A270488 -id:A270488 - cp's OEIS frontend

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.

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}]

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}]

A270920 Number of ordered ways to write n as the sum of a positive triangular number, a positive square, and a fifth power whose absolute value does not exceed n.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Mar 25 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 112, 770, 801, 1593, 1826, 2320, 2334, 2849, 7561.

(ii) Let T(x) = x*(x+1)/2 and pen(x) = x*(3x+1)/2. Any natural number n can be written as P(x,y) + z^5, where x, y and z are integers with |z^5| <= n, and the polynomial P(x,y) is either of the following ones: T(x)+2*T(y), T(x)+2*pen(y), x^2+pen(y), x^2+y(5y+1)/2, 2*T(x)+pen(y), pen(x)+pen(y), pen(x)+y(3y+j) (j = 1,2), pen(x)+6*T(y), pen(x)+y(7y+j)/2 (j = 1,3,5), pen(x)+y(4y+j) (j = 1,3), pen(x)+y(5y+j) (j = 1,2,3,4), pen(x)+y(13y+7)/2, x(5x+i)/2+y(3y+j) (i = 1,3; j = 1,2), x(5x+j)/2+y(7y+5)/2 (j = 1,3).

			a(1) = 1 since 1 = 1*2/2 + 1^2 + (-1)^5 with |(-1)^5| <= 1.
a(112) = 1 since 112 = 10*11/2 + 5^2 + 2^5.
a(770) = 1 since 770 = 28*29/2 + 11^2 + 3^5.
a(801) = 1 since 801 = 45*46/2 + 3^2 + (-3)^5 with |(-3)^5| < 801.
a(1593) = 1 since 1593 = 49*50/2 + 20^2 + (-2)^5 with |(-2)^5| < 1593.
a(1826) = 1 since 1826 = 55*56/2 + 23^2 + (-3)^5 with |(-3)^5| < 1826.
a(2320) = 1 since 2320 = 5*6/2 + 48^2 + 1^5.
a(2334) = 1 since 2334 = 11*12/2 + 45^2 + 3^5.
a(2849) = 1 since 2849 = 70*71/2 + 11^2 + 3^5.
a(7561) = 1 since 7561 = 97*98/2 + 53^2 + (-1)^5 with |(-1)^5| < 7561.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
Z.-W. Sun, On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and zT_x+by^2+f(z), preprint, arXiv:1502.03056 [math.NT], 2015.

Cf. A000217, A000290, A000584, A001318, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270921.

TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-(-1)^k*x^5-y^2],r=r+1],{k,0,1},{x,0,n^(1/5)},{y,1,Sqrt[n-(-1)^k*x^5]}];Print[n," ",r];Continue,{n,1,80}]

A266968 Number of ordered ways to write n as x^5+y^4+z^3+w*(w+1)/2, where x, y, z and w are nonnegative integers with z > 0 and w > 0.

Original entry on oeis.org

0, 0, 1, 2, 2, 2, 1, 1, 2, 2, 2, 3, 4, 2, 1, 2, 2, 2, 3, 3, 2, 1, 1, 4, 4, 2, 1, 2, 3, 4, 7, 5, 2, 2, 4, 3, 2, 5, 6, 5, 2, 1, 2, 4, 5, 5, 6, 4, 3, 4, 4, 1, 2, 4, 5, 5, 4, 4, 2, 3, 2, 4, 5, 4, 6, 5, 4, 3, 5, 6, 5, 4, 4, 3, 4, 5, 4, 3, 2, 5, 7

Offset: 0

Zhi-Wei Sun, Mar 28 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 6, 7, 14, 21, 22, 26, 41, 51, 184, 189, 206, 225, 229, 526, 708.
(ii) Any natural number can be written as 2*x^5 + y^4 + z^3 + w*(w+1)/2 with x,y,z,w nonnegative integers. Also, each natural number can be written as x^5 + 2*y^4 + z^3 + w*(w+1)/2 with x,y,z,w nonnegative integers.
(iii) For each d = 1,2, every natural number can be written as x^5 + y^4 + z^3 + w*(3w+1)/d with x,y,z nonnegative integers and w an integer.
(iv) Any natural number can be written as x^4 + y^4 + z^3 + w*(3w+1)/2 with x,y,z nonnegative integers and w an integer.
Also, for each P(w) = w(3w+1)/2, w(7w+3)/2, we can write any natural number as x^4 + y^3 + z^3 + P(w) with x,y,z nonnegative integers and w an integer.
(v) Any natural number can be written as the sum of a nonnegative cube and three pentagonal numbers. Also, every n = 0,1,2,... can be expressed as the sum of two nonnegative cubes and two pentagonal numbers.
We have verified that a(n) > 1 for all n = 2..3*10^6.
Compare this conjecture with the conjectures in A262813, A262827, A270559 and A271026.

