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 1-9 of 9 results.

A277637 Partial sums of A007004.

Original entry on oeis.org

1, 4, 34, 454, 7384, 133510, 2583958, 52468078, 1104191608, 23891534758, 528527606698, 11905777228618, 272269758961018, 6306419621308618, 147677930682023818, 3491114167267938298, 83217317955857060788, 1998209467779811473538, 48293984598611551487038, 1174012306200941229710038, 28689784424223599507417938
Offset: 0

Views

Author

Emanuele Munarini, Oct 25 2016

Keywords

Crossrefs

Cf. A007004.

Programs

  • Mathematica
    Table[Sum[Multinomial[k, k, k]/(k + 1), {k, 0, n}], {n, 0, 100}]
  • Maxima
    makelist(sum(multinomial_coeff(k,k,k)/(k+1),k,0,n),n,0,12);

Formula

a(n) = Sum_{k=0..n} multinomial(k,k,k)/(k+1).
Recurrence: (n+2)*(n+3)*a(n+2)-(28*n^2+86*n+66)*a(n+1)+3*(3*n+5)*(3*n+4)*a(n)=0.
a(n) = hypergeometric(1/3,2/3;2;27) - (multinomial(n+1,n+1,n+1)/(n+2)) * hypergeometric(1,n+4/3,n+5/3;n+2,n+3;27).
a(n) = 0 (mod 2) and a(n) = 1 (mod 3), for all natural n.
G.f.: hypergeometric(1/3,2/3;2;27*t)/(1-t).
a(n) ~ 3^(3*n+7/2) / (52*Pi*n^2). - Vaclav Kotesovec, Oct 30 2016

A277638 Binomial partial sums of sequence A007004.

Original entry on oeis.org

1, 4, 37, 520, 8803, 165292, 3320023, 69943804, 1526981575, 34271322316, 786371041603, 18372739163632, 435772652437381, 10468369504009060, 254238148448959729, 6233226769739934964, 154092763036678601551, 3837301178450916902428, 96181503100227675085675
Offset: 0

Views

Author

Emanuele Munarini, Oct 25 2016

Keywords

Crossrefs

Cf. A007004.

Programs

  • Mathematica
    Table[Sum[Binomial[n, k] Multinomial[k, k, k]/(k+1), {k, 0, n}], {n, 0, 100}]
  • Maxima
    makelist(sum(binomial(n, k)*multinomial_coeff(k, k, k)/(k+1), k, 0, n), n, 0, 12);

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * multinomial(k,k,k)/(k+1).
a(n) = hypergeometric(1/3,2/3,-n; 1,2; -27).
a(n) == 1 (mod 3) for all natural n.
E.g.f.: exp(t) * hypergeometric(1/3,2/3; 1,2; 27*t).
From Vaclav Kotesovec, Oct 26 2016: (Start)
Recurrence: n*(n+1)*a(n) = 2*(3*n-1)*(5*n-3)*a(n-1) - (n-1)*(57*n-56)*a(n-2) + 28*(n-2)*(n-1)*a(n-3).
a(n) ~ 2^(2*n+3) * 7^(n+2) / (3^(11/2) * Pi * n^2).
(End)
Diff. eq. satisfied by the ordinary g.f.: t*(1-t)^2*(1-28*t)*A''(t)+2*(1-t)*(1-2*t)*(1-28*t)*A'(t)-2*(4-29*t+28*t^2)*A(t)=0. - Emanuele Munarini, Oct 28 2016
G.f.: hypergeom([1/3, 2/3],[2],27*x/(1-x))/(1-x). - Mark van Hoeij, Nov 28 2024

A277639 Double binomial partial sums of A007004.

Original entry on oeis.org

1, 4, 43, 718, 14779, 344452, 8725093, 234594766, 6596287411, 192032529388, 5747827847545, 175986201591130, 5490952102178725, 174077883157001740, 5594651323154783515, 181946073109880839450, 5978730547304013537475, 198263347772478727193740, 6628299876919271425393105, 223211734849614639629628010, 7566093949269408444819804937
Offset: 0

Views

Author

Emanuele Munarini, Oct 25 2016

Keywords

Crossrefs

Cf. A007004.

