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.

A166830 Number of n X 3 1..2 arrays containing at least one of each value, all equal values connected, rows considered as a single number in nondecreasing order, and columns considered as a single number in nonincreasing order.

Original entry on oeis.org

2, 8, 18, 33, 54, 82, 118, 163, 218, 284, 362, 453, 558, 678, 814, 967, 1138, 1328, 1538, 1769, 2022, 2298, 2598, 2923, 3274, 3652, 4058, 4493, 4958, 5454, 5982, 6543, 7138, 7768, 8434, 9137, 9878, 10658, 11478, 12339, 13242
Offset: 1

Views

Author

R. H. Hardin, Oct 21 2009

Keywords

Examples

			All solutions for n=3
...1.1.1...1.1.1...1.1.1...1.1.1...1.1.1...1.1.1...1.1.1...1.1.1...1.1.1
...1.1.1...1.1.1...1.1.1...2.1.1...2.1.1...2.1.1...2.2.1...2.2.1...2.2.2
...2.1.1...2.2.1...2.2.2...2.1.1...2.2.1...2.2.2...2.2.1...2.2.2...2.2.2
------
...2.1.1...2.1.1...2.1.1...2.1.1...2.1.1...2.1.1...2.2.1...2.2.1...2.2.1
...2.1.1...2.1.1...2.1.1...2.2.1...2.2.1...2.2.2...2.2.1...2.2.1...2.2.2
...2.1.1...2.2.1...2.2.2...2.2.1...2.2.2...2.2.2...2.2.1...2.2.2...2.2.2
		

Programs

Formula

Empirical: a(n) = (n^3+6*n^2+11*n-6)/6.
a(n) = A167772(n+3,n). - Philippe Deléham, Nov 11 2009
a(n) = A227819(n+6,n+2). - Alois P. Heinz, Sep 22 2013
Empirical: a(n) = floor(A000292(n+1)^3/(A000292(n+1) + 1)^ 2). - Ivan N. Ianakiev, Nov 05 2013
From G. C. Greubel, May 25 2016: (Start)
Empirical G.f.: (-1 + 6*x - 6*x^2 + 2*x^3)/(1 - x)^4 + 1.
Empirical E.g.f.: (1/6)*(-6 + 18*x + 9*x^2 + x^3)*exp(x) + 1. (End)

A065602 Triangle T(n,k) giving number of hill-free Dyck paths of length 2n and having height of first peak equal to k.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 8, 6, 3, 1, 24, 18, 10, 4, 1, 75, 57, 33, 15, 5, 1, 243, 186, 111, 54, 21, 6, 1, 808, 622, 379, 193, 82, 28, 7, 1, 2742, 2120, 1312, 690, 311, 118, 36, 8, 1, 9458, 7338, 4596, 2476, 1164, 474, 163, 45, 9, 1, 33062, 25724, 16266, 8928, 4332, 1856, 692, 218, 55, 10, 1
Offset: 2

Views

Author

N. J. A. Sloane, Dec 02 2001

Keywords

Comments

A Riordan triangle.
Subtriangle of triangle in A167772. - Philippe Deléham, Nov 14 2009
Riordan array (f(x), x*g(x)) where f(x) is the g.f. of A000958 and g(x) is the g.f. of A000108. - Philippe Deléham, Jan 23 2010

Examples

			T(3,2)=1 reflecting the unique Dyck path (UUDUDD) of length 6, with no hills and height of first peak equal to 2.
Triangle begins:
     1;
     1,    1;
     3,    2,    1;
     8,    6,    3,   1;
    24,   18,   10,   4,   1;
    75,   57,   33,  15,   5,   1;
   243,  186,  111,  54,  21,   6,  1;
   808,  622,  379, 193,  82,  28,  7,  1;
  2742, 2120, 1312, 690, 311, 118, 36,  8,  1;
		

Crossrefs

Row sums give A000957 (the Fine sequence).
First column is A000958.

Programs

  • Haskell
    a065602 n k = sum
       [(k-1+2*j) * a007318' (2*n-k-1-2*j) (n-1) `div` (2*n-k-1-2*j) |
        j <- [0 .. div (n-k) 2]]
    a065602_row n = map (a065602 n) [2..n]
    a065602_tabl = map a065602_row [2..]
    -- Reinhard Zumkeller, May 15 2014
    