			a(2) = 1 since 2 = 0^5 + 0^4 + 1^3 + 1*2/2.
a(6) = 1 since 6 = 1^5 + 1^4 + 1^3 + 2*3/2.
a(7) = 1 since 7 = 0^5 + 0^4 + 1^3 + 3*4/2.
a(14) = 1 since 14 = 0^5 + 0^4 + 2^3 + 3*4/2.
a(21) = 1 since 21 = 1^5 + 2^4 + 1^3 + 2*3/2.
a(22) = 1 since 22 = 0^5 + 0^4 + 1^3 + 6*7/2.
a(26) = 1 since 26 = 1^5 + 2^4 + 2^3 + 1*2/2.
a(41) = 1 since 41 = 2^5 + 0^4 + 2^3 + 1*2/2.
a(51) = 1 since 51 = 2^5 + 1^4 + 2^3 + 4*5/2.
a(184) = 1 since 184 = 0^5 + 0^4 + 4^3 + 15*16/2.
a(189) = 1 since 189 = 1^5 + 2^4 + 1^3 + 18*19/2.
a(206) = 1 since 206 = 2^5 + 3^4 + 3^3 + 11*12/2.
a(225) = 1 since 225 = 0^5 + 3^4 + 2^3 + 16*17/2.
a(229) = 1 since 229 = 1^5 + 3^4 + 3^3 + 15*16/2.
a(526) = 1 since 526 = 3^5 + 1^4 + 6^3 + 11*12/2.
a(708) = 1 since 708 = 1^5 + 5^4 + 3^3 + 10*11/2.

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), 1367-1396.

Cf. A000217, A000326, A000578, A000583, A000584, A001318, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270920, A271026.

TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^5-y^4-z^3],r=r+1],{x,0,n^(1/5)},{y,0,(n-x^5)^(1/4)},{z,1,(n-x^5-y^4)^(1/3)}];Print[n," ",r];Continue,{n,0,80}]

A270594 Number of ordered ways to write n as the sum of a triangular number, a positive square and the square of a generalized pentagonal number (A001318).

