A230121 -id:A230121 - cp's OEIS frontend

A230747 Number of ways to write n = x + y + 2z with 0 < x <= y and z > 0 such that x^2 + y^2 + 2z^2 is a square.

0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 2, 0, 1, 4, 0, 2, 1, 3, 1, 1, 1, 1, 4, 4, 3, 1, 2, 3, 1, 8, 2, 1, 3, 4, 4, 3, 6, 3, 2, 4, 3, 3, 5, 3, 2, 7, 7, 8, 3, 7, 5, 6, 6, 1, 4, 4, 6, 8, 7, 2, 6, 14, 8, 6, 5, 7, 4, 10, 6, 4, 5, 7, 7, 6, 10, 10, 4, 14, 11, 6, 8, 11, 8, 6, 6, 3, 8, 10, 11, 9, 7, 6, 13, 19, 4, 11, 8, 16

Offset: 1

Zhi-Wei Sun, Oct 29 2013

nonn

Conjecture: a(n) > 0 for all n > 17.
Note that a(4*k) > 0 for all k > 0 since 4*k = k + k + 2*k and  k^2 + k^2 + 2*k^2 = (2*k)^2.
See also A230121 for a related conjecture.
The conjecture was confirmed by Chao Haung and Zhi-Wei Sun in 2021. - Zhi-Wei Sun, May 09 2021

			a(9) = 1 since 9 = 1 + 4 + 2*2 with 1^2 + 4^2 + 2*2^2 = 5^2.
a(21) = 1 since 21 = 5 + 8 + 2*4 with 5^2 + 8^2 + 2*4^2 = 11^2.
a(34) = 1 since 34 = 7 + 25 + 2*1 with 7^2 + 25^2 + 2*1^2 = 26^2.
a(56) = 1 since 56 = 14 + 14 + 2*14 with 14^2 + 14^2 + 2*14^2 = 28^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..4000
Chao Huang and Zhi-Wei Sun, On partitions of integers with restrictions involving squares, arXiv:2105.03416 [math.NT], 2021.
Zhi-Wei Sun, Diophantine problems involving triangular numbers and squares, a message to Number Theory List, Oct. 11, 2013.

Cf. A000290, A230121, A343950.

SQ[n_]:=IntegerQ[Sqrt[n]]
a[n_]:=Sum[If[SQ[2i^2+j^2+(n-2i-j)^2],1,0],{i,1,(n-2)/2},{j,1,(n-2i)/2}]
Table[a[n],{n,1,100}]

A229166 Number of ordered ways to write n = x(x+1)/2 + y with y(y+1)/2 + 1 prime, where x and y are nonnegative integers.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Oct 15 2013

nonn

Conjecture: a(n) > 0 for all n > 0. Moreover, if n > 0 is not among 1, 3, 60, then there are positive integers x and y with x*(x+1)/2 + y = n such that y*(y+1)/2 + 1 is prime.

			a(6) = 1 since 6 = 2*3/2 + 3 with 3*4/2 + 1 = 7 prime.
a(60) = 1 since 60 = 0*1/2 + 60 with 60*61/2 + 1 = 1831 prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On sums of primes and triangular numbers, J. Comb. Number Theory 1(2009), 65-76.
Zhi-Wei Sun, Conjectures involving primes and quadratic forms, preprint, arXiv:1211.1588 [math.NT], 2012-2017.

Cf. A000040, A000217, A132399, A208244, A230121, A230252.

T[n_]:=n(n+1)/2
a[n_]:=Sum[If[PrimeQ[T[n-T[i]]+1],1,0],{i,0,(Sqrt[8n+1]-1)/2}]
Table[a[n],{n,1,100}]

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.

Original entry on oeis.org

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]

A343862 Number of ways to write n as x + y + z with x, y, z positive integers such that x^2y^2 + 5y^2z^2 + 10z^2*x^2 is a square.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 2, 1, 1, 4, 2, 4, 2, 2, 5, 5, 3, 3, 2, 6, 4, 3, 3, 6, 6, 5, 2, 6, 4, 10, 3, 6, 4, 6, 6, 8, 5, 6, 4, 9, 7, 6, 3, 7, 9, 5, 5, 15, 5, 12, 11, 10, 5, 6, 7, 10, 8, 9, 7, 15, 7, 6, 7, 10, 10, 7, 9, 10, 10, 12, 4, 15, 9, 9, 11, 9, 7, 12, 11, 15, 8, 9, 7, 12, 10, 3, 9, 11, 11, 19, 12, 12, 9, 10, 6, 23, 11, 6, 10, 18

