A000058 -id:A000058 - cp's OEIS frontend

A051179 a(n) = 2^(2^n) - 1.

1, 3, 15, 255, 65535, 4294967295, 18446744073709551615, 340282366920938463463374607431768211455, 115792089237316195423570985008687907853269984665640564039457584007913129639935

Offset: 0

Alan DeKok (aland(AT)ox.org)

In a tree with binary nodes (0, 1 children only), the maximum number of unique child nodes at level n.
Number of binary trees (each vertex has 0, or 1 left, or 1 right, or 2 children) such that all leaves are at level n. Example: a(1) = 3 because we have (i) root with a left child, (ii) root with a right child and (iii) root with two children. a(n) = A000215(n) - 2. - Emeric Deutsch, Jan 20 2004
Similarly, this is also the number of full balanced binary trees of height n. (There is an obvious 1-to-1 correspondence between the two sets of trees.) - David Hobby (hobbyd(AT)newpaltz.edu), May 02 2010
Partial products of A000215.
The first 5 terms n (only) have the property that phi(n)=(n+1)/2, where phi(n) = A000010(n) is Euler's totient function. - Lekraj Beedassy, Feb 12 2007
If A003558(n) is of the form 2^n and A179480(n+1) is even, then (2^(A003558(n)) - 1) is in A051179. Example: A003558(25) = 8 with A179480(25) = 4, even. Then (2^8 - 1) = 255. - Gary W. Adamson, Aug 20 2012
For any odd positive a(0), the sequence defined by a(n) = a(n-1) * (a(n-1) + 2) gives a constructive proof that there exist integers with at least n distinct prime factors, e.g., a(n), since omega(a(n)) >= n. As a corollary, this gives a constructive proof of Euclid's theorem stating that there are infinitely many primes. - Daniel Forgues, Mar 07 2017
From Sergey Pavlov, Apr 24 2017: (Start)
I conjecture that, for n > 7, omega(a(n)) > omega(a(n-1)) > n.
It seems that the largest prime divisor p(n+1) of a(n+1) is always bigger than the largest prime divisor of a(n): p(n+1) > p(n). For 3 < n < 8, p(n+1) > 100 * p(n).
(End)
It appears that a(n) is the integer whose bits indicate the possible subset sums of the first n powers of two. For another example, see the calculation for primes at A368491 - Yigit Oktar, Mar 20 2025

			15 = 3*5; 255 = 3*5*17; 65535 = 3*5*17*257; ... - _Daniel Forgues_, Mar 07 2017

M. Aigner and G. M. Ziegler, Proofs from The Book, Springer-Verlag, Berlin, 1999; see p. 4.

Vincenzo Librandi, Table of n, a(n) for n = 0..11
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!)
J. H. Conway, Integral lexicographic codes, Discrete Mathematics 83.2-3 (1990): 219-235. See p. 235.
Ben Delo and Filip Saidak, Euclid's theorem redux, Fib. Q., 57:4 (2019), 331-336.
Index entries for sequences of form a(n+1)=a(n)^2 + ....

Cf. A001146, A007018, A048649.
Cf. A003558, A179480, A000215.
Cf. A368491

[2^(2^n)-1: n in [0..8]]; // Vincenzo Librandi, Jun 20 2011

A051179:=n->2^(2^n)-1; seq(A051179(n), n=0..8); # Wesley Ivan Hurt, Feb 08 2014

Table[2^(2^n)-1,{n,0,9}] (* Vladimir Joseph Stephan Orlovsky, Mar 16 2010 *)

PARI
```
a(n)=if(n<0,0,2^2^n-1)
```

def A051179(n): return (1<<(1<Chai Wah Wu, May 03 2023

a(n) = A000215(n) - 2.
a(n) = (a(n-1) + 1)^2 - 1, a(0) = 1. [ or a(n) = a(n-1)(a(n-1) + 2) ].
1 = 2/3 + 4/15 + 16/255 + 256/65535 + ... = Sum_{n>=0} A001146(n)/a(n+1) with partial sums: 2/3, 14/15, 254/255, 65534/65535, ... - Gary W. Adamson, Jun 15 2003
a(n) = b(n-1) where b(1)=1, b(n) = Product_{k=1..n-1} (b(k) + 2). - Benoit Cloitre, Sep 13 2003
A136308(n) = A007088(a(n)). - Jason Kimberley, Dec 19 2012
A000215(n) = a(n+1) / a(n). - Daniel Forgues, Mar 07 2017
Sum_{n>=0} 1/a(n) = A048649. - Amiram Eldar, Oct 27 2020

A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.

Original entry on oeis.org

3, 7, 47, 2207, 4870847, 23725150497407, 562882766124611619513723647, 316837008400094222150776738483768236006420971486980607

