A262813 -id:A262813 - cp's OEIS frontend

A275150 Number of ordered ways to write n as x^3 + 2y^2 + kz^2, where x,y,z are nonnegative integers, k is 1 or 5, and k = 1 if z = 0.

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

Offset: 0

Zhi-Wei Sun, Jul 17 2016

nonn

Conjecture 1: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 15, 79, 120, 218, 399, 454, 622, 725, 3240.
We have verified that a(n) > 0 for all n = 0..10^7.
Conjecture 2: For any positive integers a, b, c and integers i, j, k greater than one, there are infinitely many positive integers not in the set {a*x^i + b*y^j + c*z^k: x,y,z = 0,1,2,...}. - Zhi-Wei Sun, May 24 2023

			a(0) = 1 since 0 = 0^3 + 2*0^2 + 0^2.
a(15) = 1 since 15 = 2^3 + 2*1^2 + 5*1^2.
a(79) = 1 since 79 = 3^3 + 2*4^2 + 5*2^2.
a(120) = 1 since 120 = 2^3 + 2*4^2 + 5*4^2.
a(218) = 1 since 218 = 6^3 + 2*1^2 + 0^2.
a(399) = 1 since 399 = 5^3 + 2*3^2 + 16^2.
a(454) = 1 since 454 = 0^3 + 2*15^2 + 2^2.
a(622) = 1 since 622 = 2^3 + 2*17^2 + 6^2.
a(725) = 1 since 725 = 5^3 + 2*10^2 + 20^2.
a(3240) = 1 since 3240 = 7^3 + 2*38^2 + 3^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
G. Doyle and K. S. Williams, A positive-definite ternary quadratic form does not represent all positive integers, Integers 17 (2017), #A41, 19pp (electronic).
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Cf. A000290, A000578, A262813, A262941, A262954, A270488, A274274, A275083.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
TQ[n_]:=TQ[n]=SQ[n]||SQ[n/5]
Do[r=0;Do[If[TQ[n-x^3-2*y^2],r=r+1],{x,0,n^(1/3)},{y,0,Sqrt[(n-x^3)/2]}];Print[n," ",r];Continue,{n,0,80}]

A275168 Positive integers not of the form x^3 + 3*y^2 + z^2 with x,y,z nonnegative integers.

Original entry on oeis.org

6, 18, 23, 41, 42, 59, 78, 86, 96, 114, 115, 123, 142, 187, 195, 205, 213, 214, 240, 261, 262, 266, 303, 322, 329, 330, 383, 423, 478, 501, 510, 581, 610, 618, 642, 682, 690, 698, 761, 774, 807, 865, 870, 906, 959, 963, 990, 1206, 1222, 1230, 1234, 1302, 1312, 1314, 1320, 1346, 1411, 1697, 1706, 1781

Offset: 1

Zhi-Wei Sun, Jul 18 2016

nonn

Conjecture: The sequence has totally 150 terms as listed in the b-file the largest of which is 182842. Thus any integer n > 182842 can be written as x^3 + 3*y^2 + z^2 with x,y,z nonnegative integers.
We note that the sequence has no term greater than 182842 and not exceeding 10^6.
See also A275169 for a similar conjecture.
It is known that for any positive integers a,b,c there are infinitely many positive integers not of the form a*x^2 + b*y^2 + c*z^2 with x,y,z nonnegative integers.

			a(1) = 6 since 1 = 0^3 + 3*0^2 + 1^2, 2 = 1^3 + 3*0^2 + 1^2, 3 = 0^3 + 3*1^2 + 0^2, 4 = 0^3 + 3*1^2 + 1^2, 5 = 1^3 + 3*1^2 + 1^2, but 6 cannot be written as x^3 + 3*y^2 + z^2 with x,y,z nonnegative integers.

