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.

A003095 a(n) = a(n-1)^2 + 1 for n >= 1, with a(0) = 0.

Original entry on oeis.org

0, 1, 2, 5, 26, 677, 458330, 210066388901, 44127887745906175987802, 1947270476915296449559703445493848930452791205, 3791862310265926082868235028027893277370233152247388584761734150717768254410341175325352026
Offset: 0

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Comments

Number of binary trees of height less than or equal to n. [Corrected by Orson R. L. Peters, Jan 03 2020]
The rightmost digits cycle (0,1,2,5,6,7,0,1,2,5,6,7,...). - Jonathan Vos Post, Jul 21 2005
Apart from the initial term, a subsequence of A008318. - Reinhard Zumkeller, Jan 17 2008
Partial sums of A001699. - Jonathan Vos Post, Feb 17 2010
Corresponds to the second and second last diagonals of A119687. - John M. Campbell, Jul 25 2011
This is a divisibility sequence. - Michael Somos, Jan 01 2013
Sum_{n>=1} 1/a(n) = 1.739940825174794649210636285335916041018367182486941... . - Vaclav Kotesovec, Jan 30 2015
From Vladimir Vesic, Oct 03 2015: (Start)
Forming Herbrand's domains of formula: (∃x)(∀y)(∀z)(∃k)(P(x)∨Q(y)∧R(k))
where: x->a
k->f(y,z)
we get:
H0 = {a}
H1 = {a, f(a,a)}
H2 = {a, f(a,a), f(a,f(a,a)), f(f(a,a),a), f(f(a,a),f(a,a))}
...
The number of elements in each domain follows this sequence.
(End)
It is an open question whether or not this sequence satisfies Benford's law [Berger-Hill, 2017] - N. J. A. Sloane, Feb 07 2017
This is a strong divisibility sequence; see A329429. - Clark Kimberling, Nov 13 2019
From Peter Bala, Oct 31 2022: (Start)
Let k be a positive integer. Clearly, the sequence obtained by reducing a(n) modulo k is eventually periodic. Conjectures:
1) The sequence obtained by reducing a(n) modulo 2^k is eventually periodic with period 2.
2) The sequence obtained by reducing a(n) modulo 10^k is eventually periodic with period 6 (the case k = 1 is noted above).
3) The sequence obtained by reducing a(n) modulo 20^k is eventually periodic with period 6.
4) For n >= floor(k/2) and for 1 <= i <= 6, the value of a(6*n+i) mod 10^k is a constant independent of n. The digits of these 6 constant integers, when read from right to left, are the first k digits of the 10-adic numbers A318135 (i = 1), A318136 (i = 2), A318137 (i = 3), A318138 (i = 4), A318139 (i = 5) and A318140 (i = 6), respectively. An example is given below.
n a(6*n+1) mod 10^11
1 10066388901
2 72084948901
3 67988948901
4 61588948901
5 01588948901
6 01588948901
7 01588948901
... ...
A318135 begins 1, 0, 9, 8, 4, 9, 8, 8, 5, 1, 0, 2, .... (End)

References

  • Mordechai Ben-Ari, Mathematical Logic for Computer Science, Third edition, 173-203.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 443-448.
  • R. K. Guy, How to factor a number, Proc. 5th Manitoba Conf. Numerical Math., Congress. Num. 16 (1975), 49-89.
  • R. Penrose, The Emperor's New Mind, Oxford, 1989, p. 122.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A137560, which enumerates binary trees of height less than n and exactly j leaf nodes. - Robert Munafo, Nov 03 2009

Programs

Formula

a(n) = B_{n-1}(1) where B_n(x) = 1 + x*B_{n-1}(x)^2 is the generating function of trees of height <= n.
a(n) is asymptotic to c^(2^n) where c=1.2259024435287485386279474959130085213... (see A076949). - Benoit Cloitre, Nov 27 2002
c = b^(1/4) where b is the constant in Bottomley's formula in A004019. a(n) appears very asymptotic to c^(2^n) - Sum_{k>=1} A088674(k)/(2*c^(2^n))^(2*k-1). - Gerald McGarvey, Nov 17 2007
a(n) = Sum_{i=1..n} A001699(i). - Jonathan Vos Post, Feb 17 2010
G.f. = x + 2*x^2 + 5*x^3 + 26*x^4 + 677*x^5 + 458330*x^6 + 210066388901*x^7 + ... . - Michael Somos, Jan 01 2013
a(2n) mod 2 = 0 ; a(2n+1) mod 2 = 1. - Altug Alkan, Oct 04 2015
a(n) + a(n-1) = A213437(n). - Peter Bala, Feb 03 2017
0 = a(n)^2*(+a(n+1) + a(n+2)) + a(n+1)^2*(-a(n+1) - a(n+2) - a(n+3)) + a(n+2)^3 for all n>=0. - Michael Somos, Feb 10 2017
a(n) = A091980(2^(n-1)) for n > 0. - Alois P. Heinz, Jul 11 2019

Extensions

Additional comments from Cyril Banderier, Jun 05 2000
Minor edits by Vaclav Kotesovec, Oct 04 2014
Initial term clarified by Clark Kimberling, Nov 13 2019