OFFSET
1,5
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0. Also, a(n) = 1 only for those numbers 2^k (k = 0,1,2,...), 4^k*m (k = 0,1,...; m = 3, 7), 4^k*47 (k = 0,1,2,3), 4^k*s (k = 0,1,2; s = 15, 23, 31, 39, 71); 4^k*t (k =0,1; t = 87, 111, 119, 159, 191, 311).
(ii) Any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that a*x^2 + b*y^2 = q^2 for some practical number q, provided that (a,b) is among the ordered pairs (5,16), (7,36), (16,77), (36,55), (36,91), (36, -5), (64,-7).
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II: CANT, New York, NY, USA, 2015 and 2016, Springer Proc. in Math. & Stat., Vol. 220, Springer, New York, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT].)
EXAMPLE
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 and 64*1^2 + 65*0^2 = 8^2 with 8 practical.
a(71) = 1 since 71 = 3^2 + 6^2 + 1^2 + 5^2 and 64*3^2 + 65*6^2 = 54^2 with 54 practical.
a(159) = 1 since 159 = 5^2 + 10^2 + 3^2 + 5^2 and 64*5^2 + 65*10^2 = 90^2 with 90 practical.
a(311) = 1 since 311 = 1^2 + 2^2 + 9^2 + 15^2 and 64*1^2 + 65*2^2 = 18^2 with 18 practical.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
f[n_]:=f[n]=FactorInteger[n];
Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2]);
Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}];
pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0);
pQ[n_]:=pQ[n]=SQ[n]&&pr[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&pQ[64x^2+65y^2], r=r+1], {x, 1, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[(n-x^2-y^2)/2]}]; tab=Append[tab, r], {n, 1, 70}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 14 2017
STATUS
approved