A305934 Powers of 3 that have exactly one digit '0' in base 10.

59049, 14348907, 43046721, 129140163, 387420489, 3486784401, 847288609443, 68630377364883, 328256967394537077627, 26588814358957503287787, 717897987691852588770249, 6461081889226673298932241, 1144561273430837494885949696427, 22528399544939174411840147874772641, 67585198634817523235520443624317923

Offset: 1

Author

M. F. Hasler (following a suggestion by Zak Seidov), Jun 14 2018

Keywords

• nonn
• base

Comments

Motivated by A030700: decimal expansion of 3^n contains no zeros (probably finite).

It appears that this sequence is finite. Is a(15) = 3^73 the last term?

There are no more terms through at least 3^(10^7) (which is a 4771213-digit number). It seems nearly certain that no power of 3 containing this many or more decimal digits could have fewer than two '0' digits. (Among numbers of the form 3^k with 73 < k <= 10^7, the only one having fewer than two '0' digits among its final 200 digits is 3^5028978.) - Jon E. Schoenfield, Jun 24 2018

The first 6 terms coincide with A305931: powers of 3 having at least one digit 0, with complement A238939 (within A000244: powers of 3) conjectured to be finite, too. Then, a(7..8) = A305931(9..10), etc.

Crossrefs

Cf. A030700: decimal expansion of 3^n contains no zeros (probably finite), A238939: powers of 3 with no digit '0' in their decimal expansion, A000244: powers of 3.

Subsequence of A305931: powers of 3 having at least one '0'.

Cf. A305933: row n = { k | 3^k has n digits '0' }.

Programs

Mathematica

Select[3^Range[120], DigitCount[#, 10, 0] == 1 &] (* Michael De Vlieger, Jul 01 2018 *)

PARI

for(n=1,99, #select(t->!t,digits(3^n))==1&& print1(3^n","))

Formula

a(n) = 3^A305933(1,n).