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.

Showing 1-3 of 3 results.

A005558 a(n) is the number of n-step walks on square lattice such that 0 <= y <= x at each step.

Original entry on oeis.org

1, 1, 3, 6, 20, 50, 175, 490, 1764, 5292, 19404, 60984, 226512, 736164, 2760615, 9202050, 34763300, 118195220, 449141836, 1551580888, 5924217936, 20734762776, 79483257308, 281248448936, 1081724803600, 3863302870000, 14901311070000, 53644719852000
Offset: 0

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Keywords

Comments

Number of n-step walks that start at the origin, constrained to stay in the first octant (0 <= y <= x). (Conjectured) - Benjamin Phillabaum, Mar 11 2011, corrected by Robert Israel, Oct 07 2015
For n >= 1, a(n-1) is the number of Dyck Paths with semilength n having floor((n+2)/2) U's in odd numbered positions. Example: (U is in odd numbered position and u is in even numbered position) Dyck path with n=5, floor ((5+2)/2)=3: UuddUuUddd. - Roger Ford, May 27 2017
The ratio of the number of n-step walks on the octant with an equal number of North steps and South steps to the total number of n-step walks on the octant is A005817(n)/a(n). For the reduced ratio, if n is divisible by 4 or n-1 is divisible by 4 the ratio is 1:floor(n/4)+1 and for all other values of n the ratio is 2:floor(n/2)+2. Example n = 4: A005817(4) = 10; EEEE, EEEW, EEWE, EWEE, EWEW, EEWW, ENSE, ENES, ENSW, EENS; a(4) = 20; 10:20 reduces to 1:2. - Roger Ford, Nov 04 2019

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

See A138350 for a signed version.
Bisections are A000891 and A000888/2.
Cf. A000108, A005817. Column y=0 of A052174.

Programs

  • Magma
    [Binomial(n+1, Ceiling(n/2))*Binomial(n, Floor(n/2)) - Binomial(n+1, Ceiling((n-1)/2))*Binomial(n, Floor((n-1)/2)): n in [0..30]]; // Vincenzo Librandi, Sep 30 2015
    
  • Maple
    A:= proc(n,x,y) option remember;
        local j, xpyp, xp,yp, res;
        xpyp:= [[x-1,y],[x+1,y],[x,y-1],[x,y+1]];
        res:= 0;
        for j from 1 to 4 do
          xp:= xpyp[j,1];
          yp:= xpyp[j,2];
          if xp < 0 or xp > yp or xp + yp > n then next fi;
          res:= res + procname(n-1,xp,yp)
        od;
    return res
    end proc:
    A(0,0,0) := 1:
    seq(add(add(A(n,x,y), y = x .. n - x), x = 0 .. floor(n/2)), n = 0 .. 50); # Robert Israel, Oct 07 2015
  • Mathematica
    a[n_] := 1/2*Binomial[2*Floor[n/2]+1, Floor[n/2]+1]*CatalanNumber[1/2*(n+Mod[n, 2])]*(Mod[n, 2]+2); Table[a[n]//Abs, {n, 0, 27}] (* Jean-François Alcover, Mar 13 2014 *)
  • PARI
    a(n)=binomial(n+1,ceil(n/2))*binomial(n,floor(n/2)) - binomial(n+1,ceil((n-1)/2))*binomial(n,floor((n-1)/2))
    
  • Python
    from sympy import ceiling as c, binomial
    def a(n):
        return binomial(n + 1, c(n/2))*binomial(n, n//2) - binomial(n + 1, c((n - 1)/2))*binomial(n, (n - 1)//2)
    print([a(n) for n in range(51)]) # Indranil Ghosh, Jul 02 2017

