

A271026


Number of ordered ways to write n as x^7 + y^4 + z^3 + w*(3w+1)/2, where x, y, z are nonnegative integers, and w is an integer.


4



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

0,2


COMMENTS

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 47, 61, 62, 112, 175, 448, 573, 714, 1073, 1175, 1839, 2167, 8043, 13844.
(ii) Any natural number can be written as 3*x^6 + y^4 + z^3 + w*(3w+1)/2, where x, y, z are nonnegative integers and w is an integer.
(iii) For every a = 3, 4, 5, 9, 12, any natural number can be written as a*x^5 + y^4 + z^3 + w*(3w+1)/2, where x, y, z are nonnegative integers and w is an integer. Also, any natural number can be written as x^5 + 2*y^4 + 2*z^3 + w*(3w+1)/2 (or 3*x^5 + 2*y^4 + z^3 + w*(3w+1)/2), where x, y, z are nonnegative integers and w is an integer.
We have verified that a(n) > 0 for n up to 2*10^6.
See also A266968 for a related conjecture.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 0..10000
Z.W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103113.
Z.W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), 13671396.


EXAMPLE

a(47) = 1 since 47 = 1^7 + 2^4 + 2^3 + (4)*(3*(4)+1)/2.
a(61) = 1 since 61 = 1^7 + 1^4 + 2^3 + (6)*(3*(6)+1)/2.
a(62) = 1 since 62 = 0^7 + 0^4 + 3^3 + (5)*(3*(5)+1)/2.
a(112) = 1 since 112 = 1^7 + 3^4 + 2^3 + (4)*(3*(4)+1)/2.
a(175) = 1 since 175 = 1^7 + 3^4 + 1^3 + (8)*(3*(8)+1)/2.
a(448) = 1 since 448 = 2^7 + 4^4 + 4^3 + 0*(3*0+1)/2.
a(573) = 1 since 573 = 1^7 + 4^4 + 6^3 + 8*(3*8+1)/2.
a(714) = 1 since 714 = 2^7 + 4^4 + 0^3 + (15)*(3*(15)+1)/2.
a(1073) = 1 since 1073 = 0^7 + 2^4 + 10^3 + 6*(3*6+1)/2.
a(1175) = 1 since 1175 = 0^7 + 5^4 + 5^3 + (17)*(3*(17)+1)/2.
a(1839) = 1 since 1839 = 1^7 + 4^4 + 5^3 + 31*(3*31+1)/2.
a(2167) = 1 since 2167 = 1^7 + 5^4 + 11^3 + (12)*(3*(12)+1)/2.
a(8043) = 1 since 8043 = 1^7 + 2^4 + 20^3 + 4*(3*4+1)/2.
a(13844) = 1 since 13844 = 3^7 + 2^4 + 21^3 + (40)*(3*(40)+1)/2.


MATHEMATICA

pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]
Do[r=0; Do[If[pQ[nx^7y^4z^3], r=r+1], {x, 0, n^(1/7)}, {y, 0, (nx^7)^(1/4)}, {z, 0, (nx^7y^4)^(1/3)}]; Print[n, " ", r]; Continue, {n, 0, 80}]


CROSSREFS

Cf. A000326, A000578, A000583, A000584, A001015, A001318, A262813, A262815, A262816, A262827, A266968, A270469, A270488, A270516, A270533, A270559, A270566, A270920.
Sequence in context: A086775 A132266 A084104 * A093582 A275639 A201940
Adjacent sequences: A271023 A271024 A271025 * A271027 A271028 A271029


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Mar 29 2016


STATUS

approved



