cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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

Views

Author

Zhi-Wei Sun, Mar 29 2016

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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 + 3*y^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

Views

Author

Zhi-Wei Sun, Mar 30 2016

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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}]

A271076 Number of ordered ways to write n as u^5 + v^4 + x^3 + 2*y^3 + 3*z^3, where u, v , x, y and z are nonnegative integers with v > 0.

Original entry on oeis.org

1, 2, 2, 3, 3, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 4, 4, 4, 4, 3, 1, 1, 3, 4, 4, 5, 4, 2, 2, 2, 5, 4, 3, 5, 2, 1, 1, 2, 5, 4, 6, 5, 2, 3, 2, 4, 5, 4, 3, 3, 3, 2, 2, 4, 5, 4, 5, 5, 1, 2, 3, 3, 5, 2, 5, 5, 3, 3, 3, 3, 3, 4, 4, 1, 1, 2, 3, 5
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 07 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 8, 9, 10, 11, 13, 14, 15, 16, 23, 24, 38, 39, 61, 76, 77, 104, 118, 188, 214, 229.
(ii) We have {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(u,v,x,y,z) is among the following polynomials: u^5+v^4+x^3+2*y^3+5*z^3, u^5+2*v^4+x^3+2*y^3+3*z^3, u^5+3*v^4+x^3+2*y^3+3*z^3, 2*u^5+v^4+x^3+y^3+4*z^3, 2*u^5+v^4+x^3+2*y^3+4*z^3, 3*u^5+v^4+x^3+2*y^3+4*z^3, 5*u^5+v^4+x^3+2*y^3+4*z^3, u^4+2*v^4+x^3+y^3+4*z^3, u^4+2*v^4+x^3+2*y^3+3*z^3, u^4+2*v^4+x^3+2*y^3+4*z^3, u^4+2*v^4+x^3+2*y^3+6*z^3, u^4+2*v^4+x^3+3*y^3+4*z^3, u^4+2*v^4+x^3+4*y^3+5*z^3, u^4+2*v^4+x^3+4*y^3+6*z^3, u^4+2*v^4+x^3+4*y^3+10*z^3, u^4+3*v^4+x^3+2*y^3+3*z^3, u^4+3*v^4+x^3+2*y^3+4*z^3, u^4+3*v^4+x^3+2*y^3+6*z^3, u^4+4*v^4+x^3+y^3+2*z^3, u^4+4*v^4+x^3+2*y^3+3*z^3, u^4+4*v^4+x^3+2*y^3+4*z^3, u^4+5*v^4+x^3+2*y^3+4*z^3, u^4+6*v^4+x^3+2*y^3+3*z^3, u^4+7*v^4+x^3+2*y^3+3*z^3, u^4+9*v^4+x^3+2*y^3+4*z^3, 2*u^4+4*v^4+x^3+2*y^3+3*z^3,2*u^4+6*v^4+x^3+2*y^3+4*z^3, 3*u^4+6*v^4+x^3+2*y^3+4*z^3.
(iii) We have {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(u,v,x,y,z) is among the following polynomials: u^5+v^3+x^3+2*y^3+4*z^3, u^5+v^3+2*x^3+3*y^3+c*z^3 (c = 6,9), a*u^5+v^3+2*x^3+4*y^3+5*z^3 (a = 1,2,6), b*u^5+v^3+2*x^3+4*y^3+6*z^3 (b = 2,3), u^5+v^3+2*x^3+4*y^3+d*z^3 (d = 9,13).
The listed polynomials in part (ii), together with u^5+v^4+x^3+2*y^3+3*z^3, should essentially exhaust all those polynomials P(u,v,x,y,z) = s*u^k+t*v^j+a*x^3+b*y^3+c*z^3 with s,t,a,b,c positive integers and k >= j > 3, such that {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...}.
There are also finitely many (but quite a lot) polynomials P(u,v,x,y,z) of the form m*u^4+a*v^3+b*x^3+c*y^3+d*z^3 with a,b,c,d and m positive integers such that {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...}.
See also A267826, A271099 and A271237 for related comments.
Conjectures (i), (ii) and (iii) verified for n up to 10^11 for all polynomials. - Mauro Fiorentini, Sep 20 2023

Examples

			a(16) = 1 since 16 = 0^5 + 2^4 +0^3 + 2*0^3 + 3*0^3.
a(104) = 1 since 104 = 0^5 + 2^4 + 4^3 + 2*0^3 + 3*2^3.
a(188) = 1 since 188 = 2^5 + 1^4 + 3^3 + 2*4^3 + 3*0^3.
a(229) = 1 since 229 = 1^5 + 3^4 + 4^3 + 2*1^3 + 3*3^3.
		

Crossrefs

Programs

  • Mathematica
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
    Do[r=0;Do[If[CQ[n-u^5-v^4-3z^3-2y^3],r=r+1],{u,0,(n-1)^(1/5)},{v,1,(n-u^5)^(1/4)},{z,0,((n-u^5-v^4)/3)^(1/3)},{y,0,((n-u^5-v^4-3z^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

Views

Author

Zhi-Wei Sun, Apr 05 2016

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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}]
Showing 1-4 of 4 results.