|
%I #50 Aug 21 2021 16:29:00
%S 1,1,2,1,6,2,1,8,162,2,1,12,548,17538,2,1,14,980,33704228,
%T 2341874005255914498,2,1,18,1386,139300292,2661014892563136514212,
%U 87117602843745867025899669792849494278146,2
%N 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)).
%C The denominators are given by 2^(3^k)-1.
%C 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.
%C We define A014682^0(n) = n and A014682^1(n) = A014682(n), A014682^2(n) = A014682(A014682(n)).
%e Let f(n) = Sum_{j>=1} 1/2^(A014682^n(j)).
%e f(0) = 1, f(1) = 9/7, f(2) = 951/511, f(3) = 47165693/19173961, ...
%e So triangle begins:
%e 1; f(0) = 1/1
%e 1, 2; f(1) = 1/1 + 2/7
%e 1, 6, 2; f(2) = 1/1 + 6/7 + 2/511
%e 1, 8, 162, 2; f(3) = 1/1 + 8/7 + 162/511 + 2/(2^27-1)
%e 1, 12, 548, 17538, 2; f(4) = 1/1 + 12/7 + 548/511 + 17538/(2^27-1) + 2/(2^81-1)
%e ...
%e 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)).
%o (MATLAB)
%o function a = A343684( row_n )
%o a = zeros(1,1+row_n);
%o for k = 1:2^row_n
%o first = A014682_exp(k,row_n);
%o second = A014682_exp(k+2^row_n,row_n);
%o e = round(log2(second-first)/log2(3));
%o a(e+1) = a(e+1)+2^(3^e-first);
%o end
%o end
%o function [ out ] = A014682_exp( in, num )
%o out = in;
%o for n = 1:num
%o out = A014682( out );
%o end
%o end
%o function [ out ] = A014682( in )
%o if mod(in,2) == 0
%o out = in/2;
%o else
%o out = ((in*3) + 1)/2;
%o end
%o end
%Y Cf. A014682.
%K nonn,tabl,frac
%O 0,3
%A _Thomas Scheuerle_, Apr 26 2021
|
