A287616 -id:A287616 - cp's OEIS frontend

A286944 Number of ways to write n as x^2 + 15y^2 + z(3z+1)/2, where x and y are nonnegative integers and z is an integer.

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

Offset: 0

Zhi-Wei Sun, May 16 2017

nonn

Conjecture: a(n) > 0 for any nonnegative integer n, and a(n) = 1 only for n = 0, 3, 4, 7, 8, 10, 12, 13, 14, 29, 33, 34, 48, 68, 113, 129, 220.

Let a,b,c,d,e,f be nonnegative integers with a > b, c > d, e > f, a == b (mod 2), c == d (mod 2), e == f (mod 2), a >= c >= e >= 2, b >= d if a = c, and d >= f if c = e. We have shown in arXiv:1502.03056 that if the ordered tuple (a,b,c,d,e,f) is universal, i.e., each n = 0,1,2,... can be written as x(ax+b)/2 + y(cy+d)/2 + z(ez+f)/2 with x,y,z integers, then (a,b,c,d,e,f) must be among the 12082 tuples listed in the linked a-file. We also conjecture that all the listed tuples not yet proved to be universal are indeed universal. Note that those numbers x(4x+2)/2 with x integral coincide with triangular numbers.

			a(33) = 1 since 33 = 4^2 + 15*1^2 + 1*(3*1+1)/2.
a(34) = 1 since 34 = 2^2 + 15*1^2 + 3*(3*3+1)/2.
a(48) = 1 since 48 = 6^2 + 15*0^2 + (-3)*(3*(-3)+1)/2.
a(68) = 1 since 68 = 1^2 + 15*2^2 + 2*(2*3+1)/2.
a(113) = 1 since 113 = 6^2 + 15*0^2 + 7*(3*7+1)/2.
a(129) = 1 since 129 = 8^2 + 15*2^2 + (-2)*(3*(-2)+1)/2.
a(220) = 1 since 220 = 13^2 + 15*0^2 + 6*(3*6+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, List of those tuples (a,b,c,d,e,f) for which each nonnegative integer should be represented by x(ax+b)/2 + y(cy+d)/2 + z(ez+f)/2 with x,y,z integers
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
Zhi-Wei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275-283.
Zhi-Wei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 2015-2017.
Zhi-Wei Sun, Riddles of Representations of Integers, presentation to Nanjing Normal Univ. (China, 2019).
Hai-Liang Wu and Zhi-Wei Sun, Some universal quadratic sums over the integers, arXiv:1707.06223 [math.NT], 2017.

Cf. A000290, A001318, A287616.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];Do[r=0;Do[If[SQ[24(n-x^2-15y^2)+1],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[(n-x^2)/15]}];Print[n," ",r],{n,0,80}]

A290342 Number of ways to write n as x^2 + 2y^2 + z(z+1)/2, where x is a nonnegative integer, and y and z are positive integers.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Jul 27 2017

nonn

Conjecture: a(n) > 0 for all n > 2. In other words, each n = 0,1,2,... can be written as x^2 + 2y*(y+2) + z*(z+3)/2 with x,y,z nonnegative integers.
As pointed out by Sun in his 2007 paper in Acta Arith., a result of Jones and Pall implies that every n = 0,1,2,... can be written as x^2 + 2*(2y)^2 + z*(z+1)/2 with x,y,z nonnegative integers.
Let a,c,e be positive integers, and let b,d,f be nonnegative integers with a-b, c-d, e-f all even. Suppose that a|b, c|d and e|f. The author studied in arXiv:1502.03056 when each nonnegative integer can be written as x*(a*x+b)/2 + y*(c*y+d)/2 + z*(e*z+f)/2 with x,y,z nonnegative integers, and conjectured that the answer is positive if (a,b,c,d,e,f) is among the following ten tuples (4,0,2,0,1,3), (4,0,2,0,1,5), (4,0,2,6,1,1), (4,0,2,6,2,0), (4,4,2,0,1,3), (4,8,2,0,1,1), (4,8,2,0,1,3), (4,12,2,0,1,1), (6,0,2,0,1,3), (6,6,2,0,1,3).
See also A287616 and A286944 for related comments.

			a(10) = 1 since 10 = 1^2 + 2*2^2 + 1*2/2.
a(11) = 1 since 11 = 0^2 + 2*2^2 + 2*3/2.
a(16) = 1 since 16 = 2^2 + 2*1^2 + 4*5/2.
a(26) = 1 since 26 = 3^2 + 2*1^2 + 5*6/2.
a(31) = 1 since 31 = 1^2 + 2*1^2 + 7*8/2.
a(41) = 1 since 41 = 6^2 + 2*1^2 + 2*3/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
B. W. Jones and G. Pall, Regular and semi-regular positive ternary quadratic forms, Acta Math. 70 (1939), 165-191.
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 2015-2017.

