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

N. J. A. Sloane

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).

Links

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)

A002967 Egyptian fractions: number of solutions of 1 = 1/x_1 + ... + 1/x_n in positive integers.

Original entry on oeis.org

1, 1, 10, 215, 12231, 2025462, 1351857641, 6255560531733
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Solutions differing only in the order of the x_i are counted as distinct.
All denominators in the expansion 1 = 1/x_1 + ... + 1/x_n are bounded by A000058(n-1) = A129871(n). - Max Alekseyev, Dec 30 2003

Examples

			For n=3 the 10 solutions are {2,3,6} (6 ways), {2,4,4} (3 ways), {3,3,3} (1 way).

References

  • R. K. Guy, Unsolved Problems in Number Theory, D11.
  • D. Singmaster, "The number of representations of one as a sum of unit fractions," unpublished manuscript, 1972.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A002966, A006585.
Cf. A000058.

Extensions

a(7) from Jud McCranie
a(8) from John Dethridge, Jan 11 2004

A118227 Decimal expansion of Cahen's constant.

Original entry on oeis.org

6, 4, 3, 4, 1, 0, 5, 4, 6, 2, 8, 8, 3, 3, 8, 0, 2, 6, 1, 8, 2, 2, 5, 4, 3, 0, 7, 7, 5, 7, 5, 6, 4, 7, 6, 3, 2, 8, 6, 5, 8, 7, 8, 6, 0, 2, 6, 8, 2, 3, 9, 5, 0, 5, 9, 8, 7, 0, 3, 0, 9, 2, 0, 3, 0, 7, 4, 9, 2, 7, 7, 6, 4, 6, 1, 8, 3, 2, 6, 1, 0, 8, 4, 8, 4, 4, 0, 8, 9, 5, 5, 5, 0, 4, 6, 3, 4, 3, 1, 9, 5, 4, 0, 5, 3
Offset: 0

Views

Author

Eric W. Weisstein, Apr 16 2006

Keywords

Comments

Cahen proved that his constant is irrational. Davison and Shallit proved that it is transcendental and computed its simple continued fraction expansion A006280. - Jonathan Sondow, Aug 17 2014
Named after the French mathematician Eugène Cahen (1865 - 1941). - Amiram Eldar, Oct 29 2020

Examples

			0.6434105462883380261...

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 6.7, p. 436.

Links

Crossrefs

Cf. A000058, A006279, A006280, A006281, A123180, A129871, A242724.

Programs

  • Mathematica
    a[0] = 2; a[n_] := a[n] = a[n-1]^2 - a[n-1]+1; kmax = 1; FixedPoint[ RealDigits[ Sum[(-1)^k/(a[k]-1), {k, 0, kmax += 10}], 10, 105][[1]]&, kmax] (* Jean-François Alcover, Jul 28 2011, updated Jun 19 2014 *)
Most@First@RealDigits@N[x=1; 1+Sum[x=x(1+x); (-1)^k/x, {k, 1, 9}], 106] (* Oliver Seipel, Aug 25 2024, after Charles R Greathouse IV *)
Most@First@RealDigits@N[x=1; 1/2+Sum[x=x(1+x)(1+x+x^2); 1/(x+1), {k, 1, 4}], 106] (* Oliver Seipel, Aug 25 2024 *)
  • PARI
    C=1;1+suminf(k=1,C+=C^2; (-1)^k/C) \\ Charles R Greathouse IV, Jul 14 2020

Formula

Equals Sum_{k >= 0} (-1)^k/(A000058(k)-1).
Equals Sum_{n>=0} 1/A000058(2*n) = 1 - Sum_{n>=0} 1/A000058(2*n+1). - Amiram Eldar, Oct 29 2020
Equals 1 + (1/2) * Sum_{n>=0} (-1)^(n+1)/A129871(n). - Bernard Schott, Apr 06 2021

A275664 a(n) is the sum of the LCM and GCD of all previous terms, with a(0) = 2.

