1,31

Column 1 is A096433. Column 2 = [0,1,0,-1,0,0,0,...(zero for n>4)]. Column 3 is A118230.

Leroy Quet, Home Page (listed in lieu of email address)

For column k>1: Sum_{i=2..n, gcd(n,i)=1} T(i,k) = 1 when n=k+1, 0 elsewhere; for column k=1: Sum_{i=2..n, gcd(n,i)=1} T(i,1) = 1 when n=1 or 2, 0 elsewhere.

Describes a sequence transformation as follows.

Say we have the arbitrary sequence {a(k)}.

We define {b(k)}, based on {a(k)}, by:

b(n) = sum{1<=k<=n, GCD(k,n)=1} a(k).

So given {b(k)} (which must have b(1) = b(2)), how do we get the sequence {a(k)}?

If a(n) = sum{k>=2} b(k) * T(n,k), then there is a triangular array {T(n,k)} which begins:

1;

-1, 1;

-1, 0, 1;

1,-1,-1, 1;

-1, 0, 0, 0, 1;

1, 0, 0,-1,-1, 1;

1, 0,-1, 0,-1, 0, 1;

-1, 0, 2,-1, 0, 0,-1, 1;

-1, 0, 0, 0, 1, 0,-1, 0, 1;

1, 0,-1, 1, 0,-1, 1,-1,-1, 1;

-1, 0, 1, 0, 0, 0,-1, 0, 0, 0, 1;

1, 0,-1, 0, 0, 0, 1, 0, 0,-1,-1, 1;

3, 0,-2, 0,-2, 0, 2, 0,-1, 0,-1, 0, 1;

-3, 0, 1, 0, 3, 0,-1,-1, 1, 0, 0, 0,-1, 1; ...

(PARI) {T(n, k)=if(n<k|k<0, 0, (matrix(n, n, r, c, if(r>=c, if(gcd(r-c+1, c)==1, 1, 0)))^-1)[n, k])}

Cf. A054431 (matrix inverse), A096433 (column 1), A118230 (column 3).

Sequence in context: A056929 A151692 A115201 this_sequence A117201 A060953 A082858

Adjacent sequences: A118226 A118227 A118228 this_sequence A118230 A118231 A118232

sign,tabl

Leroy Quet, Paul D. Hanna (pauldhanna(AT)juno.com), Apr 16 2006