%I
%S 1,2,3,5,8,13,4,7,11,12,23,15,6,111,22,14,25,17,18,9,117,28,10,21,33,
%T 16,27,19,110,31,24,26,38,211,32,35,48,112,43,37,210,41,45,29,311,42,
%U 36,39,212,53,58,113,34,47,411,52,57,312,63,49,213,44,68,114,55,310,51,46,410,61,67,313,54,59,214,65,511,62,78,115,56,611,72,69,215,66,412,73,510,71,88,116,77,314,75,512,83,711,82,610,81,79,216,87,315,76,413
%N Start T with a(1)=1 and a(2)=2. We try to set a(n) to be the sum s of the last two decimal digits of T, except that if s is already in the sequence, we replace s with the smallest unused integer that ends with the same digit as s.
%C This is not a permutation of the natural numbers; for instance, 100 (or any number ending in "100") does not appear in this sequence since the last two digits of S will never sum to 0, 00, or 100.  _Jim Nastos_, Mar 13 2014
%H E. Angelini, <a href="/A238948/a238948_2.pdf">The sum rhymes</a> [Cached copy, with permission]
%e a(3) = 1+2 = 3;
%e a(4) = 2+3 = 5;
%e a(5) = 3+5 = 8;
%e a(6) = 5+8 = 13;
%e a(7) = 1+3 = 4;
%e a(8) = 3+4 = 7;
%e a(9) = 4+7 = 11;
%e a(10) = 12 because 1+1 = 2 is already in the sequence (as a(2)), and 12 is the smallest unused integer ending with "2";
%e a(11) = 23 because both 1+2 = 3 and 13 are already in the sequence (as a(3) and a(6), respectively);
%e a(12) = 15 because 2+3 = 5 is a(5);
%e a(13) = 1+5 = 6.
%K nonn,base
%O 1,2
%A _Eric Angelini_, Mar 14 2014
