0,2

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 (deutsch(AT)duke.poly.edu), Oct 31 2002

a(n) has at least n different prime factors [Saidak]

Subsequence of square-free numbers (A005117). - Reinhard Zumkeller (reinhard.zumkeller(AT)lhsystems.com), Nov 15 2004

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. - D. E. 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 ln(3/2)+sum_{n >= 0} ln(1+(2a_n+1)^{-2}). For example, theta^64-1/2 =~ 3263442.0000000383. - D. E. Knuth, Jun 04 2007

a(n) = A139145(2^(n+1) - 1). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Apr 10 2008

D. R. Curtiss, "On Kellogg's Diophantine Problem". The American Mathematical Monthly Vol. 29, No. 10 (1922), pp. 380-387.

R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 94.

M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see p. 577.

Filip Saidak, A New Proof of Euclid's Theorem, Amer. Math. Monthly, Dec 2006

A. Urquhart, The complexity of propositional proofs, Bull. Symbolic Logic, 1 (1995) pp. 425-, esp. p. 434.

N. J. A. Sloane, Table of n, a(n) for n = 0..12

Index entries for sequences of form a(n+1)=a(n)^2 + ...

a(n) = A000058(n)-1 = A000058(n-1)^2-A000058(n-1) = 1/(1-sum{j<n}[1/A000058(j)]) where A000058 is Sylvester's sequence. - Henry Bottomley (se16(AT)btinternet.com), Jul 23 2001

a(n) = floor(c^(2^n)) where c=1.597910218031873178338070118157... - Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 06 2002

a(1)=1, a(n) = prod(k=1, n-1, a(k)+1) - Benoit Cloitre (benoit7848c(AT)orange.fr), Sep 13 2003

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

Cf. A003687.

Sequence in context: A123137 A014117 A054377 this_sequence A100016 A000610 A023363

Adjacent sequences: A007015 A007016 A007017 this_sequence A007019 A007020 A007021

nonn,nice,easy

njas

The next term is too large to include.