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%I #101 Jan 07 2024 10:53:16
%S 1,3,15,255,65535,4294967295,18446744073709551615,
%T 340282366920938463463374607431768211455,
%U 115792089237316195423570985008687907853269984665640564039457584007913129639935
%N a(n) = 2^(2^n) - 1.
%C In a tree with binary nodes (0, 1 children only), the maximum number of unique child nodes at level n.
%C 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
%C 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
%C Partial products of A000215.
%C 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
%C 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
%C 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
%C From _Sergey Pavlov_, Apr 24 2017: (Start)
%C I conjecture that, for n > 7, omega(a(n)) > omega(a(n-1)) > n.
%C 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).
%C (End)
%D M. Aigner and G. M. Ziegler, Proofs from The Book, Springer-Verlag, Berlin, 1999; see p. 4.
%D Ben Delo and Filip Saidak, Euclid's theorem redux, Fib. Q., 57:4 (2019), 331-336.
%H Vincenzo Librandi, <a href="/A051179/b051179.txt">Table of n, a(n) for n = 0..11</a>
%H For rate of growth see A. V. Aho and N. J. A. Sloane, <a href="https://www.fq.math.ca/Scanned/11-4/aho-a.pdf">Some doubly exponential sequences</a>, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437, <a href="http://neilsloane.com/doc/doubly.html">alternative link</a>.
%H J. H. Conway, <a href="https://doi.org/10.1016/0012-365X(90)90008-6">Integral lexicographic codes</a>, Discrete Mathematics 83.2-3 (1990): 219-235. See p. 235.
%H <a href="/index/Aa#AHSL">Index entries for sequences of form a(n+1)=a(n)^2 + ...</a>
%F a(n) = A000215(n) - 2.
%F a(n) = (a(n-1) + 1)^2 - 1, a(0) = 1. [ or a(n) = a(n-1)(a(n-1) + 2) ].
%F 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
%F a(n) = b(n-1) where b(1)=1, b(n) = Product_{k=1..n-1} (b(k) + 2). - _Benoit Cloitre_, Sep 13 2003
%F A136308(n) = A007088(a(n)). - _Jason Kimberley_, Dec 19 2012
%F A000215(n) = a(n+1) / a(n). - _Daniel Forgues_, Mar 07 2017
%F Sum_{n>=0} 1/a(n) = A048649. - _Amiram Eldar_, Oct 27 2020
%e 15 = 3*5; 255 = 3*5*17; 65535 = 3*5*17*257; ... - _Daniel Forgues_, Mar 07 2017
%p A051179:=n->2^(2^n)-1; seq(A051179(n), n=0..8); # _Wesley Ivan Hurt_, Feb 08 2014
%t Table[2^(2^n)-1,{n,0,9}] (* _Vladimir Joseph Stephan Orlovsky_, Mar 16 2010 *)
%o (PARI) a(n)=if(n<0,0,2^2^n-1)
%o (Magma) [2^(2^n)-1: n in [0..8]]; // _Vincenzo Librandi_, Jun 20 2011
%o (Python)
%o def A051179(n): return (1<<(1<<n))-1 # _Chai Wah Wu_, May 03 2023
%Y Cf. A001146, A007018, A048649.
%Y Cf. A003558, A179480, A000215.
%K nonn,easy,nice
%O 0,2
%A Alan DeKok (aland(AT)ox.org)
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