cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A044888 Numbers whose base-2 run lengths alternate: even, odd, even, ...

Original entry on oeis.org

3, 6, 15, 24, 27, 30, 54, 63, 96, 99, 111, 120, 123, 126, 198, 216, 219, 222, 246, 255, 384, 387, 399, 438, 447, 480, 483, 495, 504, 507, 510, 774, 792, 795, 798, 864, 867, 879, 888, 891, 894, 966, 984, 987, 990, 1014, 1023, 1536
Offset: 1

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From Emeric Deutsch, Jan 27 2018: (Start)
Also the indices of the compositions whose parts alternate: even, odd, even, ... . For the definition of the index of a composition see A298644.
For example, 99 is in the sequence since its binary form is 1100011 and the composition [2,3,2] has parts: even, odd, even. 100 is not in the sequence since its binary form is 1100100 and the composition [2,2,1,2] has parts even, even, odd, even. The command c(n) from the Maple program yields the composition having index n. (End)

Crossrefs

Programs

  • Maple
    Runs := proc (L) local j, r, i, k: j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: A := {}: for n to 1540 do if `and`(type(c(n)[1], even) = true, type(product(c(n)[j]-c(n)[j+1], j = 1 .. nops(c(n))-1), odd)) then A := `union`(A, {n}) else  end if end do: A; # most of the Maple program is due to W. Edwin Clark. # Emeric Deutsch, Jan 27 2018
  • Mathematica
    oeQ[n_]:=Module[{idsleb=Boole[EvenQ/@(Length/@Split[IntegerDigits[ n,2]])]},idsleb == PadRight[{},Length[idsleb],{1,0}]]; Select[ Range[ 1600],oeQ] (* Harvey P. Dale, Sep 04 2020 *)
  • Python
    from itertools import groupby
    def ok(n): return all(len(list(g))%2 == i%2 for i, (k, g) in enumerate(groupby(bin(n)[2:])))
    print([i for i in range(1, 1537) if ok(i)]) # Michael S. Branicky, Jan 04 2021