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 1-5 of 5 results.

A190155 Central coefficients of the Riordan matrix (g(x),x*g(x)), where g(x) = (1-x-x^2-sqrt(1-2*x-5*x^2+2*x^3+x^4))/(2*x^2) (A132276).

Original entry on oeis.org

1, 2, 12, 64, 385, 2346, 14672, 92936, 595179, 3841970, 24959726, 162988464, 1068860884, 7034520304, 46437268905, 307351081056, 2038878634695, 13552394472612, 90242046694715, 601847594327000, 4019556724362165, 26879647264387170
Offset: 0

Views

Author

Emanuele Munarini, May 05 2011

Keywords

Crossrefs

Cf. A132276.

Programs

  • Mathematica
    Table[Sum[Binomial[n+2i,i](n+1)/(i+n+1)Sum[Binomial[2n-j,n+2i]Binomial[n-2i-j,j],{j,0,n-2i}],{i,0,n/2}],{n,0,21}]
  • Maxima
    makelist(sum(binomial(n+2*i,i)*(n+1)/(i+n+1)*sum(binomial(2*n-j,n+2*i)*binomial(n-2*i-j,j),j,0,n-2*i),i,0,n/2),n,0,21);
    
  • PARI
    for(n=0,30, print1(sum(i=0,n/2, binomial(n+2*i,i)*((n+1)/(i+n+1)) *sum(j=0, n-2*i, binomial(2*n-j,n+2*i)*binomial(n-2*i-j,j))), ", ")) \\ G. C. Greubel, Dec 28 2017

Formula

a(n) = T(2*n,n) where T(n,k) = A132276(n,k).
a(n) = Sum_{i=0..(n/2)} ( binomial(n+2*i,i)*((n+1)/(i+n+1)) * Sum_{j=0..(n-2*i)} binomial(2*n-j,n+2*i)*binomial(n-2*i-j,j) ).

A128720 Number of paths in the first quadrant from (0,0) to (n,0) using steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0).

Original entry on oeis.org

1, 1, 3, 6, 16, 40, 109, 297, 836, 2377, 6869, 20042, 59071, 175453, 524881, 1579752, 4780656, 14536878, 44394980, 136107872, 418757483, 1292505121, 4001039563, 12418772656, 38641790001, 120510911885, 376628460529, 1179376013552, 3699860515924, 11626784875214
Offset: 0

Views

Author

Emeric Deutsch, Mar 30 2007, revised Sep 03 2007

Keywords

Comments

Points of two kinds are placed on a line: light points having weight 1 and heavy points having weight 2. Number of configurations of points of total weight n, with some of the light points being paired off by nonintersecting arcs.
Number of skew Dyck paths of semilength n having no UUU's. A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of a path is defined to be the number of its steps. Example: a(3)=6 because we have UDUDUD, UDUUDD, UDUUDL, UUDDUD, UUDUDD and UUDUDL. a(n)=A128719(n,0). a(n)=A059397(n,n). a(n)=A132276(n,0).
Hankel transform is the (1,3) Somos-4 sequence A174168. - Paul Barry, Mar 10 2010
First column of the Riordan matrix A132276. - Emanuele Munarini, May 05 2011

Examples

			a(3)=6 because we have hhh, hH, Hh, hUD, UhD and UDh.
G.f. = 1 + x + 3*x^2 + 6*x^3 + 16*x^4 + 40*x^5 + 109*x^6 + 297*x^7 + ...
		

Crossrefs

Programs

  • Maple
    a[0]:=1: a[1]:=1: for n from 2 to 30 do a[n]:=a[n-1]+a[n-2]+add(a[j]*a[n-2-j], j=0..n-2) end do: seq(a[n],n=0..30); G:=((1-z-z^2-sqrt((1+z-z^2)*(1-3*z-z^2)))*1/2)/z^2: Gser:=series(G,z=0,33): seq(coeff(Gser,z,n),n=0..30);
  • Mathematica
    Table[Sum[Binomial[2k,k]/(k+1)Sum[Binomial[n-j,2k]Binomial[n-j-2k,j],{j,0,n/2}],{k,0,n/2}],{n,0,12}] (* Emanuele Munarini, May 05 2011 *)
  • Maxima
    makelist(sum(binomial(2*k,k)/(k+1)*sum(binomial(n-j,2*k)*binomial(n-j-2*k,j),j,0,n/2),k,0,n/2),n,0,12); /* Emanuele Munarini, May 05 2011 */