Zhi-Wei Sun, Table of n, a(n) for n = 1..150

Cf. A000290, A000578, A022551, A022552, A262813, A270488, A274274, A275083, A275150, A275169.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
n=0;Do[Do[If[SQ[m-x^3-3*y^2],Goto[aa]],{x,0,m^(1/3)},{y,0,Sqrt[(m-x^3)/3]}];n=n+1;Print[n," ",m];Label[aa];Continue,{m,1,1800}]

A306459 Number of ways to write n as w^3 + C(x+2,3) + C(y+2,3) + C(z+2,3), where w,x,y,z are nonnegative integers with x <= y <= z, and C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Feb 20 2019

nonn

Conjecture: a(n) > 0 for all n >= 0. In other words, each nonnegative integer can be written as the sum of a nonnegative cube and three tetrahedral numbers.
It seems that a(n) = 1 only for n = 0, 7, 17, 27, 34, 53, 54, 110, 118, 163, 207, 263, 270, 309, 362, 443, 1174, 1284.
We have verified a(n) > 0 for all n = 0..2*10^6.

			a(0) = 1 with 0 = 0^3 + C(2,3) + C(2,3) + C(2,3).
a(17) = 1 with 17 = 2^3 + C(3,3) + C(4,3) + C(4,3).
a(27) = 1 with 27 = 3^3 + C(2,3) + C(2,3) + C(2,3).
a(362) = 1 with 362 = 0^3 + C(6,3) + C(8,3) + C(13,3).
a(443) = 1 with 443 = 3^3 + C(5,3) + C(10,3) + C(13,3).
a(1174) = 1 with 1174 = 1^3 + C(9,3) + C(10,3) + C(19,3).
a(1284) = 1 with 1284 = 10^3 + C(7,3) + C(9,3) + C(11,3).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Cf. A000292, A000578, A000797, A262813, A306460, A306462, A306471, A306477.

f[n_]:=f[n]=Binomial[n+2,3];
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
tab={};Do[r=0;Do[If[f[x]>n/3,Goto[cc]];Do[If[f[y]>(n-f[x])/2,Goto[bb]];Do[If[f[z]>n-f[x]-f[y],Goto[aa]];If[CQ[n-f[x]-f[y]-f[z]],r=r+1],{z,y,n-f[x]-f[y]}];Label[aa],{y,x,(n-f[x])/2}];Label[bb],{x,0,n/3}];Label[cc];tab=Append[tab,r],{n,0,100}];Print[tab]

A270705 Number of ordered ways to write n as x^2pen(x) + pen(y) + pen(z) with pen(x) = x(3x+1)/2 and pen(y) <= pen(z), where x, y and z are integers ("pen" stands for "pentagonal").

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 21 2016

nonn

Conjecture: (i) Any natural number can be written as a*x^2*pen(x) + b*pen(y) + c*pen(z) with x, y and z integers, provided that (a,b,c) is among the following ordered triples: (j,1,k) (j = 1,2; k = 1,2,3,4), (1,2,3), (3,1,4) and (4,1,3).
(ii) Every n = 0,1,2,... can be expressed as x^2*pen(x) + T(y) + T(z) with x, y and z integers, where T(m) denotes the triangular number m*(m+1)/2. Also, for each (a,b) = (1,2),(1,4),(2,2), any natural number can be written as a*x^2*T(x) + b*T(y) + T(z) with x, y and z integers.
(iii) Each natural number can be written as x^2*P(x) + pen(y) + pen(z) with x, y and z integers, where P(x) is either of the following polynomials: a*T(x) (a = 1,2,3,4,5), x*(5x+3)/2, x*(3x+1), x*(3x+2), x*(7x+1)/2, x*(4x+1), x*(4x+3), x*(9x+5)/2, x*(5x+3), x*(11x+9)/2, x*(13x+5)/2, x*(17x+9)/2, 3x*(3x+2), x*(11x+2).
See also A270594 and A270706 for other similar conjectures.

			a(88) = 1 since 88 = 1^2*pen(1) + pen(-5) + pen(-6).

