|
%I #10 Feb 13 2022 09:25:56
%S 0,0,1,0,1,3,0,1,5,13,0,1,7,25,51,0,1,9,43,125,205,0,1,11,67,259,625,
%T 819,0,1,13,97,477,1555,3125,3277,0,1,15,133,803,3355,9331,15625,
%U 13107,0,1,17,175,1261,6505,23517,55987,78125,52429,0,1,19,223,1875,11605
%N Array read by antidiagonals: T(n, k) = ((n+4)^k-(n-1)^k)/5.
%C Consider a 5x5 matrix M =
%C [n, 1, 1, 1, 1]
%C [1, n, 1, 1, 1]
%C [1, 1, n, 1, 1]
%C [1, 1, 1, n, 1]
%C [1, 1, 1, 1, n].
%C The n-th row of the array contains the values of the non diagonal elements of M^k, k=0,1,.... (Corresponding diagonal entry = non diagonal entry + (n-1)^k.)
%C For row r we have polynomial ((r+4)^n-(r-1)^n)/5. Corresponding g.f.s: x/((1-(r-1)x)(1-(r+4)x))
%C If r(n) denotes a row sequence, r(n+1)/r(n) converges to n+4.
%C Triangle T(n, k) = (4^(n-k-1)-(-1)^(n-k-1))/5*(binomial(k+(n-k-1),n-k-1)) gives coefficients for polynomials for the columns of the array. First four polynomial are:
%C 1
%C 3 + 2*k
%C 13 + 9*k + 3*k^2
%C 51 + 52*k + 18*k^2 + 4*k^3
%C ...
%e Array begins:
%e 0, 1, 3, 13, 51, 205, ...
%e 0, 1, 5, 25, 125, 625, ...
%e 0, 1, 7, 43, 259, 1555, ...
%e 0, 1, 9, 67, 477, 3355, ...
%e 0, 1, 11, 97, 803, 6505, ...
%e ...
%o (PARI) MM(n,N)=local(M);M=matrix(n,n);for(i=1,n, for(j=1,n,if(i==j,M[i,j]=N,M[i,j]=1)));M
%o for(k=0,10, for(i=0,10,print1((MM(5,k)^i)[1,2],","));print())
%o (PARI) p(n,k)=((n+4)^k-(n-1)^k)/5
%o for(k=0,10, for(i=0,10,print1(p(k,i),","));print())
%o (PARI) for(k=0,10, for(i=0,10,print1(polcoeff(x/((1-(k-1)*x)*(1-(k+4)*x)),i),","));print())
%Y Cf. A015521 (for n=0), A000351 (for n=1), A003464 (for n=2), A016130 (for n=3), A016140 (for n=4), A016153 (for n=5), A016164 (for n=6), A016174 (for n=7), A016184 (for n=8), A015441 (for n=-1), A091005 (for n=-2).
%K nonn,tabl
%O 0,6
%A Lambert Klasen (lambert.klasen(AT)gmx.net) and _Gary W. Adamson_, Feb 10 2005
|