Programs

  • Mathematica
    Table[Sum[Binomial[n, k]^2 Multinomial[k,k,k]/(k+1), {k,0,n}], {n,0,100}]
  • Maxima
    makelist(sum(binomial(n,k)^2*multinomial_coeff(k,k,k)/(k+1),k,0,n),n,0,12);

Formula

a(n) = Sum_{k=0..n} binomial(n,k)^2*multinomial(k,k,k)/(k+1).
a(n) = hypergeometric(1/3,2/3,-n,-n;1,1,2;27).
Double e.g.f.: BesselI(0,2*sqrt(t))*hypergeometric(1/3,2/3;1,1,2;27*t).
D-finite with recurrence: n^2*(n+1)^2*(1058*n^4 - 7061*n^3 + 16158*n^2 - 14048*n + 3284)*a(n) = 2*n*(30682*n^7 - 219052*n^6 + 555798*n^5 - 545060*n^4 + 16565*n^3 + 323730*n^2 - 206943*n + 39408)*a(n-1) - (834762*n^8 - 7954803*n^7 + 30596846*n^6 - 59518007*n^5 + 57023894*n^4 - 13636388*n^3 - 20674168*n^2 + 16952656*n - 3600432)*a(n-2) + 2*(n-2)^2*(744832*n^6 - 5313736*n^5 + 13458434*n^4 - 12947434*n^3 - 64535*n^2 + 6504872*n - 2110473)*a(n-3) - 676*(n-3)^2*(n-2)^2*(1058*n^4 - 2829*n^3 + 1323*n^2 + 1317*n - 609)*a(n-4). - Vaclav Kotesovec, Oct 30 2016
a(n) ~ sqrt(205/162 + 1939/(729*sqrt(3))) * (28+6*sqrt(3))^n / (Pi^(3/2)*n^(5/2)). - Vaclav Kotesovec, Oct 30 2016

A006480 De Bruijn's S(3,n): (3n)!/(n!)^3.

Original entry on oeis.org

1, 6, 90, 1680, 34650, 756756, 17153136, 399072960, 9465511770, 227873431500, 5550996791340, 136526995463040, 3384731762521200, 84478098072866400, 2120572665910728000, 53494979785374631680, 1355345464406015082330, 34469858696831179429500, 879619727485803060256500, 22514366432046593564460000
Offset: 0

Views

Author

Keywords

Comments

Number of paths of length 3n in an n X n X n grid from (0,0,0) to (n,n,n), using steps (0,0,1), (0,1,0), and (1,0,0).
Appears in Ramanujan's theory of elliptic functions of signature 3.
S(s,n) = Sum_{k=0..2n} (-1)^(k+n) * binomial(2n, k)^s. The formula S(3,n) = (3n)!/(n!)^3 is due to Dixon (according to W. N. Bailey 1935). - Charles R Greathouse IV, Dec 28 2011
a(n) is the number of ballot results that end in a 3-way tie when 3n voters each cast two votes for two out of three candidates vying for 2 slots on a county board; in such a tie, each of the three candidates receives 2n votes. Note there are C(3n,2n) ways to choose the voters who cast a vote for the youngest candidate. The n voters who did note vote for the youngest candidate voted for the two older candidates. Then there are C(2n,n) ways to choose the other n voters who voted for both the youngest and the second youngest candidate. The remaining voters vote for the oldest candidate. Hence there are C(3n,2n)*C(2n,n)=(3n)!/(n!)^3 ballot results. - Dennis P. Walsh, May 02 2013
a(n) is the constant term of (X+Y+1/(X*Y))^(3*n). - Mark van Hoeij, May 07 2013
For n > 2 a(n) is divisible by (n+2)*(n+1)^2, a(n) = (n+1)^2*(n+2)*A161581(n). - Alexander Adamchuk, Dec 27 2013
a(n) is the number of permutations of the multiset {1^n, 2^n, 3^n}, the number of ternary words of length 3*n with n of each letters. - Joerg Arndt, Feb 28 2016
Diagonal of the rational function 1/(1 - x - y - z). - Gheorghe Coserea, Jul 06 2016
At least two families of elliptic curves, x = 2*H1 = (p^2+q^2)*(1-q) and x = 2*H2 = p^2+q^2-3*p^2*q+q^3 (0Bradley Klee, Feb 25 2018
The ordinary generating function also determines periods along a family of tetrahedral-symmetric sphere curves ("du troisième ordre"). Compare links to Goursat "Étude des surfaces..." and "Proof Certificate". - Bradley Klee, Sep 28 2018

Examples

			G.f.: 1 + 6*x + 90*x^2 + 1680*x^3 + 34650*x^4 + 756756*x^5 + 17153136*x^6 + ...
		

