A262857 -id:A262857 - cp's OEIS frontend

A262824 Number of ordered ways to write n as w^2 + x^3 + 2y^3 + 3z^3, where w, x, y and z are nonnegative integers.

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

Offset: 0

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: (i) For any m = 3, 4, 5, 6 and n >= 0, there are nonnegative integers w, x, y, z such that n = w^2 + x^3 + 2*y^3 + m*z^3.
(ii) For P(w,x,y,z) = w^2 + x^3 + 2*y^3 + z^4, w^2 + x^3 + 2*y^3 + 3*z^4, w^2 + x^3 + 2*y^3 + 6*z^4, 2*w^2 + x^3 + 4*y^3 + z^4, we have {P(w,x,y,z): w,x,y,z = 0,1,2,...} ={0,1,2,...}.
Conjectures (i) and (ii) verified up to 10^11. - Mauro Fiorentini, Jul 22 2023
See also A262827 and A262857 for similar conjectures.

			a(0) = 1 since 0 = 0^2 + 0^3 + 2*0^3 + 3*0^3.
a(8) = 2 since 8 = 2^2 + 1^3 + 2*0^3 + 3*1^3 = 0^2 + 2^3 + 2*0^3 + 3*0^3.
a(23) = 1 since 23 = 2^2 + 0^3 + 2*2^3 + 3*1^3.
a(37) = 1 since 37 = 6^2 + 1^3 + 2*0^3 + 3*0^3.
a(72) = 1 since 72 = 8^2 + 2^3 + 2*0^3 + 3*0^3.

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

Cf. A000290, A000578, A262813, A262815, A262816, A262827, A262857.

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

A274274 Number of ordered ways to write n as x^3 + y^2 + z^2, where x,y,z are nonnegative integers with y <= z.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Jul 14 2016

nonn

Conjecture: Let n be any nonnegative integer.
(i) Either a(n) > 0 or a(n-2) > 0. Also, a(n) > 0 or a(n-6) > 0. Moreover, if n has the form 2^k*(4m+1) with k and m nonnegative integers, then a(n) > 0 except for n = 813, 4404, 6420, 28804.
(ii) Either n or n-3 can be written as x^3 + y^2 + 3*z^2 with x,y,z nonnegative integers.
(iii) For each d = 4, 5, 11, 12, either n or n-d can be written as x^3 + y^2 + 2*z^2 with x,y,z nonnegative integers.
We have verified that a(n) or a(n-2) is positive for every n = 0..2*10^6. Note that for each n = 0,1,2,... either n or n-2 can be written as x^2 + y^2 + z^2 with x,y,z nonnegative integers, which follows immediately from the Gauss-Legendre theorem on sums of three squares.

			a(6) = 1 since 6 = 1^3 + 1^2 + 2^2.
a(14) = 1 since 14 = 1^3 + 2^2 + 3^2.
a(31) = 1 since 31 = 3^3 + 0^2 + 2^2.

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

Cf. A000290, A000578, A022551, A022552, A262857, A272979.

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

A272979 Number of ways to write n as x^2 + 2y^2 + 3z^3 + 4*w^4 with x,y,z,w nonnegative integers.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Jul 13 2016

nonn

Conjecture: For positive integers a,b,c,d, any natural number can be written as a*x^2 + b*y^2 + c*z^3 + d*w^4 with x,y,z,w nonnegative integers, if and only if (a,b,c,d) is among the following 49 quadruples: (1,2,1,1), (1,3,1,1), (1,6,1,1), (2,3,1,1), (2,4,1,1), (1,1,2,1), (1,4,2,1), (1,2,3,1), (1,2,4,1), (1,2,12,1), (1,1,1,2), (1,2,1,2), (1,3,1,2), (1,4,1,2), (1,5,1,2), (1,11,1,2), (1,12,1,2), (2,4,1,2), (3,5,1,2), (1,1,4,2), (1,1,1,3), (1,2,1,3), (1,3,1,3), (1,2,4,3), (1,2,1,4), (1,3,1,4), (2,3,1,4), (1,1,2,4), (1,2,2,4), (1,8,2,4), (1,2,3,4), (1,1,1,5), (1,2,1,5), (2,3,1,5), (2,4,1,5), (1,3,2,5), (1,1,1,6), (1,3,1,6), (1,1,2,6), (1,2,1,8), (1,2,4,8), (1,2,1,10), (1,1,2,10), (1,2,1,11), (2,4,1,11), (1,2,1,12), (1,1,2,13), (1,2,1,14),(1,2,1,15).

