login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A109139 Numerators associated with the continued fraction of the differences of consecutive prime numbers. 3
1, 2, 5, 12, 53, 118, 525, 1168, 5197, 32350, 69897, 451732, 1876825, 4205382, 18698353, 116395500, 717071353, 1550538206, 10020300589, 41631740562, 93283781713, 601334430840, 2498621505073, 15593063461278, 127243129195297 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
The value of the continued fraction up to n is: R(n) = A(n)/B(n) where B(0) = 1, B(1) = a(1) *B(0), B(n) = a(n)* B(n - 1) + B(n-2) (n>=2).
From theory related to the continued fractions, we have:
- the continued fraction is a simple continued fraction (i.e., generated by integer positive numbers);
- the limit C0 (for n to infinity) exists, it is greater than 1 and is R(n) = A(n)/B(n) = C0 = 1.71010202343009...;
- the limit C0 is an irrational number;
- there is a unique simple continued fraction with limit C0;
- the generating number sequences of the simple continued fraction with limit C0 is unique;
- the sequence of generating numbers of the continued fraction (i.e., the difference of consecutive prime numbers and, consequently, the prime numbers) can be evaluated from C0 by:
a(0) = floor(C0), C1 = 1/(C0-a(0)), a(1) = floor(C1), C2 = 1/(C1-a(1)), ... a(n) = floor(Cn) ...;
- C0 satisfies the inequality: A(n)/B(n) - 1/B(n)^2 < C0 < A(n)/B(n) + 1/B(n)^2;
- this inequality allows us to evaluate the range of a(n+1), given A(n) and B(n);
- knowledge of A(n)/B(n) allows us to evaluate a(0), a(1) ..., a(n), i.e., the difference of consecutive prime numbers and, consequently, the prime numbers.
- The continued fraction derived from the sequences of consecutive prime number differences performs lower gradient w.r.t. the continued fraction based on prime sequence and it is therefore computationally easier to use.
The denominators B(n) are in A109140. Related sequences are D(n) = A(n) - B(n), S(n) = A(n) + B(n).
LINKS
FORMULA
A(0) = a(0), A(1) = a(1)*A(0) + 1, A(n) = a(n)*A(n - 1) + A(n-2) (n>=2) where a(0) = p(0) - 1, a(1) = p(1) - p(0), a(2) = p(2) - p(1) ..., a(n) = p(n) - p(n-1) where p(n) is the n-th prime number.
EXAMPLE
n = 2, A(n) = A(2) = 5 because A(0) = 2-1 = 1, A(1) = (3-2) * A(0) + 1 = 2, A(2) = (5-3) * A(1) + 1 * A(0) = 5.
CROSSREFS
Sequence in context: A291484 A145997 A067578 * A038576 A002358 A083699
KEYWORD
nonn,frac
AUTHOR
Giorgio Balzarotti, Aug 18 2005
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified April 19 21:09 EDT 2024. Contains 371798 sequences. (Running on oeis4.)