A253241 The "Reverse and Add!" problem in base 12: sequence lists the final palindrome number for n, or -1 if no palindrome is ever reached. (Written in base 10.)
0, 2, 4, 6, 8, 10, 13, 39, 65, 91, 117, 143, 13, 26, 39, 52, 65, 78, 91, 104, 117, 130, 143, 169, 26, 39, 52, 65, 78, 91, 104, 117, 130, 143, 169, 169, 39, 52, 65, 78, 91, 104, 117, 130, 143, 169, 169, 507, 52, 65, 78, 91, 104, 117, 130, 143, 169, 169, 507, 676, 65, 78, 91, 104, 117
Offset: 0
Examples
a(29) = 91 since (in duodecimal) 25 (decimal 29) + 52 = 77 (decimal 91) and 77 is a palindrome. a(69) = 507 since (in duodecimal) 59 (decimal 69) + 95 = 132, 132 + 231 = 363 (decimal 507) and 363 is a palindrome. a(105) = 1885 since (in duodecimal) 89 (decimal 105) + 98 = 165, 165 + 561 = 706, 706 + 607 = 1111 (decimal 1885) and 1111 is a palindrome.
Programs
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Mathematica
tol = 1728; r[n_] := FromDigits[Reverse[IntegerDigits[n, 12]], 12]; palQ[n_] := n == r[n]; ar[n_] := n + r[n]; Table[k = 0; If[palQ[n], n = ar[n]; k = 1]; While[! palQ[n] && k < tol, n = ar[n]; k++]; If[k == tol, n = -1]; n, {n, 0, 144}]
Comments