Offset: 0

N. J. A. Sloane

Expansion of 1/phi: 1/phi = (1-1/3)*(1-1/((3-1)*7))*(1-1/(((3-1)*7-1)*47))*(1-1/((((3-1)*7-1)*47-1)*2207))... (phi being the golden ration (1+sqrt(5))/2). - Thomas Baruchel, Nov 06 2003
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
Starting with 7, the terms end with 7,47,07,47,07,..., of the form 8a+7 where a = 0,1,55,121771,... Conjecture: Every a is squarefree, every other a is divisible by 55, the a's are a subset of A046194, the heptagonal triangular numbers (the first, 2nd, 3rd, 6th, 11th, ?, ... terms). - Gerald McGarvey, Aug 08 2004
Also the reduced numerator of the convergents to sqrt(5) using Newton's recursion x = (5/x+x)/2. - Cino Hilliard, Sep 28 2008
The subsequence of primes begins a(n) for n = 0, 1, 2, 3. - Jonathan Vos Post, Feb 26 2011
We have Sum_{n=0..N} a(n)^2 = 2*(N+1) + Sum_{n=1..N+1} a(n), Sum_{n=0..N} a(n)^4 = 5*(Sum_{n=1..N+1} a(n)) + a(N+1)^2 + 6*N -3, etc. which is very interesting with respect to the fact that a(n) = Lucas(2^(n+1)); see W. Webb's problem in Witula-Slota's paper. - Roman Witula, Nov 02 2012
From Peter Bala, Nov 11 2012: (Start)
The present sequence corresponds to the case x = 3 of the following general remarks.
The recurrence a(n+1) = a(n)^2 - 2 with initial condition a(0) = x > 2 has the solution a(n) = ((x + sqrt(x^2 - 4))/2)^(2^n) + ((x - sqrt(x^2 - 4))/2)^(2^n).
We have the product expansion sqrt(x + 2)/sqrt(x - 2) = Product_{n>=0} (1 + 2/a(n)) (essentially due to Euler - see Mendes-France and van der Poorten). Another expansion is sqrt(x^2 - 4)/(x + 1) = Product_{n>=0} (1 - 1/a(n)), which follows by iterating the identity sqrt(x^2 - 4)/(x + 1) = (1 - 1/x)*sqrt(y^2 - 4)/(y + 1), where y = x^2 - 2.
The sequence b(n) := a(n) - 1 satisfies b(n+1) = b(n)^2 + 2*b(n) - 2. Cases currently in the database are A145502 through A145510. The sequence c(n) := a(n)/2 satisfies c(n+1) = 2*c(n)^2 - 1. Cases currently in the database are A002812, A001601, A005828, A084764 and A084765.
(End)
E. Lucas in Section XIX of "The Theory of Simply Periodic Numerical Functions" (page 56 of English translation) equation "(127) (1-sqrt(5))/2 = -1/1 + 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ..." - Michael Somos, Oct 11 2022
Let b(n) = a(n) - 3. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. - Peter Bala, Dec 08 2022
The number of digits of a(n) is given by A094057(n+1). - Hans J. H. Tuenter, Jul 29 2025

			From _Cino Hilliard_, Sep 28 2008: (Start)
Init x=1;
x = (5/1 + 1)/2 = 3/1;
x = (5/3 + 3)/2 = 7/3;
x = ((5/7)/3 + 7/3)/2 = 47/21;
x = ((5/47)/21 + 47/21)/2 = 2207/987;
(2207/987)^2 = 5.000004106... (End)

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 397.
E.-B. Escott, Note #1741, L'Intermédiaire des Mathématiciens, 8 (1901), page 13. - N. J. A. Sloane, Mar 02 2022
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 223.
Édouard Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 7.
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).

Vincenzo Librandi, 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!)
Pierre Liardet and Pierre Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Édouard Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
Édouard Lucas, The Theory of Simply Periodic Numerical Functions, Fibonacci Association, 1969. English translation of article "Théorie des Fonctions Numériques Simplement Périodiques, I", Amer. J. Math., 1 (1878), 184-240.
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
E. L. Roettger and H. C. Williams, Some Remarks Concerning the Lucas-Lehmer Primality Test, Journal of Integer Sequences, Vol. 28 (2025), Article 25.2.5. See p. 2.
Chance Sanford, Infinite Series Involving Fibonacci Numbers Via Apéry-Like Formulae, arXiv:1603.03765 [math.NT], 2016.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Wikipedia, Engel Expansion.
Roman Wituła and Damian Słota, delta-Fibonacci numbers, Applicable Analysis and Discrete Mathematics, Vol. 3, No. 2 (2009), pp. 310-329.
Index entries for sequences of form a(n+1)=a(n)^2 + ....

Lucas numbers (A000032) with subscripts that are powers of 2 greater than 1 (Herbert S. Wilf). Cf. A000045.
Cf. A003010 (starting with 4), A003423 (starting with 6), A003487 (starting with 5).
Cf. A058635. - Artur Jasinski, Oct 05 2008
Cf. A002812, A094874, A145502, A145274, A050614, A088334, A181393, A181419, A186750, A186751, A338304.

a:= n-> simplify(2*ChebyshevT(2^n, 3/2), 'ChebyshevT'):
seq(a(n), n=0..8);

NestList[#^2-2&,3,10] (* Harvey P. Dale, Dec 17 2014 *)
Table[LucasL[2^n], {n, 1, 8}] (* Amiram Eldar, Oct 22 2020 *)

a[0]:3$
a[n]:=a[n-1]^2-2$
A001566(n):=a[n]$
makelist(A001566(n),n,0,7); /* Martin Ettl, Nov 12 2012 */

{a(n) = if( n<1, 3*(n==0), a(n-1)^2 - 2)}; /* Michael Somos, Mar 14 2004 */

g(n,p) = x=1;for(j=1,p,x=(n/x+x)/2;print1(numerator(x)","));
g(5,8) \\ Cino Hilliard, Sep 28 2008

{a(n) = my(w = quadgen(5)); if( n<0, 0, n++; imag( (2*w - 1) * w^2^n ))}; /* Michael Somos, Nov 30 2014 */

{a(n) = my(y = x^2-x-1); if( n<0, 0, n++; for(i=1, n, y = polgraeffe(y)); -polcoeff(y, 1))}; /* Michael Somos, Nov 30 2014 */

