A231924 Working in base 5: a(0)=0, thereafter a(n+1) is the smallest number not already in the sequence such that the bits of a(n) and a(n+1) together can be rearranged to form a palindrome.

0, 11, 1, 10, 100, 12, 2, 20, 101, 22, 3, 13, 31, 103, 30, 110, 33, 4, 14, 41, 104, 40, 114, 24, 42, 112, 21, 102, 120, 201, 210, 1000, 111, 44, 113, 23, 32, 121, 200, 211, 222, 233, 244, 323, 332, 424, 442, 1001, 122, 133, 144, 212, 221, 313, 331, 414, 441, 1002, 123

Offset: 0

Author

N. J. A. Sloane and Robert G. Wilson v, Nov 15 2013

Keywords

• nonn
• base
• easy

Comments

This is a permutation of the nonnegative integers in base 5 - see the Comments in A228407 for the proof.

Crossrefs

Cf. A230891, A231920, A231922, A231926, A231928, A231930, A231932, A228407, A231925.

Programs

Mathematica

a[0] = 0; a[n_] := a[n] = Block[{k = 1, idm = IntegerDigits[ a[n - 1], 5], t = a@# & /@ Range[n - 1]}, Label[ start]; While[ MemberQ[t, k], k++];  While[ Select[ Permutations[ Join[ idm, IntegerDigits[k, 5]]], #[[1]] != 0 && # == Reverse@# &] == {}, k++; Goto[ start]]; k]; s = Array[a, 60, 0]; FromDigits@# & /@ IntegerDigits[s, 5]