A262815 -id:A262815 - cp's OEIS frontend

A262813 Number of ordered ways to write n as x^3 + y^2 + z*(z+1)/2 with x >= 0, y >=0 and z > 0.

1, 2, 2, 2, 2, 2, 3, 2, 1, 4, 5, 3, 2, 2, 5, 3, 2, 4, 4, 4, 1, 4, 4, 2, 3, 3, 5, 3, 5, 5, 4, 5, 3, 4, 1, 4, 9, 6, 4, 4, 3, 3, 3, 3, 7, 8, 4, 3, 3, 3, 3, 5, 7, 5, 5, 4, 4, 4, 4, 4, 3, 4, 3, 8, 6, 4, 8, 3, 4, 5, 8, 7, 5, 5, 5, 3, 2, 8, 8, 6, 4, 7, 8, 2, 5, 7, 4, 6, 2, 5, 7, 10, 6, 5, 7, 3, 5, 1, 6, 5

Offset: 1

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 21, 35, 98, 152, 306.
This has been verified for all n = 1..2*10^7.
Conjecture verified up to 10^11. - Mauro Fiorentini, Jul 18 2023
If z >= 0, a(n) = 1 only for n = 21, 35, 98, 306. - Mauro Fiorentini, Jul 20 2023
In contrast with the conjecture, in 2015 the author refined a result of Euler by proving that any positive integer can be written as the sum of two squares and a positive triangular number.
See also A262815, A262816 and A262941 for similar conjectures.

			a(1) = 1 since 1 = 0^3 + 0^2 + 1*2/2.
a(2) = 2 since 2 = 0^3 + 1^2 + 1*2/2 = 1^3 + 0^2 + 1*2/2.
a(6) = 2 since 6 = 0^3 + 0^2 + 3*4/2 = 1^3 + 2^2 + 1*2/2.
a(9) = 1 since 9 = 2^3 + 0^2 + 1*2/2.
a(21) = 1 since 21 = 0^3 + 0^2 + 6*7/2.
a(35) = 1 since 35 = 0^3 + 5^2 + 4*5/2.
a(98) = 1 since 98 = 3^3 + 4^2 + 10*11/2.
a(152) = 1 since 152 = 0^3 + 4^2 + 16*17/2.
a(306) = 1 since 306 = 1^3 + 13^2 + 16*17/2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
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.

Cf. A000217, A000290, A000578, A254885, A262785, A262815, A262816, A262941.

TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^3-y^2],r=r+1],{x,0,n^(1/3)},{y,0,Sqrt[n-x^3]}];Print[n," ",r];Continue,{n,1,100}]

A262827 Number of ordered ways to write n as w^2 + x^3 + y^4 + 2*z^4, where w, x, y and z are nonnegative integers.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: a(n) > 0 for all n >= 0. Also, a(n) = 1 only for the following 41 values of n: 0, 14, 15, 21, 22, 23, 62, 71, 78, 87, 136, 216, 405, 437, 448, 477, 535, 583, 591, 623, 671, 696, 885, 950, 1046, 1135, 1206, 1208, 1248, 1317, 2288, 2383, 2543, 3167, 3717, 3974, 6847, 7918, 8328, 9096, 21935.
We have verified that a(n) > 0 for all n = 0..10^7.
Conjecture verified up to 10^11. - Mauro Fiorentini, Jul 07 2023
We also conjecture that if f(w,x,y,z) is one of the 8 polynomials 2w^2+x^3+4y^3+z^4, w^2+x^3+2y^3+c*z^3 (c = 3,4,5,6) and w^2+x^3+2y^3+d*z^4 (d = 1,3,6) then each n = 0,1,2,... can be written as f(w,x,y,z) with w,x,y,z nonnegative integers. - Zhi-Wei Sun, Dec 30 2017
Conjecture verified up to 10^11 for all 8 polynomials. - Mauro Fiorentini, Jul 07 2023