a(n) = Fibonacci(2^(n+2))/Fibonacci(2^(n+1)) = A058635(n+2)/A058635(n+1). - Len Smiley, May 08 2000, and Artur Jasinski, Oct 05 2008
a(n) = ceiling(c^(2^n)) where c = (3+sqrt(5))/2 = tau^2 is the largest root of x^2-3*x+1=0. - Benoit Cloitre, Dec 03 2002
a(n) = round(G^(2^n)) where G is the golden ratio (A001622). - Artur Jasinski, Sep 22 2008
a(n) = (G^(2^(n+1))-(1-G)^(2^(n+1)))/((G^(2^n))-(1-G)^(2^n)) = G^(2^n)+(1-G)^(2^n) = G^(2^n)+(-G)^(-2^n) where G is the golden ratio. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(2^(n+1)*arccosh(sqrt(5)/2)). - Artur Jasinski, Oct 09 2008
a(n) = Fibonacci(2^(n+1)-1) + Fibonacci(2^(n+1)+1). (3-sqrt(5))/2 = 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ... (E. Lucas). - Philippe Deléham, Apr 21 2009
a(n)*(a(n+1)-1)/2 = A023039(2^n). - M. F. Hasler, Sep 27 2009
For n >= 1, a(n) = 2 + Product_{i=0..n-1} (a(i) + 2). - Vladimir Shevelev, Nov 28 2010
a(n) = 2*T(2^n,3/2) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
From Peter Bala, Oct 31 2012: (Start)
Engel expansion of 1/2*(3 - sqrt(5)). Thus 1/2*(3 - sqrt(5)) = 1/3 + 1/(3*7) + 1/(3*7*47) + ... as noted above by Deleham. See Liardet and Stambul.
sqrt(5)/4 = Product_{n>=0} (1 - 1/a(n)).
sqrt(5) = Product_{n>=0} (1 + 2/a(n)). (End)
a(n) - 1 = A145502(n+1). - Peter Bala, Nov 11 2012
a(n) == 2 (mod 9), for n > 1. - Ivan N. Ianakiev, Dec 25 2013
From Amiram Eldar, Oct 22 2020: (Start)
a(n) = A000032(2^(n+1)).
Sum_{k>=0} 1/a(k) = -1 + A338304. (End)
a(n) = (A000045(m+2^(n+2))+A000045(m))/A000045(m+2^(n+1)) for any m>=0. - Alexander Burstein, Apr 10 2021
a(n) = 2*cos(2^n*arccos(3/2)). - Peter Luschny, Oct 12 2022
a(n) == -1 ( mod 2^(n+2) ). - Peter Bala, Nov 07 2022
a(n) = 5*Fibonacci(2^n)^2+2 = 5*A058635(n)^2+2, for n>0. - Jianglin Luo, Sep 21 2023
Sum_{n>=0} a(n)/Fibonacci(2^(n+2)) = A094874 (Sanford, 2016). - Amiram Eldar, Mar 01 2024

A003010 A Lucas-Lehmer sequence: a(0) = 4; for n>0, a(n) = a(n-1)^2 - 2.

Original entry on oeis.org

4, 14, 194, 37634, 1416317954, 2005956546822746114, 4023861667741036022825635656102100994, 16191462721115671781777559070120513664958590125499158514329308740975788034

Offset: 0

N. J. A. Sloane

nonn

Albert Beiler states (page 228 of Recreations in the Theory of Numbers): D. H. Lehmer modified Lucas's test to the relatively simple form: If and only if 2^n-1 divides a(n-2) then 2^n-1 is a prime, otherwise it is composite. Since 2^3 - 1 is a factor of a(1) = 14, 2^3 - 1 = 7 is a prime. - Gary W. Adamson, Jun 07 2003

a(n) - a(n-1) divides a(n+1) - a(n). - Thomas Ordowski, Dec 24 2016

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 228.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 399.
R. K. Guy, Unsolved Problems in Number Theory, Section A3.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 78.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 205.

N. J. A. Sloane, Table of n, a(n) for n = 0..10
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!)
Larry Ericksen, Primality Testing and Prime Constellations, Šiauliai Mathematical Seminar, Vol. 3 (11), 2008.
James S. Hall, A remark on the primeness of Mersenne numbers, J. London Math. Soc. 28, (1953). 285-287.
D. H. Lehmer, On Lucas's test for the primality of Mersenne's numbers, Journal of the London Mathematical Society 1.3 (1935): 162-165. See page 162.
D. H. Lehmer. Mathematical Reviews. Review of: Hall, James S. A remark on the primeness of Mersenne numbers. (1953) [Annotated scanned copy]
Pierre Liardet and Pierre Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Mario Raso, Integer Sequences in Cryptography: A New Generalized Family and its Application, Ph. D. Thesis, Sapienza University of Rome (Italy 2025). See p. 25.
E. L. Roettger and H. C. Williams, Some Remarks Concerning the Lucas-Lehmer Primality Test, Journal of Integer Sequences, Vol. 28 (2025), Article 25.2.5. See p. 3.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Pierluigi Vellucci and Alberto Maria Bersani, The class of Lucas-Lehmer polynomials, arXiv preprint arXiv:1603.01989 [math.CA], 2016.
Pierluigi Vellucci and Alberto Maria Bersani, New formulas for pi involving infinite nested square roots and Gray code, arXiv preprint arXiv:1606.09597 [math.NT], 2016 (The OEIS is cited in version 1, but has been dropped from version 4.)
Eric Weisstein's World of Mathematics, Lucas-Lehmer Test.
Wikipedia, Engel Expansion
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A001566 (starting with 3), A003423 (starting with 6), A003487 (starting with 5).

Cf. A002812, A145503.

[n le 1 select 4 else Self(n-1)^2-2: n in [1..10]]; // Vincenzo Librandi, Aug 24 2015

a := n-> if n>0 then a(n-1)^2-2 else 4 fi: 'a(i)' $ i=0..9; # M. F. Hasler, Mar 09 2007
a := n-> simplify(2*ChebyshevT(2^n, 2), 'ChebyshevT'): seq(a(n), n=0..7);

seqLucasLehmer[0] = 4; seqLucasLehmer[n_] := seqLucasLehmer[n - 1]^2 - 2; Array[seqLucasLehmer, 8, 0] (* Robert G. Wilson v, Jun 28 2012 *)

a(n)=if(n,a(n-1)^2-2,4)
vector(10,i,a(i-1)) \\ M. F. Hasler, Mar 09 2007

from itertools import accumulate
def f(anm1, _): return anm1**2 - 2
print(list(accumulate([4]*8, f))) # Michael S. Branicky, Apr 14 2021

