A093645 (10,1) Pascal triangle.

1, 10, 1, 10, 11, 1, 10, 21, 12, 1, 10, 31, 33, 13, 1, 10, 41, 64, 46, 14, 1, 10, 51, 105, 110, 60, 15, 1, 10, 61, 156, 215, 170, 75, 16, 1, 10, 71, 217, 371, 385, 245, 91, 17, 1, 10, 81, 288, 588, 756, 630, 336, 108, 18, 1, 10, 91, 369, 876, 1344, 1386, 966, 444, 126, 19, 1

Offset: 0

Author

Wolfdieter Lang, Apr 22 2004

Keywords

• nonn
• easy
• tabl

Comments

The array F(10;n,m) gives in the columns m >= 1 the figurate numbers based on A017281, including the 12-gonal numbers A051624 (see the W. Lang link).

This is the tenth member, d=10, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-5 and A093644 for d=1..9.

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+9*z)/(1-(1+x)*z).

The SW-NE diagonals give A022100(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k), n >= 1, with n=0 value 9. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Examples

			Triangle begins
   1;
  10,  1;
  10, 11,  1;
  10, 21, 12,  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
• W. Lang, First 10 rows and array of figurate numbers .

Crossrefs

Row sums: 1 for n=0 and A005015(n-1), n >= 1, alternating row sums are 1 for n=0, 9 for n=2 and 0 otherwise.

The column sequences give for m=1..9: A017281, A051624 (12-gonal), A007587, A051799, A051880, A050406, A052254, A056125, A093646.

Programs

Haskell

a093645 n k = a093645_tabl !! n !! k
a093645_row n = a093645_tabl !! n
a093645_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [10, 1]
-- Reinhard Zumkeller, Aug 31 2014

Mathematica

t[0, 0] = 1; t[n_, k_] := Binomial[n, k] + 9*Binomial[n-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 05 2013, after Philippe Deléham *)

Formula

a(n, m) = F(10;n-m, m) for 0 <= m <= n, else 0, with F(10;0, 0)=1, F(10;n, 0)=10 if n >= 1 and F(10;n, m):=(10*n+m)*binomial(n+m-1, m-1)/m if m >= 1.

Recursion: a(n, m)=0 if m > n, a(0, 0)=1; a(n, 0)=10 if n >= 1; a(n, m) = a(n-1, m) + a(n-1, m-1).

G.f. column m (without leading zeros): (1+9*x)/(1-x)^(m+1), m >= 0.

T(n, k) = C(n, k) + 9*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)*(10 + 21*x + 12*x^2/2! + x^3/3!) = 10 + 31*x + 64*x^2/2! + 110*x^3/3! + 170*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