Offset: 1

Zhi-Wei Sun, May 02 2021

nonn

Conjecture 1: a(n) > 0 for all n > 2.
We have verified a(n) > 0 for all n = 3..10000. Conjecture 1 holds if a(p) > 0 for each odd prime p. For any n > 0 we have a(3*n) > 0 since 3*n = n + n + n and 1 + 5 + 10 = 4^2.
See also A340274 for a similar conjecture.
Conjecture 2: There are infinitely many triples (a,b,c) of positive integers such that each n = 3,4,... can be written as x + y + z with x,y,z positive integers and a*x^2*y^2 + b*y^2*z^2 + c*z^2*x^2 a square.
Such triple candidates include (21,19,9), (23,17,9), (24,16,9), (25,14,10), (29,19,16), (33,27,21), (35,9,5), (37,32,31) etc.
Conjecture 1 holds for all n < 2^15. - Martin Ehrenstein, May 02 2021

			a(4) = 1 with 4 = 2 + 1 + 1 and 2^2*1^2 + 5*1^2*1^2 + 10*1^2*2^2 = 7^2.
a(5) = 1 with 5 = 1 + 3 + 1 and 1^2*3^2 + 5*3^2*1^2 + 10*1^1*1^2 = 8^2.
a(8) = 1 with 8 = 4 + 2 + 2 and 4^2*2^2 + 5*2^2*2^2 + 10*2^2*4^2 = 28^2.
a(9) = 1 with 9 = 3 + 3 + 3 and 3^2*3^2 + 5*3^2*3^2 + 10*3^2*3^2 = 36^2.
a(19) = 2. We have 19 = 4 + 5 + 10 with 4^2*5^2 + 5*5^2*10^2 + 10*10^2*4^2 = 170^2, and 19 = 4 + 13 + 2 with 4^2*13^2 + 5*13^2*2^2 + 10*2^2*4^2 = 82^2.

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:1604.06723 [math.NT].

Cf. A000041, A000290, A230121, A230747, A231168, A262357, A271719, A273278, A340274.

SQ[n_]:=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[x^2*y^2+(n-x-y)^2*(5*y^2+10*x^2)],r=r+1],{x,1,n-2},{y,1,n-1-x}];tab=Append[tab,r],{n,1,100}];Print[tab]

A230596 Number of ways to write n = x + y + z with 0 < x <= y <= z such that xyz is a triangular number, and that x is a triangular number of the form (p^2 - 1)/8 with p an odd prime.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Oct 24 2013

nonn

Conjecture: (i) a(n) > 0 except for n = 1, 2, 4, 7.
(ii) For any integer n > 7, there are positive integers x, y, z with x + y + z = n such that x*y*z is a triangular number and x is among 1, 2, 3, 4, 5, 6.
Note that a(3k) and a(3k+2) are positive for every k = 1, 2, 3, .... In fact, 3k = 1 + k + (2k-1) with 1*k*(2k-1) = 2k*(2k-1)/2 a triangular number, and  3k+2 = 1 + k + (2k+1) with 1*k*(2k+1) = 2k(2k+1)/2 a triangular number.

			a(10) = 1 since 10 = 3 + 3 + 4, and 3 = (5^2-1)/8 with 5 an odd prime, and 3*3*4 = 8*9/2 is a triangular number.
a(31) = 1 since 31 = 3 + 11 + 17, and 3 = (5^2-1)/8 with 5 an odd prime, and 3*11*17 = 33*34/2 is a triangular number.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000217, A000040, A132399, A229166, A230121, A230451.

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
a[n_]:=Sum[If[TQ[(Prime[i]^2-1)/8*y*(n-(Prime[i]^2-1)/8-y)],1,0],{i,2,PrimePi[Sqrt[8n/3+1]]},{y,(Prime[i]^2-1)/8,(n-(Prime[i]^2-1)/8)/2}]
Table[a[n],{n,1,100}]

A227877 Number of ways to write n = x + y + z (x, y, z > 0) such that xy and xz are triangular numbers, and 6y-1 and 6z+1 are both prime.

