|
%I #6 Jan 21 2019 19:00:49
%S 1,2,3,4,5,6,7,8,12,34,56,78,123,45,67,81,23,456,781,234,567,812,345,
%T 678,1234,5678,12345,6781,2345,67812,3456,7812,34567,8123,4567,81234,
%U 56781,23456,78123,45678,123456,781234,567812,345678,1234567,812345
%N a(n) is the smallest number not yet in the sequence such that concatenation of all terms yields an infinite periodic stream of digits 1, 2, 3, ..., 8 (repeat from 1).
%e Considering a(9), having already 1,2,3,4,5,6,7,8, the next number must be 12 because after 1,2,3,4,5,6,7,8 we shall continue with a 1.
%e But 1 is already in the sequence so we need to add a 2 -> 12. And so on.
%p cyc8 := proc(n) op(n, [2, 3, 4, 5, 6, 7, 8, 1]) ; end:
%p A165306 := proc(n) option remember ; local k, prev, d, a ; if n = 1 then 1; else d := cyc8(procname(n-1) mod 10) ; a := d ; while true do prev := false; for k from 1 to n-1 do if procname(k) = a then prev := true; break; end if; end do; if not prev then return a; end if; d := cyc8(d) ; a := 10*a+d ; end do; end if ; end proc:
%p seq(A165306(n), n=1..60) ; # _R. J. Mathar_, Feb 02 2010
%Y Cf. A165300-A165305, A165307.
%K easy,nonn,base
%O 1,2
%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Sep 14 2009
%E Keyword base added by _R. J. Mathar_, Feb 02 2010
|
