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A123706
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Matrix inverse of triangle A010766, where A010766(n,k) = [n/k], for n>=k>=1.
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7
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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
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Unsigned elements consist of only 0's, 1's and 2's.
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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.
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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; ...
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PROG
| (PARI) T(n, k)=(matrix(n, n, r, c, r\c)^-1)[n, k] \\ simplified by M. F. Hasler, Feb 12 2012
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CROSSREFS
| Cf. A102309, A085411; A123707, A123708, A123709.
Sequence in context: A056980 A005094 A121372 * A194325 A025452 A194337
Adjacent sequences: A123703 A123704 A123705 * A123707 A123708 A123709
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KEYWORD
| sign,tabl,changed
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AUTHOR
| Paul D. Hanna (pauldhanna(AT)juno.com), Oct 09 2006
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