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.

A062745 Generalized Catalan array FS(3; n,r).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 3, 3, 3, 1, 3, 6, 9, 12, 12, 12, 1, 4, 10, 19, 31, 43, 55, 55, 55, 1, 5, 15, 34, 65, 108, 163, 218, 273, 273, 273, 1, 6, 21, 55, 120, 228, 391, 609, 882, 1155, 1428, 1428, 1428, 1, 7, 28, 83, 203, 431, 822, 1431, 2313, 3468, 4896, 6324, 7752, 7752
Offset: 0

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Author

Wolfdieter Lang, Jul 12 2001

Keywords

Comments

In the Frey-Sellers reference this array appears in Table 2, p. 143 and is called {n over r}_{m-1}, with m=3.
The step width sequence of this staircase array is [1,2,2,2,....], i.e., the degree of the row polynomials is [0,2,4,6,...] = A005843.
The columns r=0..5 give A000012 (powers of 1), A000027 (natural), A000217 (triangular), A062748, A005718, A062749.
Number of lattice paths from (0,0) to (r,n) using steps h=(1,0), v=(0,1) and staying on or above the line y = x/2. Example: a(3,2)=6 because from (0,0) to (2,3) we have the following valid paths: vvvhh, vvhvh, vvhhv, vhvvh, vhvhvh and vhvvh (see the Niederhausen reference). - Emeric Deutsch, Jun 24 2005

Examples

			Array begins:
  {1};
  {1,1,1};
  {1,2,3,3,3};
  {1,3,6,9,12,12,12};
  ...;
N(3; 1,x) = 3-3*x+x^2.

Links

Crossrefs

Cf. A009766, A280759 (rows reversed).
Cf. A062746, A062747, A062748, A062749.
Cf. A005843, A000012, A000027, A000217, A005718.

Programs

  • Maple
    a:=proc(n,r) if r<=2*n then binomial(n+r,r)-(-1)^(r-1)*sum(binomial(3*i,i)*binomial(i-n-1,r-1-2*i)/(2*i+1),i=0..floor((r-1)/2)) else 0 fi end: for n from 0 to 8 do seq(a(n,r),r=0..2*n) od; # yields sequence in triangular form # Emeric Deutsch, Jun 24 2005
  • Mathematica
    a[0, 0] = 1; a[, -1] = 0; a[n, r_] /; r > 2*n = 0; a[n_, r_] := a[n, r] = a[n, r-1] + a[n-1, r]; Table[a[n, r], {n, 0, 7}, {r, 0, 2*n}] // Flatten (* Jean-François Alcover, Jun 21 2013 *)

Formula

a(0,0)=1, a(n,-1)=0, n >= 1; a(n,r) = a(n, r-1) + a(n-1, r) if r <= 2n, 0 otherwise.
G.f. for column r = 2*k+j, k >= 0, j=1, 2: (x^(k+1))*N(3; k, x)/ (1-x)^(2*k+1+j), with row polynomials N(3; k, x) of array A062746; for column r=0: 1/(1-x).
a(n,r) = binomial(n+r, r) - (-1)^(r-1)*Sum_{i=0..floor((r-1)/2)} binomial(3i, i)*binomial(i-n-1, r-1-2i)/(2i+1), 0 <= r <= 2n (see the Niederhausen reference, eq. (17)). - Emeric Deutsch, Jun 24 2005

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 29 2003

A062746 Coefficient array for certain polynomials N(3; k,x) (rising powers of x).

Original entry on oeis.org

1, 3, -3, 1, 12, -29, 30, -15, 3, 55, -222, 405, -417, 252, -84, 12, 273, -1575, 4203, -6678, 6846, -4608, 1980, -495, 55, 1428, -10812, 38367, -83244, 121518, -124146, 89595, -44990, 15015, -3003, 273, 7752, -73017, 325164
Offset: 0

Views

Author

Wolfdieter Lang, Jul 12 2001

Keywords

Comments

The g.f. for the sequence of column r=2*k+1, k >= 0, of array A062745(n,r) is N(3; k,x)*(x^(k+1))/(1-x)^(2*k+2) with N(3; k,x) := sum(a(k,p)*x^p,p=0..2*k).
The m=0 column gives: A001764(n+1). The row sums give A000012 (powers of 1) and (unsigned) A062747.
The sequence of step width of this staircase array is [1,2,2,2,...], i.e. the degree of the row polynomials is [0,2,4,6,...]= A005843.

