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 1-5 of 5 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

Views

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)

A006892 Representation as a sum of squares requires n squares with greedy algorithm.

Original entry on oeis.org

1, 2, 3, 7, 23, 167, 7223, 13053767, 42600227803223, 453694852221687377444001767, 51459754733114686962148583993443846186613037940783223
Offset: 1

Views

Author

Jeffrey Shallit

Keywords

Comments

Of course Lagrange's theorem tells us that any positive integer can be written as a sum of at most four squares (cf. A004215).
Records in A053610. - Hugo van der Sanden, Jun 24 2015

Examples

			Here is why a(5) = 23: start with 23, subtract largest square <= 23, which is 16, getting 7.
Now subtract largest square <= 7, which is 4, getting 3.
Now subtract largest square <= 3, which is 1, getting 2.
Now subtract largest square <= 2, which is 1, getting 1.
Now subtract largest square <= 1, which is 1, getting 0.
Thus 23 = 16+4+1+1+1.
It took 5 steps to get to 0, and 23 is the smallest number which takes 5 steps. - _N. J. A. Sloane_, Jan 29 2014

References

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

Links

Crossrefs

Cf. A004215, A053610.

Programs

  • PARI
    a(n) = if (n <= 3, n , ((a(n-1)+3)/2)^2 - 2) \\ Michel Marcus, May 25 2013

Formula

For n >= 4, a(n) = a(n-1) + ((a(n-1)+1)/2)^2. - Joe K. Crump (joecr(AT)carolina.rr.com), Apr 16 2000
a(n) = n for n <= 3; for n > 3, a(n) = ((a(n-1)+3)/2)^2 - 2. - Arkadiusz Wesolowski, Mar 30 2013
a(n+2) = 2 * A053630(n) - 3. - Thomas Ordowski, Jul 14 2014
a(n+3) = A053630(n)^2 - 2. - Thomas Ordowski, Jul 19 2014

Extensions

Four more terms from Rick L. Shepherd, Jan 27 2014

A053631 Pythagorean spiral: a(n-1)+1, a(n) and a(n)+1 are the sides of a right triangle (a primitive Pythagorean triangle).

Original entry on oeis.org

2, 4, 12, 84, 3612, 6526884, 21300113901612, 226847426110843688722000884, 25729877366557343481074291996721923093306518970391612, 331013294649039928396936390888878360035026305412754995683702777533071737279144813617823976263475290370884
Offset: 1

Views

Author

Henry Bottomley, Mar 21 2000

Keywords

Comments

To derive a list of Pythagorean triples from this sequence, we note that the difference between the second and the third terms in the Pythagorean triple is 1 and that the last term of the previous triple gives us the first term in the next triple. Therefore the sequence is completely determined by the initial triple.
A053631 gives us a list of Pythagorean triples beginning with (3,4,5), since a(1)=2. Using any initial value h>1, (2h-1,2h^2-2h,2h^2-2h+1) forms a Pythagorean triple; we can use b(1)=2h-1 and the recursive formula b(n)=b(n-1)^2-b(n-1)+1 for n>1, we can create infinitely many of spirals of this type. - Haoqi Chen, Teena Carroll

Examples

			For n=3, a(n-1) = 4, so we want a right triangle with sides 4 + 1 = 5, a(n), and a(n)+1.  Solving (x+1)^2 = x^2 + 5^2 gives x = 12, so a(3) = 12. - _Michael B. Porter_, Jul 19 2016

Links

Crossrefs

Apart from the initial term, the sequence is the same as A127690.
Cf. A046092

Programs

  • Maple
    a[1]:= 2:
for n from 2 to 10 do a[n]:= a[n-1] + a[n-1]^2/2 od:
seq(a[i],i=1..10); # Robert Israel, Jul 08 2015
  • Mathematica
    NestList[# + #^2/2 &, 2, 9] (* Robert G. Wilson v, Dec 12 2012 *)
  • Maxima
    a[1]:2$
a[n]:=a[n-1] + (a[n-1]^2)/2$
A053631(n):=a[n]$
makelist(A053631(n),n,1,10); /* Martin Ettl, Nov 08 2012 */
  • PARI
    main(size)={v=vector(size); v[1]=2;for(n=2,size,v[n]=v[n-1]+v[n-1]^2/2);return(v)} /* Anders Hellström, Jul 08 2015 */