			a(14) = 1 since 14 = 2^2 + 2^3 + 0^4 + 2*1^4.
a(87) = 1 since 87 = 2^2 + 0^3 + 3^4 + 2*1^4.
a(216) = 1 since 216 = 0^2 + 6^3 + 0^4 + 2*0^4.
a(405) = 1 since 405 = 18^2 + 0^3 + 3^4 + 2*0^4.
a(1248) = 1 since 1248 = 31^2 + 5^3 + 0^4 + 2*3^4.
a(1317) = 1 since 1317 = 23^2 + 1^3 + 5^4 + 2*3^4.
a(2288) = 1 since 2288 = 44^2 + 4^3 + 4^4 +2*2^4.
a(2383) = 1 since 2383 = 1462 + 9^3 + 6^4 + 2*3^4.
a(2543) = 1 since 2543 = 50^2 + 3^3 + 2^4 + 2*0^4.
a(3167) = 1 since 3167 = 54^2 + 2^3 + 3^4 + 2*3^4.
a(3717) = 1 since 3717 = 18^2 + 15^3 + 2^4 + 2*1^4.
a(3974) = 1 since 3974 = 39^2 + 13^3 + 4^4 + 2*0^4.
a(6847) = 1 since 6847 = 52^2 + 15^3 + 4^4 + 2*4^4.
a(7918) = 1 since 7918 = 46^2 + 10^3 + 0^4 + 2*7^4.
a(8328) = 1 since 8328 = 42^2 + 1^3 + 9^4 + 2*1^4.
a(9096) = 1 since 9096 = 44^2 + 18^3 + 6^4 + 2*2^4.
a(21935) = 1 since 21935 = 66^2 + 26^3 + 1^4 + 2*1^4.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. (See Remark 1.1.)
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120. (See Theorem 1.1 and Conjecture 1.1.)

Cf. A000290, A000578, A000583, A262813, A262815, A262816, A262824.

SQ[n_]:=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^3-y^4-2*z^4],r=r+1],{x,0,n^(1/3)},{y,0,(n-x^3)^(1/4)},{z,0,((n-x^3-y^4)/2)^(1/4)}];Print[n," ",r];Continue,{n,0,100}]

A262816 Number of ordered ways to write n as x^3 + y^2 + z(3z-1)/2, where x and y are nonnegative integers, and z is a nonzero integer.

Original entry on oeis.org

1, 3, 3, 1, 2, 4, 3, 2, 3, 4, 4, 3, 4, 4, 3, 5, 5, 3, 4, 2, 3, 4, 4, 6, 2, 4, 6, 4, 5, 4, 6, 6, 3, 4, 5, 5, 4, 8, 6, 5, 5, 4, 7, 5, 5, 3, 2, 6, 5, 5, 8, 8, 4, 3, 4, 4, 6, 6, 8, 3, 4, 6, 3, 5, 7, 9, 6, 5, 6, 6, 8, 6, 4, 6, 6, 6, 7, 9, 9, 5, 4, 6, 7, 6, 6, 6, 11, 5, 4, 7, 5, 5, 7, 11, 4, 6, 4, 5, 3, 6

Offset: 1

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 4, 216.
(ii) Any positive integer can be written as x^3 + y*(3*y-1)/2 + z*(3*z-1)/2, where x and y are nonnegative integers, and z is a nonzero integer.
Conjectures (i) and (ii) verified for n up to 10^9. - Mauro Fiorentini, Jul 21 2023
See also A262813 and A262815 for similar conjectures.
By Theorem 1.7(ii) in the linked paper, any nonnegative integer can be written as x^2 + y^2 + z*(3*z-1)/2, where x, y and z are integers.

			a(1) = 1 since 1 = 0^3 + 0^2 + 1*(3*1-1)/2.
a(4) = 1 since 4 = 1^3 + 1^2 + (-1)*(3*(-1)-1)/2.
a(8) = 2 since 8 = 0^3 + 1^2 + (-2)*(3*(-2)-1)/2 = 1^3 + 0^2 + (-2)*(3*(-2)-1)/2.
a(216) = 1 since 216 = 2^3 + 14^2 + 3*(3*3-1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Cf. A000290, A000326, A000578, A001318, A160325, A160326, A262813, A262815.

