A257162 Number of ordered ways to write n as floor(x/5) + floor(y/6) + floor(z/7), where x is a pentagonal number, y is a hexagonal number and z is a heptagonal number.

8, 12, 18, 13, 21, 15, 16, 18, 21, 18, 13, 27, 20, 18, 19, 26, 22, 19, 30, 21, 22, 25, 31, 25, 20, 34, 23, 31, 21, 35, 24, 33, 28, 30, 30, 32, 34, 27, 33, 28, 35, 33, 38, 30, 31, 32, 37, 30, 34, 39, 42, 35, 32, 31, 39, 33, 40, 38, 41, 36, 41, 36, 37, 32, 41, 43, 34, 42, 43, 42, 37

Offset

0,1

Comments

Conjecture: a(n) > 0 for all n. Moreover, for any i,j,k = 3,4,5,... all sufficiently large integers can be written as floor(x/i) + floor(y/j) + floor(z/k), where x is an i-gonal number, y is a j-gonal number and z is a k-gonal number.

Note that if {i,j,k} is a subset of {18,19} then 48 cannot be written as floor(x/i) + floor(y/j) + floor(z/k), where x is an i-gonal number, y is a j-gonal number and z is a k-gonal number.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..3000

Zhi-Wei Sun, Natural numbers represented by floor(x^2/a)+floor(y^2/b)+floor(z^2/c), arXiv:1504.01608 [math.NT], 2015.

Example

a(0) = 8 since 0 = floor(x/5) + floor(y/6) + floor(z/7) for any x,y,z in {0,1}, and 0 and 1 are the only m-gonal numbers smaller than m.

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
p[m_, n_]:=p[m, n]=(m-2)*Binomial[n, 2]+n
PQ[m_, n_]:=SQ[8(m-2)n+(m-4)^2]&&(n==0||Mod[Sqrt[8(m-2)n+(m-4)^2]+m-4, 2m-4]==0)
Do[a=0; Do[If[PQ[7, 7(n-Floor[p[5, x]/5]-Floor[p[6, y]/6])+r], a=a+1], {x, 0, (Sqrt[24(5*n+4)+1]+1)/6}, {y, 0, (Sqrt[8*(6*(n-Floor[p[5, x]/5])+5)+1]+1)/4}, {r, 0, 6}];
Print[n, " ", a]; Continue, {n, 0, 70}]

Crossrefs

Cf. A000326, A000384, A000566.

Sequence in context: A175975 A030752 A091523 * A352880 A104201 A036327

Adjacent sequences: A257159 A257160 A257161 * A257163 A257164 A257165

Keyword

nonn

Author

Zhi-Wei Sun, Apr 16 2015

Status

approved