A329267 a(n) is the absolute difference between n and its nearest palindromic neighbor.
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 5, 4, 3, 2
Offset: 0
Examples
For 0 <= n <= 9, n is palindromic so a(n) = 0. a(10) = 10-9 = 11-10 = 1 (10 is equidistant from its two nearest palindromes). a(11) = 0 because 11 is palindromic. For 12 <= n <= 16, a(n) = n-11 because 11 is the nearest palindromic number. For 17 <= n <= 22, a(n) = 22-n because 22 is the nearest palindromic number. . n nearest palindrome difference -- ------------------ ---------- 1 1 1-1 = 0 2 2 2-2 = 0 3 3 3-3 = 0 4 4 4-4 = 0 5 5 5-5 = 0 6 6 6-6 = 0 7 7 7-7 = 0 8 8 8-8 = 0 9 9 9-9 = 0 10 9 or 11 10-9 = 11-10 = 1 11 11 11-11 = 0 12 11 12-11 = 1 13 11 13-11 = 2 14 11 14-11 = 3 15 11 15-11 = 4 16 11 16-11 = 5 17 22 22-17 = 5 18 22 22-18 = 4 19 22 22-19 = 3 20 22 22-20 = 2 21 22 22-21 = 1 22 22 22-22 = 0 23 22 23-22 = 1
Programs
-
Mathematica
palQ[n_] := Block[{d = IntegerDigits[n]}, d == Reverse@ d]; a[n_] := Block[{k=0}, While[! palQ[n+k] && ! palQ[n-k], k++]; k]; Array[a, 121] (* Giovanni Resta, Nov 12 2019 *) -
PARI
ispal(n) = my (d=digits(n)); d==Vecrev(d) a(n) = for (k=0, oo, if (ispal(n-k) || ispal(n+k), return (k))) \\ Rémy Sigrist, Dec 03 2019
Comments