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A076725
a(n) = a(n-1)^2 + a(n-2)^4, a(0) = a(1) = 1.
11
1, 1, 2, 5, 41, 2306, 8143397, 94592167328105, 13345346031444632841427643906, 258159204435047592104207508169153297050209383336364487461
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
Karl Petersen, Ibrahim Salama, Tree shift complexity, arXiv:1712.02251 [math.DS], 2017.
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 = A263719 = (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
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 *)
NestList[{#[[2]], #[[1]]^4+#[[2]]^2}&, {1, 1}, 10][[All, 1]] (* Harvey P. Dale, Jul 03 2021 *)
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
(PARI) a=vector(20); a[1]=1; a[2]=2; for(n=3, #a, a[n]=a[n-1]^2+a[n-2]^4); concat(1, a) \\ Altug Alkan, Apr 04 2018
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Michael Somos, Oct 29 2002
EXTENSIONS
Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 15 2007
STATUS
approved