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 1-7 of 7 results.

A271608 Number of ordered ways to write n as pen(u) + pen(v) + pen(x) + pen(y) + pen(z) with u,v,x,y,z nonnegative integers such that u + 2*v + 4*x + 5*y + 6*z is a pentagonal number, where pen(k) denotes the pentagonal number k*(3k-1)/2.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 10 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 2, 4, 5, 7, 9, 21, 22, 43. Also, every n = 0,1,2,... can be written as pen(u) + pen(v) + pen(x) + pen(y) + pen(z) with u,v,x,y,z nonnegative integers such that 3*u + 5*v + 11*x + 16*y + 19*z is also a pentagonal number.
(ii) Any integer n > 43 can be written as the sum of five pentagonal numbers u, v, x, y and z such that u + 2*v + 5*x + 7*y + 10*z is also a pentagonal number. Also, each integer n > 10 can be written as the sum of five pentagonal numbers u, v, x, y and z such that u + 2*v + 5*x + 7*y + 10*z is a square.
(iii) Any natural number n can be written as u^2 + v^2 + x^2 + y^2 + z^2 with u^2 + 2*v^2 + 3*x^2 + 4*y^2 + 5*z^2 a square, where u, v, x, y and z are integers.
As conjectured by Fermat and proved by Cauchy, each natural number can be written as the sum of five pentagonal numbers.
See also A271510, A271513, A271518 and A271644 for some similar conjectures refining Lagrange's four-square theorem.

Examples

			a(7) = 1 since 7 = 5 + 0 + 1 + 0 + 1 = pen(2) + pen(0) + pen(1) + pen(0) + pen(1) with 2 + 2*0 + 4*1 + 5*0 + 6*1 = 12 = pen(3).
a(9) = 1 since 9 = 1 + 1 + 5 + 1 + 1 = pen(1) + pen(1) + pen(2) + pen(1) + pen(1) with 1 + 2*1 + 4*2 + 5*1 + 6*1 = 22 = pen(4).
a(22) = 1 since 22 = 0 + 0 + 5 + 12 + 5 = pen(0) + pen(0) + pen(2) + pen(3) + pen(2) with 0 + 2*0 + 4*2 + 5*3 + 6*2 = 35 = pen(5).
a(43) = 1 since 43 = 5 + 1 + 35 + 1 + 1 = pen(2) + pen(1) + pen(5) + pen(1) + pen(1) with 2 + 2*1 + 4*5 + 5*1 + 6*1 = 35 = pen(5).
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    pen[x_]:=pen[x]=x*(3x-1)/2
    pQ[n_]:=pQ[n]=SQ[24n+1]&&(n==0||Mod[Sqrt[24n+1]+1,6]==0)
    Do[r=0;Do[If[pQ[n-pen[x]-pen[y]-pen[z]-pen[w]]&&pQ[x+2y+4z+5w+6*Floor[(Sqrt[24(n-pen[x]-pen[y]-pen[z]-pen[w])+1]+1)/6]],r=r+1],{x,0,(Sqrt[24n+1]+1)/6},{y,0,(Sqrt[24(n-pen[x])+1]+1)/6},{z,0,(Sqrt[24(n-pen[x]-pen[y])+1]+1)/6},{w,0,(Sqrt[24(n-pen[x]-pen[y]-pen[z])+1]+1)/6}];Print[n," ",r];Label[aa];Continue,{n,0,70}]

