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 1-5 of 5 results.

A145274 a(n) = A145231(n+1)/A145231(n).

Original entry on oeis.org

329, 10749959329, 13354478338703157414450712411084788083329
Offset: 1

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Author

Artur Jasinski, Oct 06 2008

Keywords

Comments

A member of the family of sequences of type:
(G^(k^(n + 1)) - (1 - G)^(k^(n + 1)))/(G^(k^n) - (1 - G)^(k^n)) where G = (1 + sqrt(5))/2.
For k=2 see A001566.
For k=3 see A002814(n+2).
For k=4 see this sequence.
For k=5 see A145275.
For k=6 see A145276.
For k=7 see A145277.

Crossrefs

Programs

  • Mathematica
    G = (1 + Sqrt[5])/2; Table[Expand[(G^(4^(n + 1)) - (1 - G)^(4^(n + 1)))/Sqrt[5]]/Expand[(G^(4^n) - (1 - G)^(4^n))/Sqrt[5]], {n, 1, 5}]

Formula

a(n) = (G^(4^(n + 1)) - (1 - G)^(4^(n + 1)))/(G^(4^n) - (1 - G)^(4^n)) where G = (1 + sqrt(5))/2.

A145277 a(n) = A145234(n+1)/A145234(n).

Original entry on oeis.org

598364773, 27692759465311176949233529747775189817301578781117871380248013
Offset: 1

Views

Author

Artur Jasinski, Oct 06 2008

Keywords

Comments

A member of the family of sequences of type:
(G^(k^(n + 1)) - (1 - G)^(k^(n + 1)))/(G^(k^n) - (1 - G)^(k^n)) where G = (1 + sqrt(5))/2
For k=2 see A001566.
For k=3 see A002814(n+2).
For k=4 see A145274.
For k=5 see A145275.
For k=6 see A145276.
For k=7 see this sequence.

Crossrefs

Programs

  • Mathematica
    G = (1 + Sqrt[5])/2; Table[Expand[(G^(7^(n + 1)) - (1 - G)^(7^(n + 1)))/Sqrt[5]]/Expand[(G^(7^n) - (1 - G)^(7^n))/Sqrt[5]], {n, 1, 5}]

Formula

a(n) = (G^(7^(n + 1)) - (1 - G)^(7^(n + 1)))/(G^(7^n) - (1 - G)^(7^n)) where G = (1 + sqrt(5))/2.

A145276 a(n) = A145233(n+1)/A145233(n).

Original entry on oeis.org

1866294, 41473935220454958813340461622291147206
Offset: 1

Views

Author

Artur Jasinski, Oct 06 2008

Keywords

Comments

A member of the family of sequences of type:
(G^(k^(n + 1)) - (1 - G)^(k^(n + 1)))/(G^(k^n) - (1 - G)^(k^n)) where G = (1 + sqrt(5))/2.
For k=2 see A001566.
For k=3 see A002814(n+2).
For k=4 see A145274.
For k=5 see A145275.
For k=6 see this sequence.
For k=7 see A145277.

Crossrefs

Programs

  • Mathematica
    G = (1 + Sqrt[5])/2; Table[Expand[(G^(6^(n + 1)) - (1 - G)^(6^(n + 1)))/Sqrt[5]]/Expand[(G^(6^n) - (1 - G)^(6^n))/Sqrt[5]], {n, 1, 5}]

Formula

a(n) = (G^(6^(n + 1)) - (1 - G)^(6^(n + 1)))/(G^(6^n) - (1 - G)^(6^n)) where G = (1 + sqrt(5))/2.

A181419 Numbers of the form Fibonacci(p^{k+1})/Fibonacci(p^k) where p are primes, k>=1.

Original entry on oeis.org

3, 7, 17, 47, 2207, 5777, 15005, 4870847, 598364773, 192900153617, 23725150497407, 792070839820228500005, 97415813466381445596089, 562882766124611619513723647, 400009475456580321242184872389193
Offset: 1

Views

Author

Vladimir Shevelev, Oct 18 2010

Keywords

Comments

The union of A001566 (p=2), A002814 except the first two terms (p=3), A145275 (p=5), A145277 (p=7), etc.

Crossrefs

Programs

  • Maple
    N:= 10^50: # for terms <= N
    S:= {}: p:= 1:
    do
     p:= nextprime(p);
     v:= combinat:-fibonacci(p);
     for k from 2 do
       w:= v;
       v:= combinat:-fibonacci(p^k);
       r:= v/w;
       if r > N then break fi;
       S:= S union {r};
     od;
     if k = 2 then break fi;
    od:
    sort(convert(S,list)); # Robert Israel, Apr 09 2024
  • Mathematica
    t = Sort@ Flatten[ Table[ {Prime[n]^(e + 1), Prime[n]^e}, {n, 8}, {e, 10}], 1]; u = Select[t, First@# < 350 &]; Sort[ Fibonacci[ #[[1]]]/Fibonacci[ #[[2]]] & /@ u] (* Robert G. Wilson v, Oct 21 2010 *)

Extensions

a(11) onwards from Robert G. Wilson v, Oct 21 2010

A219011 Denominators in a product expansion for sqrt(5).

Original entry on oeis.org

5, 15005, 792070839820228500005
Offset: 0

Views

Author

Peter Bala, Nov 09 2012

Keywords

Comments

Apart from the initial term same as A145275.
a(3) has 105 digits and a(4) has 523 digits.
The product expansion in question is sqrt(5) = Product_{n >= 0} (1 + 2*A219010(n)/A219011(n)) = (1 + 6/5)*(1 + 246/15005)*(1 + 56287506246/792070839820228500005)*....

Crossrefs

Programs

  • Mathematica
    a[n_] := LucasL[4*5^n] - LucasL[2*5^n] + 1; Array[a, 3, 0] (* Amiram Eldar, Jul 20 2025 *)
  • Maxima
    A219011(n):=fib(5^(n+1))/fib(5^n)$
    makelist(A219011(n),n,0,3);

Formula

a(n) = Fibonacci(5^(n+1))/Fibonacci(5^n).
a(n) = A219010(n)^2 - A219010(n) - 1.
Recurrence equation: a(n+1) = 5/2*(a(n)^4 - a(n)^2)*sqrt(4*a(n) + 5) + a(n)^5 + 15/2*a(n)^4 - 25/2*a(n)^2 + 5 with initial condition a(0) = 5.
a(n) = Lucas(4*5^n) - Lucas(2*5^n) + 1. - Ehren Metcalfe, Jul 29 2017
Showing 1-5 of 5 results.