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 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, A270594, A270616, A270706.

pen[x_]:=pen[x]=x(3x+1)/2
pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]
Do[r=0;Do[If[pQ[n-pen[y]-x^2*pen[x]],r=r+1],{y,-Floor[(Sqrt[12n+1]+1)/6],(Sqrt[12n+1]-1)/6},{x,-1-Floor[(2(n-pen[y])/3)^(1/4)],(2(n-pen[y])/3)^(1/4)}];Print[n," ",r];Continue,{n,0,90}]

A270706 Number of ordered ways to write n as x^2T(x) + y^2 + T(z), where x, y and z are integers with x nonzero, y positive and z nonnegative, and T(m) denotes the triangular number m(m+1)/2.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Mar 21 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 90, 438, 480, 7108.
(ii) Let pen(x) = x*(3x+1)/2. Any natural number can be written as a*f(x)*g(x) + f(y) + g(z) with x, y and z integers, whenever (a,f(x),g(x)) is among the following ordered triples: (1,T(x),x^2), (1,T(x),pen(x)), (1,T(x),x*(5x+1)/2), (1,T(x),x*(5x+3)/2), (1,T(x),x*(3x+j)) (j = 1,2), (1,pen(x),3*T(x)), (1,pen(x),x*(7x+j)/2) (j = 1,3,5), (1,pen(x),x*(4x+1)), (2,T(x),x^2), (2,T(x),pen(x)), (2,T(x),x(5x+j)/2) (j = 1,3), (2,T(x),x*(3x+j)) (j = 1,2), (2,2*T(x),pen(x)), (2,pen(x),x(7x+j)/2) (j = 3,5), (k,x^2,pen(x)) (k = 1,2,3,4,5,8,11).
(iii) Each 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)*x(5x+1)/2+T(y)+2*T(z), a*T(x)*pen(x)+pen(y)+pen(z) (a = 1,2,3,4), T(x)*pen(x)+pen(y)+3*pen(z), T(x)*pen(x)+pen(y)+4*pen(z), 2*T(x)*pen(x)+pen(y)+3*pen(z), pen(x)*x(5x+j)/2+pen(y)+3*pen(z) (j = 1,3), x(3x+2)*pen(x)+pen(y)+4*pen(z), pen(x)*x(7x+1)/2+pen(y)+pen(z), pen(x)*x(9x+7)/2+pen(y)+pen(z).
See also A270594 and A270705 for some other similar conjectures.

			a(1) = 1 since 1 = (-1)^2*T(-1) + 1^2 + T(0).
a(3) = 1 since 3 = 1^2*T(1) + 1^2 + T(1).
a(90) = 1 since 90 = 3^2*T(3) + 6^2 + T(0).
a(438) = 1 since 438 = 4^2*T(4) + 5^2 + T(22).
a(480) = 1 since 480 = 1^2*T(1) + 17^2 + T(19).
a(7108) = 1 since 1^2*T(1) + 69^2 + T(68).

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 universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
Zhi-Wei 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, A270594, A270616, A270705.

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[x!=0&&TQ[n-y^2-x^3*(x+1)/2],r=r+1],{y,1,Sqrt[n]},{x,-1-Floor[(2(n-y^2))^(1/4)],(2(n-y^2))^(1/4)}];Print[n," ",r];Continue,{n,1,90}]

A270921 Number of ordered ways to write n as x(3x+2) + y(5y+1)/2 - z^4, where x and y are integers, and z is a nonnegative integer with z^4 <= n.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 25 2016

nonn

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 0, 5, 6, 12, 13, 14, 112, 193, 194, 200, 242, 333, 345, 376, 492, 528, 550, 551, 613, 797, 1178, 1195, 1222, 1663, 3380, 3635, 6508, 8755, 9132, 12434, 20087.

