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.

2, 7, 71, 6959, 62255215, 4736981006316791, 26518805245879857416837904442871, 811438882694890436523185183518581584358651922339197834228784351

Offset: 1

Author

N. J. A. Sloane, Sep 14 2024

Keywords

• nonn
• base

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

Cf. A374663, A375516, A375531, A375532, A375781, A375522, A376048-A376055, A376057-A376061.

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.