cp's OEIS Frontend

A291150 Number of ways to write 2*n+1 as x^2 + y^2 + z^2 + 4*w^2, where x,y,z,w are nonnegative integers with x <= y <= z such that 2^x + 2^y + 2^z + 1 is prime.

%I A291150 #15 Aug 24 2017 19:50:56
%S A291150 1,1,1,1,2,1,2,1,2,2,2,1,3,2,1,1,3,2,3,2,2,1,6,1,3,2,3,2,5,2,3,3,5,2,
%T A291150 5,1,6,5,6,2,6,1,5,1,5,4,8,3,4,2,2,2,8,2,6,4,3,2,4,1,5,4,7,3,7,2,7,4,
%U A291150 5,3,10,1,7,4,5,2,13,4,6,5,5
%N A291150 Number of ways to write 2*n+1 as x^2 + y^2 + z^2 + 4*w^2, where x,y,z,w are nonnegative integers with x <= y <= z such that 2^x + 2^y + 2^z + 1 is prime.
%C A291150 Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 2, 3, 5, 7, 11, 14, 15, 21, 23, 35, 41, 43, 59, 71, 309, 435.
%C A291150 (ii) Any positive even number not divisibly by 8 and other than 6 and 14 can be written as x^2 + y^2 + z^2 + w^2, where w is a positive odd integer, and x,y,z are nonnegative integers with 2^x + 2^y + 2^z + 1 prime.
%C A291150 (iii) Let n be a positive integer. If n is not divisible by 8, then n can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2^x + 2^y + 1 is prime. If n is not a multiple of 2^7 = 128, then we can write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2^x + 2^y - 1 is prime.
%C A291150 (iv) Let n be a positive integer. If n is not divisible by 8, then n can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2^x + 2*2^y + 3*2^z + 4*2^w - 1 is prime. If n is not a multiple of 2^8 = 256, then we can write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2^x + 2*2^y +3*2^z + 4*2^w + 1 is prime.
%C A291150 I have verified that a(n) > 0 for all n = 0..10^7. - _Zhi-Wei Sun_, Aug 23 2017
%H A291150 Zhi-Wei Sun, <a href="/A291150/b291150.txt">Table of n, a(n) for n = 0..10000</a>
%H A291150 Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2016.11.008">Refining Lagrange's four-square theorem</a>, J. Number Theory 175(2017), 167-190.
%H A291150 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1701.05868">Restricted sums of four squares</a>, arXiv:1701.05868 [math.NT], 2017.
%e A291150 a(0) = 1 since 2*0+1 = 0^2 + 0^2 + 1^2 + 4*0^2 with 2^0 + 2^0 + 2^1 + 1 = 5 prime.
%e A291150 a(14) = 1 since 2*14+1 = 2^2 + 3^2 + 4^2 + 4*0^2 with 2^2 + 2^3 + 2^4 + 1 = 29 prime.
%e A291150 a(35) = 1 since 2*35+1 = 1^2 + 3^2 + 5^2 + 4*3^2 with 2^1 + 2^3 + 2^5 + 1 = 43 prime.
%e A291150 a(43) = 1 since 2*43+1 = 1^2 + 5^2 + 5^2 + 4*3^2 with 2^1 + 2^5 + 2^5 + 1 = 67 prime.
%e A291150 a(59) = 1 since 2*59+1 = 1^2 + 3^2 + 3^2 + 4*5^2 with 2^1 + 2^3 + 2^3 + 1 = 19 prime.
%e A291150 a(71) = 1 since 2*71+1 = 1^2 + 5^2 + 9^2 + 4*3^2 with 2^1 + 2^5 + 2^9 + 1 = 547 prime.
%e A291150 a(309) = 1 since 2*309+1 = 5^2 + 13^2 + 13^2 + 4*8^2 with 2^5 + 2^13 + 2^13 + 1 = 16417 prime.
%e A291150 a(435) = 1 since 2*435+1 = 13^2 + 13^2 + 23^2 + 4*1^2 with 2^13 + 2^13 + 2^23 + 1 = 8404993 prime.
%t A291150 SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t A291150 Do[r=0;Do[If[SQ[(2n+1-x^2-y^2-z^2)/4]&&PrimeQ[2^x+2^y+2^z+1],r=r+1],{x,0,Sqrt[(2n+1)/3]},{y,x,Sqrt[(2n+1-x^2)/2]},{z,y,Sqrt[2n+1-x^2-y^2]}];Print[n," ",r],{n,0,80}]
%Y A291150 Cf. A000040, A000079, A000118, A000290, A271518, A281976, A290935, A291191.
%K A291150 nonn
%O A291150 0,5
%A A291150 _Zhi-Wei Sun_, Aug 19 2017