A181435 First column in matrix inverse of a mixed convolution of A052906.

1, -4, -1, 0, -1, 1, -1, 0, 0, 1, -1, 0, -1, 1, 1, 0, -1, 0, -1, 0, 1, 1, -1, 0, 0, 1, 0, 0, -1, -1, -1, 0, 1, 1, 1, 0, -1, 1, 1, 0, -1, -1, -1, 0, 0, 1, -1, 0, 0, 0, 1, 0, -1, 0, 1, 0, 1, 1, -1, 0, -1, 1, 0, 0, 1, -1, -1, 0, 1, -1, -1, 0, -1, 1, 0, 0, 1

Offset: 1

Author

Mats Granvik, Oct 20 2010

Keywords

• sign

Comments

It appears that except for the second term, the sequence is identical to the Möbius function.

Links

• R. J. Mathar, Table of n, a(n) for n = 1..1091

Crossrefs

Cf. A003688, A006190, A013946, A008683, A178536, A181434.

Programs

Maple

b := proc(n)
    option remember;
    local c;
    c := 3;
    if n <= 3 then
        op(n,[1,c,c^2]) ;
    else
        c*procname(n-1)+procname(n-2) ;
    end if;
end proc:
A := proc(n,k)
    if n >= k then
        b(n-k+1) ;
    else
        0 ;
    end if;
end proc:
B := proc(n,k)
    if modp(n,k) = 0 then
        1;
    else
        0;
    end if;
end proc:
AB := proc(n,k)
    option remember;
    add( A(n,j)*B(j,k),j=1..n) ;
end proc:
ABinv := proc(n,k)
    option remember;
    if k > n then
        0;
    elif k = n then
        1;
    else
        -add( AB(n,j)*procname(j,k),j=k..n-1) ;
    end if;
end proc:
A181435 := proc(n)
    ABinv(n,1) ;
end proc:
for n from 1 do
    printf("%d %d\n",n,ABinv(n,1)) ;
end do: # R. J. Mathar, Oct 06 2017

Mathematica

Clear[t, n, k, nn, b, A, c]; nn = 77; c = 3; b[0] = 1; b[1] = 1; b[n_] := b[n] = c*b[n - 1] + b[n - 2]; t[n_, 1] = If[n >= 1, b[n], 0]; t[n_, k_] := t[n, k] = If[n >= k, Sum[t[n - i, k - 1], {i, 1, k - 1}] - Sum[t[n - i, k], {i, 1, k - 1}], 0]; MatrixForm[A = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}]]; Inverse[A][[All, 1]] (* Mats Granvik, Sep 16 2017 *)

Formula

From Mats Granvik, Sep 16 2017: (Start)

a(n) as the matrix inverse of a mixed convolution: Let c = 3 and let the sequence b be defined by the recurrence: b(1) = 1, b(2) = c, b(3) = c^2; for n >= 4, b(n) = c*b(n-1) + b(n-2), so b(n) = A052906(n-1), and let the lower triangular matrix A be: If n >= k then A(n,k) = b(n - k + 1) else A(n,k) = 0, and let B be the lower triangular matrix A051731. Then the matrix inverse (A.B)^-1 will have a(n) as its first column.

The matrix product T = A.B can be defined as follows: Let c = 3 and the sequence b be defined by the recurrence b(0) = 1, b(1) = 1; for b >= 2, b(n) = c*b(n - 1) + b(n - 2); and let T be the lower triangular matrix defined by the recurrence: T(n, 1) = If n >= 1 then T(n, 1) = b(n) else T(n, 1) = 0; for k >= 2, T(n, k) = If n >= k then (Sum_{i=1..k-1} T(n - i, k - 1) - T(n - i, k)) else 0. (Then the matrix inverse of T will have a(n) as its first column.)

(End)