A108319 Numbers of the form (2^i)*(3^j)*(7^k), with i, j, k >= 0.

1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 24, 27, 28, 32, 36, 42, 48, 49, 54, 56, 63, 64, 72, 81, 84, 96, 98, 108, 112, 126, 128, 144, 147, 162, 168, 189, 192, 196, 216, 224, 243, 252, 256, 288, 294, 324, 336, 343, 378, 384, 392, 432, 441, 448, 486, 504, 512, 567

Offset: 1

Author

Douglas Winston (douglas.winston(AT)srupc.com), Jun 30 2005

Keywords

• nonn
• easy

Comments

Numbers m | 42^e with integer e >= 0. - Michael De Vlieger, Aug 22 2019

Sum_{n>=1} 1/a(n) = (2*3*7)/((2-1)*(3-1)*(7-1)) = 7/2. - Amiram Eldar, Sep 24 2020

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A051037, A003586, A003591, A003594.

Programs

Mathematica

With[{n = 567}, Sort@ Flatten@ Table[2^i * 3^j * 7^k, {i, 0, Log2@ n}, {j, 0, Log[3, n/2^i]}, {k, 0, Log[7, n/(2^i*3^j)]}]] (* Michael De Vlieger, Aug 22 2019 *)

PARI

list(lim)=my(v=List(), s, t); for(i=0, logint(lim\=1, 7), t=7^i; for(j=0, logint(lim\t, 3), s=t*3^j; while(s<=lim, listput(v, s); s<<=1))); Set(v) \\ Charles R Greathouse IV, Nov 20 2024

Formula

a(n) ~ exp((6*log(2)*log(3)*log(7)*n)^(1/3)) / sqrt(42). - Vaclav Kotesovec, Sep 23 2020