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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 (list; graph; refs; listen; history; text; internal format)
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 a-file. 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

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Zhi-Wei 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

Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Zhi-Wei 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), 275-283.

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, Riddles of Representations of Integers, presentation to Nanjing Normal Univ. (China, 2019).

Hai-Liang Wu and Zhi-Wei 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(n-x^2-15y^2)+1], r=r+1], {x, 0, Sqrt[n]}, {y, 0, Sqrt[(n-x^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

Zhi-Wei Sun, May 16 2017

STATUS

approved

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Last modified June 25 03:37 EDT 2021. Contains 345449 sequences. (Running on oeis4.)