%I #16 May 11 2017 11:43:21
%S 75,92,696,794,921,6946,9211,9418,67365,69446,73515,92111,92592,
%T 584799,653597,694446,921111,5793295,6693466,6944446,7145554,7694443,
%U 9211111,58788989,61728398,66733665,69444446,72175395,76445374,88183426,89245679,91721145,92111111,92592592
%N Zeroless numbers n = concat(s,t) such that s * t is the 10’s complement of the digits of n.
%C Numbers of the form concat(92,(10^k  1)/9) and concat(69,4*(10^k  1)/9,6) for k>0 are part of the sequence.
%C The first digit of a term must be 5 or higher. Numbers of the form 925925...92592 are terms, i.e. concatenate 925 m times for m >= 0 and append 92. Numbers of the form concat(2*(10^k1)/3,9,(10^(k1)1)/3,4,2*(10^k1)/3) for k > 1 are also terms. It also works for the case k=1 if the middle 0 that results is removed. If a term n is even, then the digits of n besides the least significant digit must contain either 4 or 9. If a term n is odd and not divisible by 5, then the digits of n besides the least significant digit must contain 9.  _Chai Wah Wu_, May 10 2017
%H Chai Wah Wu, <a href="/A286353/b286353.txt">Table of n, a(n) for n = 1..51</a>
%e 75 = concat(7,5) and 7 * 5 = 35 is the 10’s complement of 75.
%e 584799 = concat(58479,9) and 58479 * 9 = 526311 is the 10’s complement of 584799.
%e 7694443 = concat(769,4443) and 769 * 4443 = 3416667 is the 10’s complement of 7694443.
%p P:=proc(q) local a,b,c,d,k,ok,n; for n from 1 to q do b:=ilog10(n)+1; c:=n; ok:=1;
%p d:=0; for k from 1 to b do if (c mod 10)>0 then a:=(10(c mod 10)); d:=d+10^(k1)*a; c:=trunc(c/10); else ok:=0; break; fi; od; if ok=1 then for k from 1 to b do
%p if d=(n mod 10^k)*trunc(n/10^k) then print(n); fi; od; fi; od; end: P(10^7);
%K nonn,base
%O 1,1
%A _Paolo P. Lava_, May 08 2017
