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.

A283239 Number of ways to write n as x^2 + y^2 + z*(3*z-1)/2 with x,y,z integers such that x + 2*y is a square.

Original entry on oeis.org

1, 2, 3, 2, 2, 5, 5, 5, 1, 2, 4, 2, 5, 3, 3, 4, 3, 7, 5, 3, 8, 3, 5, 3, 2, 6, 3, 7, 5, 3, 4, 7, 5, 3, 4, 6, 5, 3, 3, 7, 5, 5, 5, 1, 6, 7, 6, 4, 3, 2, 5, 5, 9, 6, 3, 7, 5, 7, 5, 4, 8, 5, 6, 4, 6, 5, 5, 8, 5, 6, 6, 7, 5, 5, 5, 7, 5, 6, 2, 4, 12
Offset: 0

Views

Author

Zhi-Wei Sun, Mar 03 2017

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 8, 43, 84, 133, 253, 399, 488, 523, 803, 7369.
(ii) Any integer n > 1 can be written as p + x^2 + y^2 with p prime and x + 2*y (or x + 3*y) a square, where x is an integer and y is a nonnegative integer.
Note that those numbers z*(3*z-1)/2 with z integral are called generalized pentagonal numbers (A001318). By Theorem 1.7(ii) of the linked paper in Sci. China Math., each n = 0,1,2,... can be written as the sum of two squares and a generalized pentagonal number.
Ju. V. Linnik proved in 1960 that any sufficiently large integer can be expressed as the sum of a prime and two squares.

Examples

			a(8) = 1 since 8 = 1^2 + 0^2 + (-2)*(3*(-2)-1)/2 with 1 + 2*0 = 1^2.
a(43) = 1 since 43 = 1^2 + 4^2 + (-4)*(3*(-4)-1)/2 with 1 + 2*4 = 3^2.
a(84) = 1 since 84 = 7^2 + (-3)^2 + (-4)*(3*(-4)-1)/2 with 7 + 2*(-3) = 1^2.
a(133) = 1 since 133 = 4^2 + 0^2 + 9*(3*9-1)/2 with 4 + 2*0 = 2^2.
a(253) = 1 since 253 = (-13)^2 + 7^2 + 5*(3*5-1)/2 with (-13) + 2*7 = 1^2.
a(399) = 1 since 399 = 18^2 + (-7)^2 + (-4)*(3*(-4)-1)/2 with 18 + 2*(-7) = 2^2.
a(488) = 1 since 488 = 9^2 + 20^2 + (-2)*(3*(-2)-1)/2 with 9 + 2*20 = 7^2.
a(523) = 1 since 523 = 9^2 + 0^2 + (-17)*(3*(-17)-1)/2 with 9 + 2*0 = 3^2.
a(803) = 1 since 803 = (-17)^2 + 13^2 + (-15)*(3*(-15)-1)/2 with (-17) + 2*13 = 3^2.
a(7369) = 1 since 7369 = 0^2 + 72^2 + (-38)*(3*(-38)-1)/2 with 0 + 2*72 = 12^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    PenQ[n_]:=PenQ[n]=SQ[24n+1];
    Do[r=0;Do[If[PenQ[n-x^2-y^2],Do[If[SQ[(-1)^i*x+2(-1)^j*y],r=r+1],{i,0,Min[x,1]},{j,0,Min[y,1]}]],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]}];Print[n," ",r];Continue,{n,0,80}]

A283205 Numbers of the form x^2 + y^2 with x and y integers such that x + 3*y is a square.

Original entry on oeis.org

0, 1, 2, 5, 8, 9, 10, 13, 16, 17, 25, 26, 29, 32, 34, 37, 40, 50, 53, 58, 61, 65, 73, 74, 80, 81, 85, 90, 104, 109, 117, 125, 128, 130, 136, 137, 144, 145, 146, 160, 162, 170, 178, 185, 193, 202, 208, 221, 229, 232, 241, 245, 250, 256, 257, 265, 269, 272, 274, 281
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 03 2017

Keywords

Comments

This sequence is interesting since part (ii) of the conjecture in A283170 implies that each n = 0,1,2,... can be expressed as the sum of two terms of the current sequence.
Clearly, the sequence is a subsequence of A001481.

