

A100450


Number of ordered triples (i,j,k) with i + j + k <= n and GCD{i,j,k} <= 1.


17



1, 7, 19, 51, 99, 195, 291, 483, 675, 963, 1251, 1731, 2115, 2787, 3363, 4131, 4899, 6051, 6915, 8355, 9507, 11043, 12483, 14595, 16131, 18531, 20547, 23139, 25443, 28803, 31107, 34947, 38019, 41859, 45315, 49923, 53379, 58851, 63171, 68547
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OFFSET

0,2


COMMENTS

Note that GCD{0,m} = m for any m.
From Robert Price, Mar 05 2013: (Start)
Distinct sums i+j+k with or without the GCD qualifier results in a(n)=2n+1 (A005408).
Distinct products i*j*k without the GCD qualifier is given by A213207.
Distinct products i*j*k with the GCD qualifier is given by A213208.
With the restriction i,j,k >= 0 ...
Distinct sums or products equal to n is trivial and always equals one (A000012).
Distinct sums <=n results in a(n)=n (A001477).
Distinct products <=n without the GCD qualifier is given by A213213.
Distinct products <=n with the GCD qualifier is given by A213212.
Ordered triples with the sum =n without the GCD qualifier is A000217(n+1).
Ordered triples with the sum =n with the GCD qualifier is A048240.
Ordered triples with the sum <=n without the GCD qualifier is A000292.
Ordered triples with the sum <=n with the GCD qualifier is A048241. (End)
This sequence (A100450) without the GCD qualifier results in A001845.  Robert Price, Jun 04 2013


LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..1000


FORMULA

G.f.: (3+Sum(moebius(k)*((1+x^k)/(1x^k))^3, k=1..infinity))/(1x).  Vladeta Jovovic, Nov 22 2004. [Sketch of proof: Let b(n) = number of ordered triples (i, j, k) with i + j + k = n and GCD{i, j, k}<= 1. Then a(n) = A100450(n) = partial sums of b(n) and Sum_{d divides n} b(d) = 4*n^2+2 = A005899(n) with g.f. ((1+x)/(1x))^3.]


MAPLE

f:=proc(n) local i, j, k, t1, t2, t3; t1:=0; for i from n to n do for j from n to n do t2:=gcd(i, j); for k from n to n do if abs(i) + abs(j) + abs(k) <= n then t3:=gcd(t2, k); if t3 <= 1 then t1:=t1+1; fi; fi; od: od: od: t1; end;


MATHEMATICA

f[n_] := Length[ Union[ Flatten[ Table[ If[ Abs[i] + Abs[j] + Abs[k] <= n && GCD[i, j, k] <= 1, {i, j, k}, {0, 0, 0}], {i, n, n}, {j, n, n}, {k, n, n}], 2]]]; Table[ f[n], {n, 0, 40}] (* Robert G. Wilson v, Dec 14 2004 *)


CROSSREFS

Cf. A000124, A000292, A018805, A027430, A048240, A048241, A100448, A100449, A213207, A213208, A213212, A213213.
Sequence in context: A027523 A100545 A203165 * A155423 A155347 A155385
Adjacent sequences: A100447 A100448 A100449 * A100451 A100452 A100453


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane, Nov 21 2004


EXTENSIONS

The keyword "more" refers to the fact that I would also like to get the sequences of the numbers of distinct sums i+j+k (also distinct products i*j*k) over all ordered triples (i,j,k) with i + j + k <= n; also over all ordered triples (i,j,k) with i + j + k <= n and GCD{i,j,k} <= 1.
Also the sequences of the numbers of distinct sums i+j+k (also distinct products i*j*k) over all ordered triples (i,j,k) with i >= 0, j >= 0, k >= 0 and i + j + k = n; also over all ordered triples (i,j,k) with i >= 0, j >= 0, k >= 0, i + j + k = n and GCD{i,j,k} <= 1.
Also the number of ordered triples (i,j,k) with i >= 0, j >= 0, k >= 0, i + j + k = n and GCD{i,j,k} <= 1.
See the Comments section for sequences that address these extensions.  Robert Price, Mar 05 2013


STATUS

approved



