cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A001604 Odd-indexed terms of A124297.

Original entry on oeis.org

11, 31, 151, 911, 5951, 40051, 272611, 1863551, 12760031, 87424711, 599129311, 4106261531, 28144128251, 192901135711, 1322159893351, 9062207833151, 62113268013311, 425730597768451, 2918000731816531, 20000274041790911, 137083916295800111, 939587136717207031, 6440026032054760351
Offset: 0

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Author

Keywords

Comments

Old name: Related to factors of Fibonacci numbers.

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    A001604:=-(11-90*z+173*z**2-90*z**3+11*z**4)/(z-1)/(z**2-3*z+1)/(z**2-7*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
  • Mathematica
    5 #^2 + 5 # + 1 &@ Fibonacci@ # & /@ Range[1, 45, 2] (* Michael De Vlieger, Apr 03 2017 *)

Formula

G.f.: -(11-90*x+173*x^2-90*x^3+11*x^4)/((x-1)*(x^2-3*x+1)*(x^2-7*x+1)). [After Simon Plouffe]
a(n) = (5+sqrt(5))/2*((3+sqrt(5))/2)^n+(5-sqrt(5))/2*((3-sqrt(5))/2)^n+(3+sqrt(5))/2*((7+3*sqrt(5))/2)^n+(3-sqrt(5))/2*((7-3*sqrt(5))/2)^n+3. [Tim Monahan, Aug 15 2011]

Extensions

Entry revised by Michel Marcus and N. J. A. Sloane, Jun 06 2015

A133320 Numbers k such that both A124296(k) = 5*F(k)^2 - 5*F(k) + 1 and A124297(k) = 5*F(k)^2 + 5*F(k) + 1 are prime, where F(k) = Fibonacci(k).

Original entry on oeis.org

3, 4, 5, 10, 40
Offset: 1

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Author

Alexander Adamchuk, Oct 18 2007

Keywords

Crossrefs

Cf. A124297 (5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci(n)).
Cf. A124296 (5*F(n)^2 - 5*F(n) + 1, where F(n) = Fibonacci(n)).

Programs

  • Mathematica
    Do[ F=Fibonacci[n]; f=5*F^2-5*F+1; g=5*F^2+5*F+1; If[ PrimeQ[f], If[ PrimeQ[g], Print[ {n,f,g} ] ] ], {n,1,1000} ]

A001946 a(n) = 11*a(n-1) + a(n-2).

Original entry on oeis.org

2, 11, 123, 1364, 15127, 167761, 1860498, 20633239, 228826127, 2537720636, 28143753123, 312119004989, 3461452808002, 38388099893011, 425730551631123, 4721424167835364, 52361396397820127, 580696784543856761, 6440026026380244498, 71420983074726546239
Offset: 0

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Author

Keywords

Comments

For odd n there is the Aurifeuillian factorization a(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]. The largest prime divisors of a(n) for n>0 are listed in A121171[n] = {11, 41, 31, 2161, 151, 2521, 911, ...}. - Alexander Adamchuk, Oct 25 2006
For more information about this type of recurrence follow the Khovanova link and see A086902 and A054413. - Johannes W. Meijer, Jun 12 2010

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 139.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

Formula

a(n) = Lucas(5n) = Fibonacci(5n-1) + Fibonacci(5n+1). - Alexander Adamchuk, Oct 25 2006
a(n) = ((11 + 5*sqrt(5))/2)^n + ((11 - 5*sqrt(5))/2)^n. - Tanya Khovanova, Feb 06 2007
Contribution from Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 11*A097842(n), a(2n) = A065705(n).
a(3n+1) = A041226(5n), a(3n+2) = A041226(5n+3), a(3n+3) = 2* A041226(5n+4).
Limit(a(n+k)/a(k), k=infinity) = (A001946(n) + A049666(n)*sqrt(125))/2.
Limit(A001946(n)/A049666(n), n=infinity) = sqrt(125). (End)
From Peter Bala, Mar 22 2015: (Start)
a(n) = Fibonacci(10*n)/Fibonacci(5*n) for n >= 1.
a(n) = ( Fibonacci(5*n + 2*k) - F(5*n - 2*k) )/Fibonacci(2*k) for nonzero integer k.
a(n) = ( Fibonacci(5*n + 2*k + 1) + F(5*n - 2*k - 1) )/Fibonacci(2*k + 1) for arbitrary integer k.
a(n) = Sum_{k = 0..2*n} binomial(2*n,k)*Lucas(n + k). (End)
a(n) = [x^n] ( (1 + 11*x + sqrt(1 + 22*x + 125*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 26 2015
E.g.f.: 2*cosh(5*sqrt(5)*x/2)*(cosh(11*x/2) + sinh(11*x/2)). - Stefano Spezia, Jan 18 2025

A124296 a(n) = 5*F(n)^2 - 5*F(n) + 1, where F(n) = Fibonacci(n).

Original entry on oeis.org

1, 1, 1, 11, 31, 101, 281, 781, 2101, 5611, 14851, 39161, 102961, 270281, 708761, 1857451, 4865911, 12744061, 33372361, 87382901, 228792301, 599019851, 1568309051, 4105974961, 10749725281, 28143378001, 73680695281, 192899171531
Offset: 0

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Author

Alexander Adamchuk, Oct 25 2006

Keywords

Comments

11 = Lucas(5) divides a(3+10k), a(7+10k), and a(8+10k). The last digit of a(n) is 1, so a(n) mod 10 = 1. For odd n there exists the 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).