Examples

			{1}; {3,-3,1}; {12,-29,30,-15,3}; ...; N(3; 1,x)= 3-3*x+x^2.

Crossrefs

Cf. A062991.

Formula

a(k, p) := [x^p]N(3; k, x) with N(3; k, x)=(N(3; k-1, x)-A001764(k)*(1-x)^(2*k+1))/x, N(3; 0, x) := 1.
a(n, k)= a(n-1, k+1)+((-1)^k)*binomial(2*n+1, k+1)*binomial(3*n+1, n)/(3*n+1) if k=0, .., (2*n-3); a(n, k)= ((-1)^k)*binomial(2*n+1, k+1)*binomial(3*n+1, n)/(3*n+1) if k=(2*n-2), ..., 2*n; else 0.

A333561 a(n) = Sum_{j = 0..2*n} binomial(n+j-1,j)*2^j.

Original entry on oeis.org

1, 7, 129, 2815, 65537, 1579007, 38862849, 970522623, 24494735361, 623210135551, 15956734640129, 410649406472191, 10612705274626049, 275241225206890495, 7159857331658817537, 186731505521384226815, 4880983719142471237633, 127836403093194475044863
Offset: 0

Views

Author

Peter Bala, Mar 27 2020

Keywords

Comments

Column 2 of the square array A333560. Compare with A119259(n) = Sum_{j = 0..n} binomial(n+j-1,j)*2^j.
We conjecture that this sequence satisfies the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Some examples are given below.

Examples

			Examples of supercongruences:
a(11) - a(1) = 410649406472191 - 7 = (2^3)*3*(11^3)*12855290711 == 0 ( mod 11^3 ).
a(3*7) - a(3) = 61103847305642669128888090623 - 2815 = (2^8)*(7^5)* 87326419*162627033103121 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 29754989698128108780761000609579007 - 1579007 = (2^11)*(5^6)*179*751*10267*673710468794491483 == 0 ( mod 5^6 ).

Crossrefs

Cf. A001764, A062747, A119259, A333560, A333562.

Programs

  • Maple
    seq(add( binomial(n+j-1,j)*2^j, j = 0..2*n), n = 0..25);
  • Mathematica
    Table[(-1)^n - 2^(2*n+1) * Binomial[3*n, 2*n+1] * Hypergeometric2F1[1, 3*n+1, 2*n+2, 2], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)
  • PARI
    a(n) = sum(j = 0, 2*n, binomial(n+j-1,j)*2^j); \\ Michel Marcus, Mar 28 2020

Formula

Conjectural o.g.f.: 1/(1 + x) + 8*x*f'(4*x)/(2*f(4*x) - 1), where f(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + ... is the o.g.f. of A001764.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 7*x + 89*x^2 + 1447*x^3 + ... appears to be the o.g.f. of A062747.
Conjectural recurrence: n*(n - 1)*(2*n - 1)*(3098*n - 6455)*a(n) = (n - 1)*(172988*n^3 - 585840*n^2 + 550321*n - 169824)*a(n-1) - 12*(11825*n^4 - 168518*n^3 + 627675*n^2 - 853766*n + 350744)*a(n-2) - 36*(n - 3)*(3*n - 7)*(3*n - 8)*(991*n - 724)*a(n-3) with a(1) = 7, a(2) = 129, a(3) = 2815.
From Vaclav Kotesovec, Mar 28 2020: (Start)
a(n) ~ 3^(3*n + 1/2) / (4*sqrt(Pi*n)).
Recurrence: n*(2*n - 1)*(7*n^2 - 20*n + 14)*a(n) = (364*n^4 - 1411*n^3 + 1818*n^2 - 868*n + 120)*a(n-1) + 6*(3*n - 5)*(3*n - 4)*(7*n^2 - 6*n + 1)*a(n-2). (End)
From Peter Bala, Mar 05 2022: (Start)
a(n) = Sum_{k = 0..2*n} binomial(3*n, 2*n-k)*binomial(n+k-1,k).
a(n) = [x^(2*n)] ( (1 + x^3)/(1 - x) )^n.
The o.g.f. satisfies the algebraic equation (108*x^3 + 212*x^2 + 100*x - 4)*A(x)^3 - (216*x^2 + 208*x - 8)*A(x)^2 + (48*x^2 + 155*x - 5)*A(x) + 8*x^2 - 40*x + 1 = 0. (End)
a(n) = binomial(3*n, 2*n)*hypergeom([-2*n, n], [n + 1], -1). - Peter Luschny, Mar 07 2022
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