A343862 -id:A343862 - cp's OEIS frontend

A340274 Number of ways to write n as x + y + z with x, y, z positive integers such that 3x^2y^2 + 5y^2z^2 + 8z^2x^2 is a square.

0, 0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 4, 2, 5, 2, 1, 2, 2, 2, 2, 4, 2, 4, 3, 3, 3, 4, 5, 2, 3, 5, 5, 4, 4, 2, 4, 4, 5, 3, 4, 3, 6, 3, 2, 5, 2, 2, 7, 7, 1, 3, 6, 4, 4, 3, 3, 6, 2, 5, 5, 3, 6, 5, 4, 6, 6, 6, 3, 6, 6, 4, 5, 6, 2, 6, 3, 5, 4, 5, 3, 5, 12, 4, 4, 5, 1, 6, 6, 7, 9, 3, 3, 6, 5, 6, 7, 4

Offset: 1

Zhi-Wei Sun, Apr 24 2021

nonn

Conjecture 1: a(n) > 0 for all n > 2.
We have verified a(n) > 0 for all n = 3..10000. The conjecture holds if a(p) > 0 for every odd prime p. For any n > 0 we have a(3*n) > 0, since 3*n = n + n + n and 3 + 5 + 8 = 4^2.
It seems that a(n) = 1 only for n = 3..8, 10, 11, 19, 53, 89, 127, 178, 257, 461.
See also A343862 for similar conjectures.
Conjecture 1 holds for all n < 2^15. Note a(1823) = 1. - Martin Ehrenstein, May 03 2021

			a(4) = 1 with 4 = 2 + 1 + 1 and 3*2^2*1^2 + 5*1^2*1^2 + 8*1^2*2^2 = 7^2.
a(19) = 1 with 19 = 9 + 9 + 1 and 3*9^2*9^2 + 5*9^2*1^2 + 8*1^2*9^2 = 144^2.
a(53) = 1 with 53 = 23 + 7 + 23 and 3*23^2*7^2 + 5*7^2*23^2 + 8*23^2*23^2 = 1564^2.
a(89) = 1 with 89 = 2 + 58 + 29 and 3*2^2*58^2 + 5*58^2*29^2 + 8*29^2*2^2 = 3770^2.
a(257) = 1 with 257 = 11 + 164 + 82 and 3*11^2*164^2 + 5*164^2*82^2 + 8*82^2*11^2 = 30340^2.
a(461) = 1 with 461 = 186 + 165 + 110 and 3*186^2*165^2 + 5*165^2*110^2 + 8*110^2*186^2 = 88440^2.

Martin Ehrenstein, Table of n, a(n) for n = 1..32767 (first 1500 terms from Zhi-Wei Sun)
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175 (2017), 167-190. See also arXiv version, arXiv:1604.06723 [math.NT], 2016-2017.

Cf. A000041, A000290, A230121, A230747, A231168, A262357, A271719, A273278, A343862.

SQ[n_]:=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[3x^2*y^2+(n-x-y)^2*(5*y^2+8*x^2)],r=r+1],{x,1,n-2},{y,1,n-1-x}];tab=Append[tab,r],{n,1,100}];Print[tab]

A343897 Number of ways to write n as 2x + y + z with x,y,z positive integers such that 16x^2y^2 + 19y^2z^2 + 29z^2*x^2 is a square.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 2, 4, 2, 1, 3, 1, 3, 1, 5, 2, 2, 6, 1, 1, 5, 5, 6, 2, 2, 4, 2, 4, 5, 6, 3, 2, 3, 2, 5, 2, 7, 10, 4, 1, 3, 3, 10, 9, 2, 5, 5, 10, 6, 7, 6, 7, 8, 8, 4, 7, 4, 5, 8, 2, 4, 4, 13, 9, 5, 6, 10, 11, 6, 11, 6, 6, 5, 4, 4, 10, 9, 8, 8, 8, 8, 9, 16, 5, 5, 6, 4, 7, 3, 12, 7, 11, 13

Offset: 1

Zhi-Wei Sun, May 03 2021

nonn

Conjecture: a(n) > 0 for all n > 3.
We have verified a(n) > 0 for all n = 4..10000. The conjecture holds if a(p) > 0 for each odd prime p.
It seems that a(n) = 1 only for n = 4..9, 13, 17, 19, 21, 26, 27, 47.