Formula

a(n) = Sum_{j=0..floor(n/2)} binomial(n-j, j)*m(n-2j), where m(k)=A001006(k) are the Motzkin numbers.
G.f. = G satisfies z^2*G^2 - (1-z-z^2)*G + 1 = 0.
G.f. = c(z^2/(1-z-z^2)^2)/(1-z-z^2), where c(z) = (1-sqrt(1-4z))/(2z) is the Catalan function.
a(n) = a(n-1) + a(n-2) + Sum_{j=0..n-2} a(j)*a(n-2-j), a(0) = a(1) = 1.
G.f.: (1/(1-x-x^2))*c(x^2/(1-x-x^2)^2) = (1/(1-x^2))*m(x/(1-x^2)), c(x) the g.f. of A000108, m(x) the g.f. of A001006. - Paul Barry, Mar 18 2010
Let A(x) be the g.f., then B(x) = 1 + x*A(x) = 1 + 1*x + 1*x^2 + 3*x^3 + 6*x^4 + ... = 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1+x-x^2) (continued fraction); more generally B(x)=C(x/(1+x-x^2)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
a(n) = Sum_{k=0..floor(n/2)} (binomial(2*k,k)/(k+1))*Sum_{j=0..floor(n/2)} binomial(n-j, 2*k)*binomial(n-j-2*k, j). - Emanuele Munarini, May 05 2011
D-finite with recurrence: (n+2)*a(n) + (-2*n-1)*a(n-1) + 5*(-n+1)*a(n-2) + (2*n-5)*a(n-3) + (n-4)*a(n-4) = 0. - R. J. Mathar, Dec 03 2012
G.f.: (1 - x - x^2 - sqrt(1 - 2*x - 5*x^2 + 2*x^3 + x^4))/(2*x^2) = 1/Q(0), where Q(k) = 1 - x - x^2 - x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 04 2013
a(n) ~ sqrt(78+22*sqrt(13)) * ((3+sqrt(13))/2)^n / (4 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Feb 13 2014

A059398 Row sums of triangle in A059397.

Original entry on oeis.org

1, 2, 6, 17, 51, 154, 473, 1464, 4568, 14332, 45187, 143024, 454217, 1446604, 4618576, 14777451, 47371177, 152110326, 489165277, 1575211177, 5078690936, 16392526502, 52963765321, 171282782902, 554393341371, 1795821017014
Offset: 0

Views

Author

N. J. A. Sloane, Jan 29 2001

Keywords

Comments

Number of paths in the first quadrant from (0,0) to the line x=n, consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0) (in other words, left factors of the paths in A128720). Example: a(2)=6 because we have hh, H, UD, hU, Uh and UU. Row sums of triangle in A132276. - Emeric Deutsch, Sep 03 2007
Row sums of the Riordan matrix (g(x),x*g(x)), where g(x) = (1-x-x^2-sqrt(1-2*x-5*x^2+2*x^3+x^4))/(2*x^2) (A132276). - Emanuele Munarini, May 05 2011

Crossrefs

Programs

  • Magma
    Q:=Rationals(); R:=PowerSeriesRing(Q, 40); Coefficients(R!(Sqrt((1+x-x^2)/(1-3*x-x^2))-1)/(2*x)) // G. C. Greubel, Jan 29 2018
  • Maple
    g:=(1/2)*(sqrt((1+x-x^2)/(1-3*x-x^2))-1)/x: gser:=series(g,x=0,30): seq(coeff(gser,x,n),n=0..25); # Emeric Deutsch, Sep 03 2007
  • Mathematica
    Table[Sum[Binomial[2k,k](-1)^(n-k+1)Sum[Binomial[i+k-1,i]Binomial[i,n-k-i+1],{i,0,n-k+1}],{k,0,n+1}]/2,{n,0,28}] (* Emanuele Munarini, May 05 2011 *)
    With[{nn = 50}, CoefficientList[Series[(Sqrt[(1 + x - x^2)/(1 - 3*x - x^2)] - 1)/x/2, {x, 0, nn}], x]] (* G. C. Greubel, Jan 29 2018 *)
  • Maxima
    makelist(sum(binomial(2*k,k)*(-1)^(n-k+1)*sum(binomial(i+k-1,i)*binomial(i,n-k-i+1),i,0,n-k+1),k,0,n+1)/2,n,0,28); /* Emanuele Munarini, May 05 2011 */
    
  • PARI
    x='x+O('x^30); Vec((sqrt((1+x-x^2)/(1-3*x-x^2))-1)/x/2) \\ G. C. Greubel, Jan 29 2018
    

Formula

G.f.: (sqrt((1+x-x^2)/(1-3*x-x^2))-1)/x/2. - Vladeta Jovovic, Jan 20 2004
a(n) = (1/2)*sum(binomial(2*k,k)*(-1)^(n-k+1)*sum(binomial(i+k-1,i)*binomial(i,n-k-i+1),i=0..n-k+1),k=0..n+1). - Emanuele Munarini, May 05 2011

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jan 31 2001

A059397 Triangle formed by right-bounded rhombus rule, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 7, 6, 1, 4, 12, 18, 16, 1, 5, 18, 37, 53, 40, 1, 6, 25, 64, 120, 148, 109, 1, 7, 33, 100, 227, 369, 430, 297, 1, 8, 42, 146, 385, 760, 1146, 1244, 836, 1, 9, 52, 203, 606, 1391, 2518, 3519, 3656, 2377, 1, 10, 63, 272, 903, 2346, 4900, 8188
Offset: 0

Views

Author

N. J. A. Sloane, Jan 29 2001

Keywords

Comments

T(n,n)=A128720(n). Mirror image of A132276. - Emeric Deutsch, Sep 03 2007

Examples

			If triangle is reflected in the vertical axis it looks like this:
1
1 1
3 2 1
6 7 3 1
16 18 12 4 1
and now the rhombus rule is clearly visible (e.g. 18 = 6 + 7 + 3 + 2).
		

