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A339204
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Decimal expansion of the generating constant for the Fibonacci numbers.
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2
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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FORMULA
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EXAMPLE
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2.95693889137798804983169009791120927869915823439362...
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MAPLE
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with(combinat, fibonacci): evalf(Sum((fibonacci(n) - 1)/Product(fibonacci(k), k = 2..n-1), n = 3..infinity), 120); # Vaclav Kotesovec, Nov 29 2020
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MATHEMATICA
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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 *)
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PROG
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(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)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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