205173838 = 2 * 3 * 7 * 59 * 163 * 1381 * 775807.
14737133470010574 = 2 * 3 * 7 * 71 * 103 * 67213 * 713863.
550843391309130318 = 2 * 3 * 7 * 71 * 103 * 61559 * 29133437.
244197000982499715087866346 = 2 * 3 * 11 * 23 * 31 * 47137 * 28282147 * 3892535183.
554079914617070801288578559178 = 2 * 3 * 11 * 23 * 31 * 47059 * 2259696349 * 110725121051.
1910667181420507984555759916338506 = 2 * 3 * 7 * 43 * 1831 * 138683 * 2861051 * 1456230512169437.
Another Giuga number (but possibly not the 13th) is 4200017949707747062038711509670656632404195753751630609228764416142557211582098432545190323474818 = 2 * 3 * 11 * 23 * 31 * 47059 * 2217342227 * 1729101023519 * 8491659218261819498490029296021 * 58254480569119734123541298976556403.
		

Crossrefs

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

Original entry on oeis.org

2, 2, 3, 2, 3, 7, 2, 3, 7, 43, 2, 3, 11, 23, 31, 2, 3, 11, 23, 31, 47059, 2, 3, 11, 17, 101, 149, 3109, 2, 3, 11, 23, 31, 47059, 2217342227, 1729101023519
Offset: 1

Views

Author

Jonathan Sondow, Jan 25 2014

Keywords

Comments

It is unknown whether there are infinitely many primary pseudoperfect numbers or any odd ones.
See A054377 for other comments, references, links, etc.
For prime factors of Giuga numbers, see A236434; of terms in Sylvester's sequence, see A126263.

Examples

			2 = 2.
6 = 2 * 3.
42 = 2 * 3 * 7.
1806 = 2 * 3 * 7 * 43.
47058 = 2 *  3 * 11 * 23 * 31.
2214502422 =  2 *  3 * 11 * 23 * 31 * 47059.
52495396602 =  2 * 3 * 11 * 17 * 101 * 149 * 3109.
8490421583559688410706771261086 = 2 *  3 * 11 * 23 * 31 * 47059 * 2217342227 * 1729101023519.
		

Crossrefs

Programs

  • Mathematica
    A054377 = Cases[Import["https://oeis.org/A054377/b054377.txt", "Table"], {, }][[All, 2]];
    First /@ Flatten[FactorInteger[A054377], 1] (* Robert Price, Mar 14 2020 *)

A375543 Sylvester primes. Yet another proof of the infinity of primes.

Original entry on oeis.org

2, 3, 7, 43, 13, 139, 547, 607, 1033, 181, 1987, 73, 2287, 29881, 13999, 17881, 31051, 52387, 67003, 74203, 128551, 352867, 635263, 74587, 1286773, 2271427, 27061, 164299, 20929, 1171, 298483, 1679143, 3229081, 3263443, 120823, 447841, 2408563, 333457, 30241, 4219
Offset: 1

Views

Author

Peter Luschny, Sep 02 2024

Keywords

Comments

Sylvester's sequence can be defined recursively S(n) = S(n-1)*(S(n-1) + 1) for n >= 1 starting S(0) = 1. (A000058(n) = S(n) + 1.)
Since S(n) and S(n) + 1 have no common divisors, it follows that S(n) has at least one more prime factor than S(n-1), and thus by induction, S(n) has at least n distinct prime factors. This simple and constructive form of Euclid's proof of the infinity of primes was formulated by Filip Saidak (see links).
To generate the sequence, select the smallest unchosen prime factor from all prime factors of S(0), S(1), ..., S(n-1). We call the infinite sequence constructed this way the 'Sylvester primes'. The terms, when ordered by size, can be found in A007996; any prime not a Sylvester prime can be located in A096264.
As a procedure, the sequence can hardly be described more clearly than in the Maple program below. Compared to other variants (for example A126263), it has the advantage that the primes generated start relatively small.

