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 1-3 of 3 results.

A049684 a(n) = Fibonacci(2n)^2.

Original entry on oeis.org

0, 1, 9, 64, 441, 3025, 20736, 142129, 974169, 6677056, 45765225, 313679521, 2149991424, 14736260449, 101003831721, 692290561600, 4745030099481, 32522920134769, 222915410843904, 1527884955772561, 10472279279564025, 71778070001175616, 491974210728665289
Offset: 0

Views

Author

Clark Kimberling

Keywords

Comments

This is the r=9 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.
Apparently, this sequence consists of those nonnegative integers k for which x*(x^2-1)*y*(y^2-1) = k*(k^2-1) has a solution in nonnegative integers x, y. If k = a(n), x = A000045(2*n-1) and y = A000045(2*n+1) are a solution. See A374375 for numbers k*(k^2-1) that can be written as a product of two or more factors of the form x*(x^2-1). - Pontus von Brömssen, Jul 14 2024

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 27.
  • H. J. H. Tuenter, Fibonacci summation identities arising from Catalan's identity, Fib. Q., 60:4 (2022), 312-319.

Links

Crossrefs

First differences give A033890.
First differences of A103434.
Bisection of A007598 and A064841.
a(n) = A064170(n+2) - 1 = (1/5) A081070.
Cf. A000032, A000045, A001622, A001906, A056854, A065563, A092184, A374375.

Programs

  • Mathematica
    Join[{a=0, b=1}, Table[c=7*b-1*a+2; a=b; b=c, {n, 60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
Fibonacci[Range[0, 40, 2]]^2 (* Harvey P. Dale, Mar 22 2012 *)
Table[Fibonacci[n - 1] Fibonacci[n + 1] - 1, {n, 0, 40, 2}] (* Bruno Berselli, Feb 12 2015 *)
LinearRecurrence[{8, -8, 1},{0, 1, 9},21] (* Ray Chandler, Sep 23 2015 *)
  • MuPAD
    numlib::fibonacci(2*n)^2 $ n = 0..35; // Zerinvary Lajos, May 13 2008
  • PARI
    a(n)=fibonacci(2*n)^2
  • Sage
    [fibonacci(2*n)^2 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009

Formula

G.f.: (x+x^2) / ((1-x)*(1-7*x+x^2)).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) with n>2, a(0)=0, a(1)=1, a(2)=9.
a(n) = 7*a(n-1) - a(n-2) + 2 = A001906(n)^2.
a(n) = (A000032(4*n)-2)/5. [This is in Koshy's book (reference under A065563) on p. 88, attributed to Lucas 1876.] - Wolfdieter Lang, Aug 27 2012
a(n) = 1/5*(-2 + ( (7+sqrt(45))/2 )^n + ( (7-sqrt(45))/2 )^n). - Ralf Stephan, Apr 14 2004
a(n) = 2*(T(n, 7/2)-1)/5 with twice the Chebyshev polynomials of the first kind evaluated at x=7/2: 2*T(n, 7/2)= A056854(n). - Wolfdieter Lang, Oct 18 2004
a(n) = F(2*n-1)*F(2*n+1)-1, see A064170 - Bruno Berselli, Feb 12 2015
a(n) = Sum_{i=1..n} F(4*i-2) for n>0. - Bruno Berselli, Aug 25 2015
From Peter Bala, Nov 20 2019: (Start)
Sum_{n >= 1} 1/(a(n) + 1) = (sqrt(5) - 1)/2.
Sum_{n >= 1} 1/(a(n) + 4) = (3*sqrt(5) - 2)/16. More generally, it appears that
Sum_{n >= 1} 1/(a(n) + F(2*k+1)^2) = ((2*k+1)*F(2*k+1)*sqrt(5) - Lucas(2*k+1))/ (2*F(2*k+1)*F(4*k+2)) for k = 0,1,2,....
Sum_{n >= 2} 1/(a(n) - 1) = (8 - 3*sqrt(5))/9. (End)
E.g.f.: (1/5)*(-2*exp(x) + exp((16*x)/(1 + sqrt(5))^4) + exp((1/2)*(7 + 3*sqrt(5))*x)). - Stefano Spezia, Nov 23 2019
Product_{n>=2} (1 - 1/a(n)) = phi^2/3, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 01 2021
a(n) = A092521(n-1)+A092521(n). - R. J. Mathar, Nov 22 2024

Extensions

Better description and more terms from Michael Somos

A048985 Working in base 2, replace n with the concatenation of its prime divisors in increasing order (write answer in base 10).

Original entry on oeis.org

1, 2, 3, 10, 5, 11, 7, 42, 15, 21, 11, 43, 13, 23, 29, 170, 17, 47, 19, 85, 31, 43, 23, 171, 45, 45, 63, 87, 29, 93, 31, 682, 59, 81, 47, 175, 37, 83, 61, 341, 41, 95, 43, 171, 125, 87, 47, 683, 63, 173, 113, 173, 53, 191, 91, 343, 115, 93, 59, 349, 61, 95, 127, 2730
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Examples

			15 = 3*5 -> 11.101 -> 11101 = 29, so a(15) = 29.

Links

Crossrefs

Cf. A037276, A048986, A064841.
Cf. A193652, A029744 (record values and where they occur).
Cf. A027746.

Programs

  • Haskell
    -- import Data.List (unfoldr)
a048985 = foldr (\d v -> 2 * v + d) 0 . concatMap
   (unfoldr (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2))
   . reverse . a027746_row
-- Reinhard Zumkeller, Jul 16 2012
  • Mathematica
    f[n_] := FromDigits[ Flatten[ IntegerDigits[ Flatten[ Table[ #1, {#2}] & @@@ FactorInteger@n], 2]], 2]; Array[f, 64] (* Robert G. Wilson v, Jun 02 2010 *)
  • Python
    from sympy import factorint
def a(n):
    if n == 1: return 1
    return int("".join(bin(p)[2:]*e for p, e in factorint(n).items()), 2)
print([a(n) for n in range(1, 65)]) # Michael S. Branicky, Oct 07 2022

Extensions

More terms from Sam Alexander (pink2001x(AT)hotmail.com) and Michel ten Voorde

A065016 Working in base 2, replace n with the concatenation of its prime factors (without repetition).

Original entry on oeis.org

1, 10, 11, 10, 101, 1011, 111, 10, 11, 10101, 1011, 1011, 1101, 10111, 11101, 10, 10001, 1011, 10011, 10101, 11111, 101011, 10111, 1011, 101, 101101, 11, 10111, 11101, 1011101, 11111, 10, 111011, 1010001, 101111, 1011, 100101, 1010011
Offset: 1

Views

Author

Robert G. Wilson v, Nov 01 2001

Keywords

Crossrefs

Cf. A064841.

Programs

  • Mathematica
    Prepend[ Table[ FromDigits[ Flatten[ IntegerDigits[ Transpose[ FactorInteger[n]] [[1]], 2]]], {n, 2, 50} ], 1]
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.