A099097 Riordan array (1, 3+x).

1, 0, 3, 0, 1, 9, 0, 0, 6, 27, 0, 0, 1, 27, 81, 0, 0, 0, 9, 108, 243, 0, 0, 0, 1, 54, 405, 729, 0, 0, 0, 0, 12, 270, 1458, 2187, 0, 0, 0, 0, 1, 90, 1215, 5103, 6561, 0, 0, 0, 0, 0, 15, 540, 5103, 17496, 19683, 0, 0, 0, 0, 0, 1, 135, 2835, 20412, 59049, 59049, 0, 0, 0, 0, 0, 0, 18, 945, 13608, 78732, 196830, 177147

Offset

0,3

Comments

Row sums are A006190(n+1). Diagonal sums are A052931. The Riordan array (1, s+tx) defines T(n,k) = binomial(k,n-k)*s^k*(t/s)^(n-k). The row sums satisfy a(n) = s*a(n-1) + t*a(n-2) and the diagonal sums satisfy a(n) = s*a(n-2) + t*a(n-3).

Triangle T(n,k), 0 <= k <= n, read by rows given by [0, 1/3, -1/3, 0, 0, 0, 0, 0, ...] DELTA [3, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 10 2008

Links

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Formula

Triangle: T(n, k) = binomial(k, n-k)*3^k*(1/3)^(n-k).

G.f. of column k: (3*x + x^2)^k.

G.f.: 1/(1 - 3*y*x - y*x^2). - Philippe Deléham, Nov 21 2011

Sum_{k=0..n} T(n,k)*x^k = A000007(n), A006190(n+1), A135030(n+1), A181353(n+1) for x = 0,1,2,3 respectively. - Philippe Deléham, Nov 21 2011

Example

Triangle begins:
  1;
  0, 3;
  0, 1, 9;
  0, 0, 6, 27;
  0, 0, 1, 27,  81;
  0, 0, 0,  9, 108, 243;
  ...

Mathematica

Table[3^(2*k-n)*Binomial[k, n-k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, May 19 2021 *)

Prog

(Sage) flatten([[3^(2*k-n)*binomial(k, n-k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 19 2021

Crossrefs

Cf. A027465.

Diagonals are of the form 3^n*binomial(n+m, m): A000244 (m=0), A027471 (m=1), A027472 (m=2), A036216 (m=3), A036217 (m=4), A036219 (m=5), A036220 (m=6), A036221 (m=7), A036222 (m=8), A036223 (m=9), A172362 (m=10).

Sequence in context: A020816 A174860 A157391 * A152150 A136239 A225443

Adjacent sequences: A099094 A099095 A099096 * A099098 A099099 A099100

Keyword

easy,nonn,tabl

Author

Paul Barry, Sep 25 2004

Status

approved