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 1-2 of 2 results.

A267861 Number of ways to write n as 2*t + u^4 + v^4 + 2*w^4 + 3*x^4 + 4*y^4 + 6*z^4, where t is 0 or 1, and u, v, w, x, y, z are nonnegative integers with u <= v and v > 0.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 07 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 1, 2, 111, 127, 143, 158, 221, 223, 240, 460, 463, 480, 545, 560, 561, 1455, 1695, 1776, 2175. Moreover, any integer n > 10^4 not among 10543, 17935, 37583, 40383, 78543 can be written as u^4 + v^4 + 2*w^4 + 3*x^4 + 4*y^4 + 6*z^4 with u,v,w,x,y,z nonnegative integers.
If a(1),...,a(7) are positive integers with a(1) <= a(2) <= ... <= a(7) and a(1)+...+a(7) = g(4) = 19 such that {a(1)*x(1)^4+...+a(7)*x(7)^4: x(1),...,x(7) = 0,1,2,...} = {0,1,2,...}, then the tuple (a(1),...,a(7)) must be (1,1,2,2,3,4,6) or (1,1,2,2,3,3,7). Similarly, if a(1),...,a(8) are positive integers with a(1) <= a(2) <= ... <= a(8) and a(1)+...+a(8) = g(5) = 37 such that {a(1)*x(1)^5+...+a(8)*x(8)^5: x(1),...,x(8) = 0,1,2,...} = {0,1,2,...}, then (a(1),...,a(8)) must be (1,1,2,3,4,6,8,12) or (1,1,2,3,4,5,7,14).

Examples

			a(111) = 1 since 111 = 2*1 + 2^4 + 3^4 + 2*1^4 + 3*0^4 + 4*1^4 + 6*1^4.
a(240) = 1 since 240 = 2*0 + 2^4 + 2^4 + 2*0^4 + 3*2^4 + 4*2^4 + 6*2^4.
a(1776) = 1 since 1776 = 2*0 + 4^4 + 5^4 + 2*3^4 + 3*3^4 + 4*1^4 + 6*3^4.
a(2175) = 1 since 2175 = 2*1 + 0^4 + 4^4 + 2*2^4 + 3*5^4 + 4*1^4 + 6*1^4.

Links

Crossrefs

Cf. A000583, A000584, A002804, A267826, A271099, A271169, A271237.

Programs

  • Mathematica
    QQ[n_]:=QQ[n]=n>0&&IntegerQ[n^(1/4)]
Do[r=0;Do[If[QQ[n-2t-6*z^4-4y^4-3x^4-2w^4-u^4],r=r+1],{t,0,Min[1,n/2]},{z,0,((n-2t^8)/6)^(1/4)},{y,0,((n-2t-6z^4)/4)^(1/4)},{x,0,((n-2t-6z^4-4y^4)/3)^(1/4)},
{w,0,((n-2t-6z^4-4y^4-3x^4)/2)^(1/4)},{u,0,((n-2t-6z^4-4y^4-3x^4-2w^4)/2)^(1/4)}];Print[n," ",r];Continue,{n,1,70}]

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.

Links

Crossrefs

Cf. A000578, A000583, A000584, A266968, A267826, A271099, A271237.

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}]
Showing 1-2 of 2 results.

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