A093562 (5,1) Pascal triangle.

1, 5, 1, 5, 6, 1, 5, 11, 7, 1, 5, 16, 18, 8, 1, 5, 21, 34, 26, 9, 1, 5, 26, 55, 60, 35, 10, 1, 5, 31, 81, 115, 95, 45, 11, 1, 5, 36, 112, 196, 210, 140, 56, 12, 1, 5, 41, 148, 308, 406, 350, 196, 68, 13, 1, 5, 46, 189, 456, 714, 756, 546, 264, 81, 14, 1, 5, 51, 235, 645, 1170

Offset: 0

Author

Wolfdieter Lang, Apr 22 2004

Keywords

• nonn
• easy
• tabl

Comments

This is the fifth member, d=5, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-1, for d=1..4.

This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=(1+4*z)/(1-(1+x)*z).

The SW-NE diagonals give A022095(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 4. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

The array F(5;n,m) gives in the columns m >= 1 the figurate numbers based on A016861, including the heptagonal numbers A000566 (see the W. Lang link).

For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018

Examples

			Triangle begins
  [1];
  [5,  1];
  [5,  6,  1];
  [5, 11,  7,  1];
  ...
		

References

• Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
• Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Links

• Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
• P. Bala, A note on the diagonals of a proper Riordan Array
• W. Lang, First 10 rows and array of figurate numbers .

Crossrefs

Cf. Row sums: A007283(n-1), n>=1, 1 for n=0. A082505(n+1), alternating row sums are 1 for n=0, 4 for n=2 and 0 else.

Column sequences give for m=1..9: A016861, A000566 (heptagonal), A002413, A002418, A027800, A051946, A050484, A052255, A055844.

A007318, A093563 (d=6), A228196, A228576.

Programs

Haskell

a093562 n k = a093562_tabl !! n !! k
a093562_row n = a093562_tabl !! n
a093562_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [5, 1]
-- Reinhard Zumkeller, Aug 31 2014

Python

from math import comb, isqrt
def A093562(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<2))//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

a(n, m) = F(5;n-m, m) for 0<= m <= n, otherwise 0, with F(5;0, 0)=1, F(5;n, 0)=5 if n>=1 and F(5;n, m):=(5*n+m)*binomial(n+m-1, m-1)/m if m>=1.

G.f. column m (without leading zeros): (1+4*x)/(1-x)^(m+1), m>=0.

Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=5 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).

T(n, k) = C(n, k) + 4*C(n-1, k). - Philippe Deléham, Aug 28 2005

exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(5 + 11*x + 7*x^2/2! + x^3/3!) = 5 + 16*x + 34*x^2/2! + 60*x^3/3! + 95*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014