# Find subsequence A in sequence B A007908 - OEIS

I have sequence A as A007908 - OEIS and pattern B which is a sub-sequence of that sequence A. I need to find index of first occurrence of the given pattern. For example pattern ..1010...

I've looked at this. Could anyone explain how to solve the problem?

Finding a formula for a pattern

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• Your question could be a question about the software system Mathematica, but it seems from the link like it might belong on math.SE instead. Could you say more? (See mathematica.stackexchange.com/help/on-topic) – Michael E2 Oct 2 '15 at 19:07
• Can you construct a simple example that you can include in the question instead of making folks go to some external site? – george2079 Oct 2 '15 at 19:08
• possible duplicate: mathematica.stackexchange.com/questions/941/… ( Unless the specific input sequence makes a difference ) – george2079 Oct 2 '15 at 19:17
• Flatten[ IntegerDigits[ Range@# ] & /@ Range[ n ] ] – george2079 Oct 2 '15 at 19:21

Using the findSubsequence function from the linked answer: https://mathematica.stackexchange.com/a/942/2079

 findSubsequence[list_, {ss__}] :=
ReplaceList[list, {pre___, ss, ___} :> Length[{pre}] + 1]


suppose we have b:

 b = {8, 4, 9, 1, 2}


In principle you search the full sequence:

 findSubsequence[Flatten[IntegerDigits[Range@#] & /@ Range[FromDigits[b]] ,b]
(*huge, dont do it*)


Find the first subsequence long enough to contain b:

 i = 1; While[ Length@Flatten[IntegerDigits[Range@i] ] < Length@b,
i++]; i


5

check if b occurs early..

 findSubsequence[Flatten[IntegerDigits[Range@#] & /@ Range[i]] , b]


{}

beyond here we need only look in adjacent pairs:

 While[
findSubsequence[
Flatten[Join[IntegerDigits[Range[i]],
IntegerDigits[Range[i + 1]]]], b] == {}, ++i]; i


49

now a manageable* full search up to that i:

 findSubsequence[Flatten[IntegerDigits[Range@#] & /@ Range[i + 1]] , b]


{2043}

*manageable because I picked a sequence I knew occurred pretty early..You can readily eliminate that last search by keeping a cumulative tally of all the sub-sequence lengths.

Edit: a little cleaner version tracking the lengths...

 b = {8, 4, 9, 1, 2}
i = 1; While[ Length@Flatten[IntegerDigits[Range@i] ] < Length@b,
i++]; i
early = Flatten[IntegerDigits[Range@#] & /@ Range[i]];
findSubsequence[early , b]
res = NestWhile[  {
z = Flatten@IntegerDigits[Range[#[[2]] + 1]],
#[[2]] + 1, #[[3]] + Length@z } & ,
{Flatten@IntegerDigits[Range[i]], i , Length@early} ,
findSubsequence[ Flatten[Join[#1[[1]], #2[[1]]]], b] == {} &,
2 ];
lastseq = Flatten@IntegerDigits[Range[res[[2]] - 1]]~Join~
Flatten@IntegerDigits[Range[res[[2]]]];
res[[3]] - Length@lastseq + First@findSubsequence[ lastseq, b]


2043

• Pending clarification from the OP, I don't think this answers the question, unless they're abusing terminology in subsequence... – ciao Oct 3 '15 at 7:36