Crossrefs

Programs

  • Mathematica
    Table[5*Fibonacci[n]^2-5*Fibonacci[n]+1,{n,0,50}]
    5#^2-5#+1&/@Fibonacci[Range[0,30]] (* Harvey P. Dale, Nov 29 2011 *)
  • PARI
    a(n)=subst(5*t*(t-1)+1, t, fibonacci(n)) \\ Charles R Greathouse IV, Jan 03 2013

Formula

a(n) = 5*Fibonacci(n)^2 - 5*Fibonacci(n) + 1.
G.f.: -(x^5+9*x^4-15*x^3+x^2+3*x-1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)). [Colin Barker, Jan 03 2013]

A156094 5 F(2n) (F(2n) - 1) + 1 where F(n) denotes the n-th Fibonacci number.

Original entry on oeis.org

1, 1, 31, 281, 2101, 14851, 102961, 708761, 4865911, 33372361, 228792301, 1568309051, 10749725281, 73680695281, 505017569551, 3461448647801, 23725139605861, 162614572159411, 1114576979567761, 7639424583421961, 52361395886149351
Offset: 0

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Author

Stuart Clary, Feb 04 2009

Keywords

Comments

Natural bilateral extension (brackets mark index 0): ..., 15401, 2311, 361, 61, 11, [1], 1, 31, 281, 2101, 14851, ... This is A156095-reversed followed by A156094, without repeating the central 1. That is, A156094(-n) = A156095(n).

Crossrefs

Programs

  • Mathematica
    a[n_Integer] := 5 Fibonacci[2n] (Fibonacci[2n] - 1) + 1
    5(#*(#-1))&/@Fibonacci[Range[0,40,2]]+1 (* Harvey P. Dale, Jan 06 2013 *)

Formula

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
Alternate formula: a(n) = L(4n) - 5 F(2n) - 1.
Recurrence: a(n) - 10 a(n-1) + 23 a(n-2) - 10 a(n-3) + a(n-4) = -5.
Recurrence: a(n) - 11 a(n-1) + 33 a(n-2) - 33 a(n-3) + 11 a(n-4) - a(n-5) = 0.
G.f.: A(x) = (1 - 10 x + 53 x^2 - 60 x^3 + 11 x^4)/(1 - 11 x + 33 x^2 - 33 x^3 + 11 x^4 - x^5) = (1 - 10 x + 53 x^2 - 60 x^3 + 11 x^4)/((1 - x) (1 - 7 x + x^2) (1 - 3 x + x^2)).
a(n)=A056854(n)-5*A001906(n)-1. - R. J. Mathar, Feb 23 2009
a(n)=((2*sqrt(5))/2)*(((3-sqrt(5))/2)^n-((3+sqrt(5))/2)^n)+((7+3*sqrt(5))/2)^n+((7-3*sqrt(5))/2)^n-1. - Tim Monahan, Aug 15 2011

A001603 Odd-indexed terms of A124296.

Original entry on oeis.org

1, 11, 101, 781, 5611, 39161, 270281, 1857451, 12744061, 87382901, 599019851, 4105974961, 28143378001, 192899171531, 1322154751061, 9062194370461, 62113232767531, 425730505493801, 2918000490238361, 20000273409331051, 137083914639998701, 939587132382262661
Offset: 0

Views

Author

Keywords

Comments

Old name: Related to factors of Fibonacci numbers.

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    A001603:=-(1+13*z**2+z**4)/(z-1)/(z**2-3*z+1)/(z**2-7*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
  • Mathematica
    5 #^2 - 5 # + 1 &@ Fibonacci@ # & /@ Range[1, 43, 2] (* Michael De Vlieger, Apr 03 2017 *)

Formula

G.f.: -(1+13*x^2+x^4)/((x-1)*(x^2-3*x+1)*(x^2-7*x+1)). [After Simon Plouffe]

Extensions

Entry revised by Michel Marcus and N. J. A. Sloane, Jun 06 2015

A156095 5 F(2n) (F(2n) + 1) + 1 where F(n) denotes the n-th Fibonacci number.

Original entry on oeis.org

1, 11, 61, 361, 2311, 15401, 104401, 712531, 4875781, 33398201, 228859951, 1568486161, 10750188961, 73681909211, 505020747661, 3461456968201, 23725161388951, 162614629188281, 1114577128871281, 7639424974303651, 52361396909490901
Offset: 0

Views

Author

Stuart Clary, Feb 04 2009

Keywords

Comments

Natural bilateral extension (brackets mark index 0): ..., 14851, 2101, 281, 31, 1, [1], 11, 61, 361, 2311, 15401, ... This is A156095-reversed followed by A156095, without repeating the central 1. That is, A156095(-n) = A156094(n).

Crossrefs

Programs

  • Mathematica
    a[n_Integer] := 5 Fibonacci[2n] (Fibonacci[2n] + 1) + 1

Formula

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
Alternate formula: a(n) = L(4n) + 5 F(2n) - 1.
Recurrence: a(n) - 10 a(n-1) + 23 a(n-2) - 10 a(n-3) + a(n-4) = -5.
Recurrence: a(n) - 11 a(n-1) + 33 a(n-2) - 33 a(n-3) + 11 a(n-4) - a(n-5) = 0.
G.f.: A(x) = (1 - 27 x^2 + 20 x^3 + x^4)/(1 - 11 x + 33 x^2 - 33 x^3 + 11 x^4 - x^5) = (1 - 27 x^2 + 20 x^3 + x^4)/((1 - x) (1 - 7 x + x^2) (1 - 3 x + x^2)).
a(n)=((2*sqrt(5))/2)*(((3+sqrt(5))/2)^n-((3-sqrt(5))/2)^n)+((7+3*sqrt(5))/2)^n+((7-3*sqrt(5))/2)^n-1. - Tim Monahan, Aug 15 2011
Showing 1-7 of 7 results.