PenQ[n_]:=n>0&&IntegerQ[Sqrt[24n+1]]
Do[r=0;Do[If[PenQ[n-x^3-y^2],r=r+1],{x,0,n^(1/3)},{y,0,Sqrt[n-x^3]}];Print[n," ",r];Continue,{n,1,100}]

A270488 Number of ordered ways to write n = x^2 + y(y+1) + z(z^2+1), where x, y and z are nonnegative integers.

Original entry on oeis.org

1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 3, 3, 3, 2, 2, 4, 2, 3, 1, 3, 3, 3, 3, 2, 2, 3, 2, 2, 2, 4, 6, 3, 3, 3, 1, 5, 3, 4, 4, 3, 4, 3, 2, 3, 3, 6, 2, 5, 2, 2, 5, 3, 3, 1, 4, 4, 4, 5, 3, 3, 5, 1, 1, 2, 3, 7, 4, 5, 4, 3, 3, 6, 2, 5, 4, 6, 2, 5, 4, 3

Offset: 0

Zhi-Wei Sun, Mar 17 2016

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 5, 7, 19, 35, 54, 62, 63, 197, 285, 339, 479, 505, 917. Moreover, any integer n > 2 can be written as x^2 + y*(y+1) + z*(z^2+1), where x is a positive integer, and y and z are nonnegative integers.
We also guess that each n = 0,1,2,... can be expressed as x*(x+1)/2 + P(y,z) with x, y and z nonnegative integers, where P(y,z) is any of the polynomials y(y+1) + z^2*(z+1), y^2 + z*(z^2+2), y^2 + z*(z^2+7), y^2 + z*(z^2+z+2), y^2 + z*(z^2+2z+3), y^2 + z*(2z^2+z+1).
It is known that every n = 0,1,2,... can be written as x^2 + y*(y+1) + z*(z+1), where x, y and z are nonnegative integers.

			a(35) = 1 since 35 = 5^2 + 0*1 + 2*(2^2+1).
a(54) = 1 since 54 = 2^2 + 4*5 + 3*(3^2+1).
a(62) = 1 since 62 = 2^2 + 7*8 + 1*(1^2+1).
a(63) = 1 since 63 = 7^2 + 3*4 + 1*(1^2+1).
a(197) = 1 since 197 = 5^2 + 6*7 + 5*(5^2+1).
a(285) = 1 since 285 = 15^2 + 5*6 + 3*(3^2+1).
a(339) = 1 since 339 = 17^2 + 4*5 + 3*(3^2+1).
a(479) = 1 since 479 = 7^2 + 20*21 + 2*(2^2+1).
a(505) = 1 since 505 = 13^2 + 17*18 + 3*(3^2+1).
a(917) = 1 since 917 = 15^2 + 18*19 + 7*(7^2+1).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
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.

Cf. A000217, A000290, A000578, A002378, A002522, A262813, A262815, A262816, A270469.

SQ[x_]:=SQ[x]=IntegerQ[Sqrt[x]]
Do[r=0;Do[If[SQ[n-y(y+1)-z(z^2+1)],r=r+1],{y,0,(Sqrt[4n+1]-1)/2},{z,0,(n-y(y+1))^(1/3)}];Print[n," ",r];Continue,{n,0,80}]

A270566 Number of ordered ways to write n as x^4 + y(3y+1)/2 + z(7z+1)/2, where x, y and z are integers with x nonnegative.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 19 2016

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 24, 47, 63, 78, 79, 143, 153, 325, 494, 949, 1079, 3328, 4335, 5609, 7949, 7967, 8888, 9665.

