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A095660 Pascal (1,3) triangle. 19
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 (list; table; graph; refs; listen; history; text; internal format)
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
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
KEYWORD
nonn,easy,tabl
AUTHOR
Wolfdieter Lang, May 21 2004
STATUS
approved

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Last modified March 28 05:39 EDT 2024. Contains 371235 sequences. (Running on oeis4.)