a(n) = ceiling((2 + sqrt(3))^(2^n)). - Benoit Cloitre, Nov 30 2002
More generally, if u(0) = z, integer > 2 and u(n) = a(n-1)^2 - 2 then u(n) = ceiling(c^(2^n)) where c = (1/2)*(z+sqrt(z^2-4)) is the largest root of x^2 - zx + 1 = 0. - Benoit Cloitre, Dec 03 2002
a(n) = (2+sqrt(3))^(2^n) + (2-sqrt(3))^(2^n). - John Sillcox (johnsillcox(AT)hotmail.com), Sep 20 2003
a(n) = ceiling(tan(5*Pi/12)^(2^n)). Note: 5*Pi/12 radians is 75 degrees. - Jason M. Follas (jasonfollas(AT)hotmail.com), Jan 16 2004
Sum_{n >= 0} 1/( Product_{k = 0..n} a(k) ) = 2 - sqrt(3). - Paul D. Hanna, Aug 11 2004
From Ulrich Sondermann, Sep 04 2006: (Start)
To generate the n-th number in the sequence: let x = 2^(n-1), a = 2, b = sqrt(3). Take every other term of the binomial expansion (a+b)^x times 2.
E.g., for the 4th term: x = 2^(4-1) = 8, the binomial expansion is: a^8 + 7a^7 b + 28a^6 b^2 + 56a^5 b^3 + 70a^4 b^4 + 56a^3 b^5 + 28a^2 b^6 + 7a b^7 + b^8, every other term times 2: 2(a^8 + 28a^6 b^2 + 70a^4 b^4 + 28a^2 b^6 + b^8) = 2(256 + (28)(64)(3) + (70)(16)(9) + (28)(4)(27) + 81) = 2(18817) = 37634. (End)
a(n) = 2*cosh( 2^(n-1)*log(sqrt(3)+2) ) For n > 0, a(n) = 2 + 3 * 4^n * (Product_{k=0..n-2} (a(k)/2))^2, where a(k)/2 = A002812(k) is a coprime sequence. - M. F. Hasler, Mar 09 2007
a(n) = A003500(2^n). - John Blythe Dobson, Oct 28 2007
a(n) = 2*T(2^n,2) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
Engel expansion of 2 - sqrt(3). Thus 2 - sqrt(3) = 1/4 + 1/(4*14) + 1/(4*14*194) + ... as noted by Hanna above. See Liardet and Stambul. Cf. A001566, A003423 and A003487. - Peter Bala, Oct 31 2012
From Peter Bala, Nov 11 2012: (Start)
2*sqrt(3)/5 = Product_{n = 0..oo} (1 - 1/a(n)).
sqrt(3) = Product_{n = 0..oo} (1 + 2/a(n)).
a(n) - 1 = A145503(n+1).
a(n) = 2*A002812(n). (End)
a(n+1) - a(n) = a(n)^2 - a(n-1)^2. - Thomas Ordowski, Dec 24 2016
a(n) = 2*cos(2^n * arccos(2)). - Ryan Brooks, Oct 27 2020
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2 + 2*Product_{k = 0..n-1} (a(k) + 2) for n >= 1.
Let b(n) = a(n) - 4. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. (End)

One more term from Thomas A. Rockwell (LlewkcoRAT(AT)aol.com), Jan 18 2005

A374663 Lexicographically earliest sequence of positive integers a(1), a(2), a(3), ... such that for any n > 0, Sum_{k = 1..n} 1 / (k*a(k)) < 1.

Original entry on oeis.org

2, 2, 2, 4, 10, 201, 34458, 1212060151, 1305857607493406801, 1534737681943564047120326770001682121, 2141290683979549415450148346297540185977813099483710032048213090481251382

Offset: 1

Rémy Sigrist, Aug 04 2024

nonn

The harmonic series, Sum_{k > 0} 1/k, diverges. We divide each of its terms in such a way as to have a series bounded by 1.

			The initial terms, alongside the corresponding sums, are:
  n  a(n)        Sum_{k=1..n} 1/(k*a(k))
  -  ----------  -----------------------------------------
  1           2  1/2
  2           2  3/4
  3           2  11/12
  4           4  47/48
  5          10  1199/1200
  6         201  241199/241200
  7       34458  9696481199/9696481200
  8  1212060151  11752718467440661199/11752718467440661200
...
The denominators are in A375516.

Rémy Sigrist and N. J. A. Sloane, Dampening Down a Divergent Series, Manuscript in preparation, September 2024.

N. J. A. Sloane, Table of n, a(n) for n = 1..14
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]

Cf. A000058, A001008/A002805, A002387, A374983, A375516, A375517.

s:= proc(n) s(n):= `if`(n=0, 0, s(n-1)+1/(n*a(n))) end:
a:= proc(n) a(n):= 1+floor(1/((1-s(n-1))*n)) end:
seq(a(n), n=1..11);  # Alois P. Heinz, Oct 18 2024

s[n_] := s[n] = If[n == 0, 0, s[n - 1] + 1/(n*a[n])];
a[n_] := 1 + Floor[1/((1 - s[n - 1])*n)];
Table[a[n], {n, 1, 11}] (* Jean-François Alcover, Jan 09 2025, after Alois P. Heinz *)

{ t = 0; for (n = 1, 11, for (v = ceil(1/(n*(1-t))), oo, if (t + 1/(n*v) < 1, t += 1/(n*v); print1 (v", "); break;););); }

from itertools import count, islice
from math import gcd
def A374663_gen(): # generator of terms
    p, q = 0, 1
    for k in count(1):
        yield (m:=q//(k*(q-p))+1)
        p, q = p*k*m+q, k*m*q
        p //= (r:=gcd(p,q))
        q //= r
A374663_list = list(islice(A374663_gen(),11)) # Chai Wah Wu, Aug 28 2024

The ratios a(n)^2/a(n+1) are very close to the values 2, 2, 1, 8/5, 1/2, 7/6, 48/49, 9/8, 10/9, 11/10, 24/11^2, 13/12, 56/13^2, ... So it seems that often (but not always), a(n+1) is very close to (n/(n+1))*a(n)^2. - N. J. A. Sloane, Sep 08 2024

A002065 a(n+1) = a(n)^2 + a(n) + 1.