A271714 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 such that (10*w+5*x)^2 + (12*y+36*z)^2 is a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 2, 1, 3, 1, 2, 1, 2, 3, 1, 4, 4, 2, 2, 1, 3, 3, 5, 2, 2, 5, 2, 1, 2, 3, 3, 3, 2, 3, 2, 3, 4, 4, 2, 3, 9, 2, 3, 1, 1, 6, 2, 3, 4, 6, 4, 1, 2, 5, 3, 3, 4, 3, 5, 1, 4, 5, 1, 3, 6, 6, 1, 3, 4, 5, 12, 2, 4, 6, 2, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 12 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 7, 9, 19, 49, 133, 589, 2^k, 2^k*3, 4^k*q (k = 0,1,2,... and q = 14, 67, 71, 199).
(ii) If P(y,z) is one of 2y-3z, 2y-8z and 4y-6z, then any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (w-x)^2 + P(y,z)^2 is a square.
(iii) For each triple (a,b,c) = (1,4,4), (1,12,12), (2,4,8), (2,6,6), (2,12,12), (3,4,4), (3,4,8), (3,8,8), (3,12,12), (3,12,36), (5,4,4), (5,4,8), (5,8,16), (5,36,36), (6,4,4), (7,12,12), (7,20,20), (7,24,24), (9,4,4), (9,12,12),(9,36,36), (11,12,12), (13,4,4), (15,12,12), (16,12,12), (21,20,20), (21,24,24), (23,12,12), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (w+a*x)^2 + (b*y-c*z)^2 is a square.
See also A271510, A271513, A271518, A271644, A271665, A271721 and A271724 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(2) = 1 since 2 = 1^2 + 1^2 + 0^2 + 0^2 with (10*1+5*1)^2 + (12*0+36*0)^2 = 15^2 + 0^2 = 15^2.
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with (10*1+5*1)^2 + (12*0+36*1)^2 = 15^2 + 36^2 = 39^2.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with (10*2+5*0)^2 + (12*0+36*0)^2 = 20^2 + 0^2 = 20^2.
a(6) = 1 since 6 = 2^2 + 0^2 + 1^2 + 1^2 with (10*2+5*0)^2 + (12*1+36*1)^2 = 20^2 + 48^2 = 52^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with (10*1+5*2)^2 + (12*1+36*1)^2 = 20^2 + 48^2 = 52^2.
a(9) = 1 since 9 = 3^2 + 0^2 + 0^2 + 0^2 with (10*3+5*0)^2 + (12*0+36*0)^2 = 30^2 + 0^2 = 30^2.
a(19) = 1 since 19 = 3^2 + 0^2 + 3^2 + 1^2 with (10*3+5*0)^2 + (12*3+36*1)^2 = 30^2 + 72^2 = 78^2.
a(49) = 1 since 49 = 7^2 + 0^2 + 0^2 + 0^2 with (10*7+5*0)^2 + (12*0+36*0)^2 = 70^2 + 0^2 = 70^2.
a(133) = 1 since 133 = 9^2 + 0^2 + 6^2 + 4^2 with (10*9+5*0)^2 + (12*6+36*4)^2 = 90^2 + 216^2 = 234^2.
a(589) = 1 since 589 = 17^2 + 10^2 + 2^2 + 14^2 with (10*17+5*10)^2 + (12*2+36*14)^2 = 220^2 + 528^2 = 572^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(10*Sqrt[n-x^2-y^2-z^2]+5x)^2+(12y+36z)^2],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]

