A053631 -id:A053631 - cp's OEIS frontend

A007018 a(n) = a(n-1)^2 + a(n-1), a(0)=1.

1, 2, 6, 42, 1806, 3263442, 10650056950806, 113423713055421844361000442, 12864938683278671740537145998360961546653259485195806

Offset: 0

N. J. A. Sloane

Number of ordered trees having nodes of outdegree 0,1,2 and such that all leaves are at level n. Example: a(2)=6 because, denoting by I a path of length 2 and by Y a Y-shaped tree with 3 edges, we have I, Y, I*I, I*Y, Y*I, Y*Y, where * denotes identification of the roots. - Emeric Deutsch, Oct 31 2002
Equivalently, the number of acyclic digraphs (dags) that unravel to a perfect binary tree of height n. - Nachum Dershowitz, Jul 03 2022
a(n) has at least n different prime factors. [Saidak]
Subsequence of squarefree numbers (A005117). - Reinhard Zumkeller, Nov 15 2004 [This has been questioned, see MathOverflow link. - Charles R Greathouse IV, Mar 30 2015]
For prime factors see A007996.
Curtiss shows that if the reciprocal sum of the multiset S = {x_1, x_2, ..., x_n} is 1, then max(S) <= a(n). - Charles R Greathouse IV, Feb 28 2007
The number of reduced ZBDDs for Boolean functions of n variables in which there is no zero sink. (ZBDDs are "zero-suppressed binary decision diagrams.") For example, a(2)=6 because of the 2-variable functions whose truth tables are 1000, 1010, 1011, 1100, 1110, 1111. - Don Knuth, Jun 04 2007
Using the methods of Aho and Sloane, Fibonacci Quarterly 11 (1973), 429-437, it is easy to show that a(n) is the integer just a tiny bit below the real number theta^{2^n}-1/2, where theta =~ 1.597910218 is the exponential of the rapidly convergent series Sum_{n>=0} log(1+1/a_n)/2^{n+1}. For example, theta^32 - 1/2 =~ 3263442.0000000383. - Don Knuth, Jun 04 2007 [Corrected by Darryl K. Nester, Jun 19 2017]
The next term has 209 digits. - Harvey P. Dale, Sep 07 2011
Urquhart shows that a(n) is the minimum size of a tableau refutation of the clauses of the complete binary tree of depth n, see pp. 432-434. - Charles R Greathouse IV, Jan 04 2013
For any positive a(0), the sequence a(n) = a(n-1) * (a(n-1) + 1) gives a constructive proof that there exists integers with at least n distinct prime factors, e.g. a(n). As a corollary, this gives a constructive proof of Euclid's theorem stating that there are an infinity of primes. - Daniel Forgues, Mar 03 2017
Lower bound for A100016 (with equality for the first 5 terms), where a(n)+1 is replaced by nextprime(a(n)). - M. F. Hasler, May 20 2019

R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 94.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, 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!)
David Adjiashvili, Sandro Bosio and Robert Weismantel, Dynamic Combinatorial Optimization: a complexity and approximability study, 2012.
Gilles Audemard, Steve Bellart, Louenas Bounia, Frédéric Koriche, Jean-Marie Lagniez, and Pierre Marquis, On the Explanatory Power of Decision Trees, arXiv:2108.05266 [cs.AI], 2021.
Arvind Ayyer, Anne Schilling, Benjamin Steinberg and Nicolas M. Thiéry, Markov chains, R-trivial monoids and representation theory, Int. J. Algebra Comput., Vol. 25 (2015), pp. 169-231, arXiv preprint, arXiv:1401.4250 [math.CO], 2014.
Umberto Cerruti, Percorsi tra i numeri (in Italian), page 5.
A. Yu. Chirkov, D. V. Gribanov and N. Yu. Zolotykh, On the Proximity of the Optimal Values of the Multi-Dimensional Knapsack Problem with and without the Cardinality Constraint, arXiv:2004.08589 [math.OC], 2020.
D. R. Curtiss, On Kellogg's Diophantine problem, Amer. Math. Monthly, Vol. 29, No. 10 (1922), pp. 380-387.
Christian Elsholtz and Stefan Planitzer, Sums of four and more unit fractions and approximate parametrizations, arXiv:2012.05984 [math.NT], 2020.
Steven Finch, Exercises in Iterational Asymptotics, arXiv:2411.16062 [math.NT], 2024. See p. 9.
Samuele Giraudo, The combinator M and the Mockingbird lattice, arXiv:2204.03586 [math.CO], 2022.
Murray S. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see p. 577.
Diana Maimuţ and George Teşeleanu, Inferring Bivariate Polynomials for Homomorphic Encryption Application, Cryptology ePrint Archive (2023) Art. 844. See p. 16.
MathOverflow, Is OEIS A007018 really a subsequence of squarefree numbers?.
Marko R. Riedel, Two-colorings of unordered full binary trees on n levels.
Matthew Roughan, Surreal Birthdays and Their Arithmetic, arXiv:1810.10373 [math.HO], 2018.
Filip Saidak, A New Proof of Euclid's Theorem, Amer. Math. Monthly, Vol. 113, No. 10 (Dec., 2006), pp. 937-938.
N. J. A. Sloane, A Nasty Surprise in a Sequence and Other OEIS Stories, Experimental Mathematics Seminar, Rutgers University, Oct 10 2024, Youtube video; Slides [Mentions this sequence]
Bertrand Teguia Tabuguia, Computing with D-Algebraic Sequences, arXiv:2412.20630 [math.AG], 2024. See p. 9.
Alasdair Urquhart, The complexity of propositional proofs, Bull. Symbolic Logic, Vol. 1, No. 4 (1995) pp. 425-467, esp. p. 434.
Zalman Usiskin, Letter to N. J. A. Sloane, Oct. 1991.
Index entries for sequences of form a(n+1)=a(n)^2 + ....