Original entry on oeis.org

0, 1, 3, 13, 183, 33673, 1133904603, 1285739649838492213, 1653126447166808570252515315100129583, 2732827050322355127169206170438813672515557678636778921646668538491883473

Offset: 0

N. J. A. Sloane

a(n) is the number of trees of height <= n, generated by unary and binary composition: S = x + (S) + (S,S) = x + (x) + (x,x) + (x,(x)) + ((x),x) + ((x)) + ((x),(x)) + (x,(x,x)) + ((x,x),x) + ((x),(x,x)) + ((x,x),(x)) + ((x,x)) + ((x,x),(x,x)) + ... (x is of height 1); the first difference sequence (beginning with 1), 1 2 10 170 33490 1133870930..., gives the number h(n) of these trees whose height is n, h(n + 1) = h(n) + h(n)*h(n) + 2h(n)*a(n-1), h(1) = 1; as h(n + 1)/h(n) = 1 + a(n) + a(n-1) gives sequence 1, 2, 10 (2*5), 170 (2*5*17), 33490 (2*5*17*197), 1133870930 (2*5*17*197*33877), ... - Claude Lenormand (claude.lenormand(AT)free.fr), Sep 05 2001

This is a divisibility sequence, that is, if n divides m, then a(n) divides a(m). This is a particular case of the result: if p(x) is an integral polynomial then the sequence of n-th iterates p^n(x) (:= p(p^(n-1)(x)) with p^1(x) := p(x)), n = 1,2,..., of p(x) evaluated at x = 0 is a divisibility sequence. In this case p(x) = 1 + x + x^2. - Peter Bala, Mar 28 2018

Mordechai Ben-Ari, Mathematical Logic for Computer Science, Third edition, 173-203.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 433-434.
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!)
Steven R. Finch, Lehmer's Constant [Broken link]
Steven R. Finch, Lehmer's Constant [From the Wayback machine]
Stan C. Kalman and Barry L. Kwasny, Tail-recursive distributed representations and simple recurrent networks, Connection Science, 7 (1995), 61-80.
D. H. Lehmer, A cotangent analogue of continued fractions, Duke Math. J., 4 (1935), 323-340.
D. H. Lehmer, A cotangent analogue of continued fractions, Duke Math. J., 4 (1935), 323-340. [Annotated scanned copy]
H. P. Robinson, Letter to N. J. A. Sloane, Jul 12 1971
Eric Weisstein's World of Mathematics, Lehmer's Constant
Eric Weisstein's World of Mathematics, Lehmer Cotangent Expansion
Wikipedia, Herbrand Structure
J. W. Wrench, Jr., Letters to N. J. A. Sloane, Feb 1974
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A002665, A002794, A002795, A030125, A063573, A232806.

[n le 1 select 0 else Self(n-1)^2 + Self(n-1) + 1: n in [1..15]]; // Vincenzo Librandi, Oct 05 2015

f[x_] := 1 + x + x^2 ; NestList[f, 1, 7] (* Geoffrey Critzer, May 04 2010 *)

