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.

Previous Showing 21-29 of 29 results.

A204057 Triangle derived from an array of f(x), Narayana polynomials.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 11, 14, 1, 1, 5, 19, 45, 42, 1, 1, 6, 29, 100, 197, 132, 1, 1, 7, 41, 185, 562, 903, 429, 1, 1, 8, 55, 306, 1257, 3304, 4279, 1430, 1, 1, 9, 71, 469, 2426, 8925, 20071, 20793, 4862, 1, 1, 10, 89, 680, 4237, 20076, 65445, 124996, 103049, 16796, 1
Offset: 1

Views

Author

Gary W. Adamson, Jan 09 2012

Keywords

Comments

Row sums = (1, 2, 4, 10, 31, 113, 466, 2129, 10641, 138628, 335379, 2702364,...)
Another version of triangle in A008550. - Philippe Deléham, Jan 13 2012
Another version of A243631. - Philippe Deléham, Sep 26 2014

Examples

			First few rows of the array =
  1,....1,....1,.....1,.....1,...; = A000012
  1.....2,....5,....14,....42,...; = A000108
  1,....3,...11,....45,...197,...; = A001003
  1,....4,...19,...100,...562,...; = A007564
  1,....5,...29,...185,..1257,...; = A059231
  1,....6,...41,...306,..2426,...; = A078009
  ...
First few rows of the triangle =
  1;
  1, 1;
  1, 2,  1;
  1, 3,  5,   1;
  1, 4, 11,  14,    1;
  1, 5, 19,  45,   42,    1;
  1, 6, 29, 100,  197,  132,     1;
  1, 7, 41, 185,  562,  903,   429,     1;
  1, 8, 55, 306, 1257, 3304,  4279,  1430,    1;
  1, 9, 71, 469, 2426, 8952, 20071, 20793, 4862, 1;
  ...
Examples: column 4 of the array = A090197: (1, 14, 45, 100,...) = N(4,n) where N(4,x) is the 4th Narayana polynomial.
Term (5,3) = 29 is the upper left term of M^3, where M = the infinite square production matrix:
  1, 4, 0, 0, 0,...
  1, 1, 4, 0, 0,...
  1, 1, 1, 4, 0,...
  1, 1, 1, 1, 4,...
... generating row 5, A059231: (1, 5, 29, 185,...).
		

Crossrefs

Programs

  • Magma
    A204057:= func< n, k | n eq 0 select 1 else (&+[ Binomial(n, j)^2*k^j*(n-j)/(n*(j+1)): j in [0..n-1]]) >;
    [A204057(k, n-k): k in [1..n], n in [1..12]]; // G. C. Greubel, Feb 16 2021
  • Mathematica
    Table[Hypergeometric2F1[1-k, -k, 2, n-k], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Feb 16 2021 *)
  • Sage
    def A204057(n, k): return 1 if n==0 else sum( binomial(n, j)^2*k^j*(n-j)/(n*(j+1)) for j in [0..n-1])
    flatten([[A204057(k, n-k) for k in [1..n]] for n in [1..12]]) # G. C. Greubel, Feb 16 2021
    

Formula

The triangle is the set of antidiagonals of an array in which columns are f(x) of the Narayana polynomials; with column 1 = (1, 1, 1,...) column 2 = (1, 2, 3,..), column 3 = A028387, column 4 = A090197, then A090198, A090199,...
The array by rows is generated from production matrices of the form:
1, (N-1)
1, 1, (N-1)
1, 1, 1, (N-1)
1, 1, 1, 1, (N-1)
...(infinite square matrices with the rest zeros); such that if the matrix is M, n-th term in row N is the upper left term of M^n.
From G. C. Greubel, Feb 16 2021: (Start)
T(n, k) = Hypergeometric2F1([1-k, -k], [2], n-k).
Sum_{k=1..n} T(n, k) = A132745(n) - 1. (End)

Extensions

Corrected by Philippe Deléham, Jan 13 2012

A371364 Expansion of (1/x) * Series_Reversion( x * (1-4*x)^2 / (1-3*x) ).

