A323016 -id:A323016 - cp's OEIS frontend

A323015 a(n) is the number of unordered partitions of 24*n + 4 into four squares of primes (A001248).

0, 0, 0, 0, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 2, 4, 3, 4, 3, 4, 4, 4, 4, 4, 3, 6, 5, 5, 4, 5, 6, 4, 6, 5, 2, 6, 6, 6, 5, 5, 7, 3, 7, 8, 3, 8, 7, 6, 6, 7, 9, 5, 6, 9, 4, 7, 7, 6, 7, 7, 10, 4, 5, 10, 5, 9, 7, 6, 7, 4, 10, 8, 6, 10, 5, 9, 7, 8, 10, 6, 11, 10, 8, 11

Offset: 0

Jianing Song, Jan 05 2019

nonn

Also, a(n) is the number of unordered partitions of n into four terms of A024702.
a(n) > 0 for 4 <= n <= 2*10^4. Conjecture: a(n) > 0 for all n >= 4. A stronger conjecture: lim inf a(n) = +oo.
This is a quadratic analog of Goldbach's conjecture, asking for the smallest k such that any sufficiently large number congruent to k modulo 24 can be written as the sum of k squares of primes. k = 1 is trivially false. Let n = 49*t + 2 (t > 0), then 24*n + 2 = 24*49*t + 98, which is a multiple of 7^2. If p^2 + q^2 = 24*n + 2, since the squares of primes other than 7 are congruent to 1, 2, 4 modulo 7, we must have p = q = 7, but p^2 + q^2 < n. So k = 2 is false. Let n = 245*t + 103, then 24*n + 3 = 24*245*t + 2475, which is a multiple of 5. If p^2 + q^2 + r^2 = 24*n + 3, since the squares of primes other than 5 are congruent to 1, 4 modulo 5, we must have that at least one of p, q, r is 5. Suppose that p = 5, then q^2 + r^2 = 24*245*t + 2450, which is a multiple of 7^2. As is shown above, q = r = 7, but p^2 + q^2 + r^2 < n. So k = 3 is also false. On the other hand, if the case k = 4 is true, then all the cases k >= 4 are trivially true, because we can add as many 5^2 as needed. So the case k = 4 is the most interesting.

			100 = 5^2 + 5^2 + 5^2 + 5^2.
124 = 5^2 + 5^2 + 5^2 + 7^2.
148 = 5^2 + 5^2 + 7^2 + 7^2.
172 = 5^2 + 7^2 + 7^2 + 7^2.
196 = 7^2 + 7^2 + 7^2 + 7^2 = 5^2 + 5^2 + 5^2 + 11^2.
220 = 5^2 + 5^2 + 7^2 + 11^2.
244 = 5^2 + 7^2 + 7^2 + 11^2 = 5^2 + 5^2 + 5^2 + 13^2.
268 = 7^2 + 7^2 + 7^2 + 11^2 = 5^2 + 5^2 + 7^2 + 13^2.
...

Alois P. Heinz, Table of n, a(n) for n = 0..65536

See A323016 for the ordered version.

h:= proc(n) option remember; `if`(n<1, 0, (t->
      `if`((ithprime(t+2)^2-1)/24>n, t-1, t))(1+h(n-1)))
    end:
b:= proc(n, i, c) option remember; `if`(n=0, `if`(c=0, 1, 0),
      `if`(min(i, c)<1, 0, b(n, i-1, c)+(t-> b(n-t, min(i,
         h(n-t)), c-1))((ithprime(i+2)^2-1)/24)))
    end:
a:= n-> b(n, h(n), 4):
seq(a(n), n=0..120);  # Alois P. Heinz, Jan 05 2019

h[n_] := h[n] = If[n < 1, 0, Function[t, If[(Prime[t + 2]^2 - 1)/24 > n, t - 1, t]][1 + h[n - 1]]];
b[n_, i_, c_] := b[n, i, c] = If[n == 0, If[c == 0, 1, 0], If[Min[i, c] < 1, 0, b[n, i - 1, c] + Function[t, b[n - t, Min[i, h[n - t]], c - 1]][(Prime[i + 2]^2 - 1)/24]]];
a[n_] := b[n, h[n], 4];
a /@ Range[0, 120] (* Jean-François Alcover, Nov 22 2020, after Alois P. Heinz *)

a(n) = if(n<4, 0, my(i=0, k=sqrt(24*n-71)); forprime(p=5, k, forprime(q=p, k, forprime(r=q, k, forprime(s=r, k, if(p^2+q^2+r^2+s^2==24*n+4, i++))))); i)

cp's OEIS Frontend

A323015 a(n) is the number of unordered partitions of 24*n + 4 into four squares of primes (A001248).

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