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

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

Offset: 0

Author

Alan DeKok (aland(AT)ox.org)

Keywords

• nonn
• easy
• nice

Comments

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

Examples

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

References

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

Links

• 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 + ....

Crossrefs

Cf. A001146, A007018, A048649.

Cf. A003558, A179480, A000215.

Cf. A368491

Programs

Magma

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

Maple

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

Mathematica

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)

Python

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

Formula

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