Original entry on oeis.org

1, 5, 49, 597, 8129, 118469, 1807665, 28512213, 461141761, 7606159365, 127454706609, 2163636679125, 37130370808257, 643099703566277, 11227141735655345, 197356077159062613, 3490230884900117505, 62054912214781757957, 1108568475427756051761
Offset: 0

Views

Author

Seiichi Manyama, Mar 19 2024

Keywords

Crossrefs

Programs

  • Maple
    A371364 := proc(n)
        add(3^(n-k)*binomial(2*n+k+1,k)*binomial(2*n,n-k),k=0..n) ;
        %/(n+1) ;
    end proc:
    seq(A371364(n),n=0..60) ; # R. J. Mathar, Apr 22 2024
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serreverse(x*(1-4*x)^2/(1-3*x))/x)
    
  • PARI
    a(n) = sum(k=0, n, 3^(n-k)*binomial(2*n+k+1, k)*binomial(2*n, n-k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} 3^(n-k) * binomial(2*n+k+1,k) * binomial(2*n,n-k).
D-finite with recurrence (n+1)*(2*n+1)*a(n) +3*(-6*n^2-9*n+2)*a(n-1) -27*(7*n-9)*(2*n-3)*a(n-2) -243*(n-2)*(2*n-5)*a(n-3)=0. - R. J. Mathar, Apr 22 2024

A086871 Row sums of A059450.

Original entry on oeis.org

1, 2, 10, 58, 370, 2514, 17850, 130890, 983650, 7536418, 58648810, 462306266, 3683602130, 29620138994, 240059315610, 1958940281322, 16081662931650, 132723191430210, 1100568370427850, 9164925012016506, 76612776253995570
Offset: 0

Views

Author

N. J. A. Sloane, Sep 16 2003

Keywords

Comments

Hankel transform is A165928. - Paul Barry, Sep 30 2009
Number of skew Dyck paths of semilength n with the down steps coming in two colors. - David Scambler, Jun 21 2013

Examples

			G.f. = 1 + 2*x + 10*x^2 + 58*x^3 + 370*x^4 + 2514*x^5 + 17850*x^6 + 130890*x^7 + ...
		

Programs

  • Mathematica
    Table[SeriesCoefficient[2/(1+Sqrt[(1-9*x)/(1-x)]),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
  • PARI
    {a(n) = if( n<0, 0, polcoeff( 2 / (1 + sqrt((1 - 9*x) / (1 - x) + x * O(x^n))), n))}; /* Michael Somos, Mar 06 2004 */
    
  • PARI
    {a(n) = if( n<1, n==0, n++; 2 * polcoeff( serreverse( x * (1 - 4*x) / (1 - 3*x) + x * O(x^n)), n))}; /* Michael Somos, Mar 06 2004 */

Formula

a(n) = 2*A059231(n), if n>0.
G.f.: (1 - x - sqrt((1 - x) * (1 - 9*x))) / (4*x) = 2 / (1 + sqrt((1 - 9*x) / (1 - x))) =: y satisfies 0 = (1 - x) * (1 - y) + 2*x*y^2. - Michael Somos, Mar 06 2004
Moment representation: a(n) = (1/(4*Pi))*Integral_{x=1..9} x^n*sqrt(-x^2+10x-9)/x+(1/2)*0^n. - Paul Barry, Sep 30 2009
D-finite with recurrence Recurrence: (n+1)*a(n) = 5*(2*n-1)*a(n-1) - 9*(n-2)*a(n-2). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ 3^(2*n+1)/(2*sqrt(2*Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2012
G.f.: 1/(2*x) - 1/2 + G(0) where G(k) = 1 - 1/(x + x/(1 + 1/G(k+1) )) ; (continued fraction,3-step). - Sergei N. Gladkovskii, Nov 29 2012

Extensions

More terms from Ray Chandler, Sep 17 2003

A224071 Number of Schroeder paths of semilength n in which there are no (2,0)-steps at level 1.

