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A124297
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5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci[n].
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6
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1, 11, 11, 31, 61, 151, 361, 911, 2311, 5951, 15401, 40051, 104401, 272611, 712531, 1863551, 4875781, 12760031, 33398201, 87424711, 228859951, 599129311, 1568486161, 4106261531, 10750188961, 28144128251, 73681909211, 192901135711
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OFFSET
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0,2
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COMMENTS
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11 = Lucas[5] divides a(1+10k), a(2+10k), a(9+10k). Last digit of a(n) is 1, or Mod[a(n),10] = 1. For odd n there exist so called Aurifeuillian factorization A001946[n] = Lucas[5n] = Lucas[n]*A[n]*B[n] = A000032[n]*A124296[n]*A124297[n], where A[n] = A124296[n] = 5*F(n)^2 - 5*F(n) + 1 and B[n] = A124297[n] = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci[n].
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LINKS
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Table of n, a(n) for n=0..27.
Index to sequences with linear recurrences with constant coefficients, signature (4,-2,-6,4,2,-1).
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FORMULA
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a(n) = 5*Fibonacci[n]^2 + 5*Fibonacci[n] + 1.
G.f.: -(11*x^5-21*x^4-15*x^3+31*x^2-7*x-1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)). [Colin Barker, Jan 03 2013]
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MATHEMATICA
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Table[5*Fibonacci[n]^2+5*Fibonacci[n]+1, {n, 0, 50}]
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PROG
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(PARI) a(n)=subst(5*t*(t+1)+1, t, fibonacci(n)) \\ Charles R Greathouse IV, Jan 03 2013
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CROSSREFS
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Cf. A000032, A000045, A121171, A001946, A124296.
Sequence in context: A218163 A152082 A070849 * A172507 A089766 A077699
Adjacent sequences: A124294 A124295 A124296 * A124298 A124299 A124300
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KEYWORD
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nonn,easy
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AUTHOR
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Alexander Adamchuk, Oct 25 2006
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STATUS
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approved
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