A343684 Triangle read by rows: T(n,k). Row n lists the numerators of a finite sum of fractions which results from Sum_{j>=1} 1/2^(A014682^n(j)).

1, 1, 2, 1, 6, 2, 1, 8, 162, 2, 1, 12, 548, 17538, 2, 1, 14, 980, 33704228, 2341874005255914498, 2, 1, 18, 1386, 139300292, 2661014892563136514212, 87117602843745867025899669792849494278146, 2

Offset

0,3

Comments

The denominators are given by 2^(3^k)-1.

The fact that A014682^n(m+k*2^n) is of the form r+k*3^s for m = 1..k^2-1 allows an easy development as a sum of geometric series.

We define A014682^0(n) = n and A014682^1(n) = A014682(n), A014682^2(n) = A014682(A014682(n)).

Links

Table of n, a(n) for n=0..27.

Example

Let f(n) = Sum_{j>=1} 1/2^(A014682^n(j)).
f(0) = 1, f(1) = 9/7, f(2) = 951/511, f(3) = 47165693/19173961, ...
So triangle begins:
  1;                    f(0) = 1/1
  1, 2;                 f(1) = 1/1 + 2/7
  1, 6,  2;             f(2) = 1/1 + 6/7 + 2/511
  1, 8,  162, 2;        f(3) = 1/1 + 8/7 + 162/511 + 2/(2^27-1)
  1, 12, 548, 17538, 2; f(4) = 1/1 + 12/7 + 548/511 + 17538/(2^27-1) + 2/(2^81-1)
  ...
Sum_{j>=1} 1/2^(A014682^2(j)) = Sum_{j>=1} 1/2^j + Sum_{j>=1} 1/2^(1+j*(3^1)) + Sum_{j>=1} 1/2^(2+j*(3^1)) + Sum_{j>=1} 1/2^(8+j*(3^2)).

Prog

(MATLAB)
function a = A343684( row_n )
    a = zeros(1, 1+row_n);
    for k = 1:2^row_n
        first = A014682_exp(k, row_n);
        second = A014682_exp(k+2^row_n, row_n);
        e = round(log2(second-first)/log2(3));
        a(e+1) = a(e+1)+2^(3^e-first);
    end
end
function [ out ] = A014682_exp( in, num )
    out = in;
    for n = 1:num
        out = A014682( out );
    end
end
function [ out ] = A014682( in )
    if mod(in, 2) == 0
        out = in/2;
    else
        out = ((in*3) + 1)/2;
    end
end

Crossrefs

Cf. A014682.

Sequence in context: A030304 A248779 A286030 * A208905 A208749 A208751

Adjacent sequences: A343681 A343682 A343683 * A343685 A343686 A343687

Keyword

nonn,tabl,frac

Author

Thomas Scheuerle, Apr 26 2021

Status

approved