Original entry on oeis.org

2, 4, 6, 14, 86, 3614, 6526886, 21300113901614, 226847426110843688722000886, 25729877366557343481074291996721923093306518970391614
Offset: 0

Views

Author

Andres Cicuttin, Aug 04 2016

Keywords

Comments

Starting from 1, instead of from 2, it is generated A129871 (A variant of Sylvester's sequence A000058).

Links

Crossrefs

Cf. A129871, A000058.

Programs

  • Mathematica
    a = {2}; Do[AppendTo[a, LCM @@ a + GCD @@ a], {i, 1, 10}]; Column[a]

Formula

a(0) = 2, a(n+1) = lcm(a(0),a(1),..,a(n)) + gcd(a(0),a(1),..,a(n)).

A275698 a(0) = 2, after that a(n) is 3 plus the least common multiple of previous terms.

Original entry on oeis.org

2, 5, 13, 133, 17293, 298995973, 89398590973228813, 7992108067998667938125889533702533, 63873791370569400659097694858350356285036046451665934814399129508493
Offset: 0

Views

Author

Andres Cicuttin, Aug 05 2016

Keywords

Comments

This sequence could be considered a particular case of a possible two-parameter family of sequences of the form: a(n) = k1 + lcm(a(0),a(1),..,a(n-1)), a(0) = k2, where in this case k1=3 and k2=2. With other choices of k1 and k2 it seems it is possible to generate other sequences such as
A129871 with k1 = 1 and k2 = 1,
A000058 with k1 = 1 and k2 = 2,
A082732 with k1 = 1 and k2 = 3,
A000215 with k1 = 2 and k2 = 3,
A000324 with k1 = 4 and k2 = 1,
A001543 with k1 = 5 and k2 = 1,
A001544 with k1 = 6 and k2 = 1,
A275664 with k1 = 2 and k2 = 2,
A000289 with k1 = 3 and k2 = 1.

Links

Crossrefs

Cf. A000058, A000215, A000289, A000324, A001543, A001544, A082732, A275664.

Programs

Formula

a(n) = 3 + lcm(a(0), a(1), ..., a(n - 1)), a(0) = 2.
a(n) = 3 + a(n-1)*(a(n-1)-3), for n > 1. - Christian Krause, Oct 17 2023. Proof: Follows from associativity of lcm(...) and the fact that gcd(m,m+3)=1:
a(n)-3 = lcm(a(0),a(1),...,a(n-2),a(n-1))
= lcm(lcm(a(0),a(1),...,a(n-2)),a(n-1))
= lcm(a(n-1)-3,a(n-1))
= (a(n-1)-3)*a(n-1).

A337307 a(1) = 1; a(n) = 1 + Product_{k=1..n-1} a(k) (mod n-1).

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 1, 3, 1, 1, 7, 6, 1, 12, 1, 10, 1, 12, 1, 3, 1, 1, 21, 12, 1, 6, 21, 1, 1, 15, 1, 20, 1, 31, 15, 1, 1, 32, 13, 1, 1, 18, 1, 7, 25, 1, 17, 38, 1, 1, 1, 1, 1, 26, 1, 6, 1, 1, 29, 47, 1, 42, 1, 1, 1, 1, 61, 26, 1, 25, 1, 21, 1, 64, 21, 1, 1, 29, 1, 18
Offset: 1

Views

Author

Matt Donahoe, Aug 22 2020

Keywords

Comments

Note that the running product for each a(n) is incrementally computed mod n-1.

Crossrefs

Inspired by A066910.
Cf. A129871 (without the mod operation).

Programs

  • Mathematica
    a[1] = 1; a[n_] := a[n] = 1 + Mod[Product[a[k], {k, 1, n - 1}], n - 1]; Array[a, 100] (* Amiram Eldar, Aug 22 2020 *)
  • PARI
    lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = 1 + prod(k=1, n-1, va[k]) % (n-1);); va;} \\ Michel Marcus, Aug 23 2020
  • Python
    def f(n):
    if n == 1: return 1
    a = 1
    for k in range(1, n):
        a = a * f(k) % (n - 1)
    return a + 1
Showing 1-6 of 6 results.

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