

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,nk), 0 <= k <= n, read by rows.


26



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
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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



FORMULA

Another formula: f(h,k) = binomial(h+k,k+1) + Sum{C(i+j1, j)*C(h+kij, kj+1): i=1, 2, ..., h1, 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, nk+1)*(k+(k1)/(kn2)); end;


MATHEMATICA

Flatten[Table[Binomial[n, nk+1](k+(k1)/(kn2)), {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, nk+1)*(k+(k1)/(kn2)), ", ")); print()); \\ Michel Marcus, Mar 20 2015


CROSSREFS



KEYWORD



AUTHOR



EXTENSIONS



STATUS

approved