Original entry on oeis.org

1, 2, 5, 15, 52, 201, 841, 3726, 17213, 82047, 400600, 1993377, 10071777, 51532938, 266462229, 1390174911, 7308741084, 38682855225, 205940368441, 1102091393574, 5925177392573, 31987877317887, 173337754977904
Offset: 0

Views

Author

Keywords

Comments

Hankel transform is A006215. Invert transform of A155069. - Michael Somos, Apr 02 2013

Examples

			a(2) = 5 because we have HH, UDH, HUD, UDUD and UUDD.
G.f. = 1 + 2*x + 5*x^2 + 15*x^3 + 52*x^4 + 201*x^5 + 841*x^6 + ...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[4/(3-5*x+Sqrt[x^2-6*x+1]), {x, 0, 20}], x] (* Vaclav Kotesovec, May 23 2013 *)
    a[ n_] := SeriesCoefficient[ (3 - 5 x - Sqrt[ 1 - 6 x + x^2]) / (2 - 6 x + 6 x^2), {x, 0, n}]; (* Michael Somos, Mar 28 2014 *)
  • Maxima
    a(n):=sum((k+1)*((-1)^floor((k+2)/3)+(-1)^floor((k+1)/3))*sum(binomial(n+1,n-k-i)*binomial(n+i,n),i,0,n-k),k,0,n)/(2*(n+1)); /* Vladimir Kruchinin, Mar 08 2016*/
  • PARI
    z='z+O('z^66); Vec(4/(3-5*z+sqrt(1-6*z+z^2))) /* Joerg Arndt, Mar 30 2013 */
    

Formula

G.f.: 4/(3-5*x+sqrt(1-6*x+x^2)).
Recurrence: n*a(n) = 9*(n-1)*a(n-1) - 2*(11*n-15)*a(n-2) + 3*(7*n-12)*a(n-3) - 3*(n-3)*a(n-4). - Vaclav Kotesovec, May 23 2013
a(n) ~ sqrt(884+627*sqrt(2)) * (3+2*sqrt(2))^n / (98*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, May 23 2013
0 = +a(n)*(+9*a(n+1) - 144*a(n+2) + 174*a(n+3) - 81*a(n+4) + 12*a(n+5)) + a(n+1)*(+18*a(n+1) + 399*a(n+2) - 597*a(n+3) + 318*a(n+4) - 57*a(n+5)) + a(n+2)*(-300*a(n+2) + 538*a(n+3) - 255*a(n+4) + 52*a(n+5)) + a(n+3)*(-126*a(n+3) + 73*a(n+4) - 18*a(n+5)) + a(n+4)*(+a(n+5)) if n>=0. - Michael Somos, Mar 28 2014
a(n) = Sum_{k=0..n}((k+1)*((-1)^floor((k+2)/3)+(-1)^floor((k+1)/3))*Sum_{i=0..n-k}(binomial(n+1,n-k-i)*binomial(n+i,n)))/(2*(n+1)). - Vladimir Kruchinin, Mar 08 2016

A366166 Decimal expansion of sqrt(Pi)/(3*sqrt(3))*(Gamma(1/3)/Gamma(5/6))^3.

Original entry on oeis.org

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

Views

Author

Hugo Pfoertner, Oct 13 2023

Keywords

Comments

This constant c occurs in the probability that the "big player" in a game with 3 gamblers goes broke first, although he starts with an initial capital of N-2 units, whereas the other two gamblers start with one unit each. This probability is ~ c/N^3. See Diaconis link for details.

Examples

			4.5597944999598458154817364845572481176367423801661405635...
		

Crossrefs

Programs

  • Mathematica
    First[RealDigits[Gamma[1/3]^9/(2Pi)^4,10,100]] (* Paolo Xausa, Oct 14 2023 *)
  • PARI
    sqrt(Pi)/(3*sqrt(3))*(gamma(1/3)/gamma(5/6))^3

Formula

Equals Gamma(1/3)^9 / (2*Pi)^4. - Peter Luschny, Oct 13 2023

A371365 Expansion of (1/x) * Series_Reversion( x * (1-4*x)^3 / (1-3*x) ).

