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 1-4 of 4 results.

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).

Links

Crossrefs

Cf. A000032, A000045, A121171, A001946, A124297.
Bisections: A001604, A156094.

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]

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

Cf. A124296, A124297, A001603, A001604, A156094.

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

A156194 Period 12: 1,2,7,1,7,2,1,1,4,2,4,1 repeated.

Original entry on oeis.org

1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1
Offset: 0

Views

Author

Paul Curtz, Feb 05 2009

Keywords

Comments

Also the decimal expansion of 42390704747/333333333333. - R. J. Mathar, Feb 23 2009

Links

Programs

  • Mathematica
    PadRight[{},144,{1,2,7,1,7,2,1,1,4,2,4,1}] (* Harvey P. Dale, Mar 06 2012 *)

Formula

Palindromic properties: a(12k+i) = a(12k+6-i), i=0..3. a(12k+7+i) = a(12k+11-i), i=0..2, and similarly for successive differences.
a(n) = A156095(n) mod 9.
a(n) = A156094(n+6) mod 9.
a(4n) + a(4n+1) + a(4n+2) + a(4n+3) = A010850(n).
G.f.: (1+2*x+7*x^2+x^3+7*x^4+2*x^5+x^6+x^7+4*x^8+2*x^9+4*x^10+x^11)/((1-x)*(1+x+x^2)*(1+x)*(1-x+x^2)*(1+x^2)*(x^4-x^2+1)). - R. J. Mathar, Feb 23 2009

Extensions

Edited by R. J. Mathar, Feb 23 2009
More terms from Jinyuan Wang, Feb 26 2020

A156199 Period 12: repeat 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2 .

Original entry on oeis.org

1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7
Offset: 0

Views

Author

Paul Curtz, Feb 05 2009

Keywords

Comments

The period is essentially the reversal of the period in A156194.
The first 7 elements of the period (1, 1, 4, 2, 4, 1, 1) are palindromic and the other 5, (2, 7, 1, 7, 2), too.

Links

Crossrefs

Cf. A156194.

Programs

  • Mathematica
    LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1},{1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2},105] (* Ray Chandler, Aug 08 2015 *)
PadRight[{},110,{1,1,4,2,4,1,1,2,7,1,7,2}] (* Harvey P. Dale, Sep 14 2020 *)

Formula

a(n) = A156094(n) mod 9.
a(n) = A156194(n+6).
G.f.: (1 + x + 4*x^2 + 2*x^3 + 4*x^4 + x^5 + x^6 + 2*x^7 + 7*x^8 + x^9 + 7*x^10 + 2*x^11)/( (1-x) * (1+x+x^2) * (1+x) * (1-x+x^2) * (1+x^2) * (x^4-x^2+1)). - R. J. Mathar, Nov 22 2009

Extensions

Edited by R. J. Mathar, Nov 22 2009
Showing 1-4 of 4 results.

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