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.

A242062 Expansion of x * (1 - x^12) / ((1 - x^3) * (1 - x^4) * (1 - x^7)) in powers of x.

Original entry on oeis.org

0, 1, 0, 0, 1, 1, 0, 1, 2, 1, 1, 2, 2, 1, 2, 3, 2, 2, 3, 3, 2, 3, 4, 3, 3, 4, 4, 3, 4, 5, 4, 4, 5, 5, 4, 5, 6, 5, 5, 6, 6, 5, 6, 7, 6, 6, 7, 7, 6, 7, 8, 7, 7, 8, 8, 7, 8, 9, 8, 8, 9, 9, 8, 9, 10, 9, 9, 10, 10, 9, 10, 11, 10, 10, 11, 11, 10, 11, 12, 11, 11, 12
Offset: 0

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Author

Michael Somos, Aug 13 2014

Keywords

Examples

			G.f. = x + x^4 + x^5 + x^7 + 2*x^8 + x^9 + x^10 + 2*x^11 + 2*x^12 + x^13 + ...
		

Crossrefs

Cf. A131372.

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1 -x+x^3-x^5+x^6)/(1-x-x^7+x^8))); // G. C. Greubel, Aug 05 2018
  • Maple
    a:= n -> [0, 1, 0, 0, 1, 1, 0][n mod 7 + 1] + floor(n/7):
    seq(a(n), n=0..20); # Robert Israel, Aug 13 2014
  • Mathematica
    a[ n_] := Quotient[ n+3, 7] + {1, 0, 0, 0, 0, -1, 0}[[Mod[ n, 7, 1]]];
    a[ n_] := Sign[n] * SeriesCoefficient[ x * (1 - x^12) / ((1 - x^3) * (1 - x^4) * (1 - x^7)), {x, 0, Abs[n]}];
    CoefficientList[Series[x (1 - x + x^3 - x^5 + x^6)/(1 - x - x^7 + x^8), {x, 0, 100}], x] (* Vincenzo Librandi, Aug 14 2014 *)
    LinearRecurrence[{1,0,0,0,0,0,1,-1},{0,1,0,0,1,1,0,1},90] (* Harvey P. Dale, Sep 29 2024 *)
  • PARI
    {a(n) = (n+3)\7 + (n%7==1) - (n%7==6)};
    
  • PARI
    {a(n) = sign(n) * polcoeff( x * (1 - x^12) / ((1 - x^3) * (1 - x^4) * (1 - x^7)) + x * O(x^abs(n)), abs(n))};
    

Formula

Euler transform of length 12 sequence [0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, -1].
G.f.: x * (1 - x + x^3 - x^5 + x^6) / (1 - x - x^7 + x^8).
G.f.: x / (1 - x^3 / (1 - x / (1 + x / (1 - x^5 / (1 - x / (1 + x^2 / (1 - x^2))))))).
a(n) = -a(-n) = a(n-7) + 1 = a(n-1) + a(n-7) - a(n-8) for all n in Z.
0 = 2*a(n) - a(n+1) + a(n+2) - 2*a(n+3) + (a(n+1) - a(n+2))^2 for all n in Z.
a(n+1) - a(n) = A131372(n).