

A323015


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


2



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

0,9


COMMENTS

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.


LINKS

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


EXAMPLE

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.
...


MAPLE

h:= proc(n) option remember; `if`(n<1, 0, (t>
`if`((ithprime(t+2)^21)/24>n, t1, t))(1+h(n1)))
end:
b:= proc(n, i, c) option remember; `if`(n=0, `if`(c=0, 1, 0),
`if`(min(i, c)<1, 0, b(n, i1, c)+(t> b(nt, min(i,
h(nt)), c1))((ithprime(i+2)^21)/24)))
end:
a:= n> b(n, h(n), 4):
seq(a(n), n=0..120); # Alois P. Heinz, Jan 05 2019


PROG

(PARI) a(n) = if(n<4, 0, my(i=0, k=sqrt(24*n71)); 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)


CROSSREFS

See A323016 for the ordered version.
Sequence in context: A072628 A193386 A068212 * A105194 A008335 A106031
Adjacent sequences: A323012 A323013 A323014 * A323016 A323017 A323018


KEYWORD

nonn


AUTHOR

Jianing Song, Jan 05 2019


STATUS

approved