  • Maple
    a := proc(n,k) if n=0 and k=0 then 1 elif k<2 or k>n then 0 else sum((k-1+2*j)*binomial(2*n-k-1-2*j,n-1)/(2*n-k-1-2*j),j=0..floor((n-k)/2)) fi end: seq(seq(a(n,k),k=2..n),n=1..14);
  • Mathematica
    nmax = 12; t[n_, k_] := Sum[(k-1+2j)*Binomial[2n-k-1-2j, n-1] / (2n-k-1-2j), {j, 0, (n-k)/2}]; Flatten[ Table[t[n, k], {n, 2, nmax}, {k, 2, n}]] (* Jean-François Alcover, Nov 08 2011, after Maple *)
  • SageMath
    def T(n,k): return sum( (k+2*j-1)*binomial(2*n-2*j-k-1, n-1)/(2*n-2*j-k-1) for j in (0..(n-k)//2) )
    flatten([[T(n,k) for k in (2..n)] for n in (2..12)]) # G. C. Greubel, May 26 2022

Formula

T(n, 2) = A000958(n-1).
Sum_{k=2..n} T(n, k) = A000957(n+1).
From Emeric Deutsch, Feb 23 2004: (Start)
T(n, k) = Sum_{j=0..floor((n-k)/2)} (k-1+2*j)*binomial(2*n-k-1-2*j, n-1)/(2*n-k-1-2*j).
G.f.: t^2*z^2*C/( (1-z^2*C^2)*(1-t*z*C) ), where C = (1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
T(n,k) = A167772(n-1,k-1), k=2..n. - Reinhard Zumkeller, May 15 2014
From G. C. Greubel, May 26 2022: (Start)
T(n, n-1) = A000027(n-2).
T(n, n-2) = A000217(n-2).
T(n, n-3) = A166830(n-3). (End)

Extensions

More terms from Emeric Deutsch, Feb 23 2004

A237619 Riordan array (1/(1+x*c(x)), x*c(x)) where c(x) is the g.f. of Catalan numbers (A000108).

Original entry on oeis.org

1, -1, 1, 0, 0, 1, -1, 1, 1, 1, -2, 2, 3, 2, 1, -6, 6, 8, 6, 3, 1, -18, 18, 24, 18, 10, 4, 1, -57, 57, 75, 57, 33, 15, 5, 1, -186, 186, 243, 186, 111, 54, 21, 6, 1, -622, 622, 808, 622, 379, 193, 82, 28, 7, 1, -2120, 2120, 2742, 2120, 1312, 690, 311, 118, 36, 8, 1
Offset: 0

Views

Author

Philippe Deléham, Feb 10 2014

Keywords

Examples

			Triangle begins:
    1;
   -1,  1;
    0,  0,  1;
   -1,  1,  1,  1;
   -2,  2,  3,  2,  1;
   -6,  6,  8,  6,  3,  1;
  -18, 18, 24, 18, 10,  4, 1;
  -57, 57, 75, 57, 33, 15, 5, 1;
Production matrix begins:
  -1, 1;
  -1, 1, 1;
  -1, 1, 1, 1;
  -1, 1, 1, 1, 1;
  -1, 1, 1, 1, 1, 1;
  -1, 1, 1, 1, 1, 1, 1;
  -1, 1, 1, 1, 1, 1, 1, 1;
  -1, 1, 1, 1, 1, 1, 1, 1, 1;
		

Crossrefs

Programs

  • Mathematica
    A065602[n_, k_]:= A065602[n, k]= Sum[(k-1+2*j)*Binomial[2*(n-j)-k-1, n-1]/(2*(n - j) -k-1), {j,0,(n-k)/2}];
    T[n_, k_]:= If[k==0, A065602[n, 0], If[n==1 && k==1, 1, A065602[n, k]]];
    Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 27 2022 *)
  • SageMath
    def A065602(n, k): return sum( (k+2*j-1)*binomial(2*n-2*j-k-1, n-1)/(2*n-2*j-k-1) for j in (0..(n-k)//2) )
    def A237619(n, k):
        if (n<2): return (-1)^(n-k)
        elif (k==0): return A065602(n, 0)
        else: return A065602(n, k)
    flatten([[A237619(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 27 2022

Formula

Sum_{k=0..n} T(n,k)*x^k = A126983(n), A000957(n+1), A026641(n) for x = 0, 1, 2 respectively.
T(n, k) = A167772(n-1, k-1) for k > 0, with T(n, 0) = A167772(n, 0).
T(n, 0) = A126983(n).
T(n+1, 1) = A000957(n+1).
T(n+2, 2) = A000958(n+1).
T(n+3, 3) = A104629(n) = A000957(n+3).
T(n+4, 4) = A001558(n).
T(n+5, 5) = A001559(n).
T(n, k) = A065602(n, k) for k > 0, with T(n, k) = (-1)^(n-k), for n < 2, and T(n, 0) = A065602(n, 0). - G. C. Greubel, May 27 2022
Showing 1-3 of 3 results.