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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 (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 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 Cf. A000283, A005154, 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 S. Plewe, Jun 15 2007 STATUS approved

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Last modified December 7 05:41 EST 2022. Contains 358649 sequences. (Running on oeis4.)