a(n) := if n = 0 then 1 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<1,0,a(n-1)^2+a(n-1)+1)
```

a(n) = floor(c^(2^n)) for n > 0, where c = 1.385089248334672909882206535871311526236739234374149506334120193387331772... - Benoit Cloitre, Nov 29 2002

a(n) = (A232806(n) - 1)/2 = (A232806(n-1)^2 + 3)/4. - Peter Bala, Mar 28 2018

A082732 a(1) = 1, a(2) = 3, a(n) = LCM of all the previous terms + 1.

Original entry on oeis.org

1, 3, 4, 13, 157, 24493, 599882557, 359859081592975693, 129498558604939936868397356895854557, 16769876680757063368089314196389622249367851612542961252860614401811693

Offset: 1

Amarnath Murthy, Apr 14 2003

nonn

The LCM is in fact the product of all previous terms. From a(5) onwards the terms alternately end in 57 and 93.

Cf. A000058, A004168, A144743, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

a[1] = 1; a[2] = 3; a[n_] := Apply[LCM, Table[a[i], {i, 1, n - 1}]] + 1; Table[ a[n], {n, 1, 10}]
c=1.8806785436830780944921917650127503562630617563236301969047995953391479871\
7695395204087358090874194124503892563356447954254847544689332763; Table[c^(2^n),{n,1,6}] or a = {}; k = 4; Do[AppendTo[a, k]; k = k^2 - k + 1, {n, 1, 10}]; a (* Artur Jasinski, Sep 22 2008 *)

For n>=3, a(n+1) = a(n)^2 - a(n) + 1.
For n>=3, a(n) = A004168(n-3) + 1. - Max Alekseyev, Aug 09 2019
1/3 = Sum_{n=3..oo} 1/a(n) = 1/4 + 1/13 + 1/157 + 1/24493 + ... or 1 = Sum_{n=3..oo} 3/a(n) = 3/4 + 3/13 + 3/157 + 3/24493 + .... If we take segment of length 1 and cut off in each step fragment of maximal length such that numerator of fraction is 3, denominators of such fractions will be successive numbers of this sequence. - Artur Jasinski, Sep 22 2008
a(n+2)=1.8806785436830780944921917650127503562630617563236301969047995953391\
4798717695395204087358090874194124503892563356447954254847544689332763...^(2^n). -  Artur Jasinski, Sep 22 2008

More terms from Robert G. Wilson v, Apr 15 2003

A000435 Normalized total height of all nodes in all rooted trees with n labeled nodes.

Original entry on oeis.org

0, 1, 8, 78, 944, 13800, 237432, 4708144, 105822432, 2660215680, 73983185000, 2255828154624, 74841555118992, 2684366717713408, 103512489775594200, 4270718991667353600, 187728592242564421568, 8759085548690928992256, 432357188322752488126152, 22510748754252398927872000

Offset: 1

N. J. A. Sloane

This is the sequence that started it all: the first sequence in the database!
The height h(V) of a node V in a rooted tree is its distance from the root. a(n) = Sum_{all nodes V in all n^(n-1) rooted trees on n nodes} h(V)/n.
In the trees which have [0, n-1] = (0, 1, ..., n-1) as their ordered set of nodes, the number of nodes at distance i from node 0 is f(n,i) = (n-1)...(n-i)(i+1)n^(j-1), 0 <= i < n-1, i+j = n-1 (and f(n,n-1) = (n-1)!): (n-1)...(n-i) counts the words coding the paths of length i from any node to 0, n^(j-1) counts the Pruefer codes of the rest, words build by iterated deletion of the greater node of degree 1 ... except the last one, (i+1), necessary pointing at the path. If g(n,i) = (n-1)...(n-i)n^j, i+j = n-1, f(n,i) = g(n,i) - g(n,i+1), g(n,i) = Sum_{k>=i} f(n,k), the sequence is Sum_{i=1..n-1} g(n,i). - Claude Lenormand (claude.lenormand(AT)free.fr), Jan 26 2001
If one randomly selects one ball from an urn containing n different balls, with replacement, until exactly one ball has been selected twice, the probability that this ball was also the second ball to be selected once is a(n)/n^n. See also A001865. - Matthew Vandermast, Jun 15 2004
a(n) is the number of connected endofunctions with no fixed points. - Geoffrey Critzer, Dec 13 2011
a(n) is the number of weakly connected simple digraphs on n labeled nodes where every node has out-degree 1. A digraph where all out-degrees are 1 can be called a functional digraph due to the correspondence with endofunctions. - Andrew Howroyd, Feb 06 2024

			For n = 3 there are 3^2 = 9 rooted labeled trees on 3 nodes, namely (with o denoting a node, O the root node):
   o
   |
   o     o   o
   |      \ /
   O       O
The first can be labeled in 6 ways and contains nodes at heights 1 and 2 above the root, so contributes 6*(1+2) = 18 to the total; the second can be labeled in 3 ways and contains 2 nodes at height 1 above the root, so contributes 3*2=6 to the total, giving 24 in all. Dividing by 3 we get a(3) = 24/3 = 8.
For n = 4 there are 4^3 = 64 rooted labeled trees on 4 nodes, namely (with o denoting a node, O the root node):
   o
   |
   o     o        o   o
   |     |         \ /
   o     o   o      o     o o o
   |      \ /       |      \|/
   O       O        O       O
  (1)     (2)      (3)     (4)
Tree (1) can be labeled in 24 ways and contains nodes at heights 1, 2, 3 above the root, so contributes 24*(1+2+3) = 144 to the total;
tree (2) can be labeled in 24 ways and contains nodes at heights 1, 1, 2 above the root, so contributes 24*(1+1+2) = 96 to the total;
tree (3) can be labeled in 12 ways and contains nodes at heights 1, 2, 2 above the root, so contributes 12*(1+2+2) = 60 to the total;
tree (4) can be labeled in 4 ways and contains nodes at heights 1, 1, 1 above the root, so contributes 4*(1+1+1) = 12 to the total;
giving 312 in all. Dividing by 4 we get a(4) = 312/4 = 78.

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).

Robert G. Wilson v, Table of n, a(n) for n = 1..1000 (first 100 terms from T. D. Noe)
Vijayakumar Ambat, Article in the Malayalam newspaper Ayala Manorama - Padhippura, 12 June 2015, that mentions the OEIS, and in particular this sequence.
V. I. Arnold, Topological classification of complex trigonometric polynomials and the combinatorics of graphs with the same number of edges and vertices, Functional Anal. Appl., 30 (1996), 1-17.
Shalosh B. Ekhad and Doron Zeilberger, Going Back to Neil Sloane's FIRST LOVE (OEIS Sequence A435): On the Total Heights in Rooted Labeled Trees, arXiv:1607.05776 [math.CO], 2016.