Conjecture verified for n up to 10^11. - Mauro Fiorentini, Jul 24 2023

			a(24) = 1 since 24 = 2^4 + (-2)*(3*(-2)+1)/2 + (-1)*(7*(-1)+1)/2.
a(78) = 1 since 78 = 1^4 + 7*(3*7+1)/2 + 0*(7*0+1)/2.
a(143) = 1 since 143 = 1^4 + 6*(3*6+1)/2 + (-5)*(7*(-5)+1)/2.
a(494) = 1 since 494 = 4^4 + (-7)*(3*(-7)+1)/2 + (-7)*(7*(-7)+1)/2.
a(949) = 1 since 949 = 4^4 + 0*(3*0+1)/2 + 14*(7*14+1)/2.
a(1079) = 1 since 1079 = 0^4 + 25*(3*25+1)/2 + 6*(7*6+1)/2.
a(3328) = 1 since 3328 = 0^4 + 38*(3*38+1)/2 + 18*(7*18+1)/2.
a(4335) = 1 since 4335 = 2^4 + 49*(3*49+1)/2 + 14*(7*14+1)/2.
a(5609) = 1 since 5609 = 0^4 + (-61)*(3*(-61)+1)/2 + 4*(7*4+1)/2.
a(7949) = 1 since 7949 = 3^4 + 43*(3*43+1)/2 + 38*(7*38+1)/2.
a(7967) = 1 since 7967 = 7^4 + (-61)*(3*(-61)+1)/2 + 2*(7*2+1)/2.
a(8888) = 1 since 8888 = 0^4 + (-77)*(3*(-77)+1)/2 + 3*(7*3+1)/2.
a(9665) = 1 since 9665 = 3^4 + 73*(3*73+1)/2 + 21*(7*21+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
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.

Cf. A001318, A000583, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559.

(* From Zhi-Wei Sun, Start *)
pQ[n_] := pQ[n] = IntegerQ[Sqrt[24 n + 1]];
Do[r = 0; Do[If[pQ[n - x^4 - y (7 y + 1)/2], r = r + 1], {x, 0, n^(1/4)}, {y, -Floor[(Sqrt[56 (n - x^4) + 1] + 1)/14], (Sqrt[56 (n - x^4) + 1] - 1)/14}]; Print[n, " ", r]; Continue, {n, 0, 80}]
(* From Zhi-Wei Sun, End *)
A270566[n_] := Length@Solve[x >= 0 && n == x^4 + y*(3 y + 1)/2 + z*(7 z + 1)/2, {x, y, z}, Integers];
Array[A270566, 25, 0] (* JungHwan Min, Mar 19 2016 *)

A270533 Number of ordered ways to write n = x^4 + x^3 + y^2 + z*(3z-1)/2, where x and y are nonnegative integers, and z is an integer.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 18 2016

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 20, 22, 70, 87, 167, 252, 388, 562, 636, 658, 873, 2598, 14979, 18892, 20824.
(ii) Each n = 0,1,2,... can be written as x*P(x) + y^2 + z*(3z-1)/2 with x and y nonnegative integers, and z an integer, where P(x) is either of the polynomials x^3+2, x^3+3, x^3+2x+8, x^3+x^2+4x+2, x^3+x^2+7x+6.
(iii) Any nonnegative integer can be expressed as x*(x^3+3) + y*(5y+4) + z*(3z-1)/2, where x is an nonnegative integer, and y and z are integers.
See also A270516 for a similar conjecture.

			a(20) = 1 since 20 = 1^4 + 1^3 + 4^2 + (-1)*(3*(-1)-1)/2.
a(22) = 1 since 22 = 0^4 + 0^3 + 0^2 + 4*(3*4-1)/2.
a(873) = 1 since 873 = 5^4 + 5^3 + 11^2 + (-1)*(3*(-1)-1)/2.
a(2598) = 1 since 2598 =  4^4 + 4^3 + 4^2 + 39*(3*39-1)/2.
a(14979) = 1 since 14979 = 1^4 + 1^3 + 51^2 + 91*(3*91-1)/2.
a(18892) = 1 since 18892 = 3^4 + 3^3 + 137^2 + (-3)*(3*(-3)-1)/2.
a(20824) = 1 since 20824 = 1^4 + 1^3 + 115^2 + (-71)*(3*(-71)-1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Cf. A000290, A000578, A000583, A001318, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516.

pQ[x_]:=pQ[x]=IntegerQ[Sqrt[24x+1]]
Do[r=0;Do[If[pQ[n-y^2-x^3*(x+1)],r=r+1],{y,0,Sqrt[n]},{x,0,(n-y^2)^(1/4)}];Print[n," ",r];Continue,{n,0,90}]