See also A262824, A262827, A262857 and A273917 for similar conjectures.

			a(0) = 1 since 0 = 0^2 + 2*0^2 + 3*0^3 + 4*0^4.
a(1) = 1 since 1 = 1^2 + 2*0^2 + 3*0^3 + 4*0^4.
a(2) = 1 since 2 = 0^2 + 2*1^2 + 3*0^3 + 4*0^4.
a(14) = 1 since 14 = 3^2 + 2*1^2 + 3*1^3 + 4*0^4.
a(17) = 1 since 17 = 3^2 + 2*2^2 + 3*0^3 + 4*0^4.
a(59) = 1 since 59 = 3^2 + 2*5^2 + 3*0^3 + 4*0^4.
a(63) = 1 since 63 = 3^2 + 2*5^2 + 3*0^2 + 4*1^4.
a(287) = 1 since 287 = 11^2 + 2*9^2 + 3*0^2 + 4*1^4.

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

Cf. A000290, A000578, A000583, A262824, A262827, A262857, A270969, A273429, A273915, A273917.

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

A262880 Number of ordered ways to write n as w(w+1)/2 + x^3 + y^3 + 2z^3 with w > 0, 0 <= x <= y and z >= 0.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Oct 04 2015

nonn

Conjecture: (i) Any positive integer can be written as w*(w+1)/2 + x^3 + b*y^3 + c*z^3 with w > 0 and x,y,z >= 0, provided that (b,c) is among the following ordered pairs: (1,2),(1,3),(1,4),(1,6),(2,2),(2,3),(2,4),(2,5),(2,6),(2,7),(2,20),(2,21),(2,34),(3,3),(3,4),(3,5),(3,6),(4,10).
(ii) For (b,c) = (3,4),(3,6),(4,8), we have {w*(w+1)/2 + 2*x^3 + b*y^3 + c*z^3: w,x,y,z = 0,1,2,...} = {0,1,2,...}.
See also A262813, A262824 and A262857 for similar conjectures.

			a(2) = 1 since 2 = 1*2/2 + 0^3 + 1^3 + 2*0^3.
a(34) = 2 since 34 = 4*5/2 + 0^3 + 2^3 + 2*2^3 = 3*4/2 + 1^3 + 3^3 + 2*0^3.
a(41) = 1 since 41 = 3*4/2 + 2^3 + 3^3 + 2*0^3.
a(51) = 1 since 51 = 6*7/2 + 1^3 + 3^3 + 2*1^3.
a(104) = 1 since 104 = 5*6/2 + 2^3 + 3^3 + 2*3^3.

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

Cf. A000217, A000578, A262813, A262824, A262857.

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

A273917 Number of ordered ways to write n as w^2 + 3*x^2 + y^4 + z^5, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Jun 04 2016

nonn

Conjectures:
(i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 7, 11, 12, 15, 19, 24, 27, 31, 34, 35, 43, 46, 47, 56, 70, 71, 72, 87, 88, 115, 136, 137, 147, 167, 168, 178, 207, 235, 236, 267, 286, 297, 423, 537, 747, 762, 1017.
(ii) Any positive integer n can be written as w^2 + x^4 + y^5 + pen(z), where w is a positive integer, x,y,z are nonnegative integers, and pen(z) denotes the pentagonal number z*(3*z-1)/2.
Conjectures a(n) > 0 and (ii) verified up to 10^11. - Mauro Fiorentini, Jul 19 2023
See also A262813, A262857, A270566, A271106 and A271325 for some other conjectures on representations.

