A098951 - OEIS

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

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OFFSET

1,3

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

LINKS

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

EXAMPLE

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

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 *)

CROSSREFS

Cf. A030141, A097962, A307129.