A337218 The positive integers uniquely represented by the ternary form x^2 + 2*y^2 + 2*z^2, with integers x <= 0, and 0 <= y <= z.

1, 2, 3, 5, 6, 10, 12, 13, 14, 21, 22, 30, 37, 42, 46, 48, 58, 70, 78, 93, 133, 142, 190, 192, 253, 768, 3072, 12288, 49152, 196608, 786432, 3145728, 12582912, 50331648, 201326592, 805306368, 3221225472, 12884901888

Offset: 1

Author

Wolfdieter Lang, Aug 20 2020

Keywords

• nonn
• easy

Comments

This sequence gives Theorem 2.2. of Kaplansky, p. 88, with a proof on p. 90.

This sequence is composed of two finite ones and an infinite one: (i) 2*A337217 = {2, 6, 10, 14, 22, 30, 42, 46, 58, 70, 78, 142, 190}, the even members of A094739, (ii) {1, 5, 13, 21, 37, 93, 133, 253}, the 1 (mod 4) members of A094739, and (iii) A002001(k+1) = 4^k*3, for integer k >= 0. Beginning with a(26) = 768 only the powers 4^k*3, for k >= 4 appear.

See eq. (2.2), (2,4), p. 87, of Kaplansky for the two finite sequences with 13 and 8 members, respectively.

The positive integers which have no such solution (x, y, z) are given by 4^k*(7+8*m) = A002001(k+1)*A004771(m), for k >= 0 and m >= 0. See Kaplansky, p. 88. The other missing positive integers have more than 1 solution.

Examples

			4 is not a member because (x, y, z) = (0, 1, 1) and (2, 0, 0) give both 4.
3 is a member with one solution (1, 0, 1).
5 is a member with one solutuion (1, 1, 1).
7 is not a member because there is no solution.
11 is not a member because there are two solutions (1, 1, 2) and (3, 0, 1).
		

References

• Irving Kaplansky, Integers Uniquely Represented by Certain Ternary Forms, in "The Mathematics of Paul Erdős I", Ronald. L. Graham and Jaroslav Nešetřil (Eds.), Springer, 1997, pp. 86 - 94.

Crossrefs

Cf. A002001, A004771, A094739.

Formula

See the comment for the union of the three sequences (i), (ii) and (iii).