Formula

a(1)=2; for n >= 2: a(n) = a(n-1) + a(n-1)^2/2 = A046092(a(n-1)/2).
a(n) = A053630(n) - 1. - Robert G. Wilson v, Jul 29 2014
a(n) = 2*A007018(n-1). - Ivan Neretin, Jul 26 2015

Extensions

Corrected and extended by James Sellers, Mar 22 2000
a(1) = 2 added by Zak Seidov, Apr 10 2007

A127690 a(1)=3; for n>1, a(n) is such that a(1)^2+...+a(n)^2 = (1+a(n))^2.

Original entry on oeis.org

3, 4, 12, 84, 3612, 6526884, 21300113901612, 226847426110843688722000884, 25729877366557343481074291996721923093306518970391612, 331013294649039928396936390888878360035026305412754995683702777533071737279144813617823976263475290370884
Offset: 1

Views

Author

Artur Jasinski, Jan 23 2007, Jan 29 2007

Keywords

Examples

			a(2)=4 because (3^2+4^2=5^2) and (4+1=5), a(3)=12 because (3^2+4^2+12^2=13^2) and (12+1=13) a(5)= 3612 because (3^2+4^2+12^2+84^2+3612^2=3613^2) and (3612+1=3613) etc.

Links

Crossrefs

Cf. A018930, A127689, A127691.
Apart from the initial term, the sequence is the same as A053631.

Programs

  • Mathematica
    a = {3}; For[k = 1 + a[[Length[a]]], Length[a] < 5, While[ ! ((IntegerQ[Sqrt[(k)^2 + Sum[(a[[t]])^2, {t, 1, Length[a]}]]]) && (Sqrt[(k)^2 + Sum[(a[[t]])^2, {t, 1, Length[a]}]] == k + 1)), k++ ]; AppendTo[a, k]]; a
a = {3}; For[k = 1 + a[[Length[a]]], Length[a] < 12, s2 = Plus @@ (a^2); t = Reduce[{y^2 + s2 == (y + 1)^2}, y, Integers]; t = t /. {Equal -> Rule}; k = y /. t; AppendTo[a, k]]; a (* Daniel Huber *)

Formula

For n>2, a(n) = (a(1)^2 + a(2)^2 + ... + a(n-1)^2 - 1)/2 = ((a(n-1) + 1)^2 - 1)/2. - Max Alekseyev, Nov 23 2012
a(n) = A053630(n-1)-1 for n>=2. - R. J. Mathar, Apr 23 2007

A231579 a(1) = 7; for n > 1, a(n) is the hypotenuse of the right triangle with legs a(n) - 1 and a(n-1).

Original entry on oeis.org

7, 25, 313, 48985, 1199765113, 719718163185951385, 258997117209879873736794713791709113, 33539753361514126736178628392779244498735703225085922505721228803623385
Offset: 1

Views

Author

Zak Seidov, Nov 11 2013

Keywords

Comments

Least prime factors of a(n): 7, 5, 313, 5, 1199765113, 5, 233, 5, 101, 5, 2951438416261, 5, 457, 5, 373, 5, 89, 5, 101, 5.

Examples

			25^2 = 24^2 + 7^2, 313^2 = 312^2 + 25^2.

Crossrefs

Cf. A053630 (case a(1) = 3).

Programs

  • Mathematica
    NestList[(#^2+1)/2&,7,8]

Formula

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

Extensions

b(11) = 2951438416261 (the least prime factor of a(11)) from Jon E. Schoenfield and Charles R Greathouse IV
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