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 1-3 of 3 results.

A376059 a(n) is the denominator of the sum S(n) defined in A376058.

Original entry on oeis.org

1, 2, 6, 78, 18330, 1679962830, 22578200883132834030, 6627077016548303724729207245056971365730, 922281145448518091883798423085535218757314338662318933097843039655721026758456630
Offset: 0

Views

Author

N. J. A. Sloane, Sep 14 2024

Keywords

Examples

			The first few values of S(n) are 0, 1/2, 5/6, 77/78, 18329/18330, 1679962829/1679962830, 22578200883132834029/22578200883132834030, ...
		

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[n+1] == Fibonacci[n+1]*a[n]^2 + a[n], a[0] == 1}, a, {n, 0, 8}] (* Amiram Eldar, Sep 15 2024 *)

Formula

a(n+1) = Fibonacci(n+1)*a(n)^2 + a(n), with a(0) = 1.

A376056 Lexicographically earliest sequence of positive integers a(1), a(2), a(3), ... such that for any n > 0, S(n) = Sum_{k = 1..n} (2*k-1)/a(k) < 1.

Original entry on oeis.org

2, 7, 71, 6959, 62255215, 4736981006316791, 26518805245879857416837904442871, 811438882694890436523185183518581584358651922339197834228784351
Offset: 1

Views

Author

N. J. A. Sloane, Sep 14 2024

Keywords

Comments

Theorem: Given any sequence of nonnegative integers b(1), b(2), b(3), ..., let a(1), a(2), a(3), ... be the lexicographically earliest sequence of positive integers such that for all n >= 1, S(n) = Sum_{k = 1..n} b(k)/a(k) < 1. Then S(n) = (e(n)-1)/e(n) for positive integers e(1), e(2), e(3), ....
For the present sequence the e(k) are given in A376057.

Crossrefs

Programs

  • Maple
    # Given a sequence b(1), b(2), b(3), ... of nonnegative real numbers, this program computes the first M terms of the lexicographically earliest sequence of positive integers a(1), a(2), a(3), ... with the property that for any n > 0, S(n) = Sum_{k = 1..n} b(k)/a(k) < 1.
    # For the present sequence we set b(k) = 2*k - 1.
    b := Array(0..100,-1); a := Array(0..100,-1); S := Array(0..100,-1); d := Array(0..100,-1);
    for k from 1 to 100 do b[k]:=2*k-1; od:
    M:=8;
    S[0] := 0; d[0] := 1;
    for n from 1 to M do
    a[n] := floor(b[n]/d[n-1])+1;
    S[n] := S[n-1] + b[n]/a[n];
    d[n] := 1 - S[n];
    od:
    La:=[seq(a[n],n=1..M)]; # the present sequence
    Ls:=[seq(S[n],n=1..M)]; # the sums S(n)
    Lsn:=[seq(numer(S[n]),n=1..M)];
    Lsd:=[seq(denom(S[n]),n=1..M)]; # A376057
    Lsd-Lsn; # As a check, by the above theorem, this should (and does) produce the all-1's sequence
    # Some small changes to the program are needed if the starting sequence {b(n)} has offset 0, as for example in the case of the Fibonacci or Catalan numbers (see A376058-A376061).

Formula

a(n+1) = (2*n+1)*A376057(n) + 1.

A377229 Lexicographically earliest sequence of positive integers a(1), a(2), ... such that for any n >= 0, s(n) = Sum_{k=1..n} 1/(F(k)*a(k)) < 1, F = Fibonacci.

Original entry on oeis.org

2, 3, 4, 9, 44, 1486, 1357976, 1855074754595, 2975714380792664939835466, 46528348836004781630107949818181021469921360198769
Offset: 1

Views

Author

Alois P. Heinz, Oct 20 2024

Keywords

Examples

			s(0), s(1), ... = 0, 1/2, 5/6, 23/24, 215/216, 11879/11880, 17653679/17653680, ... .
		

Crossrefs

Programs

  • Maple
    F:= combinat[fibonacci]:
    s:= proc(n) option remember; `if`(n=0, 0, s(n-1)+1/(F(n)*a(n))) end:
    a:= proc(n) option remember; 1+floor(1/((1-s(n-1))*F(n))) end:
    seq(a(n), n=1..11);
Showing 1-3 of 3 results.