A098951 Smallest available integer fitting the even/odd/even/odd/even... digit pattern (across adjacent numbers).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 21, 23, 25, 27, 29, 41, 43, 45, 47, 49, 61, 63, 65, 67, 69, 81, 83, 85, 87, 89, 210, 10, 12, 14, 16, 18, 30, 32, 34, 36, 38, 50, 52, 54, 56, 58, 70, 72, 74, 76, 78, 90, 92, 94, 96, 98, 101, 212, 103, 214, 105, 216, 107, 218, 109

Offset: 1

Author

Eric Angelini, Oct 21 2004

Keywords

• base
• easy
• nonn

Comments

Different from A097962.

Indeed, A097962 is required to be increasing. Therefore, a(31) = 210 can here be followed by a(32) = 10, while A097962(32) = 301. - M. F. Hasler, Mar 23 2019

Examples

			After a(30) = 89, a(31) must start with an even digit. A number consisting of one even digit would work, but they are all in the sequence already. A 2-digit number with first digit even and second digit odd would work, but they are also all in the sequence already. A 3-digit number would have to have even, odd, and even digits in that order. The smallest such number is 210, so a(31) = 210. - _Michael B. Porter_, Mar 25 2019
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A030141, A097962, A307129.

See also A377919.

Programs

Mathematica

altQ[n_] := n < 10 || Union[Total /@ Partition[ Mod[ IntegerDigits@n, 2], 2, 1]] == {1}; s = {0}; Do[j=1; While[ MemberQ[s, j] || EvenQ[s[[-1]] + First@ IntegerDigits@ j] || ! altQ[j], j++];  AppendTo[s, j], {68}]; s (* Giovanni Resta, Mar 05 2014 *)
(* longer, but faster *) eoQ[n_] := Module[{d = IntegerDigits[n], alt, i}, alt = Table[If[OddQ[i], -1, 1], {i, d}]; i = 1; While[i++; i <= Length[d] && alt[[i]] == alt[[1]]*(-1)^(i + 1)]; If[i <= Length[d], alt[[1]] = 0]; alt[[1]]]; nn = 10000; tev = {}; tod = {}; Do[If[eoQ[i] == -1, AppendTo[tod, i], If[eoQ[i] == 1, AppendTo[tev, i]]], {i, nn}]; t = {0}; While[tev != {} && tod != {}, If[OddQ[t[[-1]]], AppendTo[t, tev[[1]]]; tev = Rest[tev], AppendTo[t, tod[[1]]]; tod = Rest[tod]]]; t (* T. D. Noe, Mar 05 2014 *)