A000058 -id:A000058 - cp's OEIS frontend

A213437 Nonlinear recurrence: a(n) = a(n-1) + (a(n-1)+1)*Product_{j=1..n-2} a(j).

1, 3, 7, 31, 703, 459007, 210066847231, 44127887746116242376703, 1947270476915296449559747573381594836628779007

Offset: 1

N. J. A. Sloane, Jun 11 2012

nonn

This sequence was going to be included in the Aho-Sloane paper, but was omitted from the published version.

It appears that the sequence becomes periodic mod 10^k for any k, with period 3. The last digits are (1,3,7) repeated. Modulo 10^5 the sequence enters the cycle (56703, 79007, 23231) after the first 10 terms. - M. F. Hasler, Jul 23 2012. See also A214635, A214636.

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437. [Includes many similar sequences, although not this one.]
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!)

Cf. A076949, A214635, A214636, A003095, A001699.

A213437 := proc(n)
        if n = 1 then 1;
        else procname(n-1)+(1+procname(n-1))*mul(procname(j),j=1..n-2);
        end if;
end proc: # R. J. Mathar, Jul 23 2012

RecurrenceTable[{a[n] == a[n-1]+(a[n-1]+1)*(a[n-1]-a[n-2])*a[n-2]/(a[n-2]+1),a[1]==1,a[2]==3},a,{n,1,10}] (* Vaclav Kotesovec, May 06 2015 *)

a=[1];for(n=1,11,a=concat(a, a[n] + (a[n]+1) * prod(k=1,n-1, a[k] )));a \\ - M. F. Hasler, Jul 23 2012

a(n) = a(n-1)+(a(n-1)+1)*(a(n-1)-a(n-2))*a(n-2)/(a(n-2)+1). - Johan de Ruiter, Jul 23 2012
a(2+3k) = 9007 (mod 10^4) for all k>0. - M. F. Hasler, Jul 23 2012
a(n) ~ c^(2^n), where c = A076949 = 1.2259024435287485386279474959130085213212293209696612823177009... . - Vaclav Kotesovec, May 06 2015
a(n) = A001699(n)/A001699(n-1); a(n+1) - a(n) = A001699(n) + A001699(n-1); a(n) = A003095(n) + A003095(n-1). - Peter Bala, Feb 03 2017

Definition recovered by Johan de Ruiter, Jul 23 2012

A348626 Greedy Egyptian fraction representation of 1 with square denominators.

Original entry on oeis.org

2, 2, 2, 3, 3, 7, 12, 49, 340, 6153, 362275, 234314697, 4303312007019, 8064823505928103487, 21034270897045389505182033301, 13184627067084215135862894820778146400791573, 36011454158212923548860166370685543204871921069986403871775848271, 6820216143160044256325325882329406136711110111012515344838677137010956148075846307036940303634819

Offset: 1

Max Alekseyev, Oct 25 2021

nonn

Greedy representation 1 = 1/a(1)^2 + 1/a(2)^2 + ... constructed in a similar way to Sylvester's sequence (A000058). Let s(n) = Sum_{i=1..n} 1/a(i)^2, and let g(n) = 1 - s(n). Each a(n) is taken be the smallest positive integer satisfying s(n) < 1. [Revised by N. J. A. Sloane, Apr 20 2025]
Comment from David desJardins, Apr 20 2025 (Start)
We know that s(n) < 1 and s(n)-1/a(n)^2+1/(a(n)-1)^2 > 1. Then, arguing heuristically, the gap g(n) = 1-s(n) \approx 1/a(n+1)^2. This implies
   0 < g(n) < 1/(a(n)-1)^2 - 1/a(n)^2 \approx 2/a(n)^3.
So a(n)^3/a(n+1)^2 should be roughly uniform on [0,2].
Let L(n) = ln(a(n)). Then 3*L(n) - 2*L(n+1) \approx ln(2) - e(n+1), where e(i) has an exponential distribution.
So L(n+1) \approx (3/2)*L(n) + (1/2)*(e(n+1)-ln(2)).
This gives us the conjecture that L(n) = C * (3/2)^n * (1+o(1)), as n -> oo.
The plot of L(n)*(2/3)^n (see link) shows that the conjecture is plausible, with C \approx 0.15113.
(End)

David desJardins, Table of n, a(n) for n = 1..24
David desJardins, Plot of log(a(n))*(2/3)^n vs n
David desJardins, Comma-separated list of first 40 terms

Cf. A000058, A348625.

s=1; for(n=1,20, t=sqrtint(floor(1/s))+1; s-=1/t^2; print1(t,", "));

from math import isqrt
from fractions import Fraction
def A348626List():
    s = Fraction(1, 1)
    while True:
        t = isqrt(1 // s) + 1
        yield t
        s -= Fraction(1, t * t)
a = A348626List()
print([next(a) for  in range(18)])  # _Peter Luschny, Oct 26 2021

A001543 a(0) = 1, a(n) = 5 + Product_{i=0..n-1} a(i) for n > 0.