References

  • L. A. Aizenberg and A. P. Yuzhakov, "Integral representations and residues in multidimensional complex analysis", American Mathematical Society, 1983, p. 194.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 174.
  • N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row 3 of A187783.
Related to diagonal of rational functions: A268545-A268555. Elliptic Integrals: A002894, A113424, A000897. Factors: A005809, A000984. Integrals: A007004, A024486. Sphere Curves: A318245, A318495.

Programs

  • GAP
    List([0..20],n->Factorial(3*n)/Factorial(n)^3); # Muniru A Asiru, Mar 31 2018
    
  • Magma
    [Factorial(3*n)/(Factorial(n))^3: n in [0..20] ]; // Vincenzo Librandi, Aug 20 2011
    
  • Maple
    seq((3*n)!/(n!)^3, n=0..16); # Zerinvary Lajos, Jun 28 2007
  • Mathematica
    Sum [ (-1)^(k+n) Binomial[ 2n, k ]^3, {k, 0, 2n} ]
    a[ n_] := If[ n < 0, 0, (-1)^n HypergeometricPFQ[ {-2 n, -2 n, -2 n}, {1, 1}, 1]]; (* Michael Somos, Oct 22 2014 *)
    Table[Multinomial[n, n, n], {n, 0, 100}] (* Emanuele Munarini, Oct 25 2016 *)
    CoefficientList[Series[Hypergeometric2F1[1/3,2/3,1,27*x],{x,0,5}],x] (* Bradley Klee, Feb 28 2018 *)
    Table[(3n)!/(n!)^3,{n,0,20}] (* Harvey P. Dale, Mar 09 2025 *)
  • Maxima
    makelist(multinomial_coeff(n,n,n),n,0,24); /* Emanuele Munarini, Oct 25 2016 */
    
  • PARI
    {a(n) = if( n<0, 0, (3*n)! / n!^3)}; /* Michael Somos, Dec 03 2002 */
    
  • PARI
    {a(n) = my(A, m); if( n<1, n==0, m=1; A = 1 + O(x); while( m<=n, m*=3; A = subst( (1 + 2*x) * subst(A, x, (x/3)^3), x, serreverse(x * (1 + x + x^2) / (1 + 2*x)^3 / 3 + O(x^m)))); polcoeff(A, n))}; /* Michael Somos, Dec 03 2002 */
    
  • Python
    from math import factorial
    def A006480(n): return factorial(3*n)//factorial(n)**3 # Chai Wah Wu, Oct 04 2022