Formula

a(n) = C(n+1, ceiling(n/2))*C(n, floor(n/2)) - C(n+1, ceiling((n-1)/2))*C(n, floor((n-1)/2)). - Paul D. Hanna, Apr 16 2004
G.f.: (1/(4x^2))*((16*x^2-1)*(hypergeom([1/2, 1/2],[1],16*x^2)+2*x*(4*x-1)*hypergeom([3/2, 3/2],[2],16*x^2))-2*x+1). - Mark van Hoeij, Oct 13 2009
E.g.f (conjectured): BesselI(1,2*x)*(BesselI(0,2*x)+BesselI(1,2*x))/x. - Benjamin Phillabaum, Feb 25 2011
Conjecture: (2*n+1)*(n+3)*(n+2)*a(n) - 4*(2*n^2+4*n+3)*a(n-1) - 16*n*(2*n+3)*(n-1)*a(n-2) = 0. - R. J. Mathar, Apr 02 2017
Conjecture: (n+3)*(n+2)*a(n) - 4*(n^2+3*n+1)*a(n-1) + 16*(-n^2+n+1)*a(n-2) + 64*(n-1)*(n-2)*a(n-3) = 0. - R. J. Mathar, Apr 02 2017
a(n) = Sum_{k=0..floor(n/2)} n!/(k!*k!*(floor(n/2)-k)!*(floor((n+1)/2)-k)!*(k+1)) (conjectured). - Roger Ford, Aug 04 2017
a(n) = A000108(floor((n+1)/2))*A000108(floor(n/2))*(2*(floor(n/2))+1). - Roger Ford, Nov 15 2019
a(n) = Product_{k=3..n} (4*floor((k-1)/2) + 2) / (floor((k+2)/2)). - Roger Ford, Apr 29 2024

A210736 Expansion of (1 + sqrt( (1 + 2*x) / (1 - 2*x))) / 2 in powers of x.

Original entry on oeis.org

1, 1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252, 462, 924, 1716, 3432, 6435, 12870, 24310, 48620, 92378, 184756, 352716, 705432, 1352078, 2704156, 5200300, 10400600, 20058300, 40116600, 77558760, 155117520, 300540195, 601080390, 1166803110, 2333606220, 4537567650
Offset: 0

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Author

Michael Somos, May 10 2012

Keywords

Comments

Hankel transform is period 4 sequence [ 1, 0, -1, 0, ...] A056594 and the Hankel transform of sequence omitting a(0) is the all 1s sequence A000012. This is the unique sequence with that property.
Series reversion of x*A(x) apparently yields x*A036765(-x). - R. J. Mathar, Sep 24 2012
a(n) is the number of length n words on {-1,1} such that the sum of any of its prefixes is always positive. Cf. A001405 where the sum of all prefixes is nonnegative. - Geoffrey Critzer, Jul 08 2013

Examples

			G.f. = 1 + x + x^2 + 2*x^3 + 3*x^4 + 6*x^5 + 10*x^6 + 20*x^7 + 35*x^8 + 70*x^9 + ...
		

Crossrefs

Essentially the same as A001405.

Programs

  • Mathematica
    nn=36; d=(1-(1-4x^2)^(1/2))/(2x^2);CoefficientList[Series[1/(1-x d),{x,0,nn}],x] (* Geoffrey Critzer, Jul 08 2013 *)
    CoefficientList[Series[2 x / (-1 + 2 x + Sqrt[1 - 4 x^2]), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 10 2014 *)
  • PARI
    {a(n) = if( n<1, n==0, binomial( n - 1, (n - 1)\2))};
    
  • PARI
    {a(n) = polcoeff( (1 + sqrt( (1 + 2*x) / (1 - 2*x) + x * O(x^n))) / 2, n)};