A271665 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 such that w^2 + 4*x*y + 8*y*z + 32*z*x is a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 1, 1, 6, 3, 1, 3, 1, 6, 2, 1, 7, 10, 1, 1, 9, 3, 2, 6, 2, 2, 3, 3, 8, 10, 1, 1, 10, 2, 2, 3, 5, 8, 11, 1, 7, 13, 2, 6, 16, 6, 1, 2, 6, 2, 3, 1, 3, 16, 4, 7, 9, 3, 2, 10, 4, 9, 4, 1, 8, 15, 1, 1, 15, 5, 2, 9, 6, 8, 2, 3, 10, 13, 4, 2, 17, 7, 1, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 12 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*3^m, 4^k*3^m*43, 4^k*9^m*q (k,m = 0, 1, 2, ... and q = 7, 15, 79, 95, 141, 159, 183).
(ii) Any positive integer n can be written as w^2 + x^2 + y^2 + z^2 with w*x + x*y + 2*y*z + 3*z*x (or w*x + 3*x*y + 8*y*z + 5*z*x) twice a square, where w is a positive integer and x,y,z are nonnegative integers.
(iii) For each k = 1, 2, 8, any positive integer can be written as w^2 + x^2 + y^2 + z^2 with w^2 + k*(x*y+y*z) a square, where w is a positive integer and x,y,z are nonnegative integers.
(iv) For each ordered pair (b,c) = (16,4), (24,4), (32,16), any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x^2 + b*y^2 + c*x*z + c*y*z + c*z*w is a square.
We also guess that for each triple (b,c,d) = (1,3,4), (1,6,8), (1,7,24), (1,8,15), (1,10,24), (1,12,35), (1,14,48), (1,20,48), (2,1,2), (2,4,2), (2,4,7), (2,6,7), (2,8,4), (2,8,14), (2,8,31), (2,10,23), (2,12,14), (2,12,34), (2,14,47), (3,1,1), (3,1,4), (3,1,16), (3,2,14), (3,2,17), (3,2,38), (3,3,3), (3,3,13), (3,4,1), (3,4,4), (3,5,11), (3,6,3), (3,6,6), (3,6,26), (3,8,2), (3,8,13), (3,8,22), (3,9,39), (3,12,12), (3,12,33), (3,15,1), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and x^2 + b*y^2 + c*x*z + d*y*z a square.
See also A271510, A271513, A271518 and A271644 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 1^2 + 4*0*1 + 8*1*1 + 32*1*0 = 3^2.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with 2^2 + 4*2*0 + 8*0*0 + 32*0*0 = 2^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1^2 + 4*2*1 + 8*1*1 + 32*1*2 = 9^2.
a(15) = 1 since 15 = 1^2 + 2^2 + 1^2 + 3^2 with 1^2 + 4*2*1 + 8*1*3 + 32*3*2 = 15^2.
a(43) = 1 since 43 = 3^2 + 3^2 + 4^2 + 3^2 with 3^2 + 4*3*4 + 8*4*3 + 32*3*3 = 21^2.
a(79) = 1 since 79 = 5^2 + 3^2 + 6^2 + 3^2 with 5^2 + 4*3*6 + 8*6*3 + 32*3*3 = 23^2.
a(95) = 1 since 95 = 5^2 + 6^2 + 5^2 + 3^2 with 5^2 + 4*6*5 + 8*5*3 + 32*3*6 = 29^2.
a(129) = 1 since 129 = 5^2 + 6^2 + 8^2 + 2^2 with 5^2 + 4*6*8 + 8*8*2 + 32*2*6 = 27^2.
a(141) = 1 since 141 = 8^2 + 5^2 + 4^2 + 6^2 with 8^2 + 4*5*4 + 8*4*6 + 32*6*5 = 36^2.
a(159) = 1 since 159 = 11^2 + 1^2 + 6^2 + 1^2 with 11^2 + 4*1*6 + 8*6*1 + 32*1*1 = 15^2.
a(183) = 1 since 183 = 1^2 + 9^2 + 10^2 + 1^2 with 1^2 + 4*9*10 + 8*10*1 + 32*1*9 = 27^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[4x*y+8*y*z+32*z*x+(n-x^2-y^2-z^2)],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]

A271724 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w*(x+2*y+3*z) a square, where w,x,y,z are nonnegative integers with x > 0.

