0,5

Let vector R_{n} equal row n of this array; then R_{n+1} = P * R_{n} for n>=0, where triangle P = A132625 such that row n+1 of P = row n of P^(2^n) with appended '1' for n>=0.

G.f. for row n: Sum_{i>=0} (1 + 2^i*x)^(n-1) * log(1 + 2^i*x)^i / i!.

Square array begins:

1,1,3,35,1365,169911,67945521,89356415775,396861704798625,...;

1,2,6,56,1820,201376,74974368,94525795200,409663695276000,...;

1,3,10,84,2380,237336,82598880,99949406400,422825581068000,...;

1,4,15,120,3060,278256,90858768,105637584000,436355999662176,...;

1,5,21,165,3876,324632,99795696,111600996000,450263760607584,...;

1,6,28,220,4845,376992,109453344,117850651776,464557848245920,...;

1,7,36,286,5985,435897,119877472,124397910208,479247424475040,...;

1,8,45,364,7315,501942,131115985,131254487936,494341831545120,...;

...

Form column vector R_{n} out of row n of this array;

then row n+1 can be generated from row n by:

R_{n+1} = P * R_{n} for n>=0,

where triangular matrix P = A132625 begins:

1;

1, 1;

2, 1, 1;

14, 4, 1, 1;

336, 60, 8, 1, 1;

25836, 2960, 248, 16, 1, 1;

6251504, 454072, 24800, 1008, 32, 1, 1; ...

where row n+1 of P = row n of P^(2^n) with appended '1' for n>=0.

(PARI) {T(n, k)=binomial(2^k+n-1, k)} (PARI) /* Coefficient of x^k in g.f. of row n: */ {T(n, k)=polcoeff(sum(i=0, k, (1+2^i*x+x*O(x^k))^(n-1)*log((1+2^i*x)+x*O(x^k))^i/i!), k)}

Cf. A132625; rows: A136556, A014070, A136505, A136506; diagonals: A060690, A132683, A132684; A136557 (antidiagonal sums).

Sequence in context: A027555 A059481 A113592 this_sequence A063967 A059397 A071943

Adjacent sequences: A136552 A136553 A136554 this_sequence A136556 A136557 A136558

nonn,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Jan 07 2008