			a(1) = 1 since 1 = 1^2 + 3*0^2 + 0^4 + 0^5.
a(3) = 1 since 3 = 1^2 + 3*0^2 + 1^4 + 1^5.
a(7) = 1 since 7 = 2^2 + 3*1^2 + 0^4 + 0^5.
a(11) = 1 since 11 = 3^2 + 3*0^2 + 1^4 + 1^5.
a(12) = 1 since 12 = 3^2 + 3*1^2 + 0^4 + 0^5.
a(15) = 1 since 15 = 1^2 + 3*2^2 + 1^4 + 1^5.
a(19) = 1 since 19 = 4^2 + 3*1^2 + 0^4 + 0^5.
a(24) = 1 since 24 = 2^2 + 3*1^2 + 2^4 + 1^5.
a(27) = 1 since 27 = 5^2 + 3*0^2 + 1^4 + 1^5.
a(31) = 1 since 31 = 2^2 + 3*3^2 + 0^4 + 0^5.
a(34) = 1 since 34 = 1^2 + 3*0^2 + 1^4 + 2^5.
a(35) = 1 since 35 = 4^2 + 3*1^2 + 2^4 + 0^5.
a(43) = 1 since 43 = 4^2 + 3*3^2 + 0^4 + 0^5.
a(46) = 1 since 46 = 1^2 + 3*2^2 + 1^4 + 2^5.
a(47) = 1 since 47 = 2^2 + 3*3^2 + 2^4 + 0^5.
a(56) = 1 since 56 = 6^2 + 3*1^2 + 2^4 + 1^5.
a(70) = 1 since 70 = 5^2 + 3*2^2 + 1^4 + 2^5.
a(71) = 1 since 71 = 6^2 + 3*1^2 + 0^4 + 2^5.
a(72) = 1 since 72 = 6^2 + 3*1^2 + 1^4 + 2^5.
a(87) = 1 since 87 = 6^2 + 2*1^2 + 2^4 + 2^5.
a(88) = 1 since 88 = 2^2 + 3*1^2 + 3^4 + 0^5.
a(115) = 1 since 115 = 8^2 + 3*1^2 + 2^4 + 2^5.
a(136) = 1 since 136 = 10^2 + 3*1^2 + 1^4 + 2^5.
a(137) = 1 since 137 = 11^2 + 3*0^2 + 2^4 + 0^5.
a(147) = 1 since 147 = 12^2 + 3*1^2 + 0^4 + 0^5.
a(167) = 1 since 167 = 2^2 + 3*7^2 + 2^4 + 0^5.
a(168) = 1 since 168 = 2^2 + 3*7^2 + 2^4 + 1^5.
a(178) = 1 since 178 = 7^2 + 3*4^2 + 3^4 + 0^5.
a(207) = 1 since 207 = 10^2 + 3*5^2 + 0^4 + 2^5.
a(235) = 1 since 235 = 12^2 + 3*5^2 + 2^4 + 0^5.
a(236) = 1 since 236 = 12^2 + 3*5^2 + 2^4 + 1^5.
a(267) = 1 since 267 = 12^2 + 3*5^2 + 2^4 + 2^5.
a(286) = 1 since 286 = 4^2 + 3*3^2 + 0^4 + 3^5.
a(297) = 1 since 297 = 3^2 + 3*0^2 + 4^4 + 2^5.
a(423) = 1 since 423 = 11^2 + 3*10^2 + 1^4 + 1^5.
a(537) = 1 since 537 = 21^2 + 3*4^2 + 2^4 + 2^5.
a(747) = 1 since 747 = 11^2 + 3*0^2 + 5^4 + 1^5.
a(762) = 1 since 762 = 27^2 + 3*0^2 + 1^4 + 2^5.
a(1017) = 1 since 1017 = 27^2 + 3*0^2 + 4^4 + 2^5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.NT], 2016-2017.
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 Conjecture 3.2.)

Cf. A000118, A000290, A000326, A000583, A000584, A262813, A262827, A262857, A270566, A270969, A271076, A271106, A273429, A273915.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-3*x^2-y^4-z^5],r=r+1],{x,0,Sqrt[(n-1)/3]},{y,0,(n-1-3x^2)^(1/4)},{z,0,(n-1-3x^2-y^4)^(1/5)}];Print[n," ",r];Continue,{n,1,80}]

A275083 Positive integers congruent to 0 or 1 modulo 4 that cannot be written as x^3 + y^2 + z^2 with x,y,z nonnegative integers.

Original entry on oeis.org

120, 312, 813, 2136, 2680, 3224, 4404, 5340, 6420, 10060, 11320, 11824, 14008, 15856, 26544, 28804, 34392, 47984

Offset: 1

Zhi-Wei Sun, Jul 15 2016

nonn

Conjecture: (i) The sequence has totally 18 terms as listed.
(ii) For each r = 2,3 there are infinitely many positive integers n == r (mod 4) not in the form x^3 + y^2 + z^2 with x,y,z nonnegative integers.
Our computation indicates that the sequence has no other terms below 10^6.
Let d be 2 or 6. Clearly, n-d is congruent to 0 or 1 modulo 4 if n is congruent to 2 or 3 modulo 4. So part (i) of the conjecture essentially implies that for each n = 0,1,2,... either n or n-d can be written as x^3 + y^2 + z^2 with x,y,z nonnegative integers.

			a(1) = 120 since all those positive integers congruent to 0 or 1 modulo 4 and smaller than 120 can be written as x^3 + y^2 + z^2 with x,y,z nonnegative integers but 120 (divisible by 4) cannot be written in this way.