Original entry on oeis.org

1, 3, 2, 1, 4, 4, 1, 3, 4, 6, 4, 2, 4, 7, 1, 1, 10, 8, 5, 6, 8, 5, 1, 4, 7, 10, 7, 2, 11, 13, 2, 3, 8, 9, 8, 6, 7, 13, 3, 6, 15, 8, 4, 4, 13, 8, 1, 2, 8, 15, 11, 4, 14, 18, 5, 7, 6, 6, 12, 5, 12, 17, 5, 1, 16, 21, 3, 11, 16, 12, 1, 8, 8, 18, 16, 5, 16, 12, 4, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 13 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 7, 15, 47, 151, 4^k*q (k = 0,1,2,... and q = 1, 23, 71).
(ii) For positive integers a,b,c with gcd(a,b,c) squarefree, any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and w*(a*x+b*y+c*z) a square, if and only if {a,b,c} is among {1,2,3}, {1,3,6}, {1,6,9}, {5,6,9}, {18,30,114}.
(iii) For each quadruple (a,b,c,d) = (1,1,2,12), (1,2,7,60), (1,3,9,48), (1,4,11,48), (1,5,8,24), (1,8,11,24), (2,6,8,15), (3,5,6,24), (3,6,15,40), (3,6,18,40), (3,12,15,20), (4,4,8,15), (4,8,12,21), (4,8,12,45), (4,8,20,15), (4,8,36,45), (5,10,15,24), (6,9,15,20), (7,14,28,60), (7,21,28,60), (7,21,42,60), (12,36,48,55), (14,21,28,60), (3,9,18,112), (3,21,33,80), (4,5,9,120), (4,12,16,105), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (a*x+b*y+c*z)^2 + (d*w)^2 is a square.
See also A271510, A271513, A271518, A271644, A271665, A271714 and A271721 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(1) = 1 since 1 = 0^2 + 1^2 + 0^2 + 0^2 with 1 > 0 and 0*(1+2*0+3*0) = 0^2.
a(3) = 2 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 1*(1+2*0+3*1) = 2^2, and 3 = 0^2 + 1^2 + 1^2 + 1^2 with 0*(1+2*1+3*1) = 0^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1*(1+2*1+3*2) = 3^2.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^2 + 1^2 with 2*(3+2*1+3*1) = 4^2.
a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with 1*(3+2*2+3*3) = 4^2.
a(31) = 2 since 31 = 2^2 + 1^2 + 1^2 + 5^2 with 2*(1+2*1+3*5) = 6^2, and also 31 = 2^2 + 3^2 + 3^2 + 3^2 with 2*(3+2*3+3*3) = 6^2.
a(47) = 1 since 47 = 1^2 + 1^2 + 3^2 + 6^2 with 1*(1+2*3+3*6) = 5^2.
a(71) = 1 since 71 = 1^2 + 6^2 + 5^2 + 3^2 with 1*(6+2*5+3*3) = 5^2.
a(151) = 1 since 151 = 9^2 + 6^2 + 5^2 + 3^2 with 9*(6+2*5+3*3) = 15^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[Sqrt[n-x^2-y^2-z^2](x+2y+3z)],r=r+1],{x,1,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Label[aa];Continue,{n,1,80}]