A270469 Number of ordered ways to write n = x^3 + y(y+1) + z(3*z+2), where x and y are nonnegative integers and z is a nonzero integer.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 3, 1, 2, 2, 3, 2, 2, 3, 2, 1, 4, 4, 2, 2, 2, 2, 1, 5, 4, 2, 2, 2, 3, 4, 4, 5, 2, 2, 3, 3, 5, 2, 5, 3, 2, 4, 5, 4, 2, 3, 3, 3, 3, 4, 3, 1, 2, 5, 3, 4, 3, 4, 4, 4, 5, 4, 3, 4, 4, 3, 6, 5, 5, 3, 3, 3, 6, 6, 2, 4

Offset: 1

Zhi-Wei Sun, Mar 17 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 3, 4, 5, 6, 10, 12, 20, 27, 56, 101.
(ii) Each nonnegative integer can be written as x^3 + P(y,z) with x >= 0 and y and z integers, provided that P(y,z) is among y^2+z(3z+1), y(2y+1)+z(3z+1), y(2y+1)+z(3z+2), y(2y+1)+z(5z+2), y(2y+1)+z(5z+3), y(2y+1)+2z(3z+1), y(2y+1)+ 2z(3z+2), y(2y+1)+z(6z+5), y(2y+1)+z(7z+2), y(2y+1)+z(7z+6), y(3y+1)+z(4z+1), y(3y+1)+z(7z+2), y(3y+2)+z(4z+1).
(iii) Every n = 0,1,2,... can be expressed as f(x,y,z) with x >= 0 and y and z integers, provided that f(x,y,z) is among 2x^3+y^2+z(3z+1), 2x^3+y(y+1)+z(3z+2), 3x^3+y(y+1)+z(3z+2), 2x^3+y(2y+1)+z(3z+1), 2x^3+y(2y+1)+z(3z+2), 2x^3+y(2y+1)+z(5z+4), 2x^3+y(3y+1)+z(3z+2), 2x^3+y(3y+2)+z(4z+3).
Note that those y(2y+1) with y integral are just triangular numbers.
See also A262813 for a similar conjecture.

			a(10) = 1 since 10 = 0^3 + 1*2 + (-2)(3*(-2)+2).
a(12) = 1 since 12 = 1^3 + 2*3 + 1*(3*1+2).
a(20) = 1 since 20 = 0^3 + 3*4 + (-2)*(3*(-2)+2).
a(27) = 1 since 27 = 0^3 + 2*3 + (-3)*(3*(-3)+2).
a(56) = 1 since 56 = 0^3 + 0*1 + 4*(3*4+2).
a(101) = 1 since 101 = 2^3 + 8*9 + (-3)*(3*(-3)+2).

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

Cf. A000217, A000290, A000578, A001082, A002378, A262813, A262815, A262816.

pQ[x_]:=pQ[x]=x>0&&IntegerQ[Sqrt[3x+1]]
Do[r=0;Do[If[pQ[n-x^3-y(y+1)],r=r+1],{x,0,n^(1/3)},{y,0,(Sqrt[4(n-x^3)+1]-1)/2}];Print[n," ",r];Continue,{n,1,80}]

A270559 Number of ordered ways to write n as x^4 + x^3 + y^2 + z*(z+1)/2, where x, y and z are integers with x nonzero, y nonnegative and z positive.

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 3, 1, 3, 4, 2, 5, 2, 3, 4, 2, 3, 4, 5, 1, 4, 3, 3, 4, 3, 4, 5, 5, 3, 6, 5, 3, 3, 6, 2, 4, 6, 3, 9, 4, 2, 3, 4, 3, 7, 6, 3, 6, 2, 4, 2, 6, 5, 7, 6, 4, 5, 3, 6, 4, 11, 1, 5, 9, 3, 6, 5, 3, 8, 8

