A207778 Smallest multiple of 2^n using only 1's and 2's.

1, 2, 12, 112, 112, 2112, 2112, 122112, 122112, 12122112, 12122112, 12122112, 111212122112, 1111212122112, 11111212122112, 11111212122112, 11111212122112, 11111212122112, 111211111212122112, 111211111212122112, 111211111212122112, 111211111212122112

Offset: 0

Author

Lekraj Beedassy, Feb 20 2012

Keywords

• nonn
• base

Comments

An induction-based argument can be used to show that this sequence is actually infinite.

Problem 1, proposed during the 5th All-Soviet-Union Mathematical Competition in 1971 at Riga (Pertsel link), asks for a proof that this sequence is infinite. - Bernard Schott, Mar 20 2023

References

• J. B. Tabov and P. J. Taylor, Methods of Problem Solving, Book 1, Australian Mathematics Trust, 1996.

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..300
• Vladimir A. Pertsel, Problems of the All-Soviet-Union Mathematical Competitions 1961-1986, the 5th competition, Riga, 1971, problem 144.
• Index to sequences related to Olympiads and other Mathematical competitions.

Crossrefs

Cf. A000079, A007931, A023396, A053312, A126933.

Programs

Mathematica

Table[m = 1; p = 2^k; While[Total@ DigitCount[m p][[3 ;; -1]] > 0, m++]; m p, {k, 0, 11}] (* Michael De Vlieger, Mar 17 2023 *)

PARI

a(n) = my(k=1, d=digits(k*2^n)); while (!((vecmin(d)>=1) && (vecmax(d)<=2)), k++; d=digits(k*2^n)); k*2^n; \\ Michel Marcus, Mar 15 2023

Formula

a(n) <= A053312(n).

Extensions

More terms from Alois P. Heinz, Feb 20 2012