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Mar 19 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 21, 24, 48, 90, 138, 213, 283, 462, 468, 567, 573, 1998, 2068, 2488, 2687, 5208, 5547, 5638, 6093, 6492, 6548, 6717, 7538, 7731, 8522, 14763, 16222, 17143, 24958, 26148.
(ii) Let T(x) = x(x+1)/2, pen(x) = x(3x+1)/2 and hep(x) = x(5x+3)/2. Then any natural number can be written as P(x,y,z) with x, y and z integers, where P(x,y,z) is either of the following polynomials: T(x)^2+T(y)+z(5z+1)/2, T(x)^2+T(y)+z(3z+j) (j = 1,2), T(x)^2+y^2+pen(z), T(x)^2+pen(y)+hep(z), T(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), T(x)^2+pen(y)+z(4z+j) (j = 1,3), T(x)^2+pen(y)+z(5z+j) (j = 1,3,4), T(x)^2+pen(y)+z(11z+7)/2, T(x)^2+y(5y+1)/2+z(3z+2), T(x)^2+hep(y)+z(3z+2), pen(x)^2+T(y)+pen(z), pen(x)^2+T(y)+2*pen(z), pen(x)^2+T(y)+z(9z+7)/2, pen(x)^2+y^2+pen(z), pen(x)^2+2*T(y)+pen(z), pen(x)^2+pen(y)+3*T(z), pen(x)^2+pen(y)+2z^2, pen(x)^2+pen(y)+2*pen(z), pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), pen(x)^2+pen(y)+z(4z+3), pen(x)^2+pen(y)+z(9z+1)/2, pen(x)^2+pen(y)+3*pen(z), pen(x)^2+pen(y)+z(5z+j) (j = 1,2,3,4), pen(x)^2+pen(y)+z(11z+j)/2 (j = 7,9), pen(x)^2+pen(y)+z(7z+1), pen(x)^2+pen(y)+3*hep(z), pen(x)^2+y(5y+j)/2+z(3z+k) (j = 1,3; k = 1,2), pen(x)^2+hep(y)+z(7z+j)/2 (j = 1,3,5), pen(x)^2+hep(y)+z(9z+5)/2, pen(y)^2+2pen(y)+z(3z+2), pen(x)^2+2*pen(y)+3*pen(z), (x(5x+1)/2)^2+2*T(y)+pen(z), (x(5x+1)/2)^2+pen(y)+z(7z+3)/2, (x(5x+1)/2)^2+pen(y)+z(4z+1), (x(5x+1)/2)^2+hep(y)+2*pen(z), hep(x)^2+T(y)+2*pen(z), hep(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), hep(x)^2+pen(y)+z(4z+1), hep(x)^2+pen(y)+z(5z+4), 4*pen(x)^2+T(y)+hep(z), 4*pen(x)^2+T(y)+2*pen(z), 4*pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), (x(3x+2))^2+y^2+pen(z), (x(3x+2))^2+pen(y)+z(7z+j)/2 (j = 3,5), 2*T(x)^2+T(y)+z(3z+j) (j = 1,2), 2*T(x)^2+y^2+pen(z), 2*T(x)^2+2*T(y)+pen(z), 2*T(x)^2+pen(y)+z(7z+j)/2 (j = 1,5), 2*T(x)^2+pen(y)+z(5z+1), 2*pen(y)^2+T(y)+z(3z+2), 2*pen(x)^2+y^2+pen(z), 2*pen(x)^2+pen(y)+z(7z+3)/2, 2*pen(x)^2+pen(y)+z(4z+j) (j = 1,3), 2*pen(x)^2+pen(y)+z(5z+4), 2*pen(x)^2+pen(y)+z(7z+1), 2*pen(x)^2+hep(y)+2*pen(z), 2*hep(x)^2+pen(y)+z(7z+5)/2, 3*pen(x)^2+T(y)+z(3z+2), 3*pen(x)^2+y^2+pen(z), 3*pen(x)^2+2*T(y)+pen(z), 3*pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), 3*pen(x)^2+pen(y)+z(4z+1), 6*pen(x)^2+pen(y)+z(7z+3)/2.
See also A270566 for a similar conjecture involving four powers.
It is known that any positive integer can be written as the sum of a triangular number, a square and an odd square.

			a(21) = 1 since 21 = 1*2/2 + 4^2 + (1*(3*1+1)/2)^2.
a(24) = 1 since 24 = 5*6/2 + 3^2 + (0*(3*0-1)/2)^2.
a(468) = 1 since 468 = 0*1/2 + 18^2 + (3*(3*3-1)/2)^2.
a(7538) = 1 since 7538 = 64*65/2 + 47^2 + (6*(3*6+1)/2)^2.
a(7731) = 1 since 7731 = 82*83/2 + 62^2 + (4*(3*4-1)/2)^2.
a(8522) = 1 since 8522 = 127*128/2 + 13^2 + (3*(3*3+1)/2)^2.
a(14763) = 1 since 14763 = 164*165/2 + 33^2 + (3*(3*3-1)/2)^2.
a(16222) = 1 since 16222 = 168*169/2 + 45^2 + (1*(3*1-1)/2)^2.
a(17143) = 1 since 17143 = 182*183/2 + 21^2 + (2*(3*2+1)/2)^2.
a(24958) = 1 since 24958 = 216*217/2 + 39^2 + (1*(3*1-1)/2)^2.
a(26148) = 1 since 26148 = 10*11/2 + 142^2 + (7*(3*7+1)/2)^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
B. K. Oh and Z.-W. Sun, Mixed sums of squares and triangular numbers (III), J. Number Theory 129(2009), 964-969.
Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
Z.-W. Sun, On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and zT_x+by^2+f(z), preprint, arXiv:1502.03056 [math.NT], 2015.

Cf. A000217, A000290, A001318, A001082, A085787, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559, A270566.

pQ[n_]:=pQ[n]=IntegerQ[n]&&IntegerQ[Sqrt[24n+1]]
Do[r=0;Do[If[pQ[Sqrt[n-x^2-y(y+1)/2]],r=r+1],{x,1,Sqrt[n]},{y,0,(Sqrt[8(n-x^2)+1]-1)/2}];Print[n," ",r];Continue,{n,1,90}]

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.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 29 2016

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), 1367-1396.

Cf. A000326, A000578, A000583, A000584, A001015, A001318, A262813, A262815, A262816, A262827, A266968, A270469, A270488, A270516, A270533, A270559, A270566, A270920.

