%I #4 Feb 16 2023 05:40:31
%S 1,2,1,3,7,2,1,3,3,2,1,1,3,2,1,5,10,10,10,17,4,8,38,38,19,17,2,1,1,3
%N a(n) is the least positive integer k such that k*(k+1)*...*(k+n-1) does not contain the digit 2, or -1 if there is no such k.
%C a(n) is the least positive integer k such that (k+n-1)!/(k-1)! does not contain the digit 2, or -1 if there is no such k.
%C a(32) = 13.
%C Conjecture: a(n) = -1 for n = 31 and all n >= 33.
%e a(4) = 3 because 3*4*5*6 = 360 does not contain the digit 2, while 1*2*3*4 = 24 and 2*3*4*5 = 120 do.
%p f:= proc(n) local k,t;
%p t:= n!;
%p for k from 1 to 100000 do
%p if not member(2,convert(t,base,10)) then return k fi;
%p t:= t*(n+k)/k;
%p od:
%p -1
%p end proc:
%p map(f, [$1..32]);
%Y Cf. A173333.
%K nonn,base,more
%O 1,2
%A _Robert Israel_, Feb 12 2023