Examples

			a(1) = 0 since 0 = 0^2 + 0^2 with 0 + 3*0 = 0^2.
a(2) = 1 since 1 = 1^2 + 0^2 with 1 + 3*0 = 1^2.
a(3) = 2 since 2 = 1^2 + 1^2 with 1 + 3*1 = 2^2.
a(4) = 5 since 5 = (-2)^2 + 1^2 with (-2) + 3*1 = 1^2.
a(5) = 8 since 8 = (-2)^2 + 2^2 with (-2) + 3*2 = 2^2.
a(6) = 9 since 9 = 0^2 + 3^2 with 0 + 3*3 = 3^2.
a(7) = 10 since 10 = 3^2 + (-1)^2 with 3 + 3*(-1) = 0^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    n=0;Do[Do[If[SQ[m-x^2],Do[If[SQ[(-1)^i*x+3(-1)^j*Sqrt[m-x^2]],n=n+1;Print[n," ",m];Goto[aa]],{i,0,Min[x,1]},{j,0,Min[Sqrt[m-x^2],1]}]],{x,0,Sqrt[m]}];Label[aa];Continue,{m,0,281}]

A283269 Number of ways to write n as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 5*z is a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 04 2017

Keywords

Comments

Conjecture: (i) a(2^k*m) > 0 for any positive odd integers k and m. Also, a(4*n+1) = 0 only for n = 63.
(ii) For any positive odd integers k and m, we can write 2^k*m as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 5*z is twice a square.
(iii) For any positive odd integer n not congruent to 7 modulo 8, we can write n as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 4*z is a square.
(iv) Let n be any nonnegative integer. Then we can write 8*n + 1 as x^2 + y^2 + z^2 with x + 3*y a square, where x and y are integers, and z is a positive integer. Also, except for n = 2255, 4100 we can write 4*n + 2 as x^2 + y^2 + z^2 with x + 3*y a square, where x,y,z are integers.
(v) For each n = 0,1,2,..., we can write 8*n + 6 as x^2 + y^2 + z^2 with x + 2*y a square, where x,y,z are integers with y nonnegative and z odd.
The Gauss-Legendre theorem asserts that a nonnegative integer can be written as the sum of three squares if and only if it is not of the form 4^k*(8m+7) with k and m nonnegative integers. Thus a(4^k*(8m+7)) = 0 for all k,m = 0,1,2,....
See also A283273 for a similar conjecture.

Examples

			a(11) = 1 since 11 = 3^2 + 1^2 + (-1)^2 with 3 + 3*1 + 5*(-1) = 1^2.
a(43) = 1 since 43 = (-5)^2 + (-3)^2 + 3^2 with (-5) + 3*(-3) + 5*3 = 1^2.
a(75) = 1 since 75 = (-1)^2 + 5^2 + 7^2 with (-1) + 3*5 + 5*7 = 7^2.
a(262) = 1 since 262 = 1^2 + 15^2 + (-6)^2 with 1 + 3*15 + 5*(-6) = 4^2.
a(277) = 1 since 277 = (-6)^2 + 4^2 + 15^2 with (-6) + 3*4 + 5*15 = 9^2.
a(617) = 1 since 617 = 17^2 + 18^2 + 2^2 with 17 + 3*18 + 5*2 = 9^2.
a(1430) = 1 since 1430 = (-13)^2 + (-6)^2 + 35^2 with (-13) + 3*(-6) + 5*35 = 12^2.
a(5272) = 1 since 5272 = (-66)^2 + 30^2 + (-4)^2 with (-66) + 3*30 + 5*(-4) = 2^2.
a(7630) = 1 since 7630 = (-78)^2 + 39^2 + 5^2 with (-78) + 3*39 + 5*5 = 8^2.
a(7933) = 1 since 7933 = (-56)^2 + 69^2 + (-6)^2 with (-56) + 3*69 + 5*(-6) = 11^2.
a(14193) = 1 since 14193 = (-7)^2 + 112^2 + 40^2 with (-7) + 3*112 + 5*40 = 23^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Do[r=0;Do[If[SQ[n-x^2-y^2]&&SQ[(-1)^i*x+(-1)^j*3y+(-1)^k*5*Sqrt[n-x^2-y^2]],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{i,0,Min[x,1]},{j,0,Min[y,1]},{k,0,Min[Sqrt[n-x^2-y^2],1]}];Print[n," ",r];Continue,{n,0,80}]