Formula

Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 1/2 * sqrt(3) * 27^n / (Pi*n) - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
From Karol A. Penson, Nov 21 2001: (Start)
O.g.f.: hypergeom([1/3, 2/3], [1], 27*x).
E.g.f.: hypergeom([1/3, 2/3], [1, 1], 27*x).
Integral representation as n-th moment of a positive function on [0, 27]:
a(n) = int( x^n*(-1/24*(3*sqrt(3)*hypergeom([2/3, 2/3], [4/3], 1/27*x)* Gamma(2/3)^6*x^(1/3) - 8*hypergeom([1/3, 1/3], [2/3], 1/27*x)*Pi^3)/Pi^3 /x^(2/3)/Gamma(2/3)^3), x=0..27). This representation is unique. (End)
a(n) = Sum_{k=-n..n} (-1)^k*binomial(2*n, n+k)^3. - Benoit Cloitre, Mar 02 2005
a(n) = C(2n,n)*C(3n,n) = A104684(2n,n). - Paul Barry, Mar 14 2006
G.f. satisfies: A(x^3) = A( x*(1+3*x+9*x^2)/(1+6*x)^3 )/(1+6*x). - Paul D. Hanna, Oct 29 2010
D-finite with recurrence: n^2*a(n) - 3*(3*n-1)*(3*n-2)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = (n+1)^2*(n+2)*A161581(n) for n>2. - Alexander Adamchuk, Dec 27 2013
0 = a(n)^2*(472392*a(n+1)^2 - 83106*a(n+1)*a(n+2) + 3600*a(n+2)^2) + a(n)*a(n+1)*(-8748*a(n+1)^2 + 1953*a(n+1)*a(n+2) - 120*a(n+2)^2) + a(n+1)^2*(36*a(n+1)^2 - 12*a(n+1)*a(n+2) + a(n+2)^2) for all n in Z. - Michael Somos, Oct 22 2014
0 = x*(27*x-1)*y'' + (54*x-1)*y' + 6*y, where y is g.f. - Gheorghe Coserea, Jul 06 2016
From Peter Bala, Jul 15 2016: (Start)
a(n) = 3*binomial(2*n - 1,n)*binomial(3*n - 1,n) = 3*[x^n] 1/(1 - x)^n * [x^n] 1/(1 - x)^(2*n) for n >= 1.
a(n) = binomial(2*n,n)*binomial(3*n,n) = ([x^n](1 + x)^(2*n)) *([x^n](1 + x)^(3*n)) = [x^n](F(x)^(6*n)), where F(x) = 1 + x + 2*x^2 + 14*x^3 + 127*x^4 + 1364*x^5 + 16219*x^6 + ... appears to have integer coefficients. Cf. A002894.
This sequence occurs as the right-hand side of several binomial sums:
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)^3 = a(n) (Dixon's identity).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n,n - k)*binomial(3*n + k,k) = a(n) (Gould, Vol. 4, 6.86)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = a(n).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n + k,k)*binomial(3*n,n - k) = a(n).
Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = a(n).
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k,k)*binomial(4*n - k,2*n) = a(n) (see Gould, Vol.5, 9.23).
Sum_{k = 0..2*n} (-1)^k*binomial(3*n,k)*binomial(3*n - k,n)^3 = a(n) (Sprugnoli, Section 2.9, Table 10, p. 123). (End)
From Bradley Klee, Feb 28 2018: (Start)
a(n) = A005809(n)*A000984(n).
G.f.: F(x) = 1/(2*Pi) Integral_{z=0..2*Pi} 2F1(1/3,2/3; 1/2; 27*x*sin^2(z)) dz.
With G(x) = x*2F1(1/3,2/3; 2; 27*x): F(x) = d/dx G(x). (Cf. A007004) (End)
F(x)*G(1/27-x) + F(1/27-x)*G(x) = 1/(4*Pi*sqrt(3)). - Bradley Klee, Sep 29 2018
Sum_{n>=0} 1/a(n) = A091683. - Amiram Eldar, Nov 15 2020
From Peter Bala, Sep 20 2021: (Start)
a(n) = Sum_{k = n..2*n} binomial(2*n,k)^2 * binomial(k,n). Cf. A001459.
a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integers n and k (write a(n) as C(3*n,2*n)*C(2*n,n) and apply Mestrovic, equation 39, p. 12). (End)
a(n) = 6*A060542(n). - R. J. Mathar, Jun 21 2023
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^3 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^3 * binomial(2*n, n+k)^3 = x*(x + n)*(x + 2*n)*a(n) (x arbitrary). Compare with Dixon's identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^3 = a(n). - Peter Bala, Jul 31 2023
From Peter Bala, Aug 14 2023: (Start)
a(n) = (-1)^n * [x^(2*n)] ( (1 - x)^(4*n) * Legendre_P(2*n, (1 + x)/(1 - x)) ).
Row 1 of A364509. (End)
From Peter Bala, Oct 10 2024: (Start)
The following hold for n >= 1:
a(n) = Sum_{k = 0.. 2*n} (-1)^(n+k) * k/n * binomial(2*n, k)^3 = 3/2 * Sum_{k = 0.. 2*n} (-1)^(n+k) * (k/n)^2 * binomial(2*n, k)^3.
a(n) = 3/2 * Sum_{0..2*n-1} (-1)^(n+k) * k/n * binomial(2*n, k)^2*binomial(2*n-1, k).
a(n) = 3 * Sum_{0..2*n-1} (-1)^(n+k) * k/n * binomial(2*n, k)*binomial(2*n-1, k)^2. (End)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(2*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

