%I #17 Sep 26 2020 03:46:48
%S 1,7,8,49,56,64,343,392,448,512,2401,2744,3136,3584,4096,16807,19208,
%T 21952,25088,28672,32768,117649,134456,153664,175616,200704,229376,
%U 262144,823543,941192,1075648,1229312,1404928,1605632,1835008,2097152
%N Numbers of form 7^i*8^j with i, j >= 0, sorted.
%C Could be rearranged as a triangle of numbers in which i-th row is {7^(i-j)*8^j, 0<=j<=i}; i >= 0. (This would produce a different sequence, of course).
%C The sum of the reciprocals of the terms of this sequence is equal to 4/3. Brief proof: as gcd(7, 8) = 1, 1 + 1/7 + 1/8 + 1/49 + 1/56 + 1/64 + 1/343 + ... = (Sum_{k>=0} 1/7^k) * (Sum_{m>=0} 1/8^m) = (1/(1-1/7)) * (1/(1-1/8)) = (7/(7-1)) * (8/(8-1)) = 4/3. - _Bernard Schott_, Oct 24 2019
%H Robert Israel, <a href="/A036566/b036566.txt">Table of n, a(n) for n = 1..10000</a>
%H Robert Sedgewick, <a href="http://www.cs.princeton.edu/~rs/talks/shellsort.ps">Analysis of shellsort and related algorithms</a>, Fourth European Symposium on Algorithms, Barcelona, September, 1996.
%F a(n) ~ exp(sqrt(2*log(7)*log(8)*n)) / sqrt(56). - _Vaclav Kotesovec_, Sep 25 2020
%p N:= 10^7: # for all terms <= N
%p sort([seq(seq(7^i*8^j,j=0..floor(log[8](N/7^i))),i=0..floor(log[7](N)))]); # _Robert Israel_, Oct 24 2019
%t n = 10^6; Flatten[Table[7^i*8^j, {i, 0, Log[7, n]}, {j, 0, Log[8, n/7^i]}]] // Sort (* _Amiram Eldar_, Sep 26 2020 *)
%Y Subsequence of A003591.
%K easy,nonn
%O 1,2
%A _David W. Wilson_