0,4

The first column equals the factorials. Triangle A127420 is generated by a similar recurrence.

Sum_{k=0..n*(n+1)/2} k*T(n,k) = A018927(n+1) = Sum_{k=0..n} k*k!*{(k+1)^(n-k+1)-k^(n-k+1)}.

T(n,k) = (n-t)! * (n-t)^(k - t*(t+1)/2) * (n-t+1)^(t-k + t*(t+1)/2) where t=floor((sqrt(8*k+1)-1)/2). Also, Sum_{j=k*(k+1)/2..(k+1)*(k+2)/2-1} T(n,j) = A047969(n-k,k) = (n-k)!*((n-k+1)^(k+1)-(n-k)^(k+1)).

The triangle begins:

1;

1, 1;

2, 2, 1, 1;

6, 6, 4, 4, 2, 1, 1;

24, 24, 18, 18, 12, 8, 8, 4, 2, 1, 1;

120, 120, 96, 96, 72, 54, 54, 36, 24, 16, 16, 8, 4, 2, 1, 1;

720, 720, 600, 600, 480, 384, 384, 288, 216, 162, 162, 108, 72, 48, 32, 32, 16, 8, 4, 2, 1, 1;

...

The recurrence is illustrated by the following examples.

Start with a single '1' in row 0.

To get row 1, insert 0 in row 0 at position 0,

and take partial sums in reverse order:

0,_1;

1,_1;

To get row 2, insert 0 in row 1 at positions [0,2],

and take partial sums in reverse order:

0,_1,_0,_1;

2,_2,_1,_1;

To get row 3, insert 0 in row 2 at positions [0,2,5],

and take partial sums in reverse order:

0,_2,_0,_2,_1,_0,_1;

6,_6,_4,_4,_2,_1,_1;

To get row 4, insert 0 in row 3 at positions [0,2,5,9],

and take partial sums in reverse order:

_0,__6,__0,__6,__4,_0,_4,_2,_1,_0,_1;

24,_24,_18,_18,_12,_8,_8,_4,_2,_1,_1;

etc.

Continuing in this way generates the factorials in the first column.

(PARI) {T(n, k)=if(n<0|k<0, 0, if(n==0&k==0, 1, if(k==0, n!, if(issquare(8*k+1), T(n, k-1), T(n, k-1)-T(n-1, k-(sqrtint(8*k+1)+1)\2)))))}

(PARI) {T(n, k)=local(t=(sqrtint(8*k+1)-1)\2); (n-t)!*(n-t)^(k-t*(t+1)/2)*(n-t+1)^(t-k+t*(t+1)/2)}

Cf. A018927, A127420.

Cf. A047969.

Sequence in context: A114626 A124773 A129177 this_sequence A135879 A138169 A139331

Adjacent sequences: A127449 A127450 A127451 this_sequence A127453 A127454 A127455

nonn,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Jan 15 2007