Cf. A000217, A000290, A286944, A287616.

TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^2-2y^2],r=r+1],{x,0,Sqrt[n]},{y,1,Sqrt[(n-x^2)/2]}];Print[n," ",r],{n,0,70}]

A286885 Number of ways to write 6n+1 as x^2 + 3y^2 + 54*z^2 with x,y,z nonnegative integers.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 1, 3, 2, 3, 1, 1, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 4, 4, 3, 2, 2, 4, 2, 3, 3, 3, 3, 3, 2, 2, 4, 3, 4, 1, 3, 2, 3, 4, 3, 3, 3, 3, 2, 3, 3, 2, 4, 3, 2, 3, 2

Offset: 0

Zhi-Wei Sun, Aug 02 2017

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,....

In the a-file, we list the tuples (m,r,a,b,c) with 30 >= m > max{2,r} >= 0, 100 >= a >= b >= c > 0, gcd(a,b,c) = 1, and the form a*x^2+b*y^2+c*z^2 irregular, such that all the numbers m*n+r (n = 0,1,2,...) should be representable by a*x^2+b*y^2+c*z^2 with x,y,z integers.

			a(9) = 1 since 6*9 + 1 = 1^2 + 3*0^2 + 54*1^2.
a(34) = 1 since 6*34 + 1 = 2^2 + 3*7^2 + 54*1^2.
a(125) = 1 since 6*125 + 1 = 26^2 + 3*5^2 + 54*0^2.
a(130) = 1 since 6*130 + 1 = 22^2 + 3*9^2 + 54*1^2.
a(133) = 1 since 6*133 + 1 = 11^2 + 3*8^2 + 54*3^2.
a(203) = 1 since 6*203 + 1 = 25^2 + 3*6^2 + 54*3^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Tomáš Hejda and Vítezslav Kala, Ternary quadratic forms representing arithmetic progressions, arXiv:1906.02538 [math.NT], 2019.
Zhi-Wei Sun, Tuples (m,r,a,b,c) with 30 >= m > max{2,r} >= 0 and 100 >= a >= b >= c > 0, for which all the numbers m*n+r (n = 0,1,2,...) should be representable by a*x^2+b*y^2+c*z^2 with x,y,z integers.
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), 1367-1396.
Zhi-Wei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 2015-2017.
Hai-Liang Wu and Zhi-Wei Sun, Some universal quadratic sums over the integers, arXiv:1707.06223 [math.NT], 2017.
Hai-Liang Wu and Zhi-Wei Sun, Arithmetic progressions represented by diagonal ternary quadratic forms, arXiv:1811.05855 [math.NT], 2018.

Cf. A000290, A286944, A287616, A290342.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
table={};Do[r=0;Do[If[SQ[6n+1-3y^2-54z^2],r=r+1],{y,0,Sqrt[(6n+1)/3]},{z,0,Sqrt[(6n+1-3y^2)/54]}];table=Append[table,r],{n,0,70}]

A290472 Number of ways to write 6n+1 as x^2 + 3y^2 + 7*z^2, where x is a positive integer, and y and z are nonnegative integers.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 3, 3, 1, 1, 4, 1, 3, 1, 9, 1, 1, 2, 4, 3, 3, 3, 5, 1, 4, 2, 6, 3, 6, 1, 4, 2, 3, 1, 7, 3, 3, 3, 6, 2, 3, 2, 15, 2, 5, 2, 4, 2, 2, 7, 6, 3, 6, 2, 11, 3, 7, 3, 6, 4, 5, 2, 11, 4, 3, 1, 7, 3, 2, 4, 17, 2, 3, 3, 8, 2, 5, 7, 9, 4, 4, 2, 13, 1, 13, 1, 5, 4, 3, 4, 6, 7, 7, 3, 10, 4, 6, 3, 20, 3

Offset: 0

Zhi-Wei Sun, Aug 03 2017

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 2, 4, 5, 9, 10, 12, 14, 16, 17, 24, 30, 34, 66, 84, 86, 116, 124, 152, 286.
We also conjecture that {6n+5: n = 0,1,2,...} is a subset of {2x^2+3y^2+5z^2: x,y,z are nonnegative integers with y > 0}.
See A286885 for more similar conjectures.
In support of the first conjecture, a(n) > 1 for 286 < n <= 10^7. - Charles R Greathouse IV, Aug 04 2017

			a(4) = 1 since 6*4+1 = 5^2 + 3*0^2 + 7*0^2.
a(5) = 1 since 6*5+1 = 2^2 + 3*3^2 + 7*0^2.
a(9) = 1 since 6*9+1 = 6^2 + 3*2^2 + 7*1^2.
a(116) = 1 since 6*116+1 = 9^2 + 3*14^2 + 7*2^2.
a(124) = 1 since 6*124+1 = 21^2 + 3*8^2 + 7*4^2.
a(152) = 1 since 6*152+1 = 19^2 + 3*10^2 + 7*6^2.
a(286) = 1 since 6*286+1 = 11^2 + 3*14^2 + 7*12^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), 1367-1396.
Zhi-Wei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 2015-2017.
Hai-Liang Wu and Zhi-Wei Sun, Some universal quadratic sums over the integers, arXiv:1707.06223 [math.NT], 2017.

