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%I #8 Oct 27 2019 17:30:28
%S 2,72,272,2272,22272,222272,7222272,27222272,727222272,2727222272,
%T 72727222272,772727222272,7772727222272,77772727222272,
%U 277772727222272,2277772727222272,72277772727222272,272277772727222272
%N a(n) contains n digits (either '2' or '7') and is divisible by 2^n.
%H Robert Israel, <a href="/A053318/b053318.txt">Table of n, a(n) for n = 1..999</a>
%F a(n) = a(n-1) + 10^(n-1)*(2 + 5*(a(n-1)/2^(n-1) mod 2)), i.e., a(n) ends with a(n-1); if a(n-1) is divisible by 2^n then a(n) begins with a 2, if not then a(n) begins with a 7.
%p A[1]:= 2:
%p for n from 2 to 100 do
%p if A[n-1] mod 2^n = 0 then A[n]:= A[n-1]+2*10^(n-1)
%p else A[n]:= A[n-1]+7*10^(n-1)
%p fi
%p od:
%p seq(A[i],i=1..100); # _Robert Israel_, Oct 27 2019
%Y Cf. A023399, A050621, A050622, A035014.
%K base,nonn
%O 1,1
%A _Henry Bottomley_, Mar 06 2000
%E Formula corrected by _Robert Israel_, Oct 27 2019
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