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.

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A339204 Decimal expansion of the generating constant for the Fibonacci numbers.

Original entry on oeis.org

2, 9, 5, 6, 9, 3, 8, 8, 9, 1, 3, 7, 7, 9, 8, 8, 0, 4, 9, 8, 3, 1, 6, 9, 0, 0, 9, 7, 9, 1, 1, 2, 0, 9, 2, 7, 8, 6, 9, 9, 1, 5, 8, 2, 3, 4, 3, 9, 3, 6, 2, 3, 5, 3, 4, 5, 7, 2, 4, 4, 6, 2, 7, 2, 3, 7, 5, 2, 7, 4, 6, 4, 4, 6, 6, 8, 3, 4, 6, 7, 6, 9, 3, 0, 4, 1, 7, 5
Offset: 1

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Author

A.H.M. Smeets, Nov 27 2020

Keywords

Comments

Inspired by the prime generating constant A249270, but here for the Fibonacci numbers, A000045(n); generating the Fibonacci numbers for n > 2.
The producing function is given by f' = floor(f)*(f-floor(f)+1), starting with this constant, f' denoting the next f, and floor(f) being the terms of the sequence produced by this constant.
The number of correct digits obtained from the first n terms from the series expansion for this constant as given in the formula section is roughly about (n^2)/10 (~ (3/7)*(log(Fib(n))^2)) decimal digits; i.e., for a binary representation, about (n^2)/3 binary digits.

Examples

			2.95693889137798804983169009791120927869915823439362...
		

Crossrefs

Cf. A000045 (Fibonacci).
Cf. A249270 (for primes), A339203 (for Mersenne prime exponents).

Programs

  • Maple
    with(combinat, fibonacci): evalf(Sum((fibonacci(n) - 1)/Product(fibonacci(k), k = 2..n-1), n = 3..infinity), 120); # Vaclav Kotesovec, Nov 29 2020
  • Mathematica
    Quiet[First[RealDigits[NSum[(Fibonacci[n] - 1)/Fibonorial[n - 1], {n, 3, Infinity}, Method -> {"WynnEpsilon", "ExtraTerms" -> 25}, NSumTerms -> 25, VerifyConvergence -> False, WorkingPrecision -> 105], 10, 100]], General::intnm] (* Jan Mangaldan, Nov 29 2020 *)
  • PARI
    suminf(n=3, (fibonacci(n)-1)/prod(k=2, n-1, fibonacci(k))) \\ Michel Marcus, Nov 27 2020
    
  • Python
    n, sumn, sumd, termd, f0, f1 = 0, 0, 1, 1, 1, 1
    while n < 33: # enough to obtain 100 digits
        n, sumn, sumd, termd, f0, f1 = n+1, sumn*termd+sumd*(f0-1), sumd*termd, termd*f0, f0+f1, f0
    pre, sumn, i, d = sumn//sumd, sumn%sumd, 0, ""
    while i < 100:
        dig, sumn, i = (10*sumn)//sumd, (10*sumn)%sumd, i+1
        d = d+str(dig)
    print(str(pre)+"."+d)

Formula

Equals Sum_{n > 2} (A000045(n)-1)/(Product_{k = 2..n-1} A000045(k)).
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