Original entry on oeis.org

1, 6, 11, 71, 4691, 21982031, 483209576974811, 233491495280173380882643611671, 54518278368171228201482876236565907627201914279213829353891

Offset: 0

N. J. A. Sloane

nonn

This is the special case k=5 of sequences with exact mutual k-residues. In general, a(1)=k+1 and a(n)=min{m | m>a(n-1), mod(m,a(i))=k, i=1,...,n-1}. k=1 gives Sylvester's sequence A000058 and k=2 Fermat sequence A000215. - Seppo Mustonen, Sep 04 2005

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).

Alois P. Heinz, Table of n, a(n) for n = 0..12
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!)
S. W. Golomb, On certain nonlinear recurring sequences, Amer. Math. Monthly 70 (1963), 403-405.
R. Mestrovic, 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
S. Mustonen, On integer sequences with mutual k-residues
Seppo Mustonen, On integer sequences with mutual k-residues [Local copy]
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Column k=5 of A177888.

Flatten[{1,RecurrenceTable[{a[1]==6, a[n]==a[n-1]*(a[n-1]-5)+5}, a, {n, 1, 10}]}] (* Vaclav Kotesovec, Dec 17 2014 *)
Join[{1},NestList[#(#-5)+5&,6,10]] (* Harvey P. Dale, Oct 10 2016 *)

{
  print1("1, 6");
  n=6;
  m=Mod(5,6);
  for(i=2,9,
    n=m.mod+lift(m);
    m=chinese(m,Mod(5,n));
    print1(", "n)
  )
} \\ Charles R Greathouse IV, Dec 09 2011

a(n) = a(n-1) * (a(n-1) - 5) + 5. - Charles R Greathouse IV, Dec 09 2011

a(n) ~ c^(2^n), where c = 1.696053774403103324180661918166106455311376345474042496749974632237971081462... . - Vaclav Kotesovec, Dec 17 2014

New name from Alonso del Arte, Dec 09 2011

A001544 A nonlinear recurrence: a(n) = a(n-1)^2 - 6*a(n-1) + 6, with a(0) = 1, a(1) = 7.

Original entry on oeis.org

1, 7, 13, 97, 8833, 77968897, 6079148431583233, 36956045653220845240164417232897, 1365749310322943329964576677590044473746108255675592519835615233

Offset: 0

N. J. A. Sloane

nonn

This is the special case k=6 of sequences with exact mutual k-residues. In general, a(1)=k+1 and a(n)=min{m | m>a(n-1), mod(m,a(i))=k, i=1,...,n-1}. k=1 gives Sylvester's sequence A000058 and k=2 Fermat sequence A000215. - Seppo Mustonen, Sep 04 2005

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).

Indranil Ghosh, Table of n, a(n) for n = 0..11
S. W. Golomb, On certain nonlinear recurring sequences, Amer. Math. Monthly 70 (1963), 403-405.
R. Mestrovic, 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
Seppo Mustonen, On integer sequences with mutual k-residues
Seppo Mustonen, On integer sequences with mutual k-residues [Local copy]
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Column k=6 of A177888. - Alois P. Heinz, Nov 07 2012

Flatten[{1,RecurrenceTable[{a[1]==7, a[n]==a[n-1]*(a[n-1]-6)+6}, a, {n, 1, 10}]}] (* Vaclav Kotesovec, Dec 17 2014 *)
Join[{1},NestList[#^2-6#+6&,7,10]] (* Harvey P. Dale, Nov 19 2024 *)

a(n)=if(n<1, n==0, if(n==1, 7, n=a(n-1); n^2-6*n+6))

a(n) ~ c^(2^n), where c = 1.76450357631319101484804524709844019487003729926754942591419313922841785792... . - Vaclav Kotesovec, Dec 17 2014

