A305024 Minimal number of squares, not all equal to 1, having as sum prime(n), such that their squares also sum to a prime; 0 if no such decomposition exists.

0, 0, 2, 4, 3, 2, 2, 3, 4, 2, 4, 2, 2, 3, 7, 2, 5, 4, 4, 4, 2, 4, 3, 2, 3, 3, 5, 3, 4, 5, 4, 3, 2, 3, 2, 4, 2, 4, 4, 3, 3, 2, 4, 4, 4, 4, 3, 4, 3, 4, 3, 4, 3, 3, 2, 4, 2, 4, 3, 2, 3, 2, 4, 4, 2, 4, 4, 4, 3, 2, 3, 4, 4, 2, 3, 4, 3, 2, 2, 2, 3, 2, 4, 3, 4, 3, 3, 4, 2, 4, 3, 4, 4, 3, 3

Offset: 1

Author

M. F. Hasler, May 23 2018

Keywords

• nonn

Comments

It has been conjectured (cf. A126769) that any prime p >= 5 can be written in a nontrivial way as p = Sum (b_i)^2 such that Sum (b_i)^4 is also prime. This sequence lists the number of required terms b_i for each prime.

The two initial zeros say that this decomposition is not possible for prime(1) = 2 and prime(2) = 3, and are thus conjectured to be the only zeros of the sequence. Since we are interested in the minimal number of terms, we can consider only nonzero b_i >= 1 and min{b_i} >= 2 to avoid the trivial solution b_i = 1 for all i <= k = prime(n).

Examples

			The first two primes, 2 and 3, cannot be written as a sum of squares not all equal to 1, because the smallest such sum is 1^2 + 2^2 = 5. (The empty sum and a one-term sum of a square cannot be prime, either.) Therefore a(1) = a(2) = 0.
The third prime, 5, can be written in exactly one way as a nontrivial sum of two squares, 5 = 1^2 + 2^2, and the sum of the fourth powers is 1^4 + 2^4 = 17, which is again prime. Therefore, a(3) = 2.
The fourth prime, 7, cannot be written as sum of 2 or 3 squares, but only 4 squares, as 7 = 1^2 + 1^2 + 1^2 + 2^2, and it turns out that sum of the fourth powers also yields a prime,  1^4 + 1^4 + 1^4 + 2^4 = 19. Therefore, a(4) = 4.
prime(15) = 47 = 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 4^2, and the sum of the fourth powers gives the prime 467. Since no smaller number of terms has this property, a(15) = 7.
prime(18) = 61 = 2^2 + 4^2 + 4^2 +5^2, and 2^4 + 4^4 + 4^4 +5^4 = 1153, a prime, and no smaller number of terms has this property, so a(18) = 4.
prime(27) = 103 = 1^2 + 4^2 + 5^2 + 5^2 + 6^2 and 1^4 + 4^4 + 5^4 + 5^4 + 6^4 = 2803, a prime, and no smaller number of terms has this property, so a(27) = 5.
The values 2, 3, 4, 5 appear for the first time at index n = 3, 5, 4, 17, and a(15) = 7. We don't know when the first 6 occurs, nor whether this happens at all.
Conjecture: The sequence is bounded.
Is it possible to show that no term of the sequence is larger than 7?
		

Links

• Robert Israel, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A126769, A128292.

Programs

Maple

repss:= proc(n,k,i) option remember;
# lists of k squares >= i^2 summing to n
  if k = 1 then
   if issqr(n) and n >= i^2 then {[sqrt(n)]}
   else {}
   fi
  elif n < k then {}
  else
   `union`(seq(map(t -> [j,op(t)], procname(n-j^2,k-1,j)),j=i..floor(sqrt(n))))
  fi
end proc:
f:= proc(n) local p,k,i,S; global Rep;
  p:= ithprime(n);
  for k from 2 do
    S:= select(t -> isprime(convert(map(`^`,t,4),`+`)), repss(p,k,1));
    if nops(S) > 0 then Rep[n]:= S[1]; return k fi
  od
end proc:
0,0,seq(f(n),n=3..100); # Robert Israel, Dec 12 2019

Mathematica

a[n_] := Block[{p = Prime@n, c, k=2}, c = Range[Sqrt[p]]^2; While[ kGiovanni Resta, Dec 12 2019 *)

PARI

apply( A305024(n)={n=prime(n); for(k=2, n-3, my(s=sqrtint((n-k)\3+1), t);
    forvec(b=vector(k-2, i, [1,s]), t=vecsum([t^4|t<-b]);
      for(i=1,#s=sum2sqr(n-norml2(b))/* see A133388 for sum2sqr() */,
        s[i][1]>0 && isprime(s[i][1]^4+s[i][2]^4+t) && return(k))/*end for i*/
    , 1/*forvec:increasing*/))}, [1..95]) \\ Bug fixed: M. F. Hasler, Dec 12 2019

Formula

If

Extensions

Corrected by Robert Israel, Dec 12 2019