The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A282013 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and 49*x + 48*(y-z) are squares. 9
 1, 2, 3, 3, 2, 1, 3, 2, 1, 4, 4, 2, 2, 2, 1, 2, 2, 4, 8, 4, 3, 2, 3, 2, 3, 5, 5, 7, 3, 2, 5, 1, 3, 7, 6, 5, 5, 3, 5, 3, 2, 3, 9, 5, 2, 6, 3, 1, 3, 5, 5, 10, 6, 2, 8, 4, 3, 5, 6, 3, 3, 3, 4, 4, 2, 5, 9, 8, 5, 4, 6, 1, 5, 6, 5, 9, 2, 3, 7, 1, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 5, 8, 14, 31, 47, 71, 79, 143, 248, 463, 1039). The author has proved that any nonnegative integer can be written as the sum of a fourth power and three squares. We have verified a(n) > 0 for all n = 0..10^7. See also A281976, A281977 and A282014 for similar conjectures. Qing-Hu Hou at Tianjin University verified a(n) > 0 for n up to 10^9. - Zhi-Wei Sun, Jun 02 2019 LINKS Zhi-Wei Sun, Table of n, a(n) for n = 0..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. EXAMPLE a(5) = 1 since 5 = 1^2 + 0^2 + 0^2 + 2^2 with 1 = 1^2 and 49*1 + 48*(0-0) = 7^2. a(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^2 with 0 = 0^2 and 49*0 + 48*(2-2) = 0^2. a(14) = 1 since 14 = 1^2 + 2^2 + 3^2 + 0^2 with 1 = 1^2 and 49*1 + 48*(2-3) = 1^2. a(31) = 1 since 31 = 1^2 + 1^2 + 2^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(1-2) = 1^2. a(47) = 1 since 47 = 1^2 + 6^2 + 1^2 + 3^2 with 1 = 1^2 and 49*1 + 48*(6-1) = 17^2. a(71) = 1 since 71 = 1^2 + 5^2 + 6^2 + 3^2 with 1 = 1^2 and 49*1 + 48*(5-6) = 1^2. a(79) = 1 since 79 = 1^2 + 7^2 + 2^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(7-2) = 17^2. a(143) = 1 since 143 = 1^2 + 5^2 + 6^2 + 9^2 with 1 = 1^2 and 49*1 + 48*(5-6) = 1^2. a(248) = 1 since 248 = 4^2 + 6^2 + 0^2 + 14^2 with 4 = 2^2 and 49*4 + 48*(6-0) = 22^2. a(463) = 1 since 463 = 9^2 + 6^2 + 15^2 + 11^2 with 9 = 3^2 and 49*9 + 48*(6-15) = 3^2. a(1039) = 1 since 1039 = 1^2 + 22^2 + 23^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(22-23) = 1^2. MATHEMATICA SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]; Do[r=0; Do[If[SQ[n-x^4-y^2-z^2]&&SQ[49x^2+48(y-z)], r=r+1], {x, 0, n^(1/4)}, {y, 0, Sqrt[n-x^4]}, {z, 0, Sqrt[n-x^4-y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}] CROSSREFS Cf. A000118, A000290, A270969, A271518, A281939, A281941, A281975, A281976, A281977, A282014. Sequence in context: A132312 A090431 A107336 * A275299 A205100 A156613 Adjacent sequences:  A282010 A282011 A282012 * A282014 A282015 A282016 KEYWORD nonn AUTHOR Zhi-Wei Sun, Feb 04 2017 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified May 24 17:48 EDT 2020. Contains 334574 sequences. (Running on oeis4.)