A001696 a(n) = a(n-1)*(1 + a(n-1) - a(n-2)), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 4, 12, 108, 10476, 108625644, 11798392680793836, 139202068568601568785946949658348, 19377215893777651167043206536157529523359277782016064519251404524

Offset: 0

N. J. A. Sloane

Also, numbers remaining after the following sieving process: In step 1, keep all numbers of the set N={0,1,2,...}. In step 2, keep only every second number after a(2)=2: N'={0,1,2,4,6,8,10,...}. In step 3, keep every 4th of the numbers following a(3)=4, N"={0,1,2,4,12,20,28,...}. In step 4, keep every 12th of the numbers beyond a(4)=12: {0,1,2,4,12,108,204,...}. In step 5, keep every 108th of the numbers beyond a(5)=108: {0,1,2,4,12,108,10476,...}, and so on. The next "gap" a(n+1)-a(n) is always a(n) times the former gap, i.e., a(n+1)-a(n) = a(n)*(a(n)-a(n-1)). - M. F. Hasler, Oct 28 2010

Number of plane trees where the root has fewer than n children and the i-th child of any node has fewer than i children. - David Eppstein, Dec 18 2021

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).

John Cerkan, Table of n, a(n) for n = 0..13
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!)
Index entries for sequences of form a(n+1)=a(n)^2 + ...

a(n)=A039941(2*n); first difference sequence of this sequence is A001697. - Michael Somos, May 19 2000

a001696 n = a001696_list !! n
a001696_list = 0 : 1 : zipWith (-)
   (zipWith (+) a001696_list' $ map (^ 2) a001696_list')
   (zipWith (*) a001696_list a001696_list')
   where a001696_list' = tail a001696_list
-- Reinhard Zumkeller, Apr 29 2013

a[0] = 0; a[1] = 1; a[n_] := a[n] = a[n-1]*(1 + a[n-1] - a[n-2]); Table[a[n], {n, 0, 10}] (* Jean-François Alcover, Jul 02 2013 *)

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

a(n) ~ c^(2^n), where c = 1.15552822483840350150537253088299651035583896919522349372370013726451673646... . - Vaclav Kotesovec, May 21 2015

A001697 a(n+1) = a(n)(a(0) + ... + a(n)).

Original entry on oeis.org

1, 1, 2, 8, 96, 10368, 108615168, 11798392572168192, 139202068568601556987554268864512, 19377215893777651167043206536157390321290709180447278572301746176

Offset: 0

N. J. A. Sloane

Number of binary trees of height n where for each node the left subtree is at least as high as the right subtree. - Franklin T. Adams-Watters, Feb 08 2007
The next term (a(10)) has 129 digits. - Harvey P. Dale, Jan 24 2016
Number of plane trees where the root has exactly n children and the i-th child of any node has at most i-1 children. - David Eppstein, Dec 18 2021

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).

John Cerkan, Table of n, a(n) for n = 0..12
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!)
Daniel Duverney, Takeshi Kurosawa, Iekata Shiokawa, Transformation formulas of finite sums into continued fractions, arXiv:1912.12565 [math.NT], 2019.
Index entries for sequences of form a(n+1)=a(n)^2 + ...
Index to divisibility sequences

a(n) = A039941(2*n+1); first differences of A001696 give this sequence.
Cf. A002658, A001699.
Cf. A064847.

a001697 n = a001697_list !! n
a001697_list = 1 : 1 : f [1,1] where
   f xs@(x:_) = y : f (y : xs) where y = x * sum xs
-- Reinhard Zumkeller, Apr 29 2013

[n le 2 select 1 else Self(n-1)^2*(1+1/Self(n-2)): n in [1..12]]; // Vincenzo Librandi, Nov 25 2015

a[0] = 1; a[1] = 1; a[n_] := a[n] = a[n - 1]^2*(1 + 1/a[n - 2]); Table[a[n], {n, 0, 9}]  (* Jean-François Alcover, Jul 02 2013 *)
nxt[{t_,a_}]:={t+t*a,t*a}; Transpose[NestList[nxt,{1,1},10]][[2]] (* Harvey P. Dale, Jan 24 2016 *)

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

a(n) ~ c^(2^n), where c = 1.3352454783981919948826893254756974184778316104856161827213437094446034867599... . - Vaclav Kotesovec, May 21 2015

Additional comments from Michael Somos, May 19 2000

A053630 Pythagorean spiral: a(n-1), a(n)-1 and a(n) are sides of a right triangle.

