A105615 - OEIS

%I #10 Jan 09 2024 12:16:17

%S 1,2,1,10,4,1,74,26,6,1,706,226,50,8,1,8162,2426,522,82,10,1,110410,

%T 30826,6498,1010,122,12,1,1708394,451586,93666,14458,1738,170,14,1,

%U 29752066,7489426,1532970,235466,28226,2754,226,16,1,576037442

%N Triangular matrix T, read by rows, that satisfies: SHIFT_LEFT(column 0 of T^((2*p-1)/2)) = (2*p-1)*(column p of T), or [T^((2*p-1)/2)](m,0) = (2*p-1)*T(p+m,p+1) for all m>=1 and p>=0.

%C Column 0 is A000698 (related to double factorials), offset 1. Column 1 is A105616 (column 0 of T^(1/2), offset 1). The matrix logarithm divided by 2 yields the integer triangle A105629.

%C Compare with triangular matrix A107717, which satisfies: SHIFT_LEFT(column 0 of A107717^((3*k-1)/3)) = (3*k-1)*(column k of A107717).

%F T(n, k) = 2*(k+1)*T(n, k+1) + Sum_{j=1..n-k-1} T(j, 0)*T(n, j+k+1) for n>k>=0, with T(n, n) = 1 for n>=0. T(n, 0) = A000698(n+1) for n>=0.

%e SHIFT_LEFT(column 0 of T^(-1/2)) = -1*(column 0 of T);

%e SHIFT_LEFT(column 0 of T^(1/2)) = 1*(column 1 of T);

%e SHIFT_LEFT(column 0 of T^(3/2)) = 3*(column 2 of T);

%e SHIFT_LEFT(column 0 of T^(5/2)) = 5*(column 3 of T).

%e Triangle begins:

%e 1;

%e 2,1;

%e 10,4,1;

%e 74,26,6,1;

%e 706,226,50,8,1;

%e 8162,2426,522,82,10,1;

%e 110410,30826,6498,1010,122,12,1;

%e 1708394,451586,93666,14458,1738,170,14,1;

%e 29752066,7489426,1532970,235466,28226,2754,226,16,1; ...

%e Matrix square-root T^(1/2) is A105623 which begins:

%e 1;

%e 1,1;

%e 4,2,1;

%e 26,10,3,1;

%e 226,74,19,4,1;

%e 2426,706,167,31,5,1; ...

%e compare column 0 of T^(1/2) to column 1 of T;

%e also, column 1 of T^(1/2) equals column 0 of T.

%e Matrix inverse square-root T^(-1/2) is A105620 which begins:

%e 1;

%e -1,1;

%e -2,-2,1;

%e -10,-4,-3,1;

%e -74,-20,-7,-4,1;

%e -706,-148,-39,-11,-5,1; ...

%e compare column 0 of T^(-1/2) to column 0 of T.

%e Matrix inverse T^-1 is A105619 which begins:

%e 1;

%e -2,1;

%e -2,-4,1;

%e -10,-2,-6,1;

%e -74,-10,-2,-8,1;

%e -706,-74,-10,-2,-10,1;

%e -8162,-706,-74,-10,-2,-12,1; ...

%o (PARI) {T(n,k) = if(n<k||k<0,0,if(n==k,1, 2*(k+1)*T(n,k+1)+sum(j=1,n-k-1,T(j,0)*T(n,j+k+1))))}

%o for(n=0,10,for(k=0,n,print1(T(n,k),", ")); print(""))

%o (PARI) {T(n,k) = if(n<k||k<0,0,(matrix(n+1,n+1,m,j,if(m>=j,if(m==j,1,if(m==j+1,-2*j, polcoeff(1/sum(i=0,m-j,(2*i)!/i!/2^i*x^i)+O(x^m),m-j)))))^-1)[n+1,k+1])}

%o for(n=0,10,for(k=0,n,print1(T(n,k),", ")); print(""))

%Y Cf. A000698 (column 0), A105616 (column 1), A105617 (column 2), A105618 (row sums), A105619 (T^-1), A105620 (T^(-1/2)), A105623 (T^(1/2)), A105627 (T^(3/2)), A105629 (matrix log).

%Y Cf. A107717.

%K nonn,tabl

%O 0,2

%A _Paul D. Hanna_, Apr 16 2005