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A118229
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Triangle, read by rows, equal to the matrix inverse of triangle A054431; the inverse transformation that obtains {a(n)} from b(n) = sum{1<=k<=n, GCD(k,n)=1} a(k).
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2
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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
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 1,31
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COMMENTS
| Column 1 is A096433. Column 2 = [0,1,0,-1,0,0,0,...(zero for n>4)]. Column 3 is A118230.
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FORMULA
| 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.
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EXAMPLE
| 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; ...
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PROG
| (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])}
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CROSSREFS
| Cf. A054431 (matrix inverse), A096433 (column 1), A118230 (column 3).
Sequence in context: A056929 A151692 A115201 * A172250 A179229 A117201
Adjacent sequences: A118226 A118227 A118228 * A118230 A118231 A118232
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KEYWORD
| sign,tabl
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AUTHOR
| Leroy Quet, Paul D. Hanna (pauldhanna(AT)juno.com), Apr 16 2006
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