Original entry on oeis.org

1, 9, 141, 2701, 57513, 1307553, 31083925, 763267077, 19208408721, 492817411705, 12842067417309, 338956669920189, 9042967461581753, 243464712274093713, 6606427290991922277, 180492205687604057013, 4960765361688213527073
Offset: 0

Views

Author

Seiichi Manyama, Mar 19 2024

Keywords

Crossrefs

Programs

  • PARI
    my(N=20, x='x+O('x^N)); Vec(serreverse(x*(1-4*x)^3/(1-3*x))/x)
    
  • PARI
    a(n) = sum(k=0, n, 3^(n-k)*binomial(3*n+k+2, k)*binomial(3*n+1, n-k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} 3^(n-k) * binomial(3*n+k+2,k) * binomial(3*n+1,n-k).

A257363 Number of 3-Motzkin paths with no level steps at height 1.

Original entry on oeis.org

1, 3, 10, 33, 110, 369, 1247, 4248, 14603, 50724, 178314, 635526, 2300829, 8477382, 31842897, 122103276, 478372886, 1915188093, 7831613468, 32674683984, 138871668314, 600140517762, 2631926843602, 11690520554421, 52498671870181, 237966449687118, 1087246253873875, 5001141997115010, 23137102115963262
Offset: 0

Views

Author

Keywords

Comments

For n=2 we have 10 paths: H(1)H(1), H(1)H(2), H(2)H(1), H(2)H(2), H(1)H(3), H(3)H(1), H(3)H(3), H(2)H(3), H(3)H(2), UD.

Crossrefs

Programs

  • Maple
    rec:= (95+95*n)*a(n)+(-180-9*n)*a(n+1)+(-329-197*n)*a(n+2)+(369+144*n)*a(n+3)+(-117-36*n)*a(4+n)+(12+3*n)*a(n+5):
    f:= gfun:-rectoproc({rec,a(0)=1,a(1)=3,a(2)=10,a(3)=33,a(4)=110},a(n),remember):
    seq(f(n),n=0..100); # Robert Israel, Apr 28 2015
  • Mathematica
    CoefficientList[Series[2*(3+x)/(6-17*x-9*x^2+x*Sqrt[1-6*x+5*x^2]), {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 21 2015 *)

Formula

G.f.: 1/(1-3*x-x*F(x)), where F(x) is the g.f. of the sequence A117641.
G.f.: 2*(3+x)/(6-17*x-9*x^2+x*sqrt(1-6*x+5*x^2)).
a(n) ~ 5^(n+3/2)/(98*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Apr 21 2015
From Robert Israel, Apr 28 2015 (Start):
G.f.: (6-x*sqrt(1-6*x+5*x^2)-17*x-9*x^2)/(6-36*x+42*x^2+38*x^3).
3*(-n+1)*a(n) +9*(4*n-7)*a(n-1) +9*(-16*n+39)*a(n-2) +(197*n-656)*a(n-3) +9*(n+15)*a(n-4) +95*(-n+4)*a(n-5)=0. (End)

A086873 Triangle read by rows in which row n >= 1 gives coefficients in expansion of the polynomial Sum_{k=1..n} (1/n)*binomial(n,k)*binomial(n,k-1)*x^(2k)*(1+x)^(2n-2k) / x^2 in powers of x.

Original entry on oeis.org

1, 1, 2, 2, 1, 4, 9, 10, 5, 1, 6, 21, 44, 57, 42, 14, 1, 8, 38, 116, 240, 336, 308, 168, 42, 1, 10, 60, 240, 680, 1392, 2060, 2160, 1530, 660, 132, 1, 12, 87, 430, 1545, 4152, 8449, 13014, 14985, 12540, 7227, 2574, 429, 1, 14, 119, 700, 3045, 10122, 26173, 53048
Offset: 1

Views

Author

N. J. A. Sloane, Sep 16 2003

Keywords

Comments

Row n has 2n-1 terms.

Examples

			For n=3 the polynomial is 1 + 4x + 9x^2 + 10x^3 + 5x^4.
  1;
  1,  2,  2;
  1,  4,  9,  10,    5;
  1,  6, 21,  44,   57,   42,   14;
  1,  8, 38, 116,  240,  336,  308,   168,    42;
  1, 10, 60, 240,  680, 1392, 2060,  2160,  1530,   660,  132;
  1, 12, 87, 430, 1545, 4152, 8449, 13014, 14985, 12540, 7227, 2574, 429;
		

Crossrefs

A059231 gives row sums.