Examples

			The generation of the sequence starts:
  n   selected       factors of S(i), i<n       a(n)
  [1] {}             {2}                     ->   2,
  [2] {2}            {2, 3}                  ->   3,
  [3] {2, 3}         {2, 3, 7}               ->   7,
  [4] {2, 3, 7}      {2, 3, 7, 43}           ->  43,
  [5] {2, 3, 7, 43}  {2, 3, 7, 13, 43, 139}  ->  13.
		

Crossrefs

Variants: A126263, A367020.

Programs

  • Maple
    fact := n -> NumberTheory:-PrimeFactors(n):
    SylvesterPrimes := proc(len) local p, d, w, n;
    p := 1; d := {}; w := {};
    for n from 1 to len do
       p := p*(p + 1);
       d := fact(p) minus w;
       w := w union {min(d)};
    od end:
    SylvesterPrimes(8);
    isSylvesterPrime := proc(p) local s, M;
    M := NULL: s := 2:
    while not member(s, [M]) do
       M := M, s;
       s := (s^2 + s) mod p;
       if s = 0 then return true fi;
    od: false end:
  • Mathematica
    Module[{nmax = 20, a = {}, p = 1, f}, Do[p *= p + 1; f = 2; While[MemberQ[a,f] || !Divisible[p, f], f = NextPrime[f]]; AppendTo[a, f], nmax]; a] (* Paolo Xausa, Sep 03 2024 *)
  • Python
    from sympy import sieve
    from itertools import count, islice
    def smallest_new_primefactor(n, pf):
        return next(pi for i in count(1) if (pi:=sieve[i]) not in pf and n%pi==0)
    def agen(): # generator of terms
        p, d, w, pf = 1, set(), set(), set()
        while True:
            p = p*(p + 1)
            m = smallest_new_primefactor(p, w)
            w |= {m}
            yield m
    print(list(islice(agen(), 20)))
    # Michael S. Branicky, Sep 02 2024 after Peter Luschny
    
  • SageMath
    # Returns the first 24 terms in less than 60 seconds.
    def SylvesterPrimes(len: int) -> list[int]:
        M: list[int] = []
        p = q = 1
        for n in range(1, len + 1):
            p = p * (p + 1)
            pq = p // q
            for s in Primes():
                if pq % s == 0 and not s in M:
                    M.append(s)
                    q = q * s
                    print(n, s)
                    break
        return M
    SylvesterPrimes(24)  # Peter Luschny, Sep 05 2024

Extensions

a(21)-a(31) from Michael S. Branicky, Sep 03 2024
a(32) from Paolo Xausa, Sep 04 2024
a(33)-a(36) from Peter Luschny, Sep 05 2024
a(37)-a(40) from Jinyuan Wang, Jul 25 2025

A256147 First repeated number in Sylvester's sequence modulo prime(n).

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 4, 2, 7, 3, 2, 6, 2, 1, 7, 7, 7, 17, 7, 3, 1, 43, 66, 2, 72, 51, 7, 50, 32, 3, 111, 85, 26, 1, 44, 31, 7, 7, 96, 157, 23, 1, 88, 3, 97, 7
Offset: 1

Views

Author

Alonso del Arte, Mar 16 2015

Keywords

Comments

Sylvester's sequence (A000058) is an infinite coprime sequence, a fact that may lead to the incorrect intuition that all primes occur as factors of its terms. It's quite easy to check that no multiple of 5 occurs, since Sylvester's sequence modulo 5 is 2, 3, 2, 3, 2, 3, ...
If a multiple of p occurs in Sylvester's sequence at position j, then A000058(k) == 1 (mod p) for all k > j.
But if no multiple of p occurs in Sylvester's sequence at all, then Sylvester's sequence is fully periodic modulo p or it enters a cycle at some point.

Examples

			a(4) = 1, because the fourth prime is 7 and Sylvester's sequence modulo 7 is 2, 3, 0, 1, 1, 1, ...
a(5) = 3, because the fifth prime is 11 and Sylvester's sequence modulo 11 is 2, 3, 7, 10, 3, 7, 10, 3, 7, 10, ... (3 is the first number repeated).
		

References

  • J. J. Sylvester, Postscript to Note on a Point in Vulgar Fractions. American Journal of Mathematics Vol. 3, No. 4 (Dec., 1880): 388 - 389.

Crossrefs

Showing 1-6 of 6 results.