Compare this conjecture with the conjecture in A270566.

			a(5) = 1 since 5 = 1*(3*1+2) + 0*(5*0+1)/2 - 0^4.
a(6) = 1 since 6 = 1*(3*1+2) + (-1)*(5*(-1)+1)/2 - 1^4.
a(13) = 1 since 13 = 1*(3*1+2) + (-2)*(5*(-2)+1)/2 - 1^4.
a(376) = 1 since 376 = 0*(3*0+2) + (-16)*(5*(-16)+1)/2 - 4^4.
a(9132) = 1 since 9132 = (-13)*(3*(-13)+2) + 59*(5*59+1)/2 - 3^4.
a(12434) = 1 since 12434 = (-21)*(3*(-21)+2) + 78*(5*78+1)/2 - 8^4.
a(20087) = 1 since 20087 = 19*(3*19+2) + 87*(5*87+1)/2 - 0^4.
5, 6, 12, 13, 14, 112, 193, 194, 200, 242, 333, 345, 376, 492, 528, 550, 551, 613, 797, 1178, 1195, 1222, 1663, 3380, 3635, 6508, 8755, 9132, 12434, 20087

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 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. A000583, A001082, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270920.

pQ[n_]:=pQ[n]=IntegerQ[Sqrt[40n+1]]&&(Mod[Sqrt[40n+1],10]==1||Mod[Sqrt[40n+1],10]==9)
Do[r=0;Do[If[pQ[n+x^4-y(3y+2)],r=r+1],{x,0,n^(1/4)},{y,-Floor[(Sqrt[3(n+x^4)+1]+1)/3],(Sqrt[3(n+x^4)+1]-1)/3}];Print[n," ",r];Continue,{n,0,80}]

A270928 Number of ways to write n = x(x-1)/2 + y(y-1)/2 + z*(z-1)/2, where 0 < x <= y <= z, and one of x, y, z is prime.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Mar 26 2016

nonn

Conjecture: (i) a(n) > 0 for n > 0 with the only exception n = 15^2 = 225. Also, a(n) = 1 only for n = 1, 2, 4, 5, 6, 8, 11, 14, 15, 18, 20, 29, 36, 50, 53, 60, 62, 96, 117, 119, 218, 411, 540, 645, 1125, 1590, 2346, 4068.
(ii) Any positive integer can be written as p*(p-1)/2 + x*(x-1)/2 + P(y) with p prime and x and y integers, where the polynomial P(y) is either of the following ones: y*(y-1), y*(3*y+1)/2, y*(5*y+j)/2 (j = 1,3), y*(3*y+j) (j = 1,2), y*(7*y+3)/2, y*(9*y+j)/2 (j = 1,5,7), y*(5*y+j) (j = 1,3), y*(11*y+9)/2, 2*y*(3*y+j) (j = 1,2), y*(7*y+3).
(iii) Any positive integer can be written as p*(p-1)/2 + P(x,y) with p prime and x and y integers, where the polynomial P(x,y) is either of the following ones: a*x*(x-1)/2+y*(3*y+1)/2 (a = 2,3,4), x*(x-1)+y*(5*y+3)/2, b*x^2+y*(3*y+1)/2 (b = 1,2,3), x^2+y*(5*y+j)/2 (j = 1,3), x^2+y*(3*y+1), x^2+y*(7*y+j)/2 (j = 1,3,5), x^2+y*(4*y+1).
(iv) Every positive integer can be written as p*(p-1)/2+x*(3*x+1)/2+y*(3*y+1)/2 with p prime, x an nonnegative integer and y an integer. Also, for each r = 1,3, any positive integer n can be written as p*(p-1)/2+x*(3*x-1)/2+y*(5*y+r)/2, where p is a prime, and x and y are integers with x nonnegative.
Note that Gauss proved a classical assertion of Fermat which states that any natural number is the sum of three triangular numbers.
See also A270966 for a similar conjecture involving (p-1)^2 with p prime.
The conjecture that a(n) > 0 except for n = 225 appeared as Conjecture 1.2(i) of the author's JNT paper in the links.

