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!)
 A260418 Number of ways to write 12*n+5 as 4*x^4 + 4*y^2 + z^2, where x is a nonnegative integer, and y and z are positive integers. 2
 1, 2, 2, 2, 1, 3, 2, 4, 3, 2, 3, 2, 5, 2, 3, 3, 2, 4, 3, 3, 3, 2, 5, 2, 3, 3, 2, 6, 3, 4, 3, 5, 6, 3, 3, 5, 3, 5, 4, 2, 5, 4, 7, 3, 2, 7, 4, 6, 2, 2, 4, 3, 8, 4, 1, 2, 4, 8, 6, 2, 5, 2, 7, 4, 4, 3, 4, 5, 2, 4, 5, 6, 4, 3, 2, 5, 2, 7, 4, 5, 5, 2, 5, 3, 6, 5, 4, 7, 3, 4, 3, 5, 9, 3, 4, 3, 5, 11, 4, 5, 5 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 4, 54, 159, 289, 999, 1175, 1404, 16391, 39688. (ii) All the numbers 24*n+4 (n = 0,1,2,...) can be written as 3*x^4+24*y^2+z^2, where x,y,z are integers with x > 0 and z > 0. (iii) All the numbers 24*n+13 (n = 0,1,2,...) can be written as 3*x^4+9*y^2+z^2 with x,y,z positive integers. (iv) All the numbers 24*n+r (n = 0,1,2,...) can be written as a*x^4+b*y^2+c*z^2 with x an integer and y and z positive integers, provided that (r,a,b,c) is among the following quadruples: (5,3,1,1), (5,12,4,1), (7,3,6,1), (10,5,1,1), (11,3,8,3), (11,6,3,2), (17,9,4,1). See A290491 for a similar conjecture. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 0..10000 Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), 1367-1396. Zhi-Wei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 2015-2017. Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120. EXAMPLE a(4) = 1 since 12*4+5 = 4*0^4 + 4*1^2 + 7^2. a(54) = 1 since 12*54+5 = 4*0^4 + 4*11^2 + 13^2. a(159) = 1 since 12*159+5 = 4*0^4 + 4*4^2 + 43^2. a(289) = 1 since 12*289+5 = 4*1^4 + 4*19^2 + 45^2. a(999) = 1 since 12*999+5 = 4*7^4 + 4*21^2 + 25^2. a(1175) = 1 since 12*1175+5 = 4*3^4 + 4*55^2 + 41^2. a(1404) = 1 since 12*1404+5 = 4*3^4 + 4*10^2 + 127^2. a(16391) = 1 since 12*16391+5 = 4*5^4 + 4*207^2 + 151^2. a(39688) = 1 since 12*39688+5 = 4*5^4 + 4*50^2 + 681^2. MATHEMATICA SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]; Do[r=0; Do[If[SQ[12n+5-4x^4-4y^2], r=r+1], {x, 0, ((12n+5)/4)^(1/4)}, {y, 1, Sqrt[(12n+5-4x^4)/4]}]; Print[n, " ", r], {n, 0, 100}] CROSSREFS Cf. A000290, A000583, A270566, A286885, A286944, A287616, A290342, A290472, A290491. Sequence in context: A275853 A324100 A127953 * A280919 A209266 A269400 Adjacent sequences:  A260415 A260416 A260417 * A260419 A260420 A260421 KEYWORD nonn AUTHOR Zhi-Wei Sun, Aug 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 August 10 12:39 EDT 2020. Contains 336379 sequences. (Running on oeis4.)