OFFSET
0,3
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
LINKS
Robert Israel, Table of n, a(n) for n = 0..10152 (rows 0 to 141, flattened)
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
EXAMPLE
Triangle begins:
1
1,10
1,11,100
1,12,111,1000
1,13,123,1111,10000
1,14,136,1234,11111,100000
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 *)
CROSSREFS
KEYWORD
AUTHOR
Mark Dols, Aug 28 2009
EXTENSIONS
Definition clarified, more terms, and revision of Meiburg's Mathematica code by Kellen Myers, Jan 24 2010
STATUS
approved