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 1-8 of 8 results.

A279612 Number of ways to write n = x^2 + y^2 + z^2 + w^2 with x + 2*y - 2*z a power of 4 (including 4^0 = 1), where x,y,z,w are nonnegative integers.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 1, 1, 3, 5, 2, 1, 3, 4, 1, 1, 3, 5, 5, 4, 3, 2, 3, 2, 4, 5, 1, 3, 4, 4, 1, 1, 5, 7, 7, 2, 3, 7, 3, 2, 4, 3, 4, 2, 8, 5, 1, 1, 6, 8, 3, 6, 7, 8, 2, 3, 3, 6, 8, 4, 6, 5, 2, 2, 9, 7, 7, 7, 7, 12, 3, 1, 9, 10, 7, 1, 10, 10, 2, 3
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 15 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*q (k = 0,1,2,... and q = 1, 2, 3, 6, 7, 8, 12, 15, 27, 31, 47, 72, 76, 92, 111, 127).
(ii) Let a and b be positive integers with gcd(a,b) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x - b*y a power of two (including 2^0 = 1) if and only if (a,b) = (1,1), (2,1), (2,3).
(iii) Let a,b,c be positive integers with a <= b and gcd(a,b,c) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y - c*z a power of two if and only if (a,b,c) is among the triples (1,1,1), (1,1,2), (1,2,1), (1,2,2), (1,3,1), (1,3,2), (1,3,3), (1,3,4), (1,3,5), (1,4,1), (1,4,2), (1,4,3), (1,4,4), (1,5,1), (1,5,2), (1,5,4), (1,5,5,), (1,6,3), (1,7,4), (1,7,7), (1,8,1), (1,9,2), (2,3,1), (2,3,3), (2,3,4), (2,5,1), (2,5,3), (2,5,4), (2,5,5), (2,7,1), (2,7,3), (2,7,7), (2,9,3), (2,11,5), (3,4,3), (7,8,7).
(iv) Let a,b,c be positive integers with b <= c and gcd(a,b,c) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x - b*y - c*z a power of two if and only if (a,b,c) is among the triples (2,2,1), (4,2,1), (4,3,1), (4,4,3).
(v) Let a,b,c be positive integers with a <= b, c <= d, and gcd(a,b,c,d) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y - c*z -d*w a power of two if and only if (a,b,c,d) is among the quadruples (1,2,1,1), (1,2,1,2), (1,2,1,3), (1,3,1,2), (1,3,2,3), (1,3,2,4), (1,4,1,2), (1,7,2,6), (1,9,1,4), (2,2,2,3), (2,3,1,2), (2,3,1,3), (2,3,2,3), (2,3,6,1), (2,4,1,2), (2,5,1,2), (2,5,2,3), (2,5,3,4), (3,4,1,2), (3,4,1,3), (3,4,1,5), (3,4,2,5), (3,4,3,4), (3,8,1,10), (3,8,2,3), (4,5,1,5).
(vi) Let a,b,c be positive integers with a <= b <= c and gcd(a,b,c,d) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y + c*z -d*w a power of two if and only if (a,b,c,d) is among the quadruples (1,1,2,2), (1,1,2,3), (1,1,2,4), (1,2,2,3), (1,2,3,4), (1,2,4,3), (1,2,6,7), (1,3,4,4), (1,4,6,5), (2,3,5,4).
(vii) For any positive integers a,b,c,d, not all positive integers can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x - b*y - c*z -d*w a power of two.
(viii) Let a and b be positive integers, and c and d be nonnegative integers. Then, not all positive integers can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y + c*z + d*w a power of two.
We have verified a(n) > 0 for all n = 1..2*10^7. The conjecture that a(n) > 0 for all n > 0 appeared in arXiv:1701.05868.