Original entry on oeis.org

3, 5, 13, 85, 3613, 6526885, 21300113901613, 226847426110843688722000885, 25729877366557343481074291996721923093306518970391613

Offset: 1

Henry Bottomley, Mar 21 2000

Least prime factors of a(n): 3, 5, 13, 5, 3613, 5, 233, 5, 3169, 5, 101, 5, 29, 5, 695838629, 5, 1217, 5, 2557, 5, 101, 5, 769, 5. - Zak Seidov, Nov 11 2013

			a(3)=13 because 5,12,13 is a Pythagorean triple and a(2)=5.

R. Gelca and T. Andreescu, Putnam and Beyond, Springer 2007, p. 121.

Steven Finch, Exercises in Iterational Asymptotics, arXiv:2411.16062 [math.NT], 2024. See p. 10.
Miguel-Ángel Pérez García-Ortega, Capitulo 5. Catetos, El Libro de las Ternas Pitagóricas.

Cf. A000058, A001844, A006892.
See also A018928, A180313 and A239381 for similar sequences with a(n) a leg and a(n+1) the hypotenuse of a Pythagorean triangle.
Cf. A077125, A117191 (4^(1/Pi)).

A:= proc(n) option remember; (procname(n-1)^2+1)/2 end proc: A(1):= 3:
seq(A(n),n=1..10); # Robert Israel, Jul 14 2014

NestList[(#^2+1)/2&,3,10] (* Harvey P. Dale, Sep 15 2011 *)

{a(n) = if( n>1, (a(n-1)^2 + 1) / 2, 3)}; /* Michael Somos, May 15 2011 */

a(1) = 3, a(n) = (a(n-1)^2 + 1)/2 for n > 1.
a(n) = 2*A000058(n)-1 = A053631(n)+1 = floor(2 * 1.597910218031873...^(2^n)). Constructing the spiral as a sequence of triangles with one vertex at the origin, then for large n the other vertices are close to lying on the doubly logarithmic spiral r = 2*2.228918357655...^(1.5546822754821...^theta) where theta(n) = n*Pi/2 - 1.215918200344... and 1.5546822754821... = 4^(1/Pi).
a(1) = 3, a(n+1) = (1/4)*((a(n)-1)^2 + (a(n)+1)^2). - Amarnath Murthy, Aug 17 2005
a(n)^2 - (a(n)-1)^2 = a(n-1)^2, so 2*a(n)-1 = a(n-1)^2 (see the first formula). - Thomas Ordowski, Jul 13 2014
a(n) = (A006892(n+2) + 3)/2. - Thomas Ordowski, Jul 14 2014
a(n)^2 = A006892(n+3) + 2. - Thomas Ordowski, Jul 19 2014

Corrected and extended by James Sellers, Mar 22 2000

A077496 Decimal expansion of lim_{n -> infinity} A001699(n)^(1/2^n).

Original entry on oeis.org

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

Offset: 1

Benoit Cloitre, Dec 01 2002

			1.5028368010497564997529364237321694087388717439635793069906714243...

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 6.10 Quadratic recurrence constants, p. 443.

G. C. Greubel, Table of n, a(n) for n = 1..10000
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!)
Anna de Mier and Marc Noy, On the maximum number of cycles in outerplanar and series-parallel graphs, El. Notes Discr. Math., Vol. 34 (2009), pp. 489-493, Proposition 2.2.
Eric Weisstein's World of Mathematics, Quadratic Recurrence Equation.

Cf. A001699, A003095, A076949.

function A003095(n)
  if n eq 0 then return 0;
  else return 1 + A003095(n-1)^2;
  end if; return A003095;
end function;
function S(n)
  if n eq 1 then return Log(2)/2;
  else return S(n-1) + Log(1 + 1/A003095(n)^2)/2^n;
  end if; return S;
end function;
SetDefaultRealField(RealField(120)); Exp(S(12)); // G. C. Greubel, Nov 29 2022

digits = 105; Clear[b, beta]; b[0] = 1; b[n_] := b[n] = b[n-1]^2 + 1; b[10]; beta[n_] := beta[n] = b[n]^(2^(-n)); beta[5]; beta[n = 6]; While[ RealDigits[beta[n], 10, digits+5] != RealDigits[beta[n-1], 10, digits+5], Print["n = ", n]; n = n+1]; RealDigits[beta[n], 10, digits] // First (* Jean-François Alcover, Jun 18 2014 *)
(* Second program *)
A003095[n_]:= A003095[n]= If[n==0, 0, 1 + A003095[n-1]^2];
S[n_]:= S[n]= If[n==1, Log[2]/2, S[n-1] + Log[1 + 1/A003095[n]^2]/2^n];
RealDigits[Exp[S[13]], 10, 120][[1]] (* G. C. Greubel, Nov 29 2022 *)