Original entry on oeis.org

0, 0, 1, 0, 3, 2, 2, 3, 3, 7, 3, 6, 3, 3, 2, 3, 7, 6, 7, 5, 4, 5, 10, 2, 10, 4, 5, 2, 2, 9, 5, 9, 2, 4, 3, 4, 5, 7, 5, 11, 12, 5, 8, 11, 12, 5, 11, 3, 7, 11, 4, 10, 6, 2, 9, 11, 8, 7, 9, 8, 9, 4, 3, 4, 10, 6, 9, 15, 9, 17, 3, 3, 8, 12, 10, 5, 1, 7, 9, 16, 8, 17, 6, 8, 16, 6, 8, 8, 10, 1, 6, 4, 8, 5, 23, 11, 2, 9, 6, 14

Offset: 1

Zhi-Wei Sun, Oct 25 2013

nonn

Conjecture: a(n) > 0 for all n > 4.
For n = 4*k - 1, we have n = (2k-1) + k + k with (2k-1)*k = 2k*(2k-1)/2 a triangular number. For n = 4*k + 1, we have n = (2k+1) + k + k with (2k+1)*k = 2k*(2k+1)/2 a triangular number.  For n = 4*k + 2, we have n = (2k+1) + k + (k+1), and (2k+1)*k = 2k*(2k+1)/2 and (2k+1)*(k+1) = (2k+1)(2k+2)/2 are both triangular numbers.
For n = 5*k, we have n = k + (2k-1) + (2k+1), and k*(2k-1) = 2k*(2k-1)/2 and k*(2k+1) = 2k*(2k+1)/2 are both triangular numbers. For n = 5*k - 2, we have n = k + (2k-1) + (2k-1) with k*(2k-1) = 2k*(2k-1)/2 a triangular number. For n = 5*k + 2, we have n = k + (2k+1) + (2k+1) with k*(2k+1) = 2k*(2k+1)/2 a triangular number.

			a(77) = 1 since 77 = 1 + 10 + 66, and 1*10 = 4*5/2 and 1*66 = 11*12/2 are triangular numbers, and 6*10 - 1 = 59 and 6*66 + 1 = 397 are both prime.
a(90) = 1 since 90 = 45 + 22 + 23, and 45*22 = 44*45/2 and 45*23 = 45*46/2 are triangular numbers, and 6*22 - 1 = 131 and 6*23 + 1 = 139 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, A new conjecture on triangular numbers, a message to Number Theory List, Oct. 25, 2013.

Cf. A000040, A000217, A132399, A229166, A230121, A230451, A230596.

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
a[n_]:=Sum[If[PrimeQ[6j-1]&&PrimeQ[6(n-i-j)+1]&&TQ[i*j]&&TQ[i(n-i-j)],1,0],{i,1,n-2},{j,1,n-1-i}]
Table[a[n],{n,1,100}]

A230451 Number of ways to write n = x + y + z (x, y, z > 0) such that 2x + 1, 2y + 3, 2z + 5 are all prime and xy*z is a triangular number.

Original entry on oeis.org

0, 0, 1, 0, 2, 3, 0, 4, 3, 1, 7, 3, 2, 3, 7, 4, 5, 6, 3, 4, 8, 5, 8, 3, 6, 8, 9, 9, 5, 12, 2, 11, 4, 4, 4, 13, 5, 9, 13, 8, 14, 8, 3, 15, 7, 8, 10, 6, 5, 17, 15, 4, 6, 9, 8, 10, 15, 9, 7, 15, 11, 5, 6, 11, 14, 14, 7, 11, 3, 12, 23, 16, 5, 20, 14, 4, 9, 14, 5, 19, 19, 4, 3, 12, 7, 16, 5, 12, 6, 11, 12, 12, 23, 14, 23, 12, 5, 17, 14, 5

Offset: 1

Zhi-Wei Sun, Oct 19 2013

nonn

Conjecture: (i) a(n) > 0 except for n = 1, 2, 4, 7.
(ii) Any integer n > 7 can be written as x + y + z (x, y, z > 0) such that 2*x + 1, 2*y + 1, 2*x*y + 1 are primes and x*y*z is a triangular number.
(iii) Each integer n > 4 not equal to 7 or 14 can be expressed as p + q + r (p, q, r > 0) with p and 2*q + 1 both primes, and p*q*r a triangular number.
(iv) Any integer n > 6 not among 16, 20, 60 can be written as x + y + z (x, y, z > 0) such that x*y + x*z + y*z is a triangular number.
Part (i) is stronger than Goldbach's weak conjecture which was finally proved by H. Helfgott in 2013.
See also A227877 and A230596 for some related conjectures.

			a(6) = 3 since 6 = 1 + 2 + 3 = 2 + 1 + 3 = 3 + 2 + 1, and 2*1 + 1 = 3, 2*2 + 3 = 7, 2*3 + 5 = 11, 2*2 + 1 = 5, 2*1 + 3 = 5, 2*3 + 1 = 7, 2*1 + 5 = 7 are all prime.
a(10) = 1 since 10 = 3 + 4 + 3, and 2*3 + 1 = 7, 2*4 + 3 = 11, 2*3 + 5 = 11 are all prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..6000
Zhi-Wei Sun, A new conjecture on triangular numbers, a message to Number Theory List, Oct. 25, 2013.
Zhi-Wei Sun, Conjectures involving primes and quadratic forms, preprint, arXiv:1211.1588.

