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 1-6 of 6 results.

A343397 Number of ways to write n as 2^x + [y^2/3] + [z^2/4] with x,y,z positive integers, where [.] is the floor function.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 5, 5, 8, 5, 9, 5, 8, 8, 6, 9, 9, 10, 8, 11, 10, 10, 9, 9, 14, 8, 8, 10, 12, 11, 6, 14, 13, 10, 12, 13, 15, 11, 13, 9, 20, 6, 12, 17, 13, 13, 10, 11, 17, 12, 11, 13, 15, 14, 9, 13, 13, 14, 11, 18, 11, 15, 7, 12, 22, 13, 14, 17, 17, 11, 15, 13, 24, 16, 9, 17, 15, 15, 14, 18
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 13 2021

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
We have verified a(n) > 0 for all n = 2..2*10^6.
Conjecture verified up to 10^10. - Giovanni Resta, Apr 14 2021

Examples

			a(2) = 1 with 2 = 2^1 + [1^2/3] + [1^2/4].
a(3) = 2 with 3 = 2^1 + [1^2/3] + [2^2/4] = 2^1 + [2^2/3] + [1^2/4].

Links

Crossrefs

Cf. A000079, A000290, A343326, A343368, A343384, A343387, A343391, A343411.

Programs

  • Mathematica
    PowQ[n_]:=PowQ[n]=n>1&&IntegerQ[Log[2,n]];
tab={};Do[r=0;Do[If[PowQ[n-Floor[x^2/3]-Floor[y^2/4]],r=r+1],{x,1,Sqrt[3n-1]},{y,1,Sqrt[4(n-Floor[x^2/3]-1)+1]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A343387 Number of ways to write n as x^2 + [y^2/2] + [z^4/8], where [.] is the floor function, x is a nonnegative integer, and y and z are positive integers.

Original entry on oeis.org

1, 1, 2, 2, 3, 2, 3, 1, 3, 2, 3, 4, 3, 4, 4, 3, 3, 1, 6, 4, 3, 3, 4, 3, 3, 2, 4, 5, 4, 4, 3, 2, 3, 4, 5, 6, 5, 4, 6, 2, 6, 4, 4, 7, 5, 3, 4, 1, 5, 4, 8, 8, 2, 5, 5, 1, 5, 4, 3, 8, 5, 6, 2, 3, 5, 4, 6, 4, 6, 4, 5, 3, 5, 4, 4, 5, 8, 2, 7, 2, 3, 7, 6, 9, 3, 6, 10, 5, 5, 5, 5, 8, 3, 5, 3, 6, 7, 3, 9, 8, 6
Offset: 0

Views

Author

Zhi-Wei Sun, Apr 13 2021

Keywords

Comments

Conjecture: a(n) > 0 for all n >= 0.
This has been verified for all n = 0..10^5.
We also conjecture that each n = 0,1,... can be written as x^2 + [y^2/3] + [z^4/7] with x,y,z nonnegative integers.
See also A343391 for a similar conjecture.

Examples

			a(0) = 1 with 0 = 0^2 + [1^2/2] + [1^4/8].
a(47) = 1 with 47 = 5^2 + [5^2/2] + [3^4/8].
a(55) = 1 with 55 = 7^2 + [3^2/2] + [2^4/8].
a(217) = 1 with 217 = 11^2 + [6^2/2] + [5^4/8].
a(377) = 1 with 377 9^2 + [23^2/2] + [4^4/8].
a(392) = 1 with 392 = 0^2 + [28^2/2] + [1^4/8].
a(734) = 1 with 734 = 12^2 + [32^2/2] + [5^4/8].
a(1052) = 1 with 1052 = 32^2 + [6^2/2] + [3^4/8].
a(1054) = 1 with 1054 = 30^2 + [17^2/2] + [3^4/8].
a(1817) = 1 with 1817 = 39^2 + [23^2/2] + [4^4/8].

Links

Crossrefs

Cf. A000290, A000583, A343326, A343368, A343384, A343391, A343397, A343411.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[n-Floor[x^2/2]-Floor[y^4/8]],r=r+1],{x,1,Sqrt[2n+1]},{y,1,(8(n-Floor[x^2/2])+7)^(1/4)}];tab=Append[tab,r],{n,0,100}];Print[tab]

A343368 Number of ways to write n as floor((a^3+b^3)/3) + floor((c^3+d^3)/5), where a,b,c,d are nonnegative integers with a > b and c >= d.