Examples

			a(12) = 1 since 12 = 1^2 + 1^2 + 1^2 + 3^2 with 1 + 2*1 - 2*1 = 4^0.
a(15) = 1 since 15 = 3^2 + 1^2 + 2^2 + 1^2 with 3 + 2*1 - 2*2 = 4^0.
a(27) = 1 since 27 = 4^2 + 1^2 + 1^2 + 3^2 with 4 + 2*1 - 2*1 = 4.
a(31) = 1 since 31 = 3^2 + 2^2 + 3^2 + 3^2 with 3 + 2*2 - 2*3 = 4^0.
a(47) = 1 since 47 = 3^2 + 2^2 + 3^2 + 5^2 with 3 + 2*2 - 2*3 = 4^0.
a(72) = 1 since 72 = 8^2 + 0^2 + 2^2 + 2^2 with 8 + 2*0 - 2*2 = 4.
a(76) = 1 since 76 = 1^2 + 5^2 + 5^2 + 5^2 with 1 + 2*5 - 2*5 = 4^0.
a(92) = 1 since 92 = 4^2 + 6^2 + 6^2 + 2^2 with 4 + 2*6 - 2*6 = 4.
a(111) = 1 since 111 = 9^2 + 1^2 + 5^2 + 2^2 with 9 + 2*1 - 2*5 = 4^0.
a(127) = 1 since 127 = 7^2 + 2^2 + 5^2 + 7^2 with 7 + 2*2 - 2*5 = 4^0.

Links

Crossrefs

Cf. A000079, A000118, A000290, A271518, A275656, A275675, A275676, A275738, A278560, A279616.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
FP[n_]:=FP[n]=n>0&&IntegerQ[Log[4,n]];
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&FP[x+2y-2z],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]

A275648 Number of ordered ways to write n as x^2*(1+y^2+z^2)+w^2, where x is a positive integer and y,z,w are nonnegative integers with y <= z <= w.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 04 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 6, 7, 11, 14, 15, 23, 38, 47, 71, 77, 143, 152, 191, 608, 2^(2k+1) (k = 0,1,2,...).
This is stronger than Lagrange's four-square theorem.
See A275656 for a stronger conjecture.

Examples

			a(1) = 1 since 1 = 1^2*(1+0^2+0^2) + 0^2 with 0 = 0 = 0.
a(2) = 1 since 2 = 1^2*(1+0^2+0^2) + 1^2 with 0 = 0 < 1.
a(3) = 1 since 3 = 1^2*(1+0^2+1^2) + 1^2 with 0 < 1 = 1.
a(6) = 1 since 6 = 1^2*(1+0^2+1^2) + 2^2 with 0 < 1 < 2.
a(7) = 1 since 7 = 1^2*(1+1^2+1^2) + 2^2 with 1 = 1 < 2.
a(11) = 1 since 11 = 1^2*(1+0^2+1^2) + 3^2 with 0 < 1 < 3.
a(14) = 1 since 14 = 1^2*(1+0^2+2^2) + 3^2 with 0 < 2 < 3.
a(15) = 1 since 15 = 1^2*(1+1^2+2^2) + 3^2 with 1 < 2 < 3.
a(23) = 1 since 23 = 1^2*(1+2^2+3^2) + 3^2 with 2 < 3 = 3.
a(38) = 1 since 38 = 1^2*(1+0^2+1^2) + 6^2 with 0 < 1 < 6.
a(47) = 1 since 47 = 1^2*(1+1^2+3^2) + 6^2 with 1 < 3 < 6.
a(71) = 1 since 71 = 1^2*(1+3^2+5^2) + 6^2 with 3 < 5 < 6.
a(77) = 1 since 77 = 1^2*(1+2^2+6^2) + 6^2 with 2 < 6 = 6.
a(143) = 1 since 143 = 1^2*(1+5^2+6^2) + 9^2 with 5 < 6 < 9.
a(152) = 1 since 152 = 2^2*(1+0^2+1^2) + 12^2 with 0 < 1 < 12.
a(191) = 1 since 191 = 1^2*(1+3^2+9^2) + 10^2 with 3 < 9 < 10.
a(608) = 1 since 608 = 4^2*(1+0^2+1^2) + 24^2 with 0 < 1 < 24.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275656.

Programs

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

A275675 Number of ordered ways to write n as 4^k*(1+x^2+y^2)+5*z^2, where k,x,y,z are nonnegative integers with x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 04 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
This is stronger than the known fact that any natural number can be written as w^2 + x^2 + y^2 + 5*z^2 with w,x,y,z integers.
See also A275656, A275676 and A275678 for similar conjectures.

Examples

			a(43) = 1 since 43 = 4^0*(1+1^2+6^2) + 5*1^2.
a(45) = 1 since 45 = 4*(1+0^2+3^2) + 5*1^2.
a(237) = 1 since 237 = 4^3*(1+1^2+1^2) + 5*3^2.
a(561) = 1 since 561 = 4*(1+8^2+8^2) + 5*3^2.
a(9777) = 1 since 9777 = 4*(1+11^2+31^2) + 5*33^2.
a(39108) = 1 since 39108 = 4^2*(1+11^2+31^2) + 5*66^2.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275648, A275656, A275676, A275678.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[(n-4^k*(1+x^2+y^2))/5],r=r+1],{k,0,Log[4,n]},{x,0,Sqrt[(n/4^k-1)/2]},{y,x,Sqrt[n/4^k-1-x^2]}];Print[n," ",r];Continue,{n,1,80}]

A275676 Number of ordered ways to write n as 4^k*(1+5*x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 04 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Any positive integer can be written as 4^k*(1+5*x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with y <= z.
See also A275656, A275675 and A275678 for similar conjectures.