Cf. A000040, A000217, A068307, A132399, A227877, A229166, A230121, A230141, A230219, A230596.

SQ[n_]:=IntegerQ[Sqrt[n]]
TQ[n_]:=SQ[8n+1]
a[n_]:=Sum[If[PrimeQ[2i+1]&&PrimeQ[2j+3]&&PrimeQ[2(n-i-j)+5]&&TQ[i*j(n-i-j)],1,0],{i,1,n-2},{j,1,n-1-i}]
Table[a[n],{n,1,100}]

A231168 Number of ways to write n = x + y + z (x, y, z > 0) such that x^2 + y^2 + z^2 + z is a square, and 6x + 1, 6y - 1, 6*z -1 are all prime.

Original entry on oeis.org

0, 0, 1, 0, 0, 2, 2, 2, 2, 1, 3, 2, 5, 1, 4, 3, 2, 3, 1, 1, 4, 2, 5, 3, 3, 4, 4, 8, 2, 3, 8, 2, 4, 3, 4, 8, 7, 2, 2, 8, 3, 8, 6, 1, 6, 8, 4, 1, 9, 2, 4, 10, 6, 1, 7, 11, 7, 10, 2, 6, 9, 3, 6, 3, 6, 6, 6, 8, 4, 8, 4, 4, 9, 2, 11, 4, 9, 6, 1, 4, 5, 5, 10, 7, 5, 6, 6, 7, 5, 8, 17, 8, 5, 2, 7, 8, 11, 10, 6, 4

Offset: 1

Zhi-Wei Sun, Nov 04 2013

nonn

Conjecture: a(n) > 0 for all n > 5.

			a(19) = 1 since 19 = 13 + 5 + 1 with 13^2 + 5^2 + 1^2 + 1 = 14^2, and 6*13 + 1 = 79, 6*5 - 1 = 29, 6*1 - 1 = 5 are all prime.
a(444) = 1 since 444 = 76 + 28 + 340 with 76^2 + 28^2 + 340^2 + 340 = 350^2, and 6*76 + 1 = 457, 6*28 - 1 = 167, 6*340 - 1 = 2039 are all prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..5000
Zhi-Wei Sun, A conjecture involving squares and primes, a message to Number Theory List, Nov. 5, 2013.

Cf. A000040, A000290, A227877, A230121, A230596, A230747.

SQ[n_]:=IntegerQ[Sqrt[n]]
a[n_]:=Sum[If[PrimeQ[6i+1]&&PrimeQ[6j-1]&&PrimeQ[6(n-i-j)-1]&&SQ[i^2+j^2+(n-i-j)^2+(n-i-j)],1,0],{i,1,n-2},{j,1,n-1-i}]
Table[a[n],{n,1,100}]

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]

A343950 Number of ways to write n as x + y + z with x^2 + 4y^2 + 5z^2 a square, where x,y,z are positive integers with y or z a positive power of two.

Original entry on oeis.org

0, 0, 0, 1, 1, 0, 0, 3, 1, 2, 2, 2, 3, 1, 4, 3, 2, 3, 3, 4, 4, 2, 1, 4, 6, 4, 2, 3, 12, 5, 3, 5, 8, 4, 5, 5, 8, 4, 7, 4, 4, 4, 7, 5, 5, 1, 4, 6, 5, 6, 6, 10, 7, 4, 9, 5, 10, 16, 7, 7, 9, 6, 5, 5, 14, 8, 6, 6, 3, 7, 1, 5, 4, 10, 5, 7, 10, 8, 13, 10, 3, 4, 8, 5, 12, 7, 20, 9, 12, 5, 8, 1, 9, 4, 8, 9, 8, 7, 4, 10

