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A123706
Matrix inverse of triangle A010766, where A010766(n,k) = [n/k], for n>=k>=1.
9
1, -2, 1, -1, -1, 1, 1, -1, -1, 1, -1, 0, 0, -1, 1, 2, 0, -1, 0, -1, 1, -1, 0, 0, 0, 0, -1, 1, 0, 0, 1, -1, 0, 0, -1, 1, 0, 1, -1, 0, 0, 0, 0, -1, 1, 2, -1, 0, 1, -1, 0, 0, 0, -1, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, -1, 1, 1, -1, 1, -1, 0, 0, 0, 0, -1, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 2, -1, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, -1, 1, 1, 1, -1, 1, -1, 0, 0, 0
OFFSET
1,2
COMMENTS
Unsigned elements consist of only 0's, 1's and 2's.
LINKS
FORMULA
T(n,1) = +2 when n = 2*p where p is an odd prime.
T(n,1) = -2 when n is an even squarefree number with an odd number of prime divisors.
A123709(n) = number of nonzero terms in row n = 2^(m+1) - 1 when n is an odd number with exactly m distinct prime factors.
Sum_{k=1..n} T(n,k) = moebius(n).
Sum_{k=1..n} T(n,k)*k = 0 for n>1.
Sum_{k=1..n} T(n,k)*k^2 = 2*phi(n) for n>1 where phi(n)=A000010(n).
Sum_{k=1..n} T(n,k)*k^3 = 6*A102309(n) for n>1 where A102309(n)=Sum[d|n, moebius(d)*C(n/d,2) ].
Sum_{k=1..n} T(n,k)*k*2^(k-1) = A085411(n) = Sum_{d|n} mu(n/d)*(d+1)*2^(d-2) = total number of parts in all compositions of n into relatively prime parts.
T(n,k) = mu(n/k)-mu(n/(k+1)), where mu(n/k) is A008683(n/k) if k|n and 0 otherwise. - Enrique Pérez Herrero, Feb 21 2012
EXAMPLE
Triangle begins:
1;
-2, 1;
-1,-1, 1;
1,-1,-1, 1;
-1, 0, 0,-1, 1;
2, 0,-1, 0,-1, 1;
-1, 0, 0, 0, 0,-1, 1;
0, 0, 1,-1, 0, 0,-1, 1;
0, 1,-1, 0, 0, 0, 0,-1, 1;
2,-1, 0, 1,-1, 0, 0, 0,-1, 1;
-1, 0, 0, 0, 0, 0, 0, 0, 0,-1, 1;
-1, 1, 1,-1, 1,-1, 0, 0, 0, 0,-1, 1;
-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 1;
2,-1, 0, 0, 0, 1,-1, 0, 0, 0, 0, 0,-1, 1;
1, 1,-1, 1,-1, 0, 0, 0, 0, 0, 0, 0, 0,-1, 1; ...
MATHEMATICA
t[n_, k_] := If[Divisible[n, k], MoebiusMu[n/k], 0] - If[Divisible[n, k+1], MoebiusMu[n/(k+1)], 0]; Table[t[n, k], {n, 1, 15}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 29 2013, after Enrique Pérez Herrero *)
PROG
(PARI) T(n, k)=(matrix(n, n, r, c, r\c)^-1)[n, k] \\ simplified by M. F. Hasler, Feb 12 2012
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Paul D. Hanna, Oct 09 2006
STATUS
approved