This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
The solutions (if any) for n <= 20 are as follows: n = 1: n = 2: n = 3: n = 4: 1 1 1 1 n = 5: n = 6: n = 7: 1 1 1 2 n = 8: n = 9: n = 10: 1 1 2 2 n = 11: n = 12: 1 1 1 3 n = 13: 1 2 2 2 n = 14: n = 15: 1 1 2 3 n = 16: n = 17: n = 18: 1 2 2 3 n = 19: 1 1 1 4 n = 20: 1 1 3 3 From _Wolfdieter Lang_, Mar 25 2013: (Start) a(16) = 0 because 16 is not a primitive sum of four nonzero squares. The representation [2,2,2,2] of 16 is not primitive. a(40) = 0 because the only representation as sum of four nonzero squares (A025428(4) = 1) is [2,2,4,4], but this is not primitive. a(28) = 2 because the two primitive representations of 28 are [1, 1, 1, 5] and [1, 3, 3, 3]. [2, 2, 2, 4] = 2*[1, 1, 1, 2] is not primitive due to 28 = 2^2*7. (End)
b:= proc(n, i, g, t) option remember; `if`(n=0,
`if`(g=1 and t=0, 1, 0), `if`(i<1 or t=0 or i^2*tn, 0, b(n-i^2, i, igcd(g, i), t-1))))
end:
a:= n-> `if`(n<4, 0, b(n, isqrt(n-3), 0, 4)):
seq(a(n), n=1..120); # Alois P. Heinz, Apr 02 2013
Clear[b]; b[n_, i_, g_, t_] := b[n, i, g, t] = If[n == 0, If[g == 1 && t == 0, 1, 0], If[i < 1 || t == 0 || i^2*t < n, 0, b[n, i-1, g, t] + If[i^2 > n, 0, b[n-i^2, i, GCD[g, i], t-1]]]]; a[n_] := If[n < 4, 0, b[n, Sqrt[n-3] // Floor, 0, 4]]; Table[a[n], {n, 1, 99}] (* Jean-François Alcover, Apr 05 2013, translated from Alois P. Heinz's Maple program *)
When the radius of the sphere reaches 2, one cube is completely within the sphere. When the radius reaches 7^(1/2), five cubes are completely within the sphere.
65 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 6^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2 = 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2 = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 so 65 is a term.
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 for x in range(1, 1000)]
for pos in cwr(power_terms, 9):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 10])
for x in range(len(rets)):
print(rets[x])
def A346803(n): return (63, 65, 68, 71, 72, 74, 75)[n-1] if n<8 else n+69 # Chai Wah Wu, May 09 2024
N:= 10000: # to get all terms <= N
T:= Vector(N):
for a from 1 to floor(sqrt(N/4)) do
for b from a to floor(sqrt((N-a^2)/3)) do
for c from b to floor(sqrt((N-a^2-b^2)/2)) do
for d from c to floor(sqrt(N-a^2-b^2-c^2)) do
m:= a^2+b^2+c^2+d^2;
T[m]:= T[m]+1;
od od od od:
select(i -> T[i] = 2, [$1..N]); # Robert Israel, Nov 03 2017
M = 1000;
Clear[T]; T[_] = 0;
For[a = 1, a <= Floor[Sqrt[M/4]], a++,
For[b = a, b <= Floor[Sqrt[(M - a^2)/3]], b++,
For[c = b, c <= Floor[Sqrt[(M - a^2 - b^2)/2]], c++,
For[d = c, d <= Floor[Sqrt[M - a^2 - b^2 - c^2]], d++,
m = a^2 + b^2 + c^2 + d^2;
T[m] = T[m] + 1;
]]]];
Select[Range[M], T[#] == 2&] (* Jean-François Alcover, Mar 22 2019, after Robert Israel *)
prime(n)'s are 2, 3, 5, 7, 11, 13, 17, ... giving the sequence 1, 0, 1, 2, 3, 5, 7, ...
forprime(p=2,1000, k=0; for(s1=1,sqrt((p^2)/4),for(s2=s1,sqrt((p^2 - s1^2)/3), for(s3=s2,sqrt((p^2-s1^2 - s2^2)/2), if(issquare(p^2-s1^2-s2^2-s3^2),k++)))) ; f = floor((p^2+4*p+24)/48.) ; f2 = (p^2 + 4*p + (19*(5*(p%48)+2)^2)%48 - 24)/48 ; print1([p,k,f,f2]" ")) /* code above prints [p, k, f, f2] where p is the prime, k is the number of ways the square of p can be expressed as the sum of four nonzero squares, and f and f2 are the formulas derivations. f and k are observed to be the same for p from 3 to 997; f2 and k are observed to be the same for p from 5 to 997. */
A216374(n)=sum(s1=1,.5*n=prime(n+1),my(t);sum(s2=s1,sqrtint((n^2-s1^2)\3),sum(s3=s2,sqrtint((t=n^2-s1^2-s2^2)\2),issquare(t-s3^2)))) \\ M. F. Hasler, Sep 11 2012
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