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

A271106 Number of ordered ways to write n as x^6 + 3y^3 + z^3 + w(w+1)/2, where x and y are nonnegative integers, and z and w are positive integers.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 1, 2, 2, 1, 2, 3, 3, 1, 3, 3, 1, 2, 2, 2, 1, 1, 2, 2, 1, 1, 3, 2, 2, 4, 3, 3, 4, 5, 3, 2, 4, 4, 3, 2, 4, 3, 2, 2, 1, 2, 3, 4, 3, 2, 1, 1, 2, 4, 4, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 1, 5, 5, 5, 3, 4

Offset: 0

Zhi-Wei Sun, Mar 30 2016

nonn

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 6, 9, 13, 16, 20, 21, 24, 25, 44, 50, 51, 65, 84, 189, 290, 484, 616, 664, 680, 917, 1501, 1639, 3013.
Based on our computation, we also formulate the following general conjecture.
General Conjecture: Let T(w) = w*(w+1)/2. We have {P(x,y,z,w): x,y,z,w = 0,1,2,...} = {0,1,2,...} for any of the following polynomials P(x,y,z,w): x^3+y^3+c*z^3+T(w) (c = 2,3,4,6), x^3+y^3+c*z^3+2*T(w) (c = 2,3), x^3+b*y^3+3z^3+3*T(w) (b = 1,2), x^3+2y^3+3z^3+w(5w-1)/2, x^3+2y^3+3z^3+w(5w-3)/2, x^3+2y^3+c*z^3+T(w) (c = 2,3,4,5,6,7,12,20,21,34,35,40), x^3+2y^3+c*z^3+2*T(w) (c = 3,4,5,6,11), x^3+2y^3+c*z^3+w^2 (c = 3,4,5,6), x^3+2y^3+4z^3+w(3w-1)/2, x^3+2y^3+4z^3+w(3w+1)/2, x^3+2y^3+4z^3+w(2w-1), x^3+2y^3+6z^3+w(3w-1)/2, x^3+3y^3+c*z^3+T(w) (c = 3,4,5,6,10,11,13,15,16,18,20), x^3+3y^3+c*z^3+2*T(w) (c = 5,6,11), x^3+4y^3+c*z^3+T(w) (c = 5,10,12,16), x^3+4y^3+5z^3+2*T(w), x^3+5y^3+10z^3+T(w), 2x^3+3y^3+c*z^3+T(w) (c = 4,6), 2x^3+4y^3+8z^3+T(w), x^4+y^3+3z^3+w(3w-1)/2, x^4+y^3+c*z^3+T(w) (c = 2,3,4,5,7,12,13), x^4+y^3+c*z^3+2*T(w) (c = 2,3,4,5), x^4+y^3+2z^3+w^2, x^4+y^3+4z^3+2w^2, x^4+2y^3+c*z^3+T(w) (c = 4,5,12), x^4+2y^3+3z^3+2*T(w), 2x^4+y^3+2z^3+w(3w-1)/2, 2x^4+y^3+c*z^3+T(w) (c = 1,2,3,4,5,6,10,11), 2x^4+y^3+c*z^3+2*T(w) (c = 2,3,4), 2x^4+2y^3+c*z^3+T(w) (c = 3,5), 3x^4+y^3+c*z^3+T(w) (c = 1,2,3,4,5,11), 3x^4+y^3+2z^3+2*T(w), 3x^4+y^3+2z^3+w^2, 3x^4+y^3+2z^3+w(3w-1)/2, 4x^4+y^3+c*z^3+T(w) (c = 2,3,4,6), 4x^4+y^3+2z^3+2*T(w), 5x^4+y^3+c*z^3+T(w) (c = 2,4), a*x^4+y^3+2z^3+T(w) (a = 6,20,28,40), 6x^4+y^3+2z^3+2*T(w), 6x^4+y^3+2z^3+w^2, a*x^4+y^3+3z^3+T(w) (a = 6,8,11), 8x^4+2y^3+4z^3+T(w), x^5+y^3+c*z^3+T(w) (c = 2,3,4), x^5+2y^3+c*z^3+T(w) (c = 3,6,8), 2x^5+y^3+4z^3+T(w), 3x^5+y^3+2z^3+T(w), 5x^5+y^3+c*z^3+T(w) (c = 2,4), x^6+y^3+3z^3+T(w), x^7+y^3+4z^3+T(w), x^4+2y^4+z^3+w^2, x^4+2y^4+2z^3+T(w),  x^4+b*y^4+z^3+T(w) (b = 2,3,4), 2x^4+3y^4+z^3+T(w), a*x^5+y^4+z^3+T(w) (a = 1,2), x^5+2y^4+z^3+T(w).
The polynomials listed in the general conjecture should exhaust all those polynomials P(x,y,z,w) = a*x^i+b*y^j+c*z^k+w*(s*w+/-t)/2 with {P(x,y,z,w): x,y,z,w = 0,1,2,...} = {0,1,2,...}, where a,b,c,s > 0, 0 <= t <= s, s == t (mod 2), i >= j >= k >= 3, a <= b if i = j, and b <= c if j = k.

