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%I #18 Feb 03 2019 13:27:09
%S 1,20,400,13000,310000,11200000,224000000,10030000000,201100000000,
%T 4022000000000,130440000000000,3114300000000000,112341000000000000,
%U 2302320000000000000,101101400000000000000,2022033000000000000000
%N Powers of ten written in base 5.
%C The leading numbers free of the trailing end 0's in the entries of sequence a(n) are the corresponding powers of 2 written in base 5, i.e., A000866(n). - _Lekraj Beedassy_, Oct 26 2010
%C The first formula follows from the fact that the quinary representation of 10^n - 1 is equal to the concatenation of the quinary representation of 2^n - 1 with four times the n-th repunit; so the successor 10^n is the concatenation of 2^n with n zeros. See the Regan link. - _Washington Bomfim_, Dec 24 2010
%H Rick Regan, <a href="http://www.exploringbinary.com/nines-in-quinary/">Nines in quinary</a>
%F a(n) = A000866(n) followed by n zeros.
%t FromDigits[IntegerDigits[#,5]]&/@(10^Range[0,20]) (* _Harvey P. Dale_, Feb 03 2019 *)
%Y Cf. A000468, A011557.
%K base,easy,nonn
%O 0,2
%A _Henry Bottomley_, Jun 27 2000
%E More terms from _James A. Sellers_, Jul 04 2000
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