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

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

Author

A.H.M. Smeets, Nov 27 2020

Keywords

• nonn
• cons

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...
		

Links

• Dylan Friedman, Juli Garbulsky, Bruno Glecer, James Grime, and Massi Tron Florentin, A Prime-Representing Constant, 2019.

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)).