			a(9) = 1 since 9 = 0^6 + 3*0^6 + 2^3 + 1*2/2.
a(24) = 1 since 24 = 1^6 + 3*0^6 + 2^3 + 5*6/2.
a(1501) = 1 since 1501 = 2^6 + 3*5^3 + 3^3 + 45*46/2.
a(1639) = 1 since 1639 = 0^6 + 3*6^3 + 1^3 + 44*45/2.
a(3013) = 1 since 3013 = 3^6 + 3*3^3 + 13^3 + 3*4/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), 1367-1396.

Cf. A000217, A000290, A000326, A000578, A000583, A000584, A001014, A005449, A262813, A262824, A262827, A262857, A262880, A266968, A270469, A270488, A270516, A270533, A270559, A270566, A270920, A271026.

TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^6-3*y^3-z^3],r=r+1],{x,0,n^(1/6)},{y,0,((n-x^6)/3)^(1/3)},{z,1,(n-x^6-3y^3)^(1/3)}];Print[n," ",r];Continue,{n,0,70}]

A271325 Number of ordered ways to write n as x^3 + y^2 + z*(3z+1), where x, y and z are integers with x positive and y nonnegative.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Apr 04 2016

nonn

We guess that a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 3, 4, 6, 7, 8, 13, 18, 20, 23, 25, 44, 49, 55, 59, 121, 238.
Based on our computation, we propose the following general conjecture (which extends the conjectures in A262813 and A270469).
Conjecture: Let T(x) = x*(x+1)/2 and pen(x) = x*(3x+1)/2. Every natural number can be written as P(x,y,z) with x a nonnegative integer and y and z integers, where P(x,y,z) is any of the following cubic polynomials: x^3+T(y)+z^2, a*x^3+T(y)+pen(z) (a = 1,2,3,4), x^3+T(y)+z*(5z+1)/2, x^3+T(y)+z*(3z+r) (r = 1,2), x^3+T(y)+z*(7z+3)/2, x^3+T(y)+z*(9z+j)/2 (j = 5,7), x^3+T(y)+z*(5z+r) (r = 2,3), x^3+T(y)+2z*(3z+r) (r = 1,2), x^3+T(y)+z*(6z+5), x^3+T(y)+z*(13z+j)/2 (j = 3,7,9), x^3+T(y)+z*(7z+k) (k = 2,6), a*x^3+y^2+pen(z) (a = 1,2,3,4), x^3+y^2+z*(5z+3)/2, x^3+y^2+2*pen(z), x^3+2*T(y)+pen(z), x^3+2*T(y)+z(5z+j)/2 (j = 1,3), a*x^3+2*T(y)+z*(3z+2) (a = 1,2,3), x^3+2*T(y)+z*(7z+3)/2, x^3+4*T(y)+pen(z), x^3+2y^2+pen(z), x^3+pen(y)+c*pen(z) (c = 1,2,3,4), x^3+b*pen(y)+z*(5z+j)/2 (b = 1,2; j = 1,3), x^3+pen(y)+z*(7z+k)/2 (k = 1,3,5), x^3+pen(y)+z*(4z+j) (j = 1,3), x^3+pen(y)+z*(9z+5)/2, a*x^3+pen(y)+z*(9z+r)/2 (a = 1,2; r = 1,7), x^3+pen(y)+z*(5z+r) (r = 1,2,3,4), a*x^3+pen(y)+z*(11z+9)/2 (a = 1,2), x^3+pen(y)+2z*(3z+2),x^3+pen(y)+z*(13z+11)/2, x^3+pen(y)+z*(7z+k) (k = 4,5,6), x^3+pen(y)+3z*(5z+3)/2, x^3+pen(y)+z*(15z+11)/2, x^3+pen(y)+z*(8z+7), x^3+pen(y)+z*(11z+7), x^3+2*pen(y)+z*(7z+j)/2 (j = 1,5), x^3+2*pen(y)+3*pen(z), x^3+2*pen(y)+z*(4z+1), x^3+2*pen(y)+z*(7z+2), x^3+y*(5y+j)/2+z*(7z+k)/2 (j = 1,3; k = 3,5), x^3+y*(5y+3)/2+z*(9z+7)/2, x^3+y*(3y+2)+z*(4z+1), x^3+y*(3y+2)+z*(5z+1)/2, x^3+y*(7y+3)/2+z*(7z+5)/2, 2x^3+T(y)+z*(5z+3)/2, 2x^3+T(y)+z*(3z+r) (r = 1,2), 2x^3+T(y)+z*(5z+4), 2x^3+2*T(y)+z*(5z+3)/2, 2x^3+3*T(y)+pen(z), 2x^3+y^2+2*pen(z), 2x^3+pen(y)+pen(z), a*x^3+pen(y)+3*pen(z) (a = 2,3,4), a*x^3+pen(y)+z*(7z+5)/2 (a = 2,3,4), 2x^3+pen(y)+z*(5z+k) (k = 1,3), 2x^3+y*(5y+3)/2+z*(7z+5)/2, 2x^3+2*pen(y)+z*(3z+2), 2x^3+2*pen(y)+z*(7z+5)/2, 2x^3+y*(3y+2)+z*(4z+3), 3x^3+pen(y)+z*(7z+3)/2, 4x^3+y^2+z*(5z+1)/2, 4x^3+pen(y)+z*(4z+3).
The listed ternary polynomials in the conjecture should exhaust all those P(x,y,z) = a*x^3+y*(s*y+t)/2+z*(u*z+v)/2 with a,s,u > 0, 0 <= t <= s, 0 <= v <= u, s == t (mod 2), u == v (mod 2), and (s-2t)*(u-2v) nonzero, such that any natural number can be written as P(x,y,z) with x a nonnegative integer and y and z integers. Note that those numbers y*(2y+1) with y integral are just triangular numbers.
Conjecture verified for all polynomials up to 10^11. - Mauro Fiorentini, Aug 03 2023
See also A271106 for another general conjecture on universal sums.

