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A074872
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Inverse BinomialMean transform of the Fibonacci sequence A000045 (with the initial 0 omitted).
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8
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1, 1, 5, 5, 25, 25, 125, 125, 625, 625, 3125, 3125, 15625, 15625, 78125, 78125, 390625, 390625, 1953125, 1953125, 9765625, 9765625, 48828125, 48828125, 244140625, 244140625, 1220703125, 1220703125, 6103515625, 6103515625
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| See A075271 for the definition of the BinomialMean transform.
The inverse binomial transform of 2^n*c(n+1), where c(n) is the solution to c(n)=c(n-1)+kc(n-2), a(0)=0,a(1)=1 is 1,1,4k+1,4k+1,(4k+1)^2,... - Paul Barry (pbarry(AT)wit.ie), Feb 12 2004
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LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 1..2000
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FORMULA
| a(n)=5^Floor((n-1)/2). a(1)=1, a(2)=1 and, for n>2, a(n)=5*a(n-2).
G.f.: (1+x)/(1-5x^2); a(n)=(1/(2sqrt(5))((1+sqrt(5))(sqrt(5))^n-(1-sqrt(5))(-sqrt(5))^n)). Inverse binomial transform of A063727 (2^n*Fib(n+1)). - Paul Barry (pbarry(AT)wit.ie), Feb 12 2004
a(n)=(1/5)*5^[(1/2)*n]*5^[(1/4)*(-1)^n]*125^(1/4), with n>=0 [From Paolo P. Lava (paoloplava(AT)gmail.com), Oct 06 2008]
a(n+3) = a(n+2)*a(n+1)/a(n). [Reinhard Zumkeller, Mar 04 2011]
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MATHEMATICA
| a[1] := 1; a[2] := 1; a[n_] := 5a[n - 2]; Table[a[n], {n, 30}] (* From Alonso del Arte, Mar 04 2011 *)
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PROG
| (MAGMA) [5^Floor((n-1)/2): n in [1..40]]; // Vincenzo Librandi, Aug 16 2011
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CROSSREFS
| Cf. A016116, A108411.
Sequence in context: A165826 A071340 A056451 * A162962 A170834 A154630
Adjacent sequences: A074869 A074870 A074871 * A074873 A074874 A074875
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KEYWORD
| nonn,easy
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AUTHOR
| John W. Layman (layman(AT)math.vt.edu), Sep 12 2002
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