A165300 a(n) is the smallest number not already present that permits the cyclic repetition of the path 1,2 of the digits in the sequence.

1, 2, 12, 121, 21, 212, 1212, 12121, 2121, 21212, 121212, 1212121, 212121, 2121212, 12121212, 121212121, 21212121, 212121212, 1212121212, 12121212121, 2121212121, 21212121212, 121212121212, 1212121212121, 212121212121

Offset: 1

Author

Paolo P. Lava and Giorgio Balzarotti, Sep 14 2009

Keywords

• easy
• base
• nonn

Comments

Conjecture. (1) If n > 1, and a(n) begins and ends with 1, then a(n+1) is obtained by deleting the initial 1 of a(n); (2) if a(n) begins with 1 and ends with 2 then a(n+1) is obtained by adding a final 1 to a(n); (3) if a(n) begins with 2 and ends with 1 then a(n+1) is obtained by adding a final 2 to a(n); (4) if a(n) begins and ends with 2 then a(n+1) is obtained by adding an initial 1 to a(n). This has been confirmed through a(140), which has 71 digits (and should be fairly easy to prove). - John W. Layman, Sep 22 2009

Examples

			Starting from 1,2 the next number must be 12 because after 1,2 we shall continue with a 1. But 1 is already in the sequence so we need to add a 2 -> 12. And so on.
		

Crossrefs

Cf. A165301-A165307.

Programs

Maple

P:=proc(i) local a,n; a:=2; print(1);print(2); for n from 3 by 1 to i do a:=1/24*((a+10^floor(1+evalf(log10(a),100)))*(((n-2) mod 4)+((n-1) mod 4)+7*(n mod 4)-5*((n+1) mod 4))+(10*a+1)*(((n-2) mod 4)+7*((n-1) mod 4)-5*(n mod 4)+((n+1) mod 4))+(a-10^floor(evalf(log10(a),100)))*(7*((n-2) mod 4)-5*((n-1) mod 4)+(n mod 4)+((n+1) mod 4))+(10*a+2)*(-5*((n-2) mod 4)+((n-1) mod 4)+(n mod 4)+7*((n+1) mod 4))); print(a); od; end: P(200); # Paolo P. Lava, Oct 02 2009

Formula

a(n+1) = (1/24)*((a(n) + 10^floor(1 + log_10(a(n))))*(((n-2) mod 4) + ((n-1) mod 4) + 7*(n mod 4) - 5*((n+1) mod 4)) + (10*a(n)+1)*(((n-2) mod 4) + 7*((n-1) mod 4) - 5*(n mod 4) + ((n+1) mod 4)) + (a(n) - 10^floor(log_10(a(n))))*(7*((n-2) mod 4) - 5*((n-1) mod 4) + (n mod 4) + ((n+1) mod 4)) + (10*a(n) + 2)*(-5*((n-2) mod 4) + ((n-1) mod 4) + (n mod 4) + 7*((n+1) mod 4))), with n >= 3 and a(1)=1, a(2)=2. - Paolo P. Lava, Oct 02 2009

Extensions

Terms a(21) onward from John W. Layman, Sep 22 2009

Edited by N. J. A. Sloane, Oct 06 2009