A271775 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 (x >= y >= z <= w) with x - y a square, where w,x,y,z are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 13 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 3, 11, 47, 2^{4k+3}*m (k = 0,1,2,... and m = 1, 3, 7, 15, 79).
(ii) Let a and b be positive integers with a <= b and gcd(a,b) squarefree. Then any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x-b*y a square, if and only if (a,b) is among the ordered pairs (1,1), (2,1), (2,2), (4,3), (6,2). Verified for all nonnegative integers up to 10^11. - Mauro Fiorentini, Jun 14 2024
(iii) Let a and b be positive integers with gcd(a,b) squarefree. Then any natural number 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 square, if and only if {a,b} is among {1,2}, {1,3} and {1,24}. Verified for all nonnegative integers up to 10^11. - Mauro Fiorentini, Jun 14 2024
(iv) Let a,b,c be positive integers with a <= b and gcd(a,b,c) squarefree. Then, any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x+b*y-c*z a square, if and only if (a,b,c) is among the triples (1,1,1), (1,1,2), (1,2,1), (1,2,2), (1,2,3), (1,3,1), (1,3,3), (1,4,4), (1,5,1), (1,6,6), (1,8,6), (1,12,4), (1,16,1), (1,17,1), (1,18,1), (2,2,2), (2,2,4), (2,3,2), (2,3,3), (2,4,1), (2,4,2), (2,6,1), (2,6,2), (2,6,6), (2,7,4), (2,7,7), (2,8,2), (2,9,2), (2,32,2), (3,3,3), (3,4,2), (3,4,3), (3,8,3), (4,5,4), (4,8,3), (4,9,4), (4,14,14), (5,8,5), (6,8,6), (6,10,8), (7,9,7), (7,18,7), (7,18,12), (8,9,8), (8,14,14), (8,18,8), (14,32,14), (16,18,16), (30,32,30), (31,32,31), (48,49,48), (48,121,48). Verified for all nonnegative integers up to 10^11. - Mauro Fiorentini, Jun 14 2024
(v) Let a,b,c be positive integers with b <= c and gcd(a,b,c) squarefree. Then, any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x-b*y-c*z a square, if and only if (a,b,c) is among the triples (1,1,1), (2,1,1), (2,1,2), (3,1,2) and (4,1,2).
(vi) Let a,b,c,d be positive integers with a <= b, c <= d and gcd(a,b,c,d) squarefree. Then, any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x+b*y-(c*z+d*w) a square, if and only if (a,b,c,d) is among the quadruples (1,2,1,1), (1,2,1,2), (1,3,1,2), (1,4,1,3), (2,4,1,2), (2,4,2,4), (8,16,7,8), (9,11,2,9) and (9,16,2,7).
(vii) Let a,b,c,d be positive integers with a <= b <= c and gcd(a,b,c,d) squarefree. Then, any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x+b*y+c*z-d*w a square, if and only if (a,b,c,d) is among the quadruples (1,1,2,1), (1,2,3,1), (1,2,3,3), (1,2,4,2), (1,2,4,4), (1,2,5,5), (1,2,6,2), (1,2,8,1), (2,2,4,4), (2,4,6,4), (2,4,6,6), and (2,4,8,2).
It is known that any natural number not of the form 4^k*(16*m+14) (k,m = 0,1,2,...) can be written as x^2 + y^2 + 2*z^2 = x^2 + y^2 + z^2 + z^2 with x,y,z nonnegative integers.
See also A271510, A271513, A271518, A271644, A271665, A271714, A271721 and A271724 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 1 = 1 > 0 < 1 and 1 - 1 = 0^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 = 1 = 1 < 2 and 1 - 1 = 0^2.
a(8) = 1 since 8 = 2^2 + 2^2 + 0^2 + 0^2 with 2 = 2 > 0 = 0 and 2 - 2 = 0^2.
a(11) = 1 since 11 = 1^2 + 1^2 + 0^2 + 3^2 with 1 = 1 > 0 < 3 and 1 - 1 = 0^2.
a(24) = 1 since 24 = 2^2 + 2^2 + 0^2 + 4^2 with 2 = 2 > 0 < 4 and 2 - 2 = 0^2.
a(47) = 1 since 47 = 3^2 + 3^2 + 2^2 + 5^2 with 3 = 3 > 2 < 5 and 3 - 3 = 0^2.
a(53) = 2 since 53 = 3^2 + 2^2 + 2^2 + 6^2 with 3 > 2 = 2 < 6 and 3 - 2 = 1^2, and also 53 = 6^2 + 2^2 + 2^2 + 3^2 with 6 > 2 = 2 < 3 and 6 - 2 = 2^2.
a(56) = 1 since 56 = 6^2 + 2^2 + 0^2 + 4^2 with 6 > 2 > 0 < 4 and 6 - 2 = 2^2.
a(120) = 1 since 120 = 8^2 + 4^2 + 2^2 + 6^2 with 8 > 4 > 2 < 6 and 8 - 4 = 2^2.
a(632) = 1 since 632 = 16^2 + 12^2 + 6^2 + 14^2 with 16 > 12 > 6 < 14 and 16 - 12 = 2^2.
		

