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Smallest cube containing exactly n 0's.
20

%I #19 Mar 20 2020 23:32:26

%S 1,0,140608,1000,4096000,140608000,1000000,4096000000,140608000000,

%T 1000000000,4096000000000,140608000000000,1000000000000,

%U 4096000000000000,140608000000000000,1000000000000000,4096000000000000000,140608000000000000000,1000000000000000000,4096000000000000000000,140608000000000000000000,1000000000000000000000

%N Smallest cube containing exactly n 0's.

%C a(n)^(1/3) = A048365(n) is the index of the first occurrence of n in A269250. -- For n = 3k, obviously a(n) = 10^n. The first terms for indices n = 3k+1 and n = 3k+2 equals 4096*10^3k resp. 140608*10^3k. Is there an index from where on this is no longer true? - _M. F. Hasler_, Feb 20 2016

%F a(n) = A048365(n)^3; a(3n) = 10^(3n); a(3n+1) <= 4096*10^(3n) = (16*10^n)^3 for n>0; a(3n+2) <= 140608*10^(3n) = (52*10^n)^3, with equality for all known terms. - _M. F. Hasler_, Feb 20 2016

%t nsmall = Table[Infinity, 20];

%t For[i = 0, i <= 10^6, i++, n0 = Count[IntegerDigits[i^3], 0];

%t If[nsmall[[n0 + 1]] > i^3, nsmall[[n0 + 1]] = i^3]];

%t Cases[nsmall, _?NumberQ] (* _Robert Price_, Mar 20 2020 *)

%Y Cf. A269250, A086008, A048365.

%Y Cf. A036528 - A036536 for other digits 1 - 9.

%Y Analog for squares: A036507 = A048345^2.

%K nonn,base

%O 0,3

%A _David W. Wilson_

%E Extended to a(0) = 1 and three lines of data completed by _M. F. Hasler_, Feb 20 2016