Cf. A000290, A000578, A022551, A022552, A262813, A262857, A272979, A274274.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
n=0;Do[If[Mod[m,4]>1,Goto[aa]];Do[If[SQ[m-x^3-y^2],Goto[aa]],{x,0,m^(1/3)},{y,0,Sqrt[(m-x^3)/2]}];n=n+1;Print[n," ",m];Label[aa];Continue,{m,1,50000}]

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}]

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.

Original entry on oeis.org

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

Zhi-Wei Sun, Apr 05 2016

nonn

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.

			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.

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.

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

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}]

A271381 Number of ordered ways to write n as u^2 + v^2 + x^3 + y^3, where u, v, x, y are nonnegative integers with 2 | u*v, u <= v and x <= y.

Original entry on oeis.org

1, 2, 2, 1, 1, 2, 2, 1, 2, 4, 3, 1, 1, 3, 2, 1, 3, 5, 3, 1, 2, 3, 2, 0, 2, 5, 3, 3, 3, 4, 1, 2, 4, 5, 3, 2, 5, 4, 3, 2, 4, 6, 2, 3, 4, 5, 2, 2, 4, 3, 2, 2, 5, 6, 4, 3, 2, 3, 2, 2, 4, 4, 3, 3, 5, 7, 4, 5, 5, 6, 4, 2, 6, 9, 6, 2, 4, 5, 1, 3, 8

Offset: 0

Zhi-Wei Sun, Apr 06 2016

nonn

Conjecture: (i) a(n) > 0 except for n = 23, and a(n) = 1 only for n = 0, 3, 4, 7, 11, 12, 15, 19, 23, 30, 78, 203, 219.

(ii) Any natural number can be written as the sum of two squares and three fourth powers.

			a(3) = 1 since 3 = 0^2 + 1^2 + 1^3 + 1^3 with 0 even.
a(4) = 1 since 4 = 0^2 + 2^2 + 0^3 + 0^3 with 0 and 2 even.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^3 + 1^3 with 2 even.
a(11) = 1 since 11 = 0^2 + 3^2 + 1^3 + 1^3 with 0 even.
a(12) = 1 since 12 = 0^2 + 2^2 + 0^3 + 2^3 with 0 and 2 even.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^3 + 1^3 with 2 even.
a(19) = 1 since 19 = 1^2 + 4^2 + 1^3 + 1^3 with 4 even.
a(30) = 1 since 30 = 2^2 + 5^2 + 0^3 + 1^3 with 2 even.
a(78) = 1 since 78 = 2^2 + 3^2 + 1^3 + 4^3 with 2 even.
a(203) = 1 since 203 = 7^2 + 10^2 + 3^3 + 3^3 with 10 even.
a(219) = 1 since 219 = 8^2 + 8^2 + 3^3 + 4^3 with 8 even.

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

Cf. A000290, A000578, A000583, A262813, A262824, A262827, A262857, A262880, A270488, A270559, A270969, A271325.

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

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A262824 Number of ordered ways to write n as w^2 + x^3 + 2*y^3 + 3*z^3, where w, x, y and z are nonnegative integers.

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A274274 Number of ordered ways to write n as x^3 + y^2 + z^2, where x,y,z are nonnegative integers with y <= z.

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

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A272979 Number of ways to write n as x^2 + 2*y^2 + 3*z^3 + 4*w^4 with x,y,z,w nonnegative integers.

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A262880 Number of ordered ways to write n as w*(w+1)/2 + x^3 + y^3 + 2*z^3 with w > 0, 0 <= x <= y and z >= 0.

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A273917 Number of ordered ways to write n as w^2 + 3*x^2 + y^4 + z^5, where w is a positive integer and x,y,z are nonnegative integers.

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A275083 Positive integers congruent to 0 or 1 modulo 4 that cannot be written as x^3 + y^2 + z^2 with x,y,z nonnegative integers.

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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.

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A262824 Number of ordered ways to write n as w^2 + x^3 + 2y^3 + 3z^3, where w, x, y and z are nonnegative integers.

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.

A272979 Number of ways to write n as x^2 + 2y^2 + 3z^3 + 4*w^4 with x,y,z,w nonnegative integers.

A262880 Number of ordered ways to write n as w(w+1)/2 + x^3 + y^3 + 2z^3 with w > 0, 0 <= x <= y and z >= 0.