References

  • L. E. Dickson, Modern Elementary Theory of Numbers, University of Chicago Press, Chicago, 1939, pp. 112-113.

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[x-y]&&SQ[n-x^2-y^2-z^2],r=r+1],{z,0,Sqrt[n/4]},{y,z,Sqrt[(n-z^2)/2]},{x,y,Sqrt[(n-y^2-z^2)]}];Print[n," ",r];Continue,{n,0,70}]

A271721 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= z >= 0, x > 0 and w >= z such that (x-y)*(w-z) is a square.

Original entry on oeis.org

1, 2, 1, 2, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 2, 4, 5, 3, 3, 3, 2, 1, 2, 3, 5, 4, 5, 2, 2, 4, 2, 3, 5, 1, 4, 4, 5, 3, 3, 4, 5, 4, 3, 4, 2, 2, 3, 3, 5, 3, 8, 4, 6, 3, 2, 4, 6, 3, 3, 4, 4, 5, 2, 3, 7, 6, 7, 2, 3, 2, 5, 6, 8, 3, 7, 3, 2, 2, 3, 6, 11, 5, 8, 5, 8, 4, 2, 3, 8, 4, 5, 5, 3, 1, 2, 9, 10, 5, 8
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 12 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 5, 11, 15, 23, 35, 95, 4^k*190 (k = 0,1,2,...).
(ii) For each k = 4, 5, 6, 7, 8, 11, 12, 13, 15, 17, 18, 20, 22, 25, 27, 29, 33, 37, 38, 41, 50, 61, any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that (x-y)*(w-k*z) is a square.
(iii) For each triple (a,b,c) = (3,1,1), (1,2,1), (2,2,1), (3,2,1), (2,2,2), (6,2,1), (1,3,1), (3,3,1), (15,3,1), (1,4,1), (2,4,1), (1,5,1), (3,5,1), (5,5,1), (1,5,2), (1,6,1), (2,6,1), (3,6,1), (15,6,1), (1,7,1), (5,7,1), (1,8,1), (1,8,5), (3,9,1), (1,10,1), (1,12,1), (1,13,1), (3,13,1), (1,14,1), (1,15,1), (1,15,2), (6,16,1), (2,18,1), (3,18,1), (1,20,2), (1,21,1), (3,21,1), (1,23,1), (1,24,1), (1,27,1), (3,27,1), (1,34,1), (1,45,1), (3,45,1), (3,48,1), (1,55,1), (1,60,1), (5,60,1), any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that a*(x+b*y)*(w-c*z) is a square.
This is stronger than Lagrange's four-square theorem. Note that for k = 2 or 3, any natural number n can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and (x-y)*(w-k*z) = 0, for, if n cannot be represented by x^2 + y^2 + 2*z^2 then it has the form 4^k*(16*m+14) (k,m = 0,1,2,...) and hence it can be represented by x^2 + y^2 + (k^2+1)*z^2. It is known that natural numbers not represented by x^2 + y^2 + 5*z^2 have the form 4^k*(8*m+3), and that positive even numbers not represented by x^2 + y^2 + 10*z^2 have the form 4^k*(16*m+6) (as conjectured by S. Ramanujan and proved by L. E. Dickson).
See also A271510, A271513, A271518, A271644, A271714 and A271724 for other conjectures refining Lagrange's theorem.