Extensions

a(14)-a(16) from Eric W. Weisstein
Terms a(17) and beyond from T. D. Noe, Jun 29 2008

A060693 Triangle (0 <= k <= n) read by rows: T(n, k) is the number of Schröder paths from (0,0) to (2n,0) having k peaks.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 10, 6, 1, 14, 35, 30, 10, 1, 42, 126, 140, 70, 15, 1, 132, 462, 630, 420, 140, 21, 1, 429, 1716, 2772, 2310, 1050, 252, 28, 1, 1430, 6435, 12012, 12012, 6930, 2310, 420, 36, 1, 4862, 24310, 51480, 60060, 42042, 18018, 4620, 660, 45, 1, 16796
Offset: 0

Views

Author

F. Chapoton, Apr 20 2001

Keywords

Comments

The rows sum to A006318 (Schroeder numbers), the left column is A000108 (Catalan numbers); the next-to-left column is A001700, the alternating sum in each row but the first is 0.
T(n,k) is the number of Schroeder paths (i.e., consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis) from (0,0) to (2n,0), having k peaks. Example: T(2,1)=3 because we have UU*DD, U*DH and HU*D, the peaks being shown by *. E.g., T(n,k) = binomial(n,k)*binomial(2n-k,n-1)/n for n>0. - Emeric Deutsch, Dec 06 2003
A090181*A007318 as infinite lower triangular matrices. - Philippe Deléham, Oct 14 2008
T(n,k) is also the number of rooted plane trees with maximal degree 3 and k vertices of degree 2 (a node may have at most 2 children, and there are exactly k nodes with 1 child). Equivalently, T(n,k) is the number of syntactically different expressions that can be formed that use a unary operation k times, a binary operation n-k times, and nothing else (sequence of operands is fixed). - Lars Hellstrom (Lars.Hellstrom(AT)residenset.net), Dec 08 2009

Examples

			Triangle begins:
00: [    1]
01: [    1,     1]
02: [    2,     3,      1]
03: [    5,    10,      6,      1]
04: [   14,    35,     30,     10,      1]
05: [   42,   126,    140,     70,     15,      1]
06: [  132,   462,    630,    420,    140,     21,     1]
07: [  429,  1716,   2772,   2310,   1050,    252,    28,    1]
08: [ 1430,  6435,  12012,  12012,   6930,   2310,   420,   36,   1]
09: [ 4862, 24310,  51480,  60060,  42042,  18018,  4620,  660,  45,  1]
10: [16796, 92378, 218790, 291720, 240240, 126126, 42042, 8580, 990, 55, 1]
...
		

Crossrefs

Triangle in A088617 transposed.
T(2n,n) gives A007004.

Programs

  • Maple
    A060693 := (n,k) -> binomial(n,k)*binomial(2*n-k,n)/(n-k+1); # Peter Luschny, May 17 2011
  • Mathematica
    t[n_, k_] := Binomial[n, k]*Binomial[2 n - k, n]/(n - k + 1); Flatten[Table[t[n, k], {n, 0, 9}, {k, 0, n}]] (* Robert G. Wilson v, May 30 2011 *)
  • PARI
    T(n, k) = binomial(n, k)*binomial(2*n - k, n)/(n - k + 1);
    for(n=0, 10, for(k=0, n, print1(T(n, k),", ")); print); \\ Indranil Ghosh, Jul 28 2017
    
  • Python
    from sympy import binomial
    def T(n, k): return binomial(n, k) * binomial(2 * n - k, n) / (n - k + 1)
    for n in range(11): print([T(n, k) for k in range(n + 1)])  # Indranil Ghosh, Jul 28 2017