Formula

G.f.: 2 * x / (-1 + 2*x + sqrt(1 - 4*x^2)).
G.f. A(x) satisfies A(x) = A(x)^2 - x / (1 - 2*x).
G.f. A(x) satisfies A( x / (1 + x^2) ) = 1 / (1 - x).
G.f. A(x) satisfies A(1/3) = (1 + sqrt(5))/2.
G.f. A(x) = 1 + x / (1 - 2*x + x / A(x)).
G.f. A(x) = 1 + x / (1 - x / (1 - x / (1 + x / A(x)))).
G.f. A(x) = 1 + x * A001405(x). a(n+1) = A001405(n).
Convolution inverse is A210628. Partial sums is A072100.
Binomial transform with offset 1 is A211278 with offset 1. a(n+2) * a(n) - a(n+1)^2 = A138350(n-1).
a(n) = (-1)^floor(n/2)*hypergeom2F1([1-n, -n],[1],-1). - Peter Luschny, Sep 01 2012
D-finite with recurrence: n*a(n) -2*a(n-1) +4*(2-n)*a(n-2)=0. - R. J. Mathar, Sep 14 2012
G.f. A(x) = 1 / (1 - x / (1 - x^2 / (1 - x^2 / (1 - x^2 / ...)))). - Michael Somos, Jan 02 2013
G.f.: 1/(1 - x*C(x)) where C(x) is the o.g.f. for A126120. - Geoffrey Critzer, Jul 08 2013
a(n) ~ 2^(n-1/2) / sqrt(Pi*n). - Vaclav Kotesovec, Feb 01 2014
G.f.: A(x) = 1 - x/(- 1 + x/A(-x)). - Arkadiusz Wesolowski, Feb 28 2014
From Tom Copeland, Nov 07 2014: (Start)
Setting a(0)=0 here, we have a signed version in A126930 and
O.g.f. G(x)=[-1+sqrt(1+4*x/(1-2x))]/2 = x + x^2 + 2 x^3 + ... = -C[-P(P(x,-1),-1)]= -C[-P(x,-2)] where C(x)= [1-sqrt(1-4*x)]/2= x + x^2 + 2 x^3 + ... = A000108(x) with inverse Cinv(x)=x*(1-x), and P(x,t)= x/(1 + t*x) with inverse P(x,-t).
These types of arrays are from linear fractional transformations of C(x). See A091867.
Ginv(x) = P[-Cinv(-x),2] = x*(1+x)/(1+2*x*(1-x))= (x+x^2)/(1+2(x+x^2)) (see A146559). (End)

A138351 Central moment sequence of tr(A^2) in USp(4).

Original entry on oeis.org

1, 0, 2, 1, 11, 16, 95, 232, 1085, 3460, 14820, 54275, 227095, 895688, 3756688, 15462293, 65586405, 277342336, 1192038266, 5136760581, 22357937431, 97730561480, 430177280197, 1901975209706, 8454151507801, 37734802709796
Offset: 0

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Author

Andrew V. Sutherland, Mar 16 2008, Mar 31 2008

Keywords

Comments

If A is a random matrix in the compact group USp(4) (4 X 4 complex matrices which are unitary and symplectic), then a(n) = E[(tr(A^2)+1)^n] is the n-th central moment of the trace of A^2, since E[tr(A^2)] = -1 (see A138350).

Examples

			a(4) = 11 because E[(tr(A^2)+1)^4] = 11 for a random matrix A in USp(4).
a(4) = 1*A138350(0)+4*A138350(1)+6*A138350(2)+4*A138350(3)+1*A138350(4) = 1*1 + 4*(-1) + 6*3 + 4*(-6) + 1*20 = 11.
		

Crossrefs

Cf. A138350.

Programs

  • Mathematica
    a126120[n_] := If[EvenQ[n], CatalanNumber[n/2], 0];
    a138364[n_] := If[EvenQ[n], 0, Binomial[n, Floor[n/2]], 0];
    a138350[n_] := a126120[n] a138364[n+1] - a138364[n] a126120[n+1];
    a[n_] := Sum[Binomial[n, i] a138350[i], {i, 0, n}];
    Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Aug 13 2018 *)

Formula

a(n) = (1/2)Integral_{x=0..Pi,y=0..Pi}(2cos(2x)+2cos(2y)+1)^n(2cos(x)-2cos(y))^2(2/Pi*sin^2(x))(2/Pi*sin^2(y))dxdy.
a(n) = Sum_{i=0..n} binomial(n,i)*A138350(i).
Showing 1-3 of 3 results.