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.

A253918 Number of Motzkin n-paths with two kinds of level steps both of which are final steps.

Original entry on oeis.org

1, 2, 1, 4, 6, 12, 29, 56, 134, 300, 682, 1624, 3772, 9016, 21597, 51888, 126086, 306636, 750398, 1843864, 4543604, 11242088, 27887730, 69378192, 173050396, 432596216, 1083862132, 2720961520, 6843557944, 17242789104, 43514507997, 109983815264, 278385212358
Offset: 0

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Author

Michael Somos, Jan 19 2015

Keywords

Comments

Number of words of length n on alphabet {U,D,L,R} where U is upstep, D is downstep, L and R are level steps and can only be immediately followed by D or end of word. Defining equation is A = 1 + L + R + UADA.

Examples

			G.f. = 1 + 2*x + x^2 + 4*x^3 + 6*x^4 + 12*x^5 + 29*x^6 + 56*x^7 + ...
A = 1 + (L + R) + (UD) + (ULD + URD + UDR + UDL) + (UDUD + UUDD + ULDL + ULDR + URDL + URDR) + ...
		

Crossrefs

Cf. A143013.

Programs

  • Mathematica
    a[ n_] := SeriesCoefficient[ (1 - Sqrt[1 - 4 x^2 - 8x^3]) / (2 x^2), {x, 0, n}];
  • Maxima
    a(n):=sum((binomial(k+1,n-2*k)*binomial(2*k,k)*2^(n-2*k))/(k+1),k,0,n); /* Vladimir Kruchinin, Mar 11 2016 */
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 - sqrt(1 - 4*x^2 - 8*x^3  + x^3*O(x^n))) / (2*x^2), n))};
    
  • PARI
    {a(n) = my(A); if( n<0, 0, A = O(x); for(k=0, n\2, A = 1 + 2*x + sqr(x*A)); polcoeff( A, n))};
    

Formula

When convolved with itself yields sequence shifted left two places.
G.f. A(x) satisfies A(x) = 1 + 2*x + (A(x)*x)^2.
G.f.: (1 - sqrt(1 - 4*x^2 - 8*x^3)) / (2*x^2).
0 = a(n)*(8*n+4) + a(n+1)*(4*n+8) + a(n+3)*(-n-5) for n>=-1.
a(n) = Sum_{k=0..n}((binomial(k+1,n-2*k)*binomial(2*k,k)*2^(n-2*k))/(k+1)). - Vladimir Kruchinin, Mar 11 2016
a(n) ~ sqrt(3-4*c^2) * 4^n * c^n * (1+2*c)^(n+1) / (sqrt(Pi)*n^(3/2)), where c = 0.37743883312334638... is the root of the equation 4*c^2*(1 + 2*c) = 1. - Vaclav Kotesovec, Mar 10 2016