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.

A199575 a(n) = floor(Fibonacci(n)^(1/4)).

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 16, 18, 21, 23, 26, 30, 34, 38, 43, 48, 55, 62, 70, 79, 89, 100, 113, 127, 144, 162, 183, 207, 233, 263, 296, 334, 377, 426, 480, 541, 611, 689, 777, 876, 989, 1115, 1258, 1418, 1600, 1804, 2035, 2295, 2589, 2920, 3293, 3714, 4189, 4725, 5329, 6010, 6778
Offset: 0

Views

Author

N. J. A. Sloane, Nov 09 2011

Keywords

Comments

The Ferraro problem asks for a proof that, for n>=9, floor(F(n)^(1/4)) = floor(F(n-4)^(1/4)+F(n-8)^(1/4)). As of November 2005 this problem remained unsolved.

Links

Crossrefs

Cf. A061287.

Programs

  • Magma
    [Floor(Fibonacci(n)^(1/4)): n in [0..80]]; // Vincenzo Librandi, Aug 28 2016
  • Mathematica
    Table[Floor[Fibonacci[n]^(1/4)], {n, 0, 80}] (* Vincenzo Librandi, Aug 28 2016 *)
  • PARI
    a(n) = sqrtnint(fibonacci(n), 4); \\ Michel Marcus, Aug 28 2016

A350701 a(n) is the number of squares strictly between Fibonacci(n) and Fibonacci(n+1).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 1, 1, 1, 2, 2, 2, 3, 4, 5, 7, 8, 11, 14, 18, 22, 29, 36, 46, 58, 75, 95, 120, 154, 195, 248, 315, 402, 511, 649, 826, 1052, 1337, 1700, 2164, 2751, 3501, 4452, 5664, 7204, 9164, 11656, 14828, 18861, 23991, 30518, 38818, 49379, 62810, 79896
Offset: 0

Views

Author

Karl-Heinz Hofmann, Jan 24 2022

Keywords

Comments

Terms a(0..3) are of course 0, because A000045(4) = 3 and A000045(5) = 5 are the first terms which are letting room for at least 1 integer.

Examples

			Strictly between Fibonacci(9) = 34 and Fibonacci(10) = 55 are the 2 squares 36 and 49. So a(9) = 2.

Links

Crossrefs

Cf. A000045, A000290, A061287, A076777.

Programs

  • PARI
    a(n)={if(n<=1, 0, sqrtint(fibonacci(n+1)-1) - sqrtint(fibonacci(n)))} \\ Andrew Howroyd, Jan 25 2022
  • Python
    from math import isqrt
from sympy import fibonacci as fi
print([0,0] + [(isqrt(fi(k+1)-1) - isqrt(fi(k))) for k in range(2, 55)])
  • Python
    from math import isqrt
from gmpy2 import fib2
def A350701(n): return 0 if n <= 1 else (lambda x:isqrt(x[0]-1)-isqrt(x[1]))(fib2(n+1)) # Chai Wah Wu, Jan 25 2022

A354210 a(n) = floor(sqrt(Fibonacci(n+1)*Fibonacci(n))).

Original entry on oeis.org

0, 1, 1, 2, 3, 6, 10, 16, 26, 43, 69, 113, 183, 296, 479, 775, 1255, 2031, 3286, 5318, 8605, 13923, 22528, 36452, 58981, 95433, 154414, 249847, 404261, 654109, 1058371, 1712480, 2770851, 4483332, 7254184, 11737516, 18991701, 30729217, 49720919, 80450136, 130171055, 210621192
Offset: 0

Views

Author

Michel Marcus, May 19 2022

Keywords

Links

  • P. Ferraro, Problem B-1288, Fibonacci Q., 59 (No. 2, May 2021) p. 177.

Crossrefs

Cf. A000045, A001654, A061287, A199575.

Programs

  • Mathematica
    a[n_] := Floor[Sqrt[Fibonacci[n + 1] * Fibonacci[n]]]; Array[a, 42, 0] (* Amiram Eldar, May 19 2022 *)
  • PARI
    a(n) = sqrtint(fibonacci(n+1)*fibonacci(n));
  • Python
    from math import prod
from gmpy2 import isqrt, fib2
def A354210(n): return int(isqrt(prod(fib2(n+1)))) # Chai Wah Wu, May 19 2022

Formula

a(n) = floor(sqrt(A001654(n))).

A263727 Largest square number less than or equal to the n-th Fibonacci number.

Original entry on oeis.org

0, 1, 1, 1, 1, 4, 4, 9, 16, 25, 49, 81, 144, 225, 361, 576, 961, 1521, 2500, 4096, 6724, 10816, 17689, 28561, 46225, 74529, 121104, 196249, 316969, 514089, 831744, 1345600, 2175625, 3523129, 5702544, 9223369, 14922769, 24157225, 39087504, 63234304, 102333456
Offset: 0

Views

Author

Eli Jaffe, Oct 24 2015

Keywords

Examples

			For a(8), Fibonacci(8) = 21, the largest square under 21 is 16, so a(8) = 16.

Crossrefs

Cf. A000045, A048760, A061287.

Programs

  • Mathematica
    Floor[Sqrt[Fibonacci[Range[40]]]]^2 (* Alonso del Arte, Oct 24 2015 *)
  • PARI
    a(n) = sqrtint(fibonacci(n))^2; \\ Michel Marcus, Oct 25 2015

Formula

a(n) = floor(sqrt(Fibonacci(n)))^2.
a(n) = A061287(n)^2. - Michel Marcus, Oct 25 2015
a(n) = A048760(A000045(n)). - Michel Marcus, Nov 11 2015
Showing 1-4 of 4 results.

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

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