A271106 -id:A271106 - cp's OEIS frontend

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.

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}]

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}]

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}]

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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.

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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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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.

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