Original entry on oeis.org

3, 2, 3, 6, 2, 3, 1, 3, 2, 7, 6, 3, 6, 2, 7, 2, 6, 1, 2, 2, 1, 10, 6, 3, 6, 6, 5, 6, 6, 4, 4, 5, 1, 4, 9, 6, 4, 4, 1, 5, 2, 4, 7, 5, 6, 5, 13, 6, 4, 6, 6, 7, 6, 5, 6, 8, 4, 4, 4, 5, 3, 2, 2, 4, 7, 4, 4, 8, 8, 5, 6, 6, 9, 8, 7, 8, 3, 15, 2, 10, 3, 8, 4, 3, 7, 6, 8, 4, 7, 9, 5, 4, 7, 8, 6, 6, 2, 8, 10, 4, 6
Offset: 0

Views

Author

Zhi-Wei Sun, Apr 12 2021

Keywords

Comments

Conjecture 1: a(n) > 0 for all n >= 0.
Conjecture 2: Each n = 0,1,... can be written as floor((a^3+b^3)/4) + floor((c^3+d^3)/5) with a,b,c,d nonnegative integers.
Both conjectures have been verified for all n = 0..10^5.
We also conjecture that the pair (4,5) of denominators in Conjecture 2 can be replaced by some other pairs such as (4,6), (5,6), (3,7), (4,7), (5,7), (6,7).

Examples

			a(1) = 2 with 1 = floor((1^3+0^3)/3) + floor((2^3+0^3)/5) = floor((1^3+0^3)/3) + floor((2^3+1^3)/5).
a(17) = 1 with 17 = floor((2^3+1^3)/3) + floor((4^3+2^3)/5).
a(20) = 1 with 20 = floor((2^3+0^3)/3) + floor((4^3+3^3)/5).
a(38) = 1 with 38 = floor((4^3+2^3)/3) + floor((4^3+2^3)/5).
a(103) = 1 with 103 = floor((6^3+4^3)/3) + floor((3^3+3^3)/5).
a(304) = 1 with 304 = floor((2^3+0^3)/3) + floor((10^3+8^3)/5).

Links

Crossrefs

Cf. A000578, A343326.

Programs

  • Mathematica
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
tab={};Do[r=0;Do[If[CQ[5(n-Floor[(x^3+y^3)/3])+s-z^3],r=r+1],{s,0,4},{x,1,(3n+2)^(1/3)},{y,0,Min[x-1,(3n+2-x^3)^(1/3)]},{z,0,((5(n-Floor[(x^3+y^3)/3])+s)/2)^(1/3)}];tab=Append[tab,r],{n,0,100}];Print[tab]

A343391 Number of ways to write n as x^2 + [y^2/4] + [z^4/6] with x,y,z positive integers, where [.] is the floor function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 13 2021

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
We have verified a(n) > 0 for all n = 1..10^6.
See also A343387 for a similar conjecture.

Examples

			a(1) = 1 with 1 = 1^2 + [1^2/4] + [1^4/6].
a(2) = 1 with 2 = 1^2 + [2^2/4] + [1^4/6].
a(14) = 1 with 14 = 1^2 + [1^2/4] + [3^4/6].
a(32) = 1 with 32 = 4^2 + [8^2/4] + [1^4/6].
a(35) = 1 with 35 = 4^2 + [5^2/4] + [3^4/6].
a(77) = 1 with 77 = 8^2 + [1^2/4] + [3^4/6].
a(840) = 1 with 840 = 28^2 + [15^2/4] + [1^4/6].

Links

Crossrefs

Cf. A000290, A000583, A343326, A343368, A343384, A343387, A343397, A343411.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[n-Floor[x^2/4]-Floor[y^4/6]],r=r+1],{x,1,Sqrt[4n+3]},{y,1,(6(n-Floor[x^2/4])+5)^(1/4)}];tab=Append[tab,r],{n,1,100}];Print[tab]

A343411 Number of ways to write n as x^3 + [y^3/2] + [z^3/3] + 2^w, where [.] is the floor function, x,y,z are positive integers, and w is a nonnegative integer.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 2, 1, 4, 1, 4, 3, 2, 3, 4, 5, 2, 7, 2, 2, 2, 5, 4, 5, 5, 3, 3, 3, 3, 7, 6, 3, 5, 5, 6, 2, 11, 3, 6, 2, 6, 6, 8, 10, 2, 9, 2, 5, 5, 10, 5, 2, 6, 4, 4, 7, 5, 7, 2, 2, 4, 6, 7, 3, 12, 3, 7, 4, 9, 6, 5, 10, 4, 15, 4, 8, 5, 11, 4, 8, 14, 6, 4, 6, 10, 7, 8, 9, 5, 6, 4, 4, 13, 5, 7, 3, 10, 2, 7, 11
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 14 2021