Offset: 1

Zhi-Wei Sun, May 05 2021

nonn

Conjecture 1: a(n) > 0 for all n > 7.
We have verified a(n) > 0 for all n = 8..50000. Clearly, a(2*n) > 0 if a(n) > 0.
Conjecture 2: For any integer n > 7, we can write n as x + y + z with x,y,z positive integers such that x^2 + 2*y^2 + 3*z^2 is a square.
Conjecture 3: For any integer n > 4, we can write n as x + y + z with x,y,z positive integers such that 3*x^2 + 4*y^2 + 5*z^2 (or x^2 + 3*y^2 + 5*z^2) is a square.

			a(4) = 1, and 4 = 1 + 1 + 2 with 1^2 + 4*1^2 + 5*2^2 = 5^2.
a(5) = 1, and 5 = 2 + 2 + 1 with 2^2 + 4*2^2 + 5*1^2 = 5^2.
a(9) = 1, and 9 = 4 + 1 + 4 with 4^2 + 4*1^2 + 5*4^2 = 10^2.
a(14) = 1, and 14 = 7 + 5 + 2 with 7^2 + 4*5^2 + 5*2^2 = 13^2.
a(23) = 1, and 23 = 7 + 8 + 8 with 7^2 + 4*8^2 + 5*8^2 = 25^2.
a(46) = 1, and 46 = 14 + 16 + 16 with 14^2 + 4*16^2 + 5*16^2 = 50^2.
a(71) = 1, and 71 = 42 + 8 + 21 with 42^2 + 4*8^2 + 5*21^2 = 65^2.
a(92) = 1, and 92 = 28 + 32 + 32 with 28^2 + 4*32^2 + 5*32^2 = 100^2.
a(142) = 1, and 142 = 84 + 16 + 42 with 84^2 + 4*16^2 + 5*42^2 = 130^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..3200
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. See also arXiv:1604.06723 [math.NT].

Cf. A000079, A000290, A271510, A271513, A271518, A230121, A230747.

PowQ[n_]:=PowQ[n]=n>1&&IntegerQ[Log[2,n]];
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[(PowQ[y]||PowQ[n-x-y])&&SQ[x^2+4*y^2+5*(n-x-y)^2],r=r+1],{x,1,n-3},{y,1,n-1-x}];tab=Append[tab,r],{n,1,100}];Print[tab]

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A230747 Number of ways to write n = x + y + 2*z with 0 < x <= y and z > 0 such that x^2 + y^2 + 2*z^2 is a square.

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A229166 Number of ordered ways to write n = x*(x+1)/2 + y with y*(y+1)/2 + 1 prime, where x and y are nonnegative integers.

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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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A343862 Number of ways to write n as x + y + z with x, y, z positive integers such that x^2*y^2 + 5*y^2*z^2 + 10*z^2*x^2 is a square.

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A230596 Number of ways to write n = x + y + z with 0 < x <= y <= z such that x*y*z is a triangular number, and that x is a triangular number of the form (p^2 - 1)/8 with p an odd prime.

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A227877 Number of ways to write n = x + y + z (x, y, z > 0) such that x*y and x*z are triangular numbers, and 6*y-1 and 6*z+1 are both prime.

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A230451 Number of ways to write n = x + y + z (x, y, z > 0) such that 2*x + 1, 2*y + 3, 2*z + 5 are all prime and x*y*z is a triangular number.

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A231168 Number of ways to write n = x + y + z (x, y, z > 0) such that x^2 + y^2 + z^2 + z is a square, and 6*x + 1, 6*y - 1, 6*z -1 are all prime.

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A230747 Number of ways to write n = x + y + 2z with 0 < x <= y and z > 0 such that x^2 + y^2 + 2z^2 is a square.

A229166 Number of ordered ways to write n = x(x+1)/2 + y with y(y+1)/2 + 1 prime, where x and y are nonnegative integers.

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.

A343862 Number of ways to write n as x + y + z with x, y, z positive integers such that x^2y^2 + 5y^2z^2 + 10z^2*x^2 is a square.

A230596 Number of ways to write n = x + y + z with 0 < x <= y <= z such that xyz is a triangular number, and that x is a triangular number of the form (p^2 - 1)/8 with p an odd prime.

A227877 Number of ways to write n = x + y + z (x, y, z > 0) such that xy and xz are triangular numbers, and 6y-1 and 6z+1 are both prime.

A230451 Number of ways to write n = x + y + z (x, y, z > 0) such that 2x + 1, 2y + 3, 2z + 5 are all prime and xy*z is a triangular number.

A231168 Number of ways to write n = x + y + z (x, y, z > 0) such that x^2 + y^2 + z^2 + z is a square, and 6x + 1, 6y - 1, 6*z -1 are all prime.

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.

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