			a(1) = 1 since 1 = 1*(1-1)/2 + 1*(1-1)/2 + 2*(2-1)/2 with 2 prime.
a(4) = 1 since 4 = 1*(1-1)/2 + 2*(2-1)/2 + 3*(3-1)/2 with 2 and 3 prime.
a(29) = 1 since 29 = 1*(1-1)/2 + 2*(2-1)/2 + 8*(8-1)/2 with 2 prime.
a(50) = 1 since 50 = 2*(2-1)/2 + 7*(7-1)/2 + 8*(8-1)/2 with 2 and 7 prime.
a(119) = 1 since 119 = 8*(8-1)/2 + 9*(9-1)/2 + 11*(11-1)/2 with 11 prime.
a(411) = 1 since 411 = 16*(16-1)/2 + 16*(16-1)/2 + 19*(19-1)/2 with 19 prime.
a(1125) = 1 since 1125 = 3*(3-1)/2 + 34*(34-1)/2 + 34*(34-1)/2 with 3 prime.
a(1590) = 1 since 1590 = 7*(7-1)/2 + 37*(37-1)/2 + 43*(43-1)/2 with 7, 37 and 43 prime.
a(2346) = 1 since 2346 = 6*(6-1)/2 + 16*(16-1)/2 + 67*(67-1)/2 with 67 prime.
a(4068) = 1 since 4068 = 7*(7-1)/2 + 34*(34-1)/2 + 84*(84-1)/2 with 7 prime.

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 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. A000040, A000217, A000290, A001318, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270966.

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x(x-1)/2-y(y-1)/2]&&(PrimeQ[x]||PrimeQ[y]||PrimeQ[(Sqrt[8(n-x(x-1)/2-y(y-1)/2)+1]+1)/2]),r=r+1],{x,1,(Sqrt[8n/3+1]+1)/2},{y,x,(Sqrt[8(n-x(x-1)/2)/2+1]+1)/2}];Print[n," ",r];Continue,{n,1,70}]

A270966 Number of ways to write n as x^2 + y^2 + z*(3z+1)/2, where x, y and z are integers with 0 <= x <= y such that x or y has the form p-1 with p prime.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Mar 27 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 49, 608.
(ii) Let T(x) = x*(x+1)/2 and pen(x) = x*(3x+1)/2. Any positive integer can be written as (p-1)^2+P(x,y) with p prime and x and y integral, where the polynomial P(x,y) is either of the following ones: T(x)+2*pen(y), 2*T(x)+pen(y), T(x)+y*(5y+1)/2, T(x)+y*(9y+5)/2, pen(x)+y*(5y+j)/2 (j = 1,3), pen(x)+y*(7y+k)/2 (k = 3,5), pen(x)+y*(4y+j) (j = 1,3), pen(x)+y*(5y+r) (r = 1,2,3,4), pen(x)+2y*(3y+i) (i = 1,2), pen(x)+6*pen(y), x*(5x+1)/2+y*(3y+2), x*(5x+1)/2+y*(9y+7)/2, x*(5x+3)/2+y*(3y+i) (i = 1,2), x*(5x+3)/2+y*(9y+5)/2.
See also A270928 for a similar conjecture involving T(p-1) = p*(p-1)/2 with p prime.

			a(1) = 1 since 1 = 0^2 + (2-1)^2 + 0*(3*0+1)/2 with 2 prime.
a(12) = 2 since 12 = (2-1)^2 + 2^2 + 2*(2*3+1)/2 = (2-1)^2 + 3^2 + 1*(3*1+1)/2 with 2 prime.
a(49) = 1 since 49 = (2-1)^2 + 6^2 + (-3)*(3*(-3)+1)/2 with 2 prime.
a(608) = 1 since 608 = (7-1)^2 + 14^2 + (-16)*(3*(-16)+1)/2 with 7 prime.

