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%I #52 Mar 22 2023 20:34:17
%S 1,2,12,112,112,2112,2112,122112,122112,12122112,12122112,12122112,
%T 111212122112,1111212122112,11111212122112,11111212122112,
%U 11111212122112,11111212122112,111211111212122112,111211111212122112,111211111212122112,111211111212122112
%N Smallest multiple of 2^n using only 1's and 2's.
%C An induction-based argument can be used to show that this sequence is actually infinite.
%C 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
%D J. B. Tabov and P. J. Taylor, Methods of Problem Solving, Book 1, Australian Mathematics Trust, 1996.
%H Alois P. Heinz, <a href="/A207778/b207778.txt">Table of n, a(n) for n = 0..300</a>
%H Vladimir A. Pertsel, <a href="https://olympiads.win.tue.nl/imo/soviet/RusMath.html">Problems of the All-Soviet-Union Mathematical Competitions 1961-1986</a>, the 5th competition, Riga, 1971, problem 144.
%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads and other Mathematical competitions</a>.
%F a(n) <= A053312(n).
%t 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 *)
%o (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
%Y Cf. A000079, A007931, A023396, A053312, A126933.
%K nonn,base
%O 0,2
%A _Lekraj Beedassy_, Feb 20 2012
%E More terms from _Alois P. Heinz_, Feb 20 2012
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