OFFSET
0,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 11, 57, 198, 229, 232, 1168, 2624.
(ii) Any natural number can be written as x^5 + y^5 + z*(3*z+1)/2 + w*(3*w+1)/2, where x,y,z are nonnegative integers and w is an integer.
(iii) For each k = 5, 6, 7, 8, 9, any natural number can be written as x^k + y^5 + z^2 + w*(w+1)/2, where x,y,z,w are nonnegative integers.
(iv) For each b = 2, 4, 5, 7, any natural number can be written as x^5 + b*y^5 + z*(z+1) + w*(w+1)/2 with x,y,z,w nonnegative integers. Also, each n = 0,1,2,... can be written as x^6 + y^5 + z*(z+1)/2 + w*(w+1)/2, where x,y,z,w are nonnegative integers.
(v) Let k be 3 or 4, and let S be the set {x^k + y^2 + z*(z+1)/2: x,y,z = 0,1,2,...} or the set {x^k + y*(y+1)/2 + z*(z+1)/2: x,y,z = 0,1,2,...}. Then, for any positive integer n, either n or n - 1 belongs to the set S.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
EXAMPLE
a(0) = 1 since 0 = 0^5 + 2*0^5 + 0*(3*0-1)/2 + 0*(3*0+1)/2.
a(11) = 1 since 11 = 1^5 + 2*1^5 + 1*(3*1-1)/2 + 2*(3*2+1)/2.
a(57) = 1 since 57 = 0^5 + 2*0^5 + 0*(3*0-1)/2 + 6*(3*6+1)/2.
a(198) = 1 since 198 = 0^5 + 2*1^5 + 7*(3*7-1)/2 + 9*(3*9+1)/2.
a(229) = 1 since 229 = 0^5 + 2*1^5 + 2*(3*2-1)/2 + 12*(3*12+1)/2.
a(232) = 1 since 232 = 1^5 + 2*2^5 + 3*(3*3-1)/2 + 10*(3*10+1)/2.
a(1168) = 1 since 1168 = 3^5 + 2*0^5 + 25*(3*25-1)/2 + 0*(3*0+1)/2.
a(2624) = 1 since 2624 = 0^5 + 2*3^5 + 11*(3*11-1)/2 + 36*(3*36+1)/2.
MATHEMATICA
pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]&&(Mod[Sqrt[24n+1], 6]==1)
Do[r=0; Do[If[pQ[n-x^5-2y^5-z(3z-1)/2], r=r+1], {x, 0, n^(1/5)}, {y, 0, ((n-x^5)/2)^(1/5)}, {z, 0, (Sqrt[24(n-x^5-2y^5)+1]+1)/6}]; Print[n, " ", r]; Continue, {n, 0, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jun 06 2016
STATUS
approved