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A238948 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. 1

%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

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Last modified May 20 01:04 EDT 2022. Contains 353847 sequences. (Running on oeis4.)