| Displaying 1-5 of 5 results found.
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COMMENTS
Number of integer pairs (x,y) satisfying |x|^3+|y|^3 <= n, any -n <= x,y <=n. Cubic variant of A057655.
EXAMPLE
a(1) = 5 counts (x,y) = (-1,0), (0,-1), (0,0), (0,1) and (1,0).
Number of integer triples (x,y,z) satisfying |x|^3 + |y|^3 + |z|^3 = n, -n <= x,y,z <= n.
+10
5
1, 6, 12, 8, 0, 0, 0, 0, 6, 24, 24, 0, 0, 0, 0, 0, 12, 24, 0, 0, 0, 0, 0, 0, 8, 0, 0, 6, 24, 24, 0, 0, 0, 0, 0, 24, 48, 0, 0, 0, 0, 0, 0, 24, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12, 24, 0, 0, 0, 0, 0, 0, 24, 0, 6, 24, 24, 0, 0, 0, 0, 0, 24, 48, 0, 0, 0, 0, 0, 0, 24, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 24, 48, 0, 0, 0
COMMENTS
A three-dimensional variant of A175362.
FORMULA
G.f.: ( 1 + 2*Sum_{j>=1} x^(j^3) )^3.
EXAMPLE
a(2) = 12 counts (x,y,z) = (-1,-1,0), (-1,0,-1), (-1,0,1), (-1,1,0), (0,-1,-1), (0,-1,1), (0,1,-1), (0,1,1), (1,-1,0), (1,0,-1), (1,0,1) and (1,1,0).
MAPLE
N:= 100: # to get a(0) to a(N)
G:= (1+2*add(x^(j^3), j=1..floor(N^(1/3))))^3:
S:= series(G, x, N+1):
PROG
(PARI) a(n, k=3) = if(n==0, return(1)); if(k <= 0, return(0)); if(k == 1, return(ispower(n, 3))); my(count = 0); for(v = 0, sqrtnint(n, 3), count += (2 - (v == 0))*if(k > 2, a(n - v^3, k-1), if(ispower(n - v^3, 3), 2 - (n - v^3 == 0), 0))); count; \\ Daniel Suteu, Aug 15 2021
Number of integer 4-tuples (x,y,z,u) satisfying |x|^3 + |y|^3 + |z|^3 + |u|^3 = n, -n <= x,y,z,u <= n.
+10
2
1, 8, 24, 32, 16, 0, 0, 0, 8, 48, 96, 64, 0, 0, 0, 0, 24, 96, 96, 0, 0, 0, 0, 0, 32, 64, 0, 8, 48, 96, 64, 0, 16, 0, 0, 48, 192, 192, 0, 0, 0, 0, 0, 96, 192, 0, 0, 0, 0, 0, 0, 64, 0, 0, 24, 96, 96, 0, 0, 0, 0, 0, 96, 192, 8, 48, 96, 64, 0, 0, 96, 0, 48, 192, 192, 0, 0, 0, 0, 0, 96, 224, 64, 0
COMMENTS
A variant of A000118 with cubes instead of squares.
FORMULA
Conjectured g.f.: (1 + 2*Sum_{j>=1} x^(j^3))^4.
EXAMPLE
a(1) = 8 counts (x,y,z,u) = (-1,0,0,0), (0,-1,0,0), (0,0,-1,0), (0,0,0,-1) and 4 more tuples with -1 replaced by +1.
a(2) = 24 counts (x,y,z,u) = (-1,-1,0,0), (-1,0,-1,0), (-1,0,0,-1), (-1,0,0,1) etc, all variants where two of the 4 values are zero and the other two +1 or -1.
PROG
(PARI) a(n, k=4) = if(n==0, return(1)); if(k <= 0, return(0)); if(k == 1, return(ispower(n, 3))); my(count = 0); for(v = 0, sqrtnint(n, 3), count += (2 - (v == 0))*if(k > 2, a(n - v^3, k-1), if(ispower(n - v^3, 3), 2 - (n - v^3 == 0), 0))); count; \\ Daniel Suteu, Aug 15 2021
Number of integer pairs (x,y) satisfying x^4 + y^4 = n.
+10
2
1, 4, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0
FORMULA
G.f.: (1 + 2*Sum_{j>=1} x^(j^4))^2.
MAPLE
seq(coeff(series((1+2*add(x^(j^4), j=1..n))^2, x, n+1), x, n), n = 0 .. 120); # Muniru A Asiru, Oct 07 2018
MATHEMATICA
CoefficientList[Series[(1 + 2*Sum[x^(j^4), {j, 1, 100}])^2, {x, 0, 120}], x] (* G. C. Greubel, Oct 06 2018 *)
PROG
(PARI) x='x+O('x^120); Vec((1+2*sum(j=1, 50, x^(j^4)))^2) \\ G. C. Greubel, Oct 06 2018 (Magma) m:=120; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*(&+[x^(j^4): j in [1..50]]))^2)); // G. C. Greubel, Oct 06 2018
Number of integer solutions to x^3 + y^3 = n.
+10
0
2, 1, 0, 0, 0, 0, 2, 2, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 2, 2, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 2, 0, 2, 2, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
MATHEMATICA
a[n_]:=Length[Solve[x^3 + y^3 == n, {x, y}, Integers]]; Table[a[i], {i, 10000}]
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