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Search: id:A084595
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| A084595 |
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For n>0: a(n)=Sum(binomial[2^n,2r+1]3^r,r=0,..,2^(n-1)-1). |
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+0
2
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| 1, 2, 16, 896, 2781184, 26794772135936, 2487085750646543836443049984, 21427531469765285263614058238314319540132878612321796096
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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A084594(n)/a(n) converges to sqrt(3). Related to Newton's iteration. a(n) is divisible by 2^n.
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LINKS
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A. V. Aho and N. J. A. Sloane, Some doubly exponentialsequences, Fib. Quart., 11 (1973), 429-437.
Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
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FORMULA
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With a=1+sqrt(3), b=1-sqrt(3): a(n)=(a^(2^n)-b^(2^n))/(2sqrt(3)). For n>1: a(n)=2a(n-1)sqrt(3a(n-1)^2+A001146(n-1)). a(n)=2a(n-1)*A084594(n-1). a(n)=A080953(2^n).
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MATHEMATICA
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For n>0: Table[Sum[Binomial[2^n, 2 r + 1]3^r, {r, 0, 2^(n - 1) - 1}], {n, 1, 8}]
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CROSSREFS
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Adjacent sequences: A084592 A084593 A084594 this_sequence A084596 A084597 A084598
Sequence in context: A013005 A013000 A013178 this_sequence A002543 A125791 A102103
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KEYWORD
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easy,nonn
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AUTHOR
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Mario Catalani (mario.catalani(AT)unito.it), May 31 2003
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