cp's OEIS Frontend

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

Denominator of b(n) where b(n) = 1/2*(b(n-1) + 1/b(n-1)), b(0)=2. - Vladeta Jovovic, Aug 15 2002

Decimal conversions: 0, 1, 2, 6, 54, 3510, 14380470, 241264265751990, 67909853583655146508751957430, ... . (See A267348.) - Michael De Vlieger, Jan 06 2016

After 10, a(n) is '110' repeated 2^(n-3) times. Therefore, for n>3, a(n) is the concatenation of a(n-1) with itself.

After 1, each term with the 0's omitted is a member of A136308.

The number of digits in a(n) is A098011(n+1).

The number of digits in a(n+2)/a(n+1) gives A103204 with 2 repeated.