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A098951
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Smallest available integer fitting the even/odd/even/odd/even... digit pattern (across adjacent numbers).
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3
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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
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1,3
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COMMENTS
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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
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LINKS
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EXAMPLE
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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
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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base,easy,nonn
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AUTHOR
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STATUS
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approved
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