A232806 -id:A232806 - cp's OEIS frontend

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

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

A166105 Quadratic recurrence from Sylvester's sequence, but starting with a(0)=1 and a(1)=2.

Original entry on oeis.org

1, 2, 4, 14, 184, 33674, 1133904604, 1285739649838492214, 1653126447166808570252515315100129584, 2732827050322355127169206170438813672515557678636778921646668538491883474

Offset: 0

Jaume Oliver Lafont, Oct 06 2009

nonn

a(n) is the size of the set S(n) constructed recursively as follows: Let S(1) = {a,b} and let P(S) be the set of pairs (s,t) where s,t are members of S and s not equal to t. We define S(n+1) as the union of S(n) and P(S(n)). - David M. Cerna, Feb 07 2018

David M. Cerna, Table of n, a(n) for n = 0..12

Cf. A000058.

a:= [1, 2];; for n in [3..13] do a[n]:= a[n-1]^2 - a[n-2]^2 + a[n-2]; od; a; # Muniru A Asiru, Feb 07 2018

a := proc(n) option remember: if n=0 then 1 elif n=1 then 2 elif n>=2 then procname(n-1)^2 - procname(n-2)^2 + procname(n-2) fi; end:
seq(a(n), n = 0..10); # Muniru A Asiru, Feb 07 2018
a:=1:A:=a : to 10 do a:=a*(a-1)+2 : A:=A,a od:
print(A); # Robert FERREOL, May 05 2020

RecurrenceTable[{a[n]==a[n-1]^2-a[n-2]^2+a[n-2],a[0]==1,a[1]==2}, a, {n,0,10}] (* Vaclav Kotesovec, Jan 19 2015 *)

a(n)=if(n<2,[1,2][n+1],a(n-1)^2-a(n-2)^2+a(n-2));

Sum_{n>=0} 1/a(n) = 1.82689305142092757947757234878575... (compare with Sum_{n>=0} 1/A000058(n) = 1).
a(n) ~ c^(2^n), where c = 1.385089248334672909882206535871311526236739234374149506334120193387331772... . - Vaclav Kotesovec, Jan 19 2015
Sum_{n>=1} arctan(1/a(n)) = Pi/4. - Carmine Suriano, Apr 07 2015
a(0)=1, a(n+1) = a(n)*(a(n)-1) + 2. - Robert FERREOL, May 05 2020
a(n) = A002065(n) + 1 = (A232806(n) + 1)/2. - Robert FERREOL, May 31 2020

cp's OEIS Frontend

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

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