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Irregular triangle read by rows where row n starts with n and each further term is the sum of the distinct palindromes in the concatenation of the decimal digits of preceding terms.
1

%I #50 Sep 26 2023 19:16:11

%S 1,1,12,125,463,476,483,491,500,500,6055,6170,2,2,24,250,497,513,517,

%T 3,3,36,375,750,2082,2112,4258,4504,4504,4548,5002,4,4,48,500,505,

%U 6065,62742,63407,63410,63411,63422,63444,5,5,60,66,738,756

%N Irregular triangle read by rows where row n starts with n and each further term is the sum of the distinct palindromes in the concatenation of the decimal digits of preceding terms.

%C Palindromes are substrings and 0s are included so that for instance 00500 is a palindrome, and 0 and 00 are distinct (but both value 0).

%C The first term in the first row is 1, which is a palindrome, so each subsequent term is the sum of all distinct palindromes in the concatenation of the sequence. So far we have 1,1,... This gives us one new palindrome (11), so our new sum is 1+11=12. Now we have 1,1,12,... This added two palindromes (111 and 2) to make the new total 125. Continue until you reach a term which repeats endlessly without adding any new palindromes. For n=1, this term is 6170.

%H Neal Gersh Tolunsky, <a href="/A356822/b356822.txt">Table of n, a(n) for n = 1..9324</a> (first 1000 rows)

%e Irregular array begins:

%e 1, 1, 12, 125, 463, 476, 483, 491, 500, 500, 6055, 6170;

%e 2, 2, 24, 250, 497, 513, 517;

%e 3, 3, 36, 375, 750, 2082, 2112, 4258, 4504, 4504, 4548, 5002;

%e 4, 4, 48, 500, 505, 6065, 62742, 63407, 63410, 63411, 63422, 63444;

%e 5, 5, 60, 66, 738, 756;

%e 6, 6, 72, 81, 90, 99, 1107, 1118, 1229, 1432, 1439;

%e 7, 7, 84, 96, 111, 234, 239;

%e 8, 8, 96, 111, 234, 243, 34910, 39277, 39361, 39754, 39759, 39759, 40698;

%e 9, 9, 108, 117, 135, 314, 14202, 14961, 14967;

%e 10, 1, 102, 225, 474, 959, 1927, 2846, 3132, 3448, 3815, 4653

%K nonn,tabf,look,base

%O 1,3

%A _Neal Gersh Tolunsky_, Sep 17 2022