A164844 Generalized Pascal Triangle - satisfying the same recurrence as Pascal's triangle, but with a(n,0)=1 and a(n,n)=10^n (instead of both being 1).

1, 1, 10, 1, 11, 100, 1, 12, 111, 1000, 1, 13, 123, 1111, 10000, 1, 14, 136, 1234, 11111, 100000, 1, 15, 150, 1370, 12345, 111111, 1000000, 1, 16, 165, 1520, 13715, 123456, 1111111, 10000000, 1, 17, 181, 1685, 15235, 137171, 1234567, 11111111, 100000000, 1, 18, 198, 1866, 16920, 152406, 1371738, 12345678, 111111111, 1000000000, 1, 19, 216, 2064, 18786, 169326, 1524144, 13717416, 123456789, 1111111111, 10000000000

Offset: 0

Author

Mark Dols, Aug 28 2009

Keywords

• nonn
• tabl
• easy

Comments

Like with Pascal's triangle, the columns grown polynomially. For example, a(n,1)=10+n, a(n,2)=(1/2)*(180+19n+n^2), a(n,3)=(1/6)*(5400 + 569n + 30n^2 + n^3). Likewise, diagonals grow exponentially: a(n,n)=10^n, a(n,n-1) = (10^n-1) / 9. - Kellen Myers, Jan 24 2010

Examples

			Triangle begins:
  1
  1,10
  1,11,100
  1,12,111,1000
  1,13,123,1111,10000
  1,14,136,1234,11111,100000
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10152 (rows 0 to 141, flattened)

Crossrefs

Cf. A007318, A093645, A011557, A228196.

Programs

Maple

f:= proc(n,k) option remember;
if k=n then 10^n elif k=0 then 1 else procname(n-1,k-1)+procname(n-1,k) fi
end proc:
seq(seq(f(n,k),k=0..n),n=0..10); # Robert Israel, Jul 01 2016

Mathematica

f[r_, k_] := Sum[10^i*Binomial[r - i - 1, r - k - 1], {i, 0, k}]; TableForm[Table[f[n, k], {n, 0, 15}, {k, 0, n}]] (* Alex Meiburg, Aug 21 2010 *)
a[n_, k_] := a[n, k] = Piecewise[{{0, k > n || k < 0}, {1, k == 0}, {10^n, k == n}}, a[n - 1, k - 1] + a[n - 1, k]]; TableForm[Table[a[n, k], {n, 0, 10}, {k, 0, n}]] (* Kellen Myers, Jan 24 2010 *)

Formula

From Kellen Myers, Jan 24 2010: (Start)

a(n,k) = Sum_{i = 0..k} 10^i * binomial(n-i-1, n-k-1), for 0<=k<=n.

a(n,0) = 1, a(n,n) = 10^n, a(n,k) = a(n-1,k-1)+a(n-1,k). (End)

T(n,k) = T(n-1,k)+11*T(n-1,k-1)-10*T(n-2,k-1)-10*T(n-2,k-2), T(0,0)=1, T(1,0)=1, T(1,1)=10, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013

G.f. of triangle: g(x,y) = (1-xy)/((1-10xy)(1-x-xy)). - Robert Israel, Jul 01 2016

Extensions

Definition clarified, more terms, and revision of Meiburg's Mathematica code by Kellen Myers, Jan 24 2010