

A286944


Number of ways to write n as x^2 + 15*y^2 + z*(3z+1)/2, where x and y are nonnegative integers and z is an integer.


7



1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 6, 3, 2, 2, 2, 4, 2, 3, 3, 2, 5, 3, 2, 1, 3, 6, 2, 1, 1, 2, 4, 3, 4, 2, 3, 5, 3, 2, 2, 2, 2, 2, 1, 2, 3, 7, 3, 2, 2, 3, 4, 2, 3, 2, 3, 4, 4, 2, 5, 5, 9, 3, 1, 4, 3, 8, 2, 4, 2, 4, 9, 3, 2, 5, 2
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OFFSET

0,2


COMMENTS

Conjecture: a(n) > 0 for any nonnegative integer n, and a(n) = 1 only for n = 0, 3, 4, 7, 8, 10, 12, 13, 14, 29, 33, 34, 48, 68, 113, 129, 220.
Let a,b,c,d,e,f be nonnegative integers with a > b, c > d, e > f, a == b (mod 2), c == d (mod 2), e == f (mod 2), a >= c >= e >= 2, b >= d if a = c, and d >= f if c = e. We have shown in arXiv:1502.03056 that if the ordered tuple (a,b,c,d,e,f) is universal, i.e., each n = 0,1,2,... can be written as x(ax+b)/2 + y(cy+d)/2 + z(ez+f)/2 with x,y,z integers, then (a,b,c,d,e,f) must be among the 12082 tuples listed in the linked afile. We also conjecture that all the listed tuples not yet proved to be universal are indeed universal. Note that those numbers x(4x+2)/2 with x integral coincide with triangular numbers.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 0..10000
ZhiWei Sun, List of those tuples (a,b,c,d,e,f) for which each nonnegative integer should be represented by x(ax+b)/2 + y(cy+d)/2 + z(ez+f)/2 with x,y,z integers
ZhiWei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 13671396.
ZhiWei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275283.
ZhiWei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 20152017.
ZhiWei Sun, Riddles of Representations of Integers, presentation to Nanjing Normal Univ. (China, 2019).
HaiLiang Wu and ZhiWei Sun, Some universal quadratic sums over the integers, arXiv:1707.06223 [math.NT], 2017.


EXAMPLE

a(33) = 1 since 33 = 4^2 + 15*1^2 + 1*(3*1+1)/2.
a(34) = 1 since 34 = 2^2 + 15*1^2 + 3*(3*3+1)/2.
a(48) = 1 since 48 = 6^2 + 15*0^2 + (3)*(3*(3)+1)/2.
a(68) = 1 since 68 = 1^2 + 15*2^2 + 2*(2*3+1)/2.
a(113) = 1 since 113 = 6^2 + 15*0^2 + 7*(3*7+1)/2.
a(129) = 1 since 129 = 8^2 + 15*2^2 + (2)*(3*(2)+1)/2.
a(220) = 1 since 220 = 13^2 + 15*0^2 + 6*(3*6+1)/2.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]; Do[r=0; Do[If[SQ[24(nx^215y^2)+1], r=r+1], {x, 0, Sqrt[n]}, {y, 0, Sqrt[(nx^2)/15]}]; Print[n, " ", r], {n, 0, 80}]


CROSSREFS

Cf. A000290, A001318, A287616.
Sequence in context: A332384 A275885 A199010 * A026605 A331104 A270646
Adjacent sequences: A286941 A286942 A286943 * A286945 A286946 A286947


KEYWORD

nonn


AUTHOR

ZhiWei Sun, May 16 2017


STATUS

approved