Formula

Triangle T(n, k) (0 <= k <= n) read by rows; given by [1, 1, 1, 1, 1, ...] DELTA [1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 12 2003
If C_n(x) is the g.f. of row n of the Narayana numbers (A001263), C_n(x) = Sum_{k=1..n} binomial(n,k-1)*(binomial(n-1,k-1)/k) * x^k and T_n(x) is the g.f. of row n of T(n,k), then T_n(x) = C_n(x+1), or T(n,k) = [x^n]Sum_{k=1..n}(A001263(n,k)*(x+1)^k). - Mitch Harris, Jan 16 2007, Jan 31 2007
G.f.: (1 - t*y - sqrt((1-y*t)^2 - 4*y)) / 2.
T(n, k) = binomial(2n-k, n)*binomial(n, k)/(n-k+1). - Philippe Deléham, Dec 07 2003
A060693(n, k) = binomial(2*n-k, k)*A000108(n-k); A000108: Catalan numbers. - Philippe Deléham, Dec 30 2003
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000108(n), A006318(n), A047891(n+1), A082298(n), A082301(n), A082302(n), A082305(n), A082366(n), A082367(n), for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively. - Philippe Deléham, Apr 01 2007
T(n,k) = Sum_{j>=0} A090181(n,j)*binomial(j,k). - Philippe Deléham, May 04 2007
Sum_{k=0..n} T(n,k)*x^(n-k) = (-1)^n*A107841(n), A080243(n), A000007(n), A000012(n), A006318(n), A103210(n), A103211(n), A133305(n), A133306(n), A133307(n), A133308(n), A133309(n) for x = -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively. - Philippe Deléham, Oct 18 2007
From Paul Barry, Jan 29 2009: (Start)
G.f.: 1/(1-xy-x/(1-xy-x/(1-xy-x/(1-xy-x/(1-xy-x/(1-.... (continued fraction);
G.f.: 1/(1-(x+xy)/(1-x/(1-(x+xy)/(1-x/(1-(x+xy)/(1-.... (continued fraction). (End)
T(n,k) = [k<=n]*(Sum_{j=0..n} binomial(n,j)^2*binomial(j,k))/(n-k+1). - Paul Barry, May 28 2009
T(n,k) = A104684(n,k)/(n-k+1). - Peter Luschny, May 17 2011
From Tom Copeland, Sep 21 2011: (Start)
With F(x,t) = (1-(2+t)*x-sqrt(1-2*(2+t)*x+(t*x)^2))/(2*x) an o.g.f. (nulling the n=0 term) in x for the A060693 polynomials in t,
G(x,t) = x/(1+t+(2+t)*x+x^2) is the compositional inverse in x.
Consequently, with H(x,t) = 1/(dG(x,t)/dx) = (1+t+(2+t)*x+x^2)^2 / (1+t-x^2), the n-th A060693 polynomial in t is given by (1/n!)*((H(x,t)*d/dx)^n) x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*d/d) u, evaluated at u = 0.
Also, dF(x,t)/dx = H(F(x,t),t). (End)
See my 2008 formulas in A033282 to relate this entry to A088617, A001263, A086810, and other matrices. - Tom Copeland, Jan 22 2016
Rows of this entry are non-vanishing antidiagonals of A097610. See p. 14 of Agapito et al. for a bivariate generating function and its inverse. - Tom Copeland, Feb 03 2016
From Werner Schulte, Jan 09 2017: (Start)
T(n,k) = A126216(n,k-1) + A126216(n,k) for 0 < k < n;
Sum_{k=0..n} (-1)^k*(1+x*(n-k))*T(n,k) = x + (1-x)*A000007(n).
(End)
Conjecture: Sum_{k=0..n} (-1)^k*T(n,k)*(n+1-k)^2 = 1+n+n^2. - Werner Schulte, Jan 11 2017

Extensions

More terms from Vladeta Jovovic, Apr 21 2001
New description from Philippe Deléham, Aug 12 2003
New name using a comment by Emeric Deutsch from Peter Luschny, Jul 26 2017

A215561 Number A(n,k) of permutations of k indistinguishable copies of 1..n with every partial sum <= the same partial sum averaged over all permutations; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 5, 30, 7, 1, 1, 1, 14, 420, 403, 35, 1, 1, 1, 42, 6930, 40350, 18720, 139, 1, 1, 1, 132, 126126, 5223915, 19369350, 746192, 1001, 1, 1, 1, 429, 2450448, 783353872, 27032968200, 9212531290, 71892912, 5701, 1
Offset: 0

