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 A276533 Least prime p with A271518(p) = n. 2
 5, 2, 19, 127, 17, 67, 163, 41, 89, 101, 131, 313, 257, 211, 227, 461, 241, 401, 613, 337, 433, 353, 577, 467, 863, 887, 617, 787, 601, 569, 761, 641, 823, 673, 857, 1217, 881, 1091, 1289, 977, 1427, 1097, 1801, 929, 1153, 953, 1321, 1049, 1747, 1409 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Conjecture: a(n) exists for any positive integer n. In contrast, it is known that for each prime p the number of ordered integral solutions to the equation x^2 + y^2 + z^2 + w^2 = p is 8*(p+1). In 1998 J. Friedlander and H. Iwaniec proved that there are infinitely many primes p of the form w^2 + x^4 = w^2 + (x^2)^2 + 0^2 + 0^2 with w and x nonnegative integers. Since x^2 + 3*0 + 5*0 is a square, we see that A271518(p) > 0 for infinitely many primes p. REFERENCES J. Friedlander and H. Iwaniec, The polynomial x^2 + y^4 captures its primes, Ann. of Math. 148 (1998), 945-1040. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..500 Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016. EXAMPLE a(1) = 5 since 5 is the first prime which can be written in a unique way as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integer and x + 3*y + 5*z a square; in fact, 5 = 1^2 + 0^2 + 0^2 + 2^2 with 1 + 3*0 + 5*0 = 1^2. a(2) = 2 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 1 + 3*0 + 5*0 = 1^2, and 2 = 1^2 + 1^2 + 0^2 + 0^2 with 1 + 3*1 + 5*0 = 2^2. MATHEMATICA SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]; Do[m=0; Label[aa]; m=m+1; r=0; Do[If[SQ[Prime[m]-x^2-y^2-z^2]&&SQ[x+3y+5z], r=r+1; If[r>n, Goto[aa]]], {x, 0, Sqrt[Prime[m]]}, {y, 0, Sqrt[Prime[m]-x^2]}, {z, 0, Sqrt[Prime[m]-x^2-y^2]}]; If[r

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Last modified August 25 13:50 EDT 2019. Contains 326324 sequences. (Running on oeis4.)