			a(13) = 1 since 13 = 2^3 + 1^2 + 1*(3*1+1).
a(18) = 1 since 18 = 2^3 + 0^2 + (-2)*(3*(-2)+1).
a(20) = 1 since 20 = 1^3 + 3^2 + (-2)*(3*(-2)+1).
a(23) = 1 since 23 = 2^3 + 1^2 + 2*(3*2+1).
a(25) = 1 since 25 = 1^3 + 0^2 + (-3)*(3*(-3)+1).
a(44) = 1 since 44 = 2^3 + 6^2 + 0*(3*0+1).
a(49) = 1 since 49 = 1^3 + 2^2 + (-4)*(3*(-4)+1).
a(55) = 1 since 55 = 3^3+ 2^2 + (-3)*(3*(-3)+1).
a(59) = 1 since 59 = 2^3 + 7^2 + (-1)*(3*(-1)+1).
a(121) = 1 since 121 = 3^3 + 8^2 + 3*(3*3+1).
a(238) = 1 since 238 = 4^3 + 12^2 + 3*(3*3+1).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), 1367-1396.
Z.-W. Sun, On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and zT_x+by^2+f(z), preprint, arXiv:1502.03056 [math.NT], 2015.

Cf. A000217, A000290, A000326, A000578, A001318, A160325, A160326, A262813, A262815, A262816, A270488, A270469, A270516, A270533, A270559, A270566, A271106.

pQ[n_]:=pQ[n]=IntegerQ[Sqrt[12n+1]]
Do[r=0;Do[If[pQ[n-x^3-y^2],r=r+1],{x,1,n^(1/3)},{y,0,Sqrt[n-x^3]}];Print[n," ",r];Label[aa];Continue,{n,1,70}]

cp's OEIS Frontend

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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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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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.

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A270920 Number of ordered ways to write n as the sum of a positive triangular number, a positive square, and a fifth power whose absolute value does not exceed n.

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A266968 Number of ordered ways to write n as x^5+y^4+z^3+w*(w+1)/2, where x, y, z and w are nonnegative integers with z > 0 and w > 0.

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A270594 Number of ordered ways to write n as the sum of a triangular number, a positive square and the square of a generalized pentagonal number (A001318).

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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.

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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.

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.

A271106 Number of ordered ways to write n as x^6 + 3y^3 + z^3 + w(w+1)/2, where x and y are nonnegative integers, and z and w are positive integers.