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 universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Cf. A000040, A000217, A000290, A001318, A160326, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270928.

pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]
Do[r=0;Do[If[(PrimeQ[x+1]||PrimeQ[y+1])&&pQ[n-x^2-y^2],r=r+1],{x,0,Sqrt[n/2]},{y,x,Sqrt[n-x^2]}];Print[n," ",r];Continue,{n,1,70}]

A280153 Number of ways to write n as x^3 + 2*y^3 + z^2 + 4^k, where x is a positive integer and y,z,k are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 1, 2, 2, 1, 3, 2, 3, 2, 2, 2, 1, 5, 2, 3, 4, 2, 3, 1, 4, 3, 4, 5, 4, 5, 2, 4, 4, 6, 3, 1, 6, 1, 2, 4, 3, 4, 3, 6, 3, 3, 4, 3, 5, 2, 3, 1, 5, 3, 2, 5, 2, 3, 3, 6, 3, 1, 5, 3, 4, 6, 6, 8, 7, 4, 5, 6, 3, 5, 7, 5, 3, 3, 5, 4

Offset: 1

Zhi-Wei Sun, Dec 27 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 7, 10, 17, 24, 36, 38, 52, 62, 115, 136, 185, 990.
(ii) Each positive integer n can be written as x^2 + 2*y^2 + z^3 + 8^k with x,y,z,k nonnegative integers.
(iii) Let a,b,c be positive integers with b <= c. Then any positive integer can be written as a*x^3 + b*y^2 + c*z^2 + 4^k with x,y,z,k nonnegative integers, if and only if (a,b,c) is among the following triples: (1,1,1), (1,1,2), (1,1,3), (1,1,5), (1,1,6), (1,2,3), (1,2,5), (2,1,1), (2,1,2), (2,1,3), (2,1,6), (4,1,2), (5,1,2), (8,1,2), (9,1,2).
We have verified that a(n) > 0 for all n = 2..10^6, and that part (ii) of the conjecture holds for all n = 1..10^6.
For any positive integer n, it is easy to see that at least one of n-1, n-8, n-64 is not of the form 4^k*(8m+7) with k and m nonnegative integers, thus, by the Gauss-Legendre theorem on sums of three squares, n = x^2 + y^2 + z^2 + 8^k for some nonnegative integers x,y,z and k < 3.
See also A280356 for a similar conjecture involving powers of two.

			a(7) = 1 since 7 = 1^3 + 2*1^3 + 0^2 + 4^1.
a(10) = 1 since 10 = 2^3 + 2*0^3 + 1^2 + 4^0.
a(17) = 1 since 17 = 1^3 + 2*0^3 + 0^2 + 4^2.
a(24) = 1 since 24 = 2^3 + 2*0^3 + 0^2 + 4^2.
a(36) = 1 since 36 = 2^3 + 2*1^3 + 5^2 + 4^0.
a(38) = 1 since 38 = 1^3 + 2*0^3 + 6^2 + 4^0.
a(52) = 1 since 52 = 3^3 + 2*0^3 + 3^2 + 4^2.
a(62) = 1 since 62 = 2^3 + 2*1^3 + 6^2 + 4^2.
a(115) = 1 since 115 = 2^3 + 2*3^3 + 7^2 + 4^1.
a(136) = 1 since 136 = 2^3 + 2*0^3 + 8^2 + 4^3.
a(185) = 1 since 185 = 3^3 + 2*3^3 + 10^2 + 4^1.
a(990) = 1 since 990 = 7^3 + 2*3^3 + 23^2 + 4^3.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Cf. A000290, A000302, A000578, A262813, A262827, A262857, A270533, A270559, A280356.

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

A306240 Number of ways to write n as x^9 + y^3 + z(z+1) + w(w+1), where x,y,z,w are nonnegative integers with x <= 2 and z <= w.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Jan 31 2019

nonn

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 11, 17, 18, 47, 108, 109, 234, 359. Also, any nonnegative integer can be written as x^6 + y^3 + z*(z+1) + w*(w+1), where x,y,z,w are nonnegative integers with x <= 2.

