A036566 Numbers of form 7^i*8^j with i, j >= 0, sorted.

1, 7, 8, 49, 56, 64, 343, 392, 448, 512, 2401, 2744, 3136, 3584, 4096, 16807, 19208, 21952, 25088, 28672, 32768, 117649, 134456, 153664, 175616, 200704, 229376, 262144, 823543, 941192, 1075648, 1229312, 1404928, 1605632, 1835008, 2097152

Offset: 1

Author

David W. Wilson

Keywords

• easy
• nonn

Comments

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).

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

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Robert Sedgewick, Analysis of shellsort and related algorithms, Fourth European Symposium on Algorithms, Barcelona, September, 1996.

Crossrefs

Subsequence of A003591.

Programs

Maple

N:= 10^7: # for all terms <= N
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

Mathematica

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 *)

Formula

a(n) ~ exp(sqrt(2*log(7)*log(8)*n)) / sqrt(56). - Vaclav Kotesovec, Sep 25 2020

a(n) = 7^A025669(n) * 8^A025675(n). - R. J. Mathar, Jul 06 2025