			a(4) = 1 with 4 = 2*1 + 1 + 1 and 16*1^2*1^2 + 19*1^2*1^2 + 29*1^2*1^2 = 8^2.
a(6) = 1 with 6 = 2*1 + 2 + 2 and 16*1^2*2^2 + 19*2^2*2^2 + 29*2^2*1^2 = 22^2.
a(9) = 1 with 9 = 2*2 + 4 + 1 and 16*2^2*4^2 + 19*4^2*1^2 + 29*1^2*2^2 = 38^2.
a(13) = 1 with 13 = 2*5 + 2 + 1 and 16*5^2*2^2 + 19*2^2*1^2 + 29*1^2*5^2 = 49^2.
a(19) = 1 with 19 = 2*2 + 14 + 1 and 16*2^2*14^2 + 19*14^2*1^2 + 29*1^2*2^2 = 128^2.
a(47) = 1 with 47 = 2*13 + 13 + 8 and 16*13^2*13^2 + 19*13^2*8^2 + 29*8^2*13^2 = 988^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..1500

Cf. A000041, A000290, A230121, A230747, A340274, A343862.

SQ[n_]:=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[16(x*y)^2+(n-2x-y)^2*(19*y^2+29x^2)],r=r+1],{x,1,(n-2)/2},{y,1,n-1-2x}];tab=Append[tab,r],{n,1,100}];Print[tab]

A343917 Positive integers m with 2*m^2 - 2^4 = x^4 + y^4 for some nonnegative integers x and y with |x-y| > 2.

Original entry on oeis.org

284, 1388, 2139, 4772, 8556, 8971, 10836, 21163, 28847, 45707, 54507, 71292, 73348, 95127, 101503, 104228, 131388, 136148, 263172, 350076, 638164, 982292, 1532148, 1687828, 1705407, 1958924, 2082188, 2299364, 2360347, 2728379, 3202356, 4042799, 5046771, 5165332, 5235323, 5560627, 7191079, 7740547, 8041364

Offset: 1

Zhi-Wei Sun, May 04 2021

nonn

Conjecture: The sequence has infinitely many terms.
It is easy to see that no term is divisible by 5. In the b-file we list all the 62 terms not exceeding 10^8.
Note that 2*(n^2+3)^2 - 2^4 = (n+1)^4 + (n-1)^4 with (n+1) - (n-1) = 2. This implies that any integer n > 4 can be written as  x + y + 2^(z-1) with x,y,z positive integers such that x^4 + y^4 + (2^z)^4 is twice a square.
See also A343913 for a similar conjecture.

			a(1) = 284, and 2*284^2 - 2^4 = 20^4 + 6^4 with |20-6| > 2.
a(62) = 97077407, and 2*97077407^2 - 2^4 = 18848045899687282 = 11563^4 + 5583^4 with |11563-5583| > 2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..62

Cf. A000290, A000583, A003336, A230121, A230747, A340274, A343862, A343897, A343913.

QQ[n_]:=IntegerQ[n^(1/4)];
n=0;Do[Do[If[QQ[2*m^2-16-x^4]&&(2*m^2-16-x^4)^(1/4)-x>2,n=n+1;Print[n," ",m];Goto[aa]],{x,0,(m^2-8)^(1/4)}];Label[aa],{m,3,8041364}]

A344058 Number of ways to write n as x + y + z with xy + 2y*z + 3zx 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

Zhi-Wei Sun, May 08 2021

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..2000
Chao Huang and Zhi-Wei Sun, On partitions of integers with restrictions involving squares, arXiv:2105.03416 [math.NT], 2021.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. See also arXiv:1604.06723 [math.NT].

Cf. A000290, A271644, A340274, A343862, A343897, A343950.

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]

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A340274 Number of ways to write n as x + y + z with x, y, z positive integers such that 3*x^2*y^2 + 5*y^2*z^2 + 8*z^2*x^2 is a square.

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A343897 Number of ways to write n as 2*x + y + z with x,y,z positive integers such that 16*x^2*y^2 + 19*y^2*z^2 + 29*z^2*x^2 is a square.

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A343917 Positive integers m with 2*m^2 - 2^4 = x^4 + y^4 for some nonnegative integers x and y with |x-y| > 2.

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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).

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A340274 Number of ways to write n as x + y + z with x, y, z positive integers such that 3x^2y^2 + 5y^2z^2 + 8z^2x^2 is a square.

A343897 Number of ways to write n as 2x + y + z with x,y,z positive integers such that 16x^2y^2 + 19y^2z^2 + 29z^2*x^2 is a square.

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