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A076725 a(n) = a(n-1)^2 + a(n-2)^4, a(0) = a(1) = 1. 8
1, 1, 2, 5, 41, 2306, 8143397, 94592167328105, 13345346031444632841427643906, 258159204435047592104207508169153297050209383336364487461 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

a(n) and a(n+1) are relatively prime for n >= 0.

The number of independent sets on a complete binary tree with 2^(n-1)-1 nodes. - Jonathan S. Braunhut (jonbraunhut(AT)usa.net), May 04 2004. For example, when n=3, the complete binary tree with 2 levels has 2^2-1 nodes and has 5 independent sets so a(3)=5. The recursion for number of independent sets splits in two cases, with or without the root node being in the set.

a(10) has 113 digits and is too large to include.

LINKS

Robert Israel, Table of n, a(n) for n = 0..13

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

FORMULA

If b(n) = 1 + 1/b(n-1)^2, b(1)=1, then b(n) = a(n)/a(n-1)^2.

Lim_{n->inf} a(n)/a(n-1)^2 = A092526 (constant).

a(n) is asymptotic to c1^(2^n) * c2.

c1 = 1.2897512927198122075..., c2 = 1/A092526 = (1/6)*(108 + 12*sqrt(93))^(1/3) - 2/(108 + 12*sqrt(93))^(1/3) = 0.682327803828019327369483739711... is the root of the equation c2*(1 + c2^2) = 1. - Vaclav Kotesovec, Dec 18 2014

EXAMPLE

a(2) = a(1)^2 + a(0)^4 = 1^2 + 1^4 = 2.

a(3) = a(2)^2 + a(1)^4 = 2^2 + 1^4 = 5.

a(4) = a(3)^2 + a(2)^4 = 5^2 + 2^4 = 41.

a(5) = a(4)^2 + a(3)^4 = 41^2 + 5^4 = 2306.

a(6) = a(5)^2 + a(4)^4 = 2306^2 + 41^4 = 8143397.

a(7) = a(6)^2 + a(5)^4 = 8143397^2 + 2306^4 = 94592167328105.

MAPLE

A[0]:= 1: A[1]:= 1:

for n from 2 to 10 do

  A[n]:= A[n-1]^2 + A[n-2]^4;

od:

seq(A[i], i=0..10); # Robert Israel, Aug 21 2017

MATHEMATICA

f[n_]:=(n+1/n)/n; Prepend[Prepend[Numerator[NestList[f, 2, 8]], 1], 1] (* Vladimir Joseph Stephan Orlovsky, Nov 19 2010 *)

RecurrenceTable[{a[n] == a[n-1]^2 + a[n-2]^4, a[0] ==1, a[1] == 1}, a, {n, 0, 10}] (* Vaclav Kotesovec, Dec 18 2014 *)

PROG

(PARI) {a(n) = if( n<2, 1, a(n-1)^2 + a(n-2)^4)}

(PARI) {a=[0, 0]; for(n=1, 99, iferr(a=[a[2], log(exp(a*[4, 0; 0, 2])*[1, 1]~)], E, return([n, exp(a[2]/2^n)])))} \\ To compute an approximation of the constant c1 = exp(lim_{n->oo} (log a(n))/2^n). \\ M. F. Hasler, May 21 2017

CROSSREFS

Cf. A000283, A112969, A114793.

Sequence in context: A218057 A126469 A054859 * A059917 A255963 A093625

Adjacent sequences:  A076722 A076723 A076724 * A076726 A076727 A076728

KEYWORD

nonn,nice

AUTHOR

Michael Somos, Oct 29 2002

EXTENSIONS

Edited by N. J. A. Sloane at the suggestion of Andrew Plewe, Jun 15 2007

STATUS

approved

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Last modified November 18 12:20 EST 2017. Contains 294891 sequences.