Examples

			a(4) = 1 since 4 = 4*(1+5*0^2+0^2) + 0^2 with 0 = 0.
a(259) = 1 since 259 = 4^0*(1+5*4^2+13^2) + 3^2 with 4 < 13.
a(333) = 1 since 333 = 4*(1+5*3^2+5^2) + 7^2 with 3 < 5.
a(621) = 1 since 621 = 4*(1+5*0^2+8^2) + 19^2 with 0 < 8.
a(717) = 1 since 717 = 4*(1+5*3^2+11^2) + 7^2 with 3 < 11.
a(1581) = 1 since 1581 = 4*(1+5*1^2+3^2) + 39^2 with 1 < 3.
a(2541) = 1 since 2541 = 4*(1+5*3^2+13^2) + 41^2 with 3 < 13.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275648, A275656, A275675, A275678

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-4^k*(1+5x^2+y^2)],r=r+1],{k,0,Log[4,n]},{x,0,Sqrt[(n/4^k-1)/6]},{y,x,Sqrt[n/4^k-1-5x^2]}];Print[n," ",r];Continue,{n,1,80}]

A275678 Number of ordered ways to write n as 4^k*(1+4*x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 05 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Any positive integer can be written as 4^k*(1+4*x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with x <= z.
This is stronger than Lagrange's four-square theorem. We have shown that each n = 1,2,3,... can be written as 4^k*(1+4*x^2+y^2) + z^2 with k,x,y,z nonnegative integers.
See also A275656, A275675 and A275676 for similar conjectures.

Examples

			a(12) = 1 since 12 = 4*(1+4*0^2+1^2) + 2^2 with 0 < 1.
a(19) = 1 since 19 = 4^0*(1+4*0^2+3^2) + 3^2 with 0 < 3.
a(61) = 1 since 61 = 4*(1+4*1^2+2^2) + 5^2 with 1 < 2.
a(125) = 1 since 125 = 4*(1+4*0^2+0^2) + 11^2 with 0 = 0.
a(359) = 1 since 359 = 4^0*(1+4*7^2+9^2) + 9^2 with 7 < 9.
a(196253) = 1 since 196253 = 4*(1+4*0^2+0^2) + 443^2 with 0 = 0.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275648, A275656, A275675, A275676.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-4^k*(1+4x^2+y^2)],r=r+1],{k,0,Log[4,n]},{x,0,Sqrt[(n/4^k-1)/5]},{y,x,Sqrt[n/4^k-1-4x^2]}];Print[n," ",r];Continue,{n,1,80}]

A275738 Number of ordered ways to write n as w^2 + x^2*(1+y^2+z^2), where w,x,y,z are nonnegative integers with x > 0, y <= z and y == z (mod 2).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 07 2016

Keywords

Comments

Conjecture: For any n > 0, we have a(n) > 0, i.e., n can be written as w^2 + x^2*(1+(z-y)^2+(y+z)^2) = w^2 + x^2*(1+2*y^2+2*z^2), where w,x,y,z are nonnegative integers with x > 0 and y <= z. Moreover, any positive integer n not equal to 449 can be written as 4^k*(1+x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with x == y (mod 2).
This is stronger than Lagrange's four-square theorem, and we have verified it for n up to 10^6.
See also A275648, A275656, A275675, A275676 and A275678 for similar conjectures.

