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 1-3 of 3 results.

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

Views

Author

Zhi-Wei Sun, Jun 04 2016

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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

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}]

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

Views

Author

Zhi-Wei Sun, Apr 06 2016

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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}]
Showing 1-3 of 3 results.