Programs

  • Maple
    j := 0:f := n->sum(binomial(n,k)*binomial(n,k-1)/n*x^(2*k)*(1+x)^(2*n-2*k),k=1..n): for n from 1 to 15 do p := expand(f(n)/x^2):for l from 0 to 2*n-2 do j := j+1: a[j] := coeff(p,x,l):od:od:seq(a[l],l=1..j); # Sascha Kurz
  • PARI
    for(n=1,8,p=sum(k=1,n,(1/n)*binomial(n,k)*binomial(n,k-1)*x^(2*k)*(1+x)^(2*n-2*k))/x^2; for(i=1,2*n-1,print1(polcoeff(p,i-1) ","); ); print; ); \\ Ray Chandler, Sep 17 2003

Extensions

More terms from Vladeta Jovovic and Ray Chandler, Sep 17 2003

A257072 Number of 3-colored Schroeder paths of semilength n in which there are no (2,0)-steps at level 1.

Original entry on oeis.org

1, 4, 17, 77, 374, 1959, 11085, 67500, 438485, 3004985, 21485222, 158744467, 1202966761, 9297312916, 72981656937, 580105886517, 4658713796790, 37736326098735, 307913254091925, 2528335636842300, 20875157745756429
Offset: 0

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Author

Keywords

Examples

			a(1) = 4  because we have H1, H2, H2, UD.
		

Crossrefs

Programs

  • Mathematica
    Table[8 DifferenceRoot[Function[{y, n}, {3645 (1 + n) (2 + n) y[n] + (-20736 - 21384 n - 5508 n^2) y[1 + n] + (6120 + 7857 n + 2061 n^2) y[2 + n] + (1330 - 165 n - 175 n^2) y[3 + n] + (-664 - 270 n - 26 n^2) y[4 + n] + 3 (4 + n) (5 + n) y[5 + n] == 0, y[0] == 1/8, y[1] == 1/2, y[2] == 17/8, y[3] == 77/8, y[4] == 187/4}]][k], {k, 0, 20}] (* Benedict W. J. Irwin, May 29 2016 *)
    CoefficientList[Series[8/(7 -27*x +Sqrt[1 -10*x +9*x^2]), {x,0,50}], x] (* G. C. Greubel, May 29 2016 *)
  • Maxima
    a(n):=(sum(m*sum(((sum(binomial(j+m,k)*binomial(j-1,j-k)*4^(j-k),k,0,j))*3^(n-j-m)*binomial(n-j,m))/(j+m),j,0,n-m),m,1,n))+3^n; /* Vladimir Kruchinin, Mar 13 2016 */

Formula

G.f.: 8/(7-27*z+sqrt(1-10*z+9*z^2))=1/(1-3*z-z*F(z)), where F(z) is the g.f. of the sequence A059231.
a(n) = (3^n+Sum_{m=1..n}(m*Sum_{j=0..n-m}(((Sum_{k=0..j}(binomial(j+m,k)*binomial(j-1,j-k)*4^(j-k)))*3^(n-j-m)*binomial(n-j,m))/(j+m)))). - Vladimir Kruchinin, Mar 13 2016
a(n) ~ sqrt(2)*3^(2*n-1) / (2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 13 2016
From Benedict W. J. Irwin, May 29 2016: (Start)
Let y(0)=1/8, y(1)=1/2, y(2)=17/8, y(3)=77/8, y(4)=187/4,
Let 3645*(n+1)*(n+2)*y(n)-(5508n^2+21384n+20736)*y(n+1)+(2061n^2+7857n+6120)*y(n+2)-(175n^2+165n-1330)*y(n+3)-(26n^2+270n+664)*y(n+4)+3*(n+4)*(n+5)*y(n+5) = 0,
a(n) = 8*y(n).
(End)
Conjecture: 3*n*a(n) +(-53*n+45)*a(n-1) +2*(151*n-213)*a(n-2) +9*(-73*n+144)*a(n-3) +405*(n-3)*a(n-4)=0. - R. J. Mathar, Sep 24 2016
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