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A076725
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a(n) = a(n-1)^2 + a(n-2)^4, a(0) = a(1) = 1.
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11
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OFFSET
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0,3
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COMMENTS
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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.
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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MAPLE
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A[0]:= 1: A[1]:= 1:
for n from 2 to 10 do
A[n]:= A[n-1]^2 + A[n-2]^4;
od:
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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