Cf. A000058, A003687, A004168, A011782, A077125, A117805, A118227, A371321.
Lower bound for A100016.
Row sums of A122888.

a007018 n = a007018_list !! n
a007018_list = iterate a002378 1  -- Reinhard Zumkeller, Dec 18 2013

[n eq 1 select 1 else Self(n-1)^2 + Self(n-1): n in [1..10]]; // Vincenzo Librandi, May 19 2015

A007018 := proc(n)
    option remember;
    local aprev;
    if n = 0 then
        1;
    else
        aprev := procname(n-1) ;
        aprev*(aprev+1) ;
    end if;
end proc: # R. J. Mathar, May 06 2016

FoldList[#^2 + #1 &, 1, Range@ 8] (* Robert G. Wilson v, Jun 16 2011 *)
NestList[#^2 + #&, 1, 10] (* Harvey P. Dale, Sep 07 2011 *)

a[1]:1$
a[n]:=(a[n-1] + (a[n-1]^2))$
A007018(n):=a[n]$
makelist(A007018(n),n,1,10); /* Martin Ettl, Nov 08 2012 */

a(n)=if(n>0,my(x=a(n-1));x^2+x,1) \\ Edited by M. F. Hasler, May 20 2019 and Jason Yuen, Mar 01 2025

from itertools import islice
def A007018_gen(): # generator of terms
    a = 1
    while True:
        yield a
        a *= a+1
A007018_list = list(islice(A007018_gen(),9)) # Chai Wah Wu, Mar 19 2024

a(n) = A000058(n)-1 = A000058(n-1)^2 - A000058(n-1) = 1/(1-Sum_{jA000058(j)) where A000058 is Sylvester's sequence. - Henry Bottomley, Jul 23 2001
a(n) = floor(c^(2^n)) where c = A077125 = 1.597910218031873178338070118157... - Benoit Cloitre, Nov 06 2002
a(1)=1, a(n) = Product_{k=1..n-1} (a(k)+1). - Benoit Cloitre, Sep 13 2003
a(n) = A139145(2^(n+1) - 1). - Reinhard Zumkeller, Apr 10 2008
If an (additional) initial 1 is inserted, a(n) = Sum_{kFranklin T. Adams-Watters, Jun 11 2009
a(n+1) = a(n)-th oblong (or promic, pronic, or heteromecic) numbers (A002378). a(n+1) = A002378(a(n)) = A002378(a(n-1)) * (A002378(a(n-1)) + 1). - Jaroslav Krizek, Sep 13 2009
a(n) = A053631(n)/2. - Martin Ettl, Nov 08 2012
Sum_{n>=0} (-1)^n/a(n) = A118227. - Amiram Eldar, Oct 29 2020
Sum_{n>=0} 1/a(n) = A371321. - Amiram Eldar, Mar 19 2024

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

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

Artur Jasinski, Jan 23 2007, Jan 29 2007

nonn

			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.

Sierpinski W., Elementary theory of numbers, Monografie Matematyczne 42 (1964), Chapter II, p. 63.

Cf. A018930, A127689, A127691.

Apart from the initial term, the sequence is the same as A053631.

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 *)

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

A118017 Largest denominators in even Egyptian fraction representation of 1.

Original entry on oeis.org

12, 84, 3612, 6526884, 21300113901612

Offset: 1

Teena Carroll, Jul 06 2011

nonn

An Egyptian Fraction representation of a rational number a/b is a list of distinct unit fractions with sum a/b. We will call it an even Egyptian Fraction representation if only even integers are used as denominators. The n-th term of this sequence gives the largest denominators that appear in an Egyptian Fraction Representation of one with length n+3. For instance, 1/2 + 1/4 + 1/6 + 1/12 gives a 4-term even representation of one, which is the shortest possible Egyptian even fraction representation of one.
This sequence can be derived from the Sylvester sequence (A000058). If s(n) represents the Sylvester sequence, s(n)-1 is the largest denominator appearing in an n term Egyptian fraction representation of 1. There is a one-to-one correspondence between k-term representations and (k+1)-term even representation for k<12. An even representation has to have at least 4 terms, thus a(1) is related to s(3). a(1) = 2*(s(3) - 1), etc.
Excluding the first two terms of the Pythagorean spiral sequence (A053631) yields this sequence.

			a(2) = 1/2*12^2 + 12 so a(2) = 84.

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330. Solution published in Vol. 43, No. 4, September 2012, pp. 340-342

Cf. A000058, A007018, A053631.

A053631 and A127690 are very similar to this sequence.

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

cp's OEIS Frontend

A007018 a(n) = a(n-1)^2 + a(n-1), a(0)=1.

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

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

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A118017 Largest denominators in even Egyptian fraction representation of 1.

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