Views

Author

Alois P. Heinz, Aug 16 2012

Keywords

Comments

"Late-growing permutations" were first defined by R. H. Hardin in A147681 and 18 related sequences. David Scambler observed that the set of orthogonal sequences includes A000108 and A007004, and he asked for the other orthogonal sequences, see link below.
"Early-growing permutations" with every partial sum >= the same partial sum averaged over all permutations define the same sequences.
Conjecture: Row r > 1 is asymptotic to c(r) * r^(r*n) / (Pi^((r-1)/2) * n^((r+1)/2)), where c(r) are a constants. - Vaclav Kotesovec, Sep 07 2016

Examples

			A(2,2) = 2: (1,1,2,2), (1,2,1,2).
A(2,3) = 5: (1,1,1,2,2,2), (1,1,2,1,2,2), (1,1,2,2,1,2), (1,2,1,1,2,2), (1,2,1,2,1,2).
A(3,1) = 3: (1,2,3), (1,3,2), (2,1,3).
a(4,1) = 7: (1,2,3,4), (1,2,4,3), (1,3,2,4), (1,4,2,3), (2,1,3,4), (2,1,4,3), (2,3,1,4).
Square array A(n,k) begins:
  1,   1,     1,        1,           1,              1, ...
  1,   1,     1,        1,           1,              1, ...
  1,   1,     2,        5,          14,             42, ...
  1,   3,    30,      420,        6930,         126126, ...
  1,   7,   403,    40350,     5223915,      783353872, ...
  1,  35, 18720, 19369350, 27032968200, 44776592395920, ...
		

Crossrefs

Programs

  • Maple
    b:= proc(l) option remember; local m, n, g;
          m, n:= nops(l), add(i, i=l);
          g:= add(i*l[i], i=1..m)-(m+1)/2*(n-1);
         `if`(n<2, 1, add(`if`(l[i]>0 and i<=g,
            b(subsop(i=l[i]-1, l)), 0), i=1..m))
        end:
    A:= (n, k)-> b([k$n]):
    seq(seq(A(n, d-n), n=0..d), d=0..10);
  • Mathematica
    b[l_] := b[l] = Module[{m, n, g}, {m, n} = {Length[l], Total[l]}; g = Sum[i*l[[i]], {i, 1, m}] - (m+1)/2*(n-1); If[n < 2, 1, Sum[If[l[[i]] > 0 && i <= g, b[ReplacePart[l, i -> l[[i]] - 1]], 0], {i, 1, m}]]]; a[n_, k_] := b[Array[k&, n]]; Table [Table [a[n, d-n], {n, 0, d}], {d, 0, 9}] // Flatten (* Jean-François Alcover, Dec 06 2013, translated from Maple *)

A365771 a(n) = binomial(2*n+1, n)/(2*n+1) * binomial(3*n-1, n) for n >= 0.

Original entry on oeis.org

1, 2, 20, 280, 4620, 84084, 1633632, 33256080, 701149020, 15191562100, 336424047960, 7584833081280, 173575987821600, 4022766574898400, 94247674040476800, 2228957491057276320, 53150802525726081660, 1276661433215969608500, 30863850087221160009000
Offset: 0

Views

Author

Paul D. Hanna, Oct 10 2023

Keywords

Comments

Equals the central terms of triangle A365770.
Conjectures: given A033042 is the sums of distinct powers of 5, then
(1) a(5*A033042(n)) == 4 (mod 5) for n > 0,
(2) a(5*A033042(n) + 1) == 2 (mod 5) for n > 0,
(3) a(n) == 0 (mod 5) for n > 0 except when n or n-1 equals 5*A033042(k) for some k >= 0.