@CachedFunction
def A003095(n): return 0 if (n==0) else 1 + A003095(n-1)^2
@CachedFunction
def S(n): return log(2)/2 if (n==1) else S(n-1) + log(1 + 1/(A003095(n))^2)/2^n
numerical_approx( exp(S(12)), digits=120) # G. C. Greubel, Nov 29 2022

Equals A076949^2. - Vaclav Kotesovec, Dec 17 2014

Equals exp(Sum_{k>=1} log(1+1/A003095(k)^2)/2^k) (Aho and Sloane, 1973). - Amiram Eldar, Feb 02 2022

A144744 Recurrence sequence a(n)=a(n-1)^2-a(n-1)-1, a(0)=4.

Original entry on oeis.org

4, 11, 109, 11771, 138544669, 19194625169774891, 368433635408155743950638444286989, 135743343700069833946317076518699443524748244656296738254150399131

Offset: 0

Artur Jasinski, Sep 20 2008

nonn

a(0)=3 is the smallest integer generating an increasing sequence of the form a(n)=a(n-1)^2-a(n-1)-1.

Cf. A000058, A082732, A144743, A144745, A144746, A144747, A144748.

a = {}; k = 4; Do[k = k^2 - k - 1; AppendTo[a, k], {n, 1, 10}]; a

a(n, s=4)={for(i=1, n, s=s^2-s-1); s} \\ M. F. Hasler, Oct 06 2014

a(n)=a(n-1)^2-a(n-1)-1 and a(0)=4.

a(n) ~ c^(2^n), where c = 3.22737450272053234771396610986262048906046050824600724014923334412606964... . - Vaclav Kotesovec, May 06 2015

Edited by M. F. Hasler, Oct 06 2014

A144745 Recurrence sequence a(n)=a(n-1)^2-a(n-1)-1, a(0)=9.

Original entry on oeis.org

9, 71, 4969, 24685991, 609398126966089, 371366077149776919833628989831, 137912763257614063309949706968500684963726537144819872418729

Offset: 0

Artur Jasinski, Sep 20 2008

nonn

The original version of this sequence had a(0)=5=A144743(1) and therefore was essentially the same as that sequence A144743.

The next term a(8) has 119 digits.

Cf. A000058, A082732, A144743, A144744, A144746, A144747, A144748.

k = 9; a = {k}; Do[k = k^2 - k - 1; AppendTo[a, k], {n, 1, 10}]; a
NestList[#^2 - # - 1 &, 9, 7]  (* Harvey P. Dale, Feb 04 2011 *)

a(n,s=9)=for(i=1,n,s=s^2-s-1);s \\ M. F. Hasler, Oct 06 2014

a(n) = a(n-1)^2-a(n-1)-1 and a(0)=9.

a(n) ~ c^(2^n), where c = 8.395688554881795978328174160925857176207363473280394010762212170489... . - Vaclav Kotesovec, May 06 2015

New initial value a(0)=9 from M. F. Hasler, Oct 20 2014

cp's OEIS Frontend

A213437 Nonlinear recurrence: a(n) = a(n-1) + (a(n-1)+1)*Product_{j=1..n-2} a(j).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A348626 Greedy Egyptian fraction representation of 1 with square denominators.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Python

A001543 a(0) = 1, a(n) = 5 + Product_{i=0..n-1} a(i) for n > 0.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A001544 A nonlinear recurrence: a(n) = a(n-1)^2 - 6*a(n-1) + 6, with a(0) = 1, a(1) = 7.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A001696 a(n) = a(n-1)*(1 + a(n-1) - a(n-2)), a(0) = 0, a(1) = 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Formula

A001697 a(n+1) = a(n)(a(0) + ... + a(n)).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

PARI

Formula

Extensions

A053630 Pythagorean spiral: a(n-1), a(n)-1 and a(n) are sides of a right triangle.

Views