A283273 Number of ways to write n as x^2 + y^2 + z^2 with x,y integers and z a nonnegative integer such that x + 2*y + 3*z is a square or twice a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 04 2017

Keywords

Comments

Conjecture: (i) For any nonnegative integer n not of the form 4^k*(8*m+7) (k,m = 0,1,2,...), we have a(n) > 0.
(ii) Any nonnegative integer not of the form 4^k*(8*m+7) (k,m = 0,1,2,...) can be written as x^2 + y^2 + z^2 with x,y,z integers such that a*x + b*y + c*z is a square or twice a square, whenever (a,b,c) is among the triples (1,2,5), (1,3,4), (1,4,7), (1,6,7), (2,3,4), (2,3,9), (3,4,7), (3,4,9).
The Gauss-Legendre theorem states that a nonnegative integer can be written as the sum of three squares if and only if it is not of the form 4^k*(8*m+7) with k and m nonnegative integers. Thus a(4^k*(8*m+7)) = 0 for all k,m = 0,1,2,..., and part (i) of our conjecture (verified for n up to 1.6*10^6) is stronger than the Gauss-Legendre theorem.
See also A283269 for a similar conjecture.

Examples

			a(82) = 1 since 82 = 0^2 + (-1)^2 + 9^2 with 0 + 2*(-1) + 3*9 = 5^2.
a(328) = 1 since 328 = 0^2 + (-2)^2 + 18^2 with 0 + 2*(-2) + 3*18 = 2*5^2.
a(330) = 1 since 330 = 5^2 + 4^2 + 17^2 with 5 + 2*4 + 3*17 = 8^2.
a(466) = 1 since 466 = 21^2 + 0^2 + 5^2 with 21 + 2*0 + 3*5 = 6^2.
a(1320) = 1 since 1320 = 10^2 + 8^2 + 34^2 with 10 + 2*8 + 3*34 = 2*8^2.
a(1387) = 1 since 1387 = 33^2  + (-17)^2 + 3^2 with 33 + 2*(-17) + 3*3 = 2*2^2.
a(1857) = 1 since 1857 = (-37)^2 + (-2)^2 + 22^2 with (-37) + 2*(-2) + 3*22 = 5^2.
a(1864) = 1 since 1864 = 42^2 + 0^2 + 10^2 with 42 + 2*0 + 3*10 = 2*6^2.
a(2386) = 1 since 2386 = (-44)^2 + 3^2 + 21^2 with (-44) + 2*3 + 3*21 = 5^2.
a(5548) = 1 since 5548 = 66^2 + (-34)^2 + 6^2 with 66 + 2*(-34) + 3*6 = 4^2.
a(7428) = 1 since 7428 = (-74)^2 + (-4)^2 + 44^2 with (-74) + 2*(-4) + 3*44 = 2*5^2.
a(9544) = 1 since 9544 = (-88)^2 + 6^2 + 42^2 with (-88) + 2*6 + 3*42 = 2*5^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    TQ[n_]:=TQ[n]=SQ[n]||SQ[2n];
    Do[r=0;Do[If[SQ[n-x^2-y^2]&&TQ[(-1)^i*x+(-1)^j*2y+3*Sqrt[n-x^2-y^2]],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{i,0,Min[x,1]},{j,0,Min[y,1]}];Print[n," ",r];Continue,{n,0,80}]
Showing 1-4 of 4 results.