Shalosh B. Ekhad and Doron Zeilberger, Going Back to Neil Sloane's FIRST LOVE (OEIS Sequence A435): On the Total Heights in Rooted Labeled Trees, Version on DZ's home page with more links; Local copy, pdf file only, no active links
I. P. Goulden and D. M. Jackson, A proof of a conjecture for the number of ramified coverings of the sphere by the torus, arXiv:math/9902009 [math.AG], 1999.
I. P. Goulden, D. M. Jackson, and A. Vainshtein, The number of ramified coverings of the sphere by the torus and surfaces of higher genera, arXiv:math/9902125 [math.AG], 1999.
I. P. Goulden, D. M. Jackson, and A. Vainshtein, The number of ramified coverings of the sphere by the torus and surfaces of higher genera Ann. Comb. 4 (2000), no. 1, 27-46. (See Theorem 1.1.)
Brady Haran, The Number Collector (with Neil Sloane), Numberphile Podcast (2019)
Andrew Lohr and Doron Zeilberger, On the limiting distributions of the total height on families of trees, Integers (2018) 18, Article #A32.
T. Kyle Petersen, Exponential generating functions and Bell numbers, Inquiry-Based Enumerative Combinatorics (2019) Chapter 7, Undergraduate Texts in Mathematics, Springer, Cham, 98-99.
A. Rényi and G. Szekeres, On the height of trees, Journal of the Australian Mathematical Society 7.04 (1967): 497-507. See (4.7).
Marko Riedel et al., Connected endofunctions with no fixed points, Mathematics Stack Exchange, Dec 2014.
J. Riordan, Letter to N. J. A. Sloane, Aug. 1970
J. Riordan and N. J. A. Sloane, Enumeration of rooted trees by total height, J. Austral. Math. Soc., vol. 10 pp. 278-282, 1969.
N. J. A. Sloane, Page from 1964 notebook showing start of OEIS [includes A000027, A000217, A000292, A000332, A000389, A000579, A000110, A007318, A000058, A000215, A000289, A000324, A234953 (= A001854(n)/n), A000435, A000169, A000142, A000272, A000312, A000111]
N. J. A. Sloane, Cover of same notebook
N. J. A. Sloane, Lengths of Cycle Times in Random Neural Networks, Ph. D. Dissertation, Cornell University, February 1967; also Report No. 10, Cognitive Systems Research Program, Cornell University, 1967. This sequence appears on page 119.
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
N. J. A. Sloane, Illustration of a(3) and a(4)
Yukun Yao and Doron Zeilberger, An Experimental Mathematics Approach to the Area Statistic of Parking Functions, arXiv:1806.02680 [math.CO], 2018.
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 3.
Index entries for sequences related to rooted trees
Index entries for sequences related to trees

Cf. A001863, A001864, A001854, A002862 (unlabeled version), A234953, A259334.

Column k=1 of A350452.

A000435 := n-> (n-1)!*add (n^k/k!, k=0..n-2);
seq(simplify((n-1)*GAMMA(n-1,n)*exp(n)),n=1..20); # Vladeta Jovovic, Jul 21 2005

f[n_] := (n - 1)! Sum [n^k/k!, {k, 0, n - 2}]; Array[f, 18] (* Robert G. Wilson v, Aug 10 2010 *)
nx = 18; Rest[ Range[0, nx]! CoefficientList[ Series[ LambertW[-x] - Log[1 + LambertW[-x]], {x, 0, nx}], x]] (* Robert G. Wilson v, Apr 13 2013 *)

x='x+O('x^30); concat(0, Vec(serlaplace(lambertw(-x)-log(1+lambertw(-x))))) \\ Altug Alkan, Sep 05 2018

A000435(n)=(n-1)*A001863(n) \\ M. F. Hasler, Dec 10 2018

from math import comb
def A000435(n): return ((sum(comb(n,k)*(n-k)**(n-k)*k**k for k in range(1,(n+1>>1)))<<1) + (0 if n&1 else comb(n,m:=n>>1)*m**n))//n # Chai Wah Wu, Apr 25-26 2023

a(n) = (n-1)! * Sum_{k=0..n-2} n^k/k!.
a(n) = A001864(n)/n.
E.g.f.: LambertW(-x) - log(1+LambertW(-x)). - Vladeta Jovovic, Apr 10 2001
a(n) = A001865(n) - n^(n-1).
a(n) = A001865(n) - A000169(n). - Geoffrey Critzer, Dec 13 2011
a(n) ~ sqrt(Pi/2)*n^(n-1/2). - Vaclav Kotesovec, Aug 07 2013
a(n)/A001854(n) ~ 1/2 [See Renyi-Szekeres, (4.7)]. Also a(n) = Sum_{k=0..n-1} k*A259334(n,k). - David desJardins, Jan 20 2017
a(n) = (n-1)*A001863(n). - M. F. Hasler, Dec 10 2018

Additional references from Valery A. Liskovets
Editorial changes by N. J. A. Sloane, Feb 03 2012
Edited by M. F. Hasler, Dec 10 2018

A001699 Number of binary trees of height n; or products (ways to insert parentheses) of height n when multiplication is non-commutative and non-associative.

Original entry on oeis.org

1, 1, 3, 21, 651, 457653, 210065930571, 44127887745696109598901, 1947270476915296449559659317606103024276803403, 3791862310265926082868235028027893277370233150300118107846437701158064808916492244872560821

Offset: 0

N. J. A. Sloane and Jeffrey Shallit

Approaches 1.5028368...^(2^n), see A077496. Row sums of A065329 as square array. - Henry Bottomley, Oct 29 2001. Also row sum of square array A073345 (AK).

			G.f. = 1 + x + 3*x^2 + 21*x^3 + 651*x^4 + 457653*x^5 + ... - _Michael Somos_, Jun 02 2019

Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 307.
I. M. H. Etherington, On non-associative combinations, Proc. Royal Soc. Edinburgh, 59 (Part 2, 1938-39), 153-162.
I. M. H. Etherington, Some problems of non-associative combinations (I), Edinburgh Math. Notes, 32 (1940), pp. i-vi. Part II is by A. Erdelyi and I. M. H. Etherington, and is on pages vii-xiv of the same issue.
T. K. Moon, Enumerations of binary trees, types of trees and the number of reversible variable length codes, submitted to Discrete Applied Mathematics, 2000.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

David Wasserman, Table of n, a(n) for n = 0..12 [Shortened file because terms grow rapidly: see Wasserman link below for an additional term]
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!)
Mayfawny Bergmann, Efficiency of Lossless Compression of a Binary Tree via its Minimal Directed Acyclic Graph Representation. Rose-Hulman Undergraduate Mathematics Journal: Vol. 15 : Iss. 2, Article 1. (2014).
Henry Bottomley, Illustration of initial terms
I. M. H. Etherington, Non-associate powers and a functional equation, Math. Gaz. 21 (1937), 36-39; addendum 21 (1937), 153.
I. M. H. Etherington, On non-associative combinations, Proc. Royal Soc. Edinburgh, 59 (Part 2, 1938-39), 153-162. [Annotated scanned copy]
Samuele Giraudo, The combinator M and the Mockingbird lattice, arXiv:2204.03586 [math.CO], 2022.
Claude Lenormand, Arbres et permutations II, see p. 6
