0,2

Note that GCD{0,m} = m for any m.

G.f.: (3+Sum(moebius(k)*((1+x^k)/(1-x^k))^3, k=1..infinity))/(1-x). - Vladeta Jovovic (vladeta(AT)Eunet.yu), 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)/(1-x))^3.]

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;

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}] (from Robert G. Wilson v Dec 14 2004)

Cf. A100448, A100449, A027430, etc., also A018805.

Sequence in context: A018728 A027523 A100545 this_sequence A092053 A072630 A002714

Adjacent sequences: A100447 A100448 A100449 this_sequence A100451 A100452 A100453

nonn,easy

njas, Nov 21 2004

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.