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
We have verified a(n) > 0 for all 1 < n <= 3*10^5.
Conjecture verified up to 10^10. - Giovanni Resta, Apr 14 2021

Examples

			a(2) = 1 with 2 = 1^3 + [1^3/2] + [1^3/3] + 2^0.
a(3) = 1 with 3 = 1^3 + [1^3/2] + [1^3/3] + 2^1.
a(4) = 1 with 4 = 1^3 + [1^3/2] + [2^3/3] + 2^0.
a(6) = 1 with 6 = 1^3 + [2^3/2] + [1^3/3] + 2^0.
a(8) = 1 with 8 = 1^3 + [2^3/2] + [2^3/3] + 2^0.
a(10) = 1 with 10 = 2^3 + [1^3/2] + [1^3/3] + 2^1.
a(103) = 1 with 103 = 3^3 + [1^3/2] + [6^3/3] + 2^2.

Links

Crossrefs

Cf. A000079, A000578, A343326, A343368, A343384, A343387, A343391, A343397.

Programs

  • Mathematica
    CQ[n_]:=CQ[n]=n>0&&IntegerQ[n^(1/3)];
tab={};Do[r=0;Do[If[CQ[n-Floor[x^3/2]-Floor[y^3/3]-2^z],r=r+1],{x,1,(2n-1)^(1/3)},{y,1,(3(n-Floor[x^3/2])-1)^(1/3)},{z,0,Log[2,n-Floor[x^3/2]-Floor[y^3/3]]}];tab=Append[tab,r],{n,1,100}];Print[tab]

A343384 Number of ways to write n as [a^3/3] + [b^3/4] + [c^3/5] + [d^6/6] with a,b,c,d positive integers, where [.] is the floor function.

Original entry on oeis.org

1, 1, 2, 2, 1, 2, 1, 3, 1, 3, 2, 3, 4, 3, 4, 3, 5, 4, 3, 4, 3, 5, 3, 4, 4, 3, 6, 5, 5, 2, 4, 5, 3, 6, 3, 3, 4, 6, 5, 2, 4, 5, 4, 7, 3, 5, 4, 4, 5, 3, 3, 4, 7, 6, 3, 6, 4, 5, 6, 5, 1, 3, 7, 3, 5, 3, 5, 3, 8, 4, 3, 2, 6, 3, 6, 4, 6, 4, 6, 5, 5, 1, 5, 5, 7, 4, 7, 6, 4, 6, 5, 2, 2, 5, 5, 5, 5, 6, 3, 7, 7
Offset: 0

Views

Author

Zhi-Wei Sun, Apr 13 2021

Keywords

Comments

3-4-5-6 Conjecture: a(n) > 0 for all n >= 0.
We have verified a(n) > 0 for all n = 0..10^6.
Conjecture verified up to 2*10^9. - Giovanni Resta, Apr 28 2021

Examples

			a(0) = 1 with 0 = [1^3/3] + [1^3/4] + [1^3/5] + [1^6/6].
a(1) = 1 with 1 = [1^3/3] + [1^3/4] + [2^3/5] + [1^6/6].
a(4) = 1 with 4 = [2^3/3] + [2^3/4] + [1^3/5] + [1^6/6].
a(6) = 1 with 6 = [1^3/3] + [3^3/4] + [1^3/5] + [1^6/6].
a(8) = 1 with 8 = [2^3/3] + [3^3/4] + [1^3/5] + [1^6/6].
a(60) = 1 with 60 = [3^3/3] + [4^3/4] + [5^3/5] + [2^6/6].
a(81) = 1 with 81 = [2^3/3] + [6^3/4] + [5^3/5] + [1^6/6].
a(300) = 1 with 300 = [7^3/3] + [5^3/4] + [9^3/5] + [2^6/6].
a(4434) = 1 with 4434 = [11^3/3] + [4^3/4] + [19^3/5] + [5^6/6].

Links

Crossrefs

Cf. A000578, A001014, A343326, A343368, A343387.

Programs

  • Mathematica
    CQ[n_]:=CQ[n]=n>0&&IntegerQ[n^(1/3)];
tab={};Do[r=0;Do[If[CQ[3(n-Floor[x^6/6]-Floor[y^3/5]-Floor[z^3/4])+s],r=r+1],{s,0,2},{x,1,(6n+5)^(1/6)},{y,1,(5(n-Floor[x^6/6])+4)^(1/3)},{z,1,(4(n-Floor[x^6/6]-Floor[y^3/5])+3)^(1/3)}];tab=Append[tab,r],{n,0,100}];Print[tab]
Showing 1-6 of 6 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.