A034261 Infinite square array f(a,b) = C(a+b,b+1)*(a*b+a+1)/(b+2), a, b >= 0, read by antidiagonals. Equivalently, triangular array T(n,k) = f(k,n-k), 0 <= k <= n, read by rows.

0, 0, 1, 0, 1, 3, 0, 1, 5, 6, 0, 1, 7, 14, 10, 0, 1, 9, 25, 30, 15, 0, 1, 11, 39, 65, 55, 21, 0, 1, 13, 56, 119, 140, 91, 28, 0, 1, 15, 76, 196, 294, 266, 140, 36, 0, 1, 17, 99, 300, 546, 630, 462, 204, 45, 0, 1, 19, 125, 435, 930, 1302, 1218, 750, 285, 55

Offset

0,6

Comments

f(h,k) = number of paths consisting of steps from (0,0) to (h,k) using h unit steps right, k+1 unit steps up and 1 unit step down, in some order, with first step not down and no repeated points.

Links

Table of n, a(n) for n=0..65.

Formula

Another formula: f(h,k) = binomial(h+k,k+1) + Sum{C(i+j-1, j)*C(h+k-i-j, k-j+1): i=1, 2, ..., h-1, j=1, 2, ..., k+1}

Example

Triangle begins:
  0;
  0, 1;
  0, 1, 3;
  0, 1, 5,  6;
  0, 1, 7, 14, 10;
  ...
As a square array,
  [ 0  0  0   0   0 ...]
  [ 1  1  1   1   1 ...]
  [ 3  5  7   9  11 ...]
  [ 6 14 25  39  56 ...]
  [10 30 65 119 196 ...]
  [...      ...     ...]

Maple

A034261 := proc(n, k) binomial(n, n-k+1)*(k+(k-1)/(k-n-2)); end;

Mathematica

Flatten[Table[Binomial[n, n-k+1](k+(k-1)/(k-n-2)), {n, 0, 15}, {k, 0, n}]] (* Harvey P. Dale, Jan 11 2013 *)

Prog

(PARI) f(h, k)=binomial(h+k, k+1)*(k*h+h+1)/(k+2)
(PARI) tabl(nn) = for (n=0, nn, for (k=0, n, print1(binomial(n, n-k+1)*(k+(k-1)/(k-n-2)), ", ")); print()); \\ Michel Marcus, Mar 20 2015

Crossrefs

Cf. A001787 (row sums), A000330(n) = f(n,1).

Cf. A034263, A034264, A034265, A034267 - A034275 for diagonals n -> f(n,n+k), for several fixed k.

Sequence in context: A356777 A254295 A143626 * A046778 A119925 A210663

Adjacent sequences: A034258 A034259 A034260 * A034262 A034263 A034264

Keyword

nonn,tabl,easy,nice

Author

Clark Kimberling

Extensions

Entry revised by N. J. A. Sloane, Apr 21 2000. The formula for f in the definition was found by Michael Somos.

Edited by M. F. Hasler, Nov 08 2017

Status

approved