Offset: 1

Zhi-Wei Sun, Mar 18 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0. In other words, for each n = 1,2,3,... there are integers x and y such that n-(x^4+x^3+y^2) is a positive triangular number.
(ii) a(n) = 1 only for n = 1, 2, 8, 20, 62, 97, 296, 1493, 4283, 4346, 5433.
In contrast, the author conjectured in A262813 that any positive integer can be expressed as the sum of a nonnegative cube, a square and a positive triangular number.

			a(1) = 1 since 1 = (-1)^4 + (-1)^3 + 0^2 + 1*2/2.
a(2) = 1 since 2 = (-1)^4 + (-1)^3 + 1^2 + 1*2/2.
a(8) = 1 since 8 = 1^4 + 1^3 + 0^2 + 3*4/2.
a(20) = 1 since 20 = (-2)^4 + (-2)^3 + 3^2 + 2*3/2.
a(62) = 1 since 62 = (-2)^4 + (-2)^3 + 3^2 + 9*10/2.
a(97) = 1 since 97 = 1^4 + 1^3 + 2^2 + 13*14/2.
a(296) = 1 since 296 = (-4)^4 + (-4)^3 + 7^2 + 10*11/2.
a(1493) = 1 since 1493 = (-2)^4 + (-2)^3 + 0^2 + 54*55/2.
a(4283) = 1 since 4283 = (-6)^4 + (-6)^3 + 50^2 + 37*38/2.
a(4346) = 1 since 4346 = (-3)^4 + (-3)^3 + 49^2 + 61*62/2.
a(5433) = 1 since 5433 = (-8)^4 + (-8)^3 + 14^2 + 57*58/2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Cf. A000217, A000290, A000578, A000583, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533.

TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[x!=0&&TQ[n-y^2-x^4-x^3],r=r+1],{y,0,Sqrt[n]},{x,-1-Floor[(n-y^2)^(1/4)],(n-y^2)^(1/4)}];Print[n," ",r];Continue,{n,1,10000}]

A270516 Number of ordered ways to write n = x^3(x+1) + y(y+1)/2 + z*(3z+2), where x and y are nonnegative integers, and z is an integer.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 18 2016

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and the only values of n > 1428 with a(n) = 1 are 2205, 2259, 3556, 4107, 4337, 5387, 9331, 16561, 22237, 27569, 63947, 78610.
(ii) Any natural number can be written as x*(x^3+2) + y*(y+1)/2 + z*(3z+1), where x and y are nonnegative integers, and z is an integer.
(iii) Every n = 0,1,2,... can be written as x*(x^3+x^2+6) + y*(y+1)/2 + z*(3z+2) (or x*(x^3+x^2+4x+1) + y*(y+1)/2 + z*(3z+1)), where x and y are nonnegative integers, and z is an integer.
See also A270533 for a similar conjecture.

			a(72) = 1 since 72 = 2^3*3 + 5*6/2 + 3*(3*3+2).
a(75) = 1 since 75 = 0^3*1 + 4*5/2 + (-5)*(3*(-5)+2).
a(5387) = 1 since 5387 = 7^3*8 + 2*3/2 + (-30)*(3*(-30)+2).
a(9331) = 1 since 9331 = 8^3*9 + 2*3/2 + (-40)*(3*(-40)+2).
a(16561) = 1 since 16561 = 1^3*2 + 101*102/2 + (-62)*(3*(-62)+2).
a(22237) = 1 since 22237 = 6^3*7 + 104*105/2 + 71*(3*71+2).
a(27569) = 1 since 27569 = 2^3*3 + 49*50/2 + (-94)*(3*(-94)+2).
a(63947) = 1 since 63947 = 0^3*1 + 173*174/2 + (-128)*(3*(-128)+2).
a(78610) = 1 since 78610 = 16^3*17 + 52*53/2 + 50*(3*50+2).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.

Cf. A000217, A000578, A001082, A001318, A262813, A262815, A262816, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270533.