Crossrefs

A variation on A059317. Row sums give A059398.

Programs

  • Maple
    g:=proc(z) options operator, arrow: (1/2-(1/2)*z-(1/2)*z^2-(1/2)*sqrt((1+z-z^2)*(1-3*z-z^2)))/z^2 end proc: G:=simplify(g(t*z)/(1-z*g(t*z))): Gser:=simplify(series(G,z=0,13)): for n from 0 to 10 do P[n]:=sort(coeff(Gser,z,n)) end do: for n from 0 to 10 do seq(coeff(P[n],t,j),j=0..n) end do; # yields sequence in triangular form - Emeric Deutsch, Sep 03 2007
  • Mathematica
    max = 10; g[z_] := (1 - z - z^2 - Sqrt[(1 + z - z^2)*(1 - 3*z - z^2)])/(2 z^2); s = Series[g[t*z]/(1 - z*g[t*z]), {z, 0, max}, {t, 0, max}] // Normal; t[n_, k_] := SeriesCoefficient[s, {z, 0, n}, {t, 0, k}]; t[0, 0] = 1; Table[t[n, k], {n, 0, max}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 16 2014, after Emeric Deutsch *)

Formula

Each entry is sum of 3 entries above it in previous row and the entry directly above two rows back (provided the entries are properly aligned).
G.f.=G(t,z)=g(tz)/(1-zg(tz)), where g(z)=(1-z-z^2-sqrt((1+z-z^2)(1-3z-z^2)))/(2z^2). - Emeric Deutsch, Sep 03 2007

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jan 31 2001

A190156 Expansion of (1-x-3*x^2-sqrt(1-2*x-5*x^2+2*x^3+x^4))/(2*x^3*(1+2*x)).

Original entry on oeis.org

1, 1, 4, 8, 24, 61, 175, 486, 1405, 4059, 11924, 35223, 105007, 314867, 950018, 2880620, 8775638, 26843704, 82420464, 253916555, 784672011, 2431695541, 7555381574, 23531026853, 73448858179, 229730744171, 719914525210, 2260031465504, 7106721944206
Offset: 0

Views

Author

Emanuele Munarini, May 05 2011

Keywords

Comments

Diagonal sums of the Riordan matrix (g(x),x*g(x)), where g(x) = (1-x-x^2-sqrt(1-2*x-5*x^2+2*x^3+x^4))/(2*x^2) (A132276).

Crossrefs

Cf. A132276.

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1-x-3*x^2-Sqrt(1-2*x-5*x^2+2*x^3+x^4))/(2*x^3*(1+2*x)))); // G. C. Greubel, Oct 22 2018
    
  • Maple
    T := (n, k) -> simplify(2^k*binomial(n-k, k)*hypergeom([-k, k-n-1], [2], 1/2)):
    seq(add(T(n, k), k=0..floor(n/2)), n=0..28); # Peter Luschny, Oct 19 2020
  • Mathematica
    CoefficientList[Series[(1-x-3x^2-Sqrt[1-2x-5x^2+2x^3+x^4])/(2x^3(1+2x)),{x,0,28}],x]
  • Maxima
    a(n):=sum(sum(2^(m-j+1)*binomial(n-m,j-1)*binomial(n-m+2,j)*binomial(n-m-j+1,m-j+1),j,0,n-m+2)/(n-m+2),m,0,n); /* Vladimir Kruchinin, Oct 19 2020 */
  • PARI
    x='x+O('x^66); Vec((1-x-3*x^2-sqrt(1-2*x-5*x^2+2*x^3+x^4))/(2*x^3*(1+2*x))) /* Joerg Arndt, May 15 2011 */
    

Formula

G.f.: (1-x-3*x^2-sqrt(1-2*x-5*x^2+2*x^3+x^4))/(2*x^3*(1+2*x)).
D-finite with recurrence: (n+3)*a(n) +3*a(n-1) +3*(-3*n-2)*a(n-2) +(-8*n-3)*a(n-3) +(5*n-9) *a(n-4) +2*(n-3)*a(n-5)=0. - R. J. Mathar, Oct 08 2016
a(n) = Sum_{m=0..n} Sum_{j=0..n-m+2} 2^(m-j+1)*C(n-m,j-1)*C(n-m+2,j)*C(n-m-j+1,m-j+1)/(n-m+2). - Vladimir Kruchinin, Oct 19 2020
a(n) = Sum_{k=0..floor(n/2)} 2^k*binomial(n-k, k)*hypergeom([-k, k-n-1], [2], 1/2). - Peter Luschny, Oct 19 2020
Showing 1-5 of 5 results.