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A305024
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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.
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1
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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
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OFFSET
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1,3
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COMMENTS
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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).
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LINKS
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FORMULA
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If
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EXAMPLE
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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?
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MAPLE
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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:
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MATHEMATICA
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a[n_] := Block[{p = Prime@n, c, k=2}, c = Range[Sqrt[p]]^2; While[ k<p, If[ Select[ IntegerPartitions[ p, {k}, c], PrimeQ@ Total[ #^2] &, 1] != {}, Break[]]; k++]; If[k < p, k, 0]]; Array[a, 95] (* Giovanni Resta, Dec 12 2019 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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