A095660 Pascal (1,3) triangle.

3, 1, 3, 1, 4, 3, 1, 5, 7, 3, 1, 6, 12, 10, 3, 1, 7, 18, 22, 13, 3, 1, 8, 25, 40, 35, 16, 3, 1, 9, 33, 65, 75, 51, 19, 3, 1, 10, 42, 98, 140, 126, 70, 22, 3, 1, 11, 52, 140, 238, 266, 196, 92, 25, 3, 1, 12, 63, 192, 378, 504, 462, 288, 117, 28, 3, 1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3

Offset

0,1

Comments

This is the third member, q=3, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with T(0,0)=2, not 1).

This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column nr. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} T(n,m)*x^m is G(z,x) = g(z)/(1-x*z*f(z)). Here: g(x) = (3-2*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (3-2*z)/(1-(1+x)*z).

The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} T(n-1-k,k) = A000285(n-2), n>=2, with n=1 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Central terms: T(2*n,n) = A028329(n) = A100320(n) for n > 0, A028329 are the central terms of triangle A028326. - Reinhard Zumkeller, Apr 08 2012

Let P be Pascal's triangle, A007318 and R the Riordan array, A097805. Then Pascal triangle (1,q) = ((q-1) * R) + P. Example: Pascal triangle (1,3) = (2 * R) + P. - Gary W. Adamson, Sep 12 2015

Links

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened

W. Lang, First 10 rows.

Formula

Recursion: T(n, m)=0 if m>n, T(0, 0)= 3; T(n, 0)=1 if n>=1; T(n, m) = T(n-1, m) + T(n-1, m-1).

G.f. column m (without leading zeros): (3-2*x)/(1-x)^(m+1), m>=0.

T(n,k) = (1+2*k/n) * binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012

Closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013

Example

Triangle starts:
  3;
  1,  3;
  1,  4,  3;
  1,  5,  7,   3;
  1,  6, 12,  10,   3;
  1,  7, 18,  22,  13,   3;
  1,  8, 25,  40,  35,  16,   3;
  1,  9, 33,  65,  75,  51,  19,   3;
  1, 10, 42,  98, 140, 126,  70,  22,   3;
  1, 11, 52, 140, 238, 266, 196,  92,  25,   3;
  1, 12, 63, 192, 378, 504, 462, 288, 117,  28,  3;
  1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3;

Maple

T(n, k):=piecewise(n=0, 3, 0<n, (1+2*k/n)*binomial(n, k)): # Mircea Merca, Apr 08 2012

Mathematica

{3}~Join~Table[(1 + 2 k/n) Binomial[n, k], {n, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 14 2015 *)

Prog

(Haskell)
a095660 n k = a095660_tabl !! n !! k
a095660_row n = a095660_tabl !! n
a095660_tabl = [3] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1, 3]
-- Reinhard Zumkeller, Apr 08 2012
(Magma) A095660:= func< n, k | n eq 0 select 3 else (1+2*k/n)*Binomial(n, k) >;
[A095660(n, k): k in [0..n], n in [1..12]]; // G. C. Greubel, May 02 2021
(Sage)
def A095660(n, k): return 3 if n==0 else (1+2*k/n)*binomial(n, k)
flatten([[A095660(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 02 2021

Crossrefs

Row sums: A000079(n+1), n>=1, 3 if n=0. Alternating row sums are [3, -2, followed by 0's].

Column sequences (without leading zeros) give for m=1..9 with n>=0: A000027(n+3), A055998(n+1), A006503(n+1), A095661, A000574, A095662, A095663, A095664, A095665.

Cf. A097805.

Sequence in context: A280526 A335552 A306841 * A290080 A289617 A035648

Adjacent sequences: A095657 A095658 A095659 * A095661 A095662 A095663

Keyword

nonn,easy,tabl

Author

Wolfdieter Lang, May 21 2004

Status

approved