Examples

			a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 0^2 with 1 = 1 > 0 = 0 and (1-1)*(0-0) = 0^2.
a(5) = 1 since 5 = 2^2 + 1^2 + 0^2 + 0^2 with 2 > 1 > 0 = 0 and (2-1)*(0-0) = 0^2.
a(11) = 1 since 11 = 1^2 + 1^2 + 0^2 + 3^2 with 1 = 1 > 0 < 3 and (1-1)*(3-0) = 0^2.
a(14) = 2 since 14 = 3^2 + 1^2 + 0^2 + 2^2 with 3 > 1 > 0 < 2 and (3-1)*(2-0) = 2^2, and also 14 = 3^2 + 2^2 + 0^2 + 1^2 with 3 > 2 > 0 < 1 and (3-2)*(1-0) = 1^2.
a(15) = 1 since 15 = 3^2 + 2^2 + 1^2 + 1^2 with 3 > 2 > 1 = 1 and (3-2)*(1-1) = 0^2.
a(23) = 1 since 23 = 3^2 + 3^2 + 1^2 + 2^2 with 3 = 3 > 1 < 2 and (3-3)*(2-1) = 0^2.
a(35) = 1 since 35 = 3^2 + 3^2 + 1^2 + 4^2 with 3 = 3 > 1 < 4 and (3-3)*(4-1) = 0^2.
a(95) = 1 since 95 = 5^2 + 5^2 + 3^2 + 6^2 with 5 = 5 > 3 < 6 and (5-5)*(6-3) = 0^2.
a(190) = 1 since 190 = 13^2 + 4^2 + 1^2 + 2^2 with 13 > 4 > 1 < 2 and (13-4)*(2-1) = 3^2.
		

References

  • L. E. Dickson, Integers represented by positive ternary quadratic forms, Bull. Amer. Math. Soc. 33(1927), 63-70.
  • L. E. Dickson, Modern Elementary Theory of Numbers, University of Chicago Press, Chicago, 1939, pp. 112-113.

Crossrefs

Programs

  • Mathematica
      SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(Sqrt[n-x^2-y^2-z^2]-z)*(x-y)],r=r+1],{z,0,Sqrt[n/4]},{y,z,Sqrt[(n-z^2)/2]},{x,Max[1,y],Sqrt[(n-y^2-2z^2)]}];Print[n," ",r];Continue,{n,1,100}]

A344058 Number of ways to write n as x + y + z with x*y + 2*y*z + 3*z*x a square, where x,y,z are positive integers with x or y a power of two (including 2^0 = 1).

Original entry on oeis.org

0, 0, 0, 1, 2, 1, 1, 1, 3, 2, 2, 5, 3, 2, 5, 1, 5, 5, 2, 8, 5, 3, 9, 5, 3, 8, 4, 7, 7, 6, 11, 1, 8, 5, 4, 14, 6, 2, 5, 8, 9, 6, 8, 11, 8, 10, 5, 5, 13, 5, 7, 18, 17, 6, 9, 7, 5, 7, 6, 14, 11, 12, 7, 1, 12, 10, 14, 9, 13, 6, 10, 14, 14, 11, 10, 9, 7, 6, 10, 8, 8, 12, 7, 12, 12, 10, 11, 11, 8, 10, 10, 25, 15, 7, 18, 5, 11, 13, 13, 12
Offset: 1

Views

Author

Zhi-Wei Sun, May 08 2021

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
We have verified a(n) > 0 for all n = 4..10^5. Clearly, a(2*n) > 0 if a(n) > 0.

Examples

			a(6) = 1 with 6 = 3 + 2^0 + 2 and 3*2^0 + 2*2^0*2 + 3*2*3 = 5^2.
a(7) = 1 with 7 = 3 + 2^0 + 3 and 3*2^0 + 2*2^0*3 + 3*3*3 = 6^2.
For each k > 1, we have a(2^k) = 1 with 2^k = 2^(k-2) + 2^(k-1) + 2^(k-2) and 2^(k-2)*2^(k-1) + 2*2^(k-1)*2^(k-2) + 3*2^(k-2)*2^(k-2) = (3*2^(k-2))^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]];
    Pow[x_]:=x>0&&IntegerQ[Log[2,x]];
    tab={};Do[r=0;Do[If[(Pow[x]||Pow[y])&&SQ[x*y+(2y+3x)*(n-x-y)],r=r+1],{x,1,n-2},{y,1,n-1-x}];tab=Append[tab,r],{n,1,100}];Print[tab]
Showing 1-7 of 7 results.