David E. Narváez, Proof of polynomial recursive equation of order 2, Jun 05 2025.
David Wasserman, Table of n, a(n) for n = 0..13
Eric Weisstein's World of Mathematics, Binary Tree
Index entries for "core" sequences
Index entries for sequences related to rooted trees
Index entries for sequences related to trees
Index entries for sequences related to parenthesizing
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Row sums of A065329.
Column sums of A335919, A335920.
Cf. A002658, A056207, A002449, A003095.
Cf. A004019, A077496, A076949, A213437.

s := proc(n) local i,j,ans; ans := [ 1 ]; for i to n do ans := [ op(ans),2*(add(j,j=ans)-ans[ i ])*ans[ i ]+ans[ i ]^2 ] od; RETURN(ans); end; s(10);

a[0] = 1; a[n_] := a[n] = 2*a[n-1]*Sum[a[k], {k, 0, n-2}] + a[n-1]^2; Table[a[n], {n, 0, 9}] (* Jean-François Alcover, May 16 2012 *)
a[ n_] := If[ n < 2, Boole[n >= 0], With[{u = a[n - 1], v = a[n - 2]}, u (u + v + u/v)]]; (* Michael Somos, Jun 02 2019 *)

{a(n) = if( n<=1, n>=0, a(n-1) * (a(n-1) + a(n-2) + a(n-1) / a(n-2)))}; /* Michael Somos, 2000 */

from functools import lru_cache
@lru_cache(maxsize=None)
def a(n): return 1 if n <= 1 else a(n-1) * (a(n-1) + a(n-2) + a(n-1)//a(n-2))
print([a(n) for n in range(10)]) # Michael S. Branicky, Nov 10 2022 after Michael Somos

a(n+1) = 2*a(n)*(a(0) + ... + a(n-1)) + a(n)^2.
a(n+1) = a(n)^2 + a(n) + a(n)*sqrt(4*a(n)-3), if n > 0.
a(n) = A003095(n+1) - A003095(n) = A003095(n)^2 - A003095(n) + 1. - Henry Bottomley, Apr 26 2001; offset of LHS corrected by Anindya Bhattacharyya, Jun 21 2013
a(n) = A059826(A003095(n-1)).
From Peter Bala, Feb 03 2017: (Start)
a(n) = Product_{k = 1..n} A213437(k).
a(n) + a(n-1) = A213437(n+1) - A213437(n). (End)
a(n) = -a(n-2)^3 + a(n-1)^2 + 3*a(n-1)*a(n-2) + 2*a(n-2)^2 + 2*a(n-1) - 4*a(n-2) (see Narváez link for proof). - Boštjan Gec, Oct 10 2024

Minor edits by Vaclav Kotesovec, Oct 04 2014

A004019 a(0) = 0; for n > 0, a(n) = (a(n-1) + 1)^2.

Original entry on oeis.org

0, 1, 4, 25, 676, 458329, 210066388900, 44127887745906175987801, 1947270476915296449559703445493848930452791204, 3791862310265926082868235028027893277370233152247388584761734150717768254410341175325352025

Offset: 0

N. J. A. Sloane

Take the standard rooted binary tree of depth n, with 2^(n+1) - 1 labeled nodes. Here is a picture of the tree of depth 3:
             R
            / \
           /   \
          /     \
         /       \
        /         \
       o           o
      / \         / \
     /   \       /   \
    o     o     o     o
   / \   / \   / \   / \
  o   o o   o o   o o   o
Let the number of rooted subtrees be s(n). For example, for n = 1 the s(2) = 4 subtrees are:
  R   R   R      R
     /     \    / \
    o       o  o   o
Then s(n+1) = 1 + 2*s(n) + s(n)^2 = (1+s(n))^2 and so s(n) = a(n+1).

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Mordechai Ben-Ari, Mathematical Logic for Computer Science, Third edition, 173-203.

Reinhard Zumkeller, Table of n, a(n) for n = 0..11 (shortened by _N. J. A. Sloane_, Jan 13 2019)
Geir Agnarsson, Elie Alhajjar, and Aleyah Dawkins, On locally finite ordered rooted trees and their rooted subtrees, arXiv:2312.11379 [math.CO], 2023.
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!)
Andressa Paola Cordeiro, Alexandre Tavares Baraviera, and Alex Jenaro Becker, Entropy for k-trees defined by k transition matrices, arXiv:2307.05850 [math.DS], 2023. See p. 15.
F. Disanto and N. A. Rosenberg, Enumeration of ancestral configurations for matching gene trees and species trees, J. Comput. Biol. 24 (2017), 831-850.
E. Lappo and N. A. Rosenberg, A lattice structure for ancestral configurations arising from the relationship between gene trees and species trees, Adv. Appl. Math. 343 (2024), 65-81.
R. P. Stanley, Letter to N. J. A. Sloane, c. 1991
Elmar Teufl and Stephan Wagner, Enumeration problems for classes of self-similar graphs, Journal of Combinatorial Theory, Series A, Volume 114, Issue 7, October 2007, Pages 1254-1277.
Wikipedia, Herbrand Structure
Damiano Zanardini, Computational Logic, UPM European Master in Computational Logic (EMCL) School of Computer Science Technical University of Madrid.
Index entries for sequences of form a(n+1)=a(n)^2 + ...
Index entries for sequences related to rooted trees

Cf. A001699, A056207.
Cf. A000290.
Cf. A003095, A091980, A355108.

a004019 n = a004019_list !! n
a004019_list = iterate (a000290 . (+ 1)) 0
-- Reinhard Zumkeller, Feb 01 2013

[n le 1 select 0 else  (Self(n-1)+1)^2: n in [1..15]]; // Vincenzo Librandi, Oct 05 2015

Table[Nest[(1 + #)^2 &, 0, n], {n, 0, 12}] (* Vladimir Joseph Stephan Orlovsky, Jul 20 2011 *)
NestList[(#+1)^2&,0,10] (* Harvey P. Dale, Oct 08 2011 *)

a(n) = if(n==0, 0, (a(n-1) + 1)^2);
vector(20, n, a(n-1)) \\ Altug Alkan, Oct 06 2015

a(n) = A003095(n)^2 = A003095(n+1) - 1 = A056207(n+1) + 1.
It follows from Aho and Sloane that there is a constant c such that a(n) is the nearest integer to c^(2^n). In fact a(n+1) = nearest integer to b^(2^n) - 1 where b = 2.25851845058946539883779624006373187243427469718511465966.... - Henry Bottomley, Aug 30 2005
a(n) is the number of root ancestral configurations for fully symmetric matching gene trees and species trees with 2^n leaves, a(n) = A355108(2^n). - Noah A Rosenberg, Jun 22 2022

One more term from Henry Bottomley, Jul 24 2000

Additional comments from Max Alekseyev, Aug 30 2005

cp's OEIS Frontend

A051179 a(n) = 2^(2^n) - 1.

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A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.

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A003010 A Lucas-Lehmer sequence: a(0) = 4; for n>0, a(n) = a(n-1)^2 - 2.

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A374663 Lexicographically earliest sequence of positive integers a(1), a(2), a(3), ... such that for any n > 0, Sum_{k = 1..n} 1 / (k*a(k)) < 1.

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A002065 a(n+1) = a(n)^2 + a(n) + 1.

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