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A323015
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a(n) is the number of unordered partitions of 24*n + 4 into four squares of primes (A001248).
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2
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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
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OFFSET
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0,9
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
...
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MAPLE
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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):
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MATHEMATICA
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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];
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PROG
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(PARI) 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)
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CROSSREFS
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See A323016 for the ordered version.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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