Crossrefs

Programs

  • Mathematica
    A365771[n_] := Binomial[2*n + 1, n]/(2*n + 1)*Binomial[3*n - 1, n];
    Array[A365771, 20, 0] (* Paolo Xausa, Oct 12 2024 *)
  • PARI
    {a(n) = binomial(2*n+1, n)/(2*n+1) * binomial(3*n-1, n)}
    for(n=0,30,print1(a(n),", "))
    
  • Python
    from math import comb
    def A365771(n): return comb(m:=(n<<1)+1,n)*comb(m+n-2,n)//m if n else 1 # Chai Wah Wu, Oct 11 2023

Formula

a(n) = A365770(2*n,n) for n >= 0.
a(n) = A000108(n) * A165817(n) for n >= 0.
a(n) = 2*A319578(n) = (2/3) * A007004(n) for n >= 1. - Peter Bala, Aug 25 2025

A385299 Number of Schröder paths of semilength 2n and having n valleys.

Original entry on oeis.org

1, 1, 11, 155, 2554, 46377, 899107, 18269407, 384577010, 8321452706, 184074021999, 4145999605431, 94799675260406, 2195442934642375, 51402741095491155, 1214975868437406375, 28956949406425290114, 695214262740084758154, 16800125921481031616230, 408354422827279445763942
Offset: 0

Views

Author

Alois P. Heinz, Jun 24 2025

Keywords

Comments

A Schröder path of semilength n is a lattice path starting from (0,0), ending at (2n,0), consisting only of steps U=(1,1) (up steps), D=(1,-1) (down steps) and H=(2,0) (level steps) and never going below the x-axis.

Examples

			a(0) = 1: the empty path.
a(1) = 1: UDUD.
a(2) = 11: HUDUDUD, UUDDUDUD, UHDUDUD, UDUUDDUD, UUDUDDUD, UDUHDUD, UDUDUUDD, UDUUDUDD, UUDUDUDD, UDUDUHD, UDUDUDH.
		

Crossrefs

Cf. A006318, A007004 (the same for peaks), A101282.

Programs

  • Maple
    b:= proc(x, y, t, c) option remember; `if`(y<0 or y>x, 0,
         `if`(x=0, `if`(c=0, 1, 0), b(x-1, y-1, 1, c+1)+
            b(x-1, y+1, 0, c+1-4*t)+b(x-2, y, 0, c+2)))
        end:
    a:= n-> b(4*n, 0$3):
    seq(a(n), n=0..19);

Formula

a(n) = A101282(2n,n).
a(n) = Sum_{k=0..n-1} (k+1)/(2*n-k)*C(2*n-k,n)*C(3*n-k,2*n+1) for n>=1, a(0) = 1.
a(n) ~ 3^(3*n + 5/2) / (50*Pi*n^2). - Vaclav Kotesovec, Jun 27 2025

A319578 a(n) = (1/3)*(n+2)^2*(3*n+3)!/(n+2)!^3.

Original entry on oeis.org

1, 10, 140, 2310, 42042, 816816, 16628040, 350574510, 7595781050, 168212023980, 3792416540640, 86787993910800, 2011383287449200, 47123837020238400, 1114478745528638160, 26575401262863040830, 638330716607984804250, 15431925043610580004500, 375239440534109892741000
Offset: 0

Views

Author

Peter Luschny, Sep 30 2018

Keywords

Comments

Number of Schröder paths of length 2n+1 having n peaks.

Crossrefs

Programs

  • Magma
    [(1/3)*(n+2)^2*Factorial(3*n+3)/Factorial(n+2)^3: n in [0..20]]; // Vincenzo Librandi, Oct 01 2018
  • Maple
    a := n -> (n+2)*(3*n+2)!/((n+2)!^2*n!): seq(a(n), n = 0..18);
  • Mathematica
    Table[(n+2) (3*n+2)! / ((n+2)!^2 n!), {n, 0, 30}] (* Vincenzo Librandi, Oct 01 2018 *)
  • PARI
    a(n) = (1/3)*(n+2)^2*(3*n+3)!/(n+2)!^3; \\ Michel Marcus, Oct 01 2018
    

Formula

a(n) = (n+2)*(3*n+2)!/((n+2)!^2*n!).
a(n) = A060693(2n+1,n).
G.f.: (hypergeom([1/3, 2/3], [2], 27*x) - 1)/(3*x). - Stefano Spezia, Aug 25 2025
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