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%I #15 May 10 2021 09:59:34
%S 0,1,10,20,1101,1121,11311,31101,40210,340210,4620020,5431101,7211311,
%T 12040210,24120020,151651121,165631101,1135531101,8084220020,
%U 9117311311,894105331101
%N Numbers k such that k mod (2, 3, 4, ... , i+1) = (d_i, d_i-1, ..., d_1), where d_1, d_2, ..., d_i are the digits of k, with MSD(k) = d_1 and LSD(k) = d_i.
%e a(11) = 5431101 because:
%e 5431101 mod 2 = 1, 5431101 mod 3 = 0, 5431101 mod 4 = 1,
%e 5431101 mod 5 = 1, 5431101 mod 6 = 3, 5431101 mod 7 = 4,
%e 5431101 mod 8 = 5.
%p P:=proc(q) local a,i,j,n,ok; print(0); print(1); for n from 1 to q do
%p for i from 0 to 1 do a:=10*n+i; ok:=1; for j from 1 to ilog10(a)+1 do
%p if (a mod 10)<>((10*n+i) mod (j+1)) then ok:=0; break; else
%p a:=trunc(a/10); fi; od; if ok=1 then print(10*n+i); break; fi;
%p od; od; end: P(10^12);
%Y Cf. A266181, A284815.
%K nonn,base,more
%O 0,3
%A _Paolo P. Lava_, Sep 24 2018
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