Examples

			a(2) = 1 since 2 = 1^2 + 1^2*(1+0^2+0^2) with 0 + 0 even.
a(7) = 1 since 7 = 2^2 + 1^2*(1+1^2+1^2) with 1 + 1 even.
a(59) = 1 since 59 = 0^2 + 1^2*(1+3^2+7^2) with 3 + 7 even.
a(71) = 1 since 71 = 6^2 + 1^2*(1+3^2+5^2) with 3 + 5 even.
a(113) = 2 since 113 = 7^2 + 8^2*(1+0^2+0^2) = 8^2 + 7^2*(1+0^2+0^2) with 0 + 0 even.
a(143) = 1 since 143 = 6^2 + 1^2*(1+5^2+9^2) with 5 + 9 even.
a(191) = 1 since 191 = 10^2 + 1^2*(1+3^2+9^2) with 3 + 9 even.
a(449) = 3 since 449 = 18^2 + 5^2*(1+0^2+2^2) with 0 + 2 even, and 449 = 7^2 + 20^2*(1+0^2+0^2) = 20^2 + 7^2*(1+0^2+0^2) with 0 + 0 even.
a(497) = 1 since 497 = 15^2 + 4^2*(1+0^2+4^2) with 0 + 4 even.
a(2033) = 1 since 2033 = 33^2 + 4^2*(1+3^2+7^2) with 3 + 7 even.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275648, A275656, A275675, A275676, A275678.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2*(1+2y^2+2z^2)],r=r+1],{x,1,Sqrt[n]},{y,0,Sqrt[(n/x^2-1)/4]},{z,y,Sqrt[(n/x^2-1-2y^2)/2]}];Print[n," ",r];Continue,{n,1,80}]

A279616 Numbers of the form x^2 + y^2 + z^2 with x + 2*y - 2*z a power of four (including 4^0 = 1), where x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 4, 5, 9, 10, 14, 16, 17, 18, 19, 20, 22, 24, 29, 33, 34, 35, 37, 41, 45, 48, 49, 50, 51, 52, 53, 58, 59, 61, 64, 65, 66, 68, 69, 70, 73, 74, 77, 78, 80, 82, 84, 88, 89, 90, 94, 97, 98, 99, 100, 104, 106, 107, 109, 113, 114, 116, 117, 121, 122, 125, 129, 130, 132, 133, 138, 139, 141, 144
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 15 2016

Keywords

Comments

Part (i) of the conjecture in A279612 implies that any positive integer can be written as the sum of a square and a term of the current sequence.
It seems that a(n)/n has the limit 2 as n tends to the infinity.

Examples

			a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 + 2*0 - 2*0 = 4^0.
a(2) = 3 since 3 = 1^2 + 1^2 + 1^2 + 0^2 with 1 + 2*1 - 2*1 = 4^0.
a(4) = 5 since 5 = 2^2 + 1^2 + 0^2 + 0^2 with 2 + 2*1 - 2*0 = 4.

Links

Crossrefs

Cf. A000079, A000290, A271518, A275656, A275675, A275676, A275738, A278560, A279612.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
FP[n_]:=FP[n]=n>0&&IntegerQ[Log[4,n]];
ex={};Do[Do[If[SQ[m-x^2-y^2]&&FP[x+2y-2*Sqrt[m-x^2-y^2]],ex=Append[ex,m];Goto[aa]],{x,0,Sqrt[m]},{y,0,Sqrt[m-x^2]}];Label[aa];Continue,{m,1,144}]

A275471 Number of ordered ways to write n as 4^k*(1+x^2+y^2)+z^2, where k,x,y,z are nonnegative integers with x <= y and x == y (mod 2).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 11 2016

Keywords

Comments

Conjecture: a(n) > 0 except for n = 449.
See also A275656, A275678 and A275738 for related conjectures.
As x^2 + y^2 = 2*((x+y)/2)^2 + 2*((x-y)/2)^2, we see that {x^2 + y^2: x and y are integers with x == y (mod 2)} = {2*x^2 + 2*y^2: x and y are integers}.

Examples

			a(8) = 1 since 8 = 4*(1+0^2+0^2) + 2^2 with 0+0 even.
a(31) = 1 since 31 = 4^0*(1+1^2+5^2) + 2^2 with 1+5 even.
a(47) = 1 since 47 = 4^0*(1+1^2+3^2) + 6^2 with 1+3 even.
a(79) = 1 since 79 = 4^0*(1+5^2+7^2)+2^2 with 5+7 even.
a(1009) = 1 since 1009 = 4^2*(1+1^2+1^2) + 31^2 with 1+1 even.
a(7793) = 1 since 7793 = 4^2*(1+12^2+18^2) + 17^2 with 12+18 even.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275648, A275656, A275675, A275676, A275678, A275738.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-4^k*(1+2x^2+2y^2)],r=r+1],{k,0,Log[4,n]},{x,0,Sqrt[(n/4^k-1)/4]},{y,x,Sqrt[(n/4^k-1-2x^2)/2]}];Print[n," ",r];Continue,{n,1,80}]
Showing 1-8 of 8 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.