A271365 Number of ordered ways to write n as u^2 + v^3 + x^4 + y^5 + z^6, where u is a positive integer, and v, x, y, z are nonnegative integers.

1, 4, 6, 5, 5, 6, 4, 1, 2, 7, 9, 6, 4, 3, 1, 1, 6, 12, 10, 4, 3, 3, 1, 1, 6, 12, 11, 7, 6, 4, 2, 4, 9, 12, 8, 5, 10, 12, 6, 2, 5, 9, 8, 8, 10, 6, 2, 3, 8, 13, 10, 8, 11, 8, 2, 1, 6, 10, 8, 7, 6, 2, 2, 7, 15, 20, 14, 9, 13, 11

Offset

1,2

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 8, 15, 16, 23, 24, 56. Moreover, the only positive integers not represented by u^2+v^3+x^4+y^5 (u > 0 and v,x,y >= 0) are 8, 15, 23, 55, 62, 71, 471, 478, 510, 646, 806, 839, 879, 939, 1023, 1063, 1287, 2127, 5135, 6811, 7499, 9191, 26471.

Note that 1/2+1/3+1/4+1/5+1/6 = 29/20 < 3/2.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.

Example

a(1) = 1 since 1 = 1^2 + 0^3 + 0^4 + 0^5 + 0^6.
a(8) = 1 since 8 = 2^2 + 1^3 + 1^4 + 1^5 + 1^6.
a(15) = 1 since 15 = 2^2 + 2^3 + 1^4 + 1^5 + 1^6.
a(16) = 1 since 16 = 4^2 + 0^3 + 0^4 + 0^5 + 0^6.
a(23) = 1 since 23 = 2^2 + 1^3 + 2^4 + 1^5 + 1^6.
a(24) = 1 since 24 = 4^2 + 2^3 + 0^4 + 0^5 + 0^6.
a(56) = 1 since 56 = 4^2 + 2^3 + 0^4 + 2^5 + 0^6.

Mathematica

SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x1^6-x2^5-x3^4-x4^3], r=r+1], {x1, 0, n^(1/6)}, {x2, 0, (n-x1^6)^(1/5)}, {x3, 0, (n-x1^6-x2^5)^(1/4)}, {x4, 0, (n-x1^6-x2^5-x3^4)^(1/3)}]; Print[n, " ", r]; Continue, {n, 1, 70}]

Crossrefs

Cf. A000290, A000578, A000583, A000584, A001014, A262813, A262824, A262827, A262857, A266968, A270488, A270516, A270533, A270559, A270566, A270920, A271026, A271106, A271325.

Sequence in context: A002245 A062117 A200497 * A088559 A092526 A307463

Adjacent sequences: A271362 A271363 A271364 * A271366 A271367 A271368

Keyword

nonn

Author

Zhi-Wei Sun, Apr 05 2016

Status

approved