We have verified a(n) > 0 for all n = 0..2*10^7.

			a(11) = 1 with 11 = 1^9 + 2^3 + 0*1 + 1*2.
a(18) = 1 with 18 = 0^9 + 0^3 + 2*3 + 3*4.
a(109) = 1 with 109 = 1^9 + 4^3 + 1*2 + 6*7.
a(234) = 1 with 234 = 0^9 + 6^3 + 2*3 + 3*4.
a(359) = 1 with 359 = 1^9 + 2^3 + 10*11 + 15*16.
a(1978) = 3 with 1978 = 2^9 + 2^3 + 26*27 + 27*28 = 2^9 + 6^3 + 19*20 + 29*30 = 2^9 + 6^3 + 24*25 + 25*26.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Cf. A000578, A001014, A001017, A002378, A262813, A262815, A270488, A306225, A306227, A306239.

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[4n+1]];
tab={};Do[r=0;Do[If[TQ[n-x^9-y^3-z(z+1)],r=r+1],{x,0,Min[2,n^(1/9)]},{y,0,(n-x^9)^(1/3)},{z,0,(Sqrt[2(n-x^9-y^3)+1]-1)/2}];tab=Append[tab,r],{n,0,100}];Print[tab]

cp's OEIS Frontend

A275150 Number of ordered ways to write n as x^3 + 2*y^2 + k*z^2, where x,y,z are nonnegative integers, k is 1 or 5, and k = 1 if z = 0.

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A275168 Positive integers not of the form x^3 + 3*y^2 + z^2 with x,y,z nonnegative integers.

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A306459 Number of ways to write n as w^3 + C(x+2,3) + C(y+2,3) + C(z+2,3), where w,x,y,z are nonnegative integers with x <= y <= z, and C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

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A270705 Number of ordered ways to write n as x^2*pen(x) + pen(y) + pen(z) with pen(x) = x*(3x+1)/2 and pen(y) <= pen(z), where x, y and z are integers ("pen" stands for "pentagonal").

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A270706 Number of ordered ways to write n as x^2*T(x) + y^2 + T(z), where x, y and z are integers with x nonzero, y positive and z nonnegative, and T(m) denotes the triangular number m*(m+1)/2.

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A270921 Number of ordered ways to write n as x*(3x+2) + y*(5y+1)/2 - z^4, where x and y are integers, and z is a nonnegative integer with z^4 <= n.

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A270928 Number of ways to write n = x*(x-1)/2 + y*(y-1)/2 + z*(z-1)/2, where 0 < x <= y <= z, and one of x, y, z is prime.

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A270966 Number of ways to write n as x^2 + y^2 + z*(3z+1)/2, where x, y and z are integers with 0 <= x <= y such that x or y has the form p-1 with p prime.

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A275150 Number of ordered ways to write n as x^3 + 2y^2 + kz^2, where x,y,z are nonnegative integers, k is 1 or 5, and k = 1 if z = 0.

A270705 Number of ordered ways to write n as x^2pen(x) + pen(y) + pen(z) with pen(x) = x(3x+1)/2 and pen(y) <= pen(z), where x, y and z are integers ("pen" stands for "pentagonal").

A270706 Number of ordered ways to write n as x^2T(x) + y^2 + T(z), where x, y and z are integers with x nonzero, y positive and z nonnegative, and T(m) denotes the triangular number m(m+1)/2.

A270921 Number of ordered ways to write n as x(3x+2) + y(5y+1)/2 - z^4, where x and y are integers, and z is a nonnegative integer with z^4 <= n.

A270928 Number of ways to write n = x(x-1)/2 + y(y-1)/2 + z*(z-1)/2, where 0 < x <= y <= z, and one of x, y, z is prime.

A306240 Number of ways to write n as x^9 + y^3 + z(z+1) + w(w+1), where x,y,z,w are nonnegative integers with x <= 2 and z <= w.