OQ[x_]:=OQ[x]=IntegerQ[Sqrt[3x+1]]
Do[r=0;Do[If[OQ[n-y(y+1)/2-x^3*(x+1)],r=r+1],{y,0,(Sqrt[8n+1]-1)/2},{x,0,(n-y(y+1)/2)^(1/4)}];Print[n," ",r];Continue,{n,0,80}]

A262944 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a pentagonal number.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 4, 3, 1, 3, 5, 3, 2, 2, 5, 5, 3, 3, 5, 5, 3, 6, 6, 3, 3, 8, 6, 5, 5, 3, 7, 5, 5, 3, 4, 4, 8, 9, 3, 5, 7, 6, 3, 5, 5, 7, 5, 3, 4, 5, 6, 6, 9, 4, 5, 7, 7, 5, 4, 4, 7, 6, 1, 5, 5, 7, 7, 7, 1, 6, 10, 8, 6, 3, 4, 3, 6, 4, 6, 9, 5, 7, 9, 3, 5, 8, 9, 8, 3, 3, 11, 10, 6, 6, 8, 12, 5, 6, 4, 7

Offset: 1

Zhi-Wei Sun, Oct 05 2015

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 63, 69, 489, 714, 1089.
(ii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is twice a square or twice a pentagonal number.
(iii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - 2*x^4 - y*(y+1)/2 is a square or a pentagonal number.
See also A262941 and A262945 for similar conjectures.

			a(1) = 1 since 1 = 0^4 + 1*2/2 + p_5(0), where p_5(n) denotes the pentagonal number n*(3*n-1)/2.
a(9) = 1 since 9 = 1^4 + 2*3/2 + p_5(2).
a(63) = 1 since 63 = 0^4 + 7*8/2 + p_5(5).
a(69) = 1 since 69 = 2^4 + 7*8/2 + 5^2.
a(489) = 1 since 489 = 3^4 + 12*13/2 + p_5(15).
a(714) = 1 since 714 = 4^4 + 18*19/2 + p_5(14).
a(1089) = 1 since 1089 = 4^4 + 38*39/2 + p_5(8).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Cf. A000217, A000290, A000583, A001318, A160325, A254885, A262813, A262815, A262816, A262941, A262945.

SQ[n_]:=IntegerQ[Sqrt[n]]||(IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1]+1,6]==0)
Do[r=0;Do[If[SQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]

cp's OEIS Frontend

A262813 Number of ordered ways to write n as x^3 + y^2 + z*(z+1)/2 with x >= 0, y >=0 and z > 0.

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A262827 Number of ordered ways to write n as w^2 + x^3 + y^4 + 2*z^4, where w, x, y and z are nonnegative integers.

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A262816 Number of ordered ways to write n as x^3 + y^2 + z*(3*z-1)/2, where x and y are nonnegative integers, and z is a nonzero integer.

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A270488 Number of ordered ways to write n = x^2 + y*(y+1) + z*(z^2+1), where x, y and z are nonnegative integers.

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A270566 Number of ordered ways to write n as x^4 + y*(3y+1)/2 + z*(7z+1)/2, where x, y and z are integers with x nonnegative.

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A270533 Number of ordered ways to write n = x^4 + x^3 + y^2 + z*(3z-1)/2, where x and y are nonnegative integers, and z is an integer.

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A270469 Number of ordered ways to write n = x^3 + y*(y+1) + z*(3*z+2), where x and y are nonnegative integers and z is a nonzero integer.

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A270559 Number of ordered ways to write n as x^4 + x^3 + y^2 + z*(z+1)/2, where x, y and z are integers with x nonzero, y nonnegative and z positive.

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A262816 Number of ordered ways to write n as x^3 + y^2 + z(3z-1)/2, where x and y are nonnegative integers, and z is a nonzero integer.

A270488 Number of ordered ways to write n = x^2 + y(y+1) + z(z^2+1), where x, y and z are nonnegative integers.

A270566 Number of ordered ways to write n as x^4 + y(3y+1)/2 + z(7z+1)/2, where x, y and z are integers with x nonnegative.

A270469 Number of ordered ways to write n = x^3 + y(y+1) + z(3*z+2), where x and y are nonnegative integers and z is a nonzero integer.

A270516 Number of ordered ways to write n = x^3(x+1) + y(y+1)/2 + z*(3z+2), where x and y are nonnegative integers, and z is an integer.