Cf. A000290, A286885, A286944, A287616, A290342.

SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[6n+1-3y^2-7z^2],r=r+1],{y,0,Sqrt[(6n+1)/3]},{z,0,Sqrt[(6n+1-3y^2)/7]}];Print[n," ",r],{n,0,100}]

a(n)=my(s=6*n+1,t); sum(z=0,sqrtint((s-1)\7), t=s-7*z^2; sum(y=0,sqrtint((t-1)\3), issquare(t-3*y^2))) \\ Charles R Greathouse IV, Aug 03 2017

first(n)=my(v=vector(n+1),mx=6*n+1,s,t,u); for(x=1,sqrtint(mx), s=x^2; for(y=0,sqrtint((mx-s)\3), t=s+3*y^2; for(z=0,sqrtint((mx-t)\7), u=t+7*z^2; if(u%6==1, v[u\6+1]++)))); v \\ Charles R Greathouse IV, Aug 03 2017

A290491 Number of ways to write 12n+1 as x^2 + 4y^2 + 8*z^4, where x and y are positive integers and z is a nonnegative integer.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Aug 03 2017

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4, 10, 18, 28, 39, 49, 98, 142, 163, 184, 208, 320, 382, 408, 814, 910, 1414, 2139, 2674, 3188, 3213, 4230, 6279, 25482.
(ii) All the numbers 16*n+5 (n = 0,1,2,...) can be written as x^4 + 4*y^2 + z^2, where x,y,z are integers with y > 0 and z > 0.
(iii) All the numbers 24*n+1 (n = 0,1,2,...) can be written as 12*x^4 + 4*y^2 + z^2 with x,y,z integers. Also, all the numbers 24*n+9 (n = 0,1,2,...) can be written as 2*x^4 + 6*y^2 + z^2 with x,y,z positive integers.
(iv) All the numbers 24*n+2 (n = 0,1,2,...) can be written as x^4 + 9*y^2 + z^2, where x,y,z are integers with z > 0. Also, all the numbers 24*n+17 (n = 0,1,2,...) can be written as x^4 + 16*y^2 + z^2, where x,y,z are integers with y > 0 and z > 0.
(v) All the numbers 30*n+3 (n = 1,2,3,...) can be written as 2*x^4 + 3*y^2 + z^2 with x,y,z positive integers. Also, all the numbers 30*n+21 (n = 0,1,2,...) can be written as 2*x^4 + 3*y^2 + z^2, where x,y,z are integers with z > 0.

			a(10) = 1 since 12*10+1 = 7^2 + 4*4^2 + 8*1^4.
a(28) = 1 since 12*28+1 = 9^2 + 4*8^2 + 8*0^4.
a(49) = 1 since 12*49+1 = 19^2 + 4*5^2 + 8*2^4.
a(3188) = 1 since 12*3188+1 = 103^2 + 4*80^2 + 8*4^4.
a(3213) = 1 since 12*3213+1 = 91^2 + 4*87^2 + 8*0^4.
a(4230) = 1 since 12*4230+1 = 223^2 + 4*16^2 + 8*1^4.
a(6279) = 1 since 12*6279+1 = 19^2 + 4*75^2 + 8*9^4.
a(25482) = 1 since 12*25482+1 = 531^2 + 4*58^2 + 8*6^4.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), 1367-1396.
Zhi-Wei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 2015-2017.
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34 (2017), No. 2, 97-120.

Cf. A000290, A000583, A270566, A286885, A286944, A287616, A290342, A290472.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[12n+1-8x^4-4y^2],r=r+1],{x,0,((12n+1)/8)^(1/4)},{y,1,Sqrt[(12n+1-8x^4)/4]}];Print[n," ",r],{n,1,100}]

A260418 Number of ways to write 12n+5 as 4x^4 + 4*y^2 + z^2, where x is a nonnegative integer, and y and z are positive integers.

Original entry on oeis.org

1, 2, 2, 2, 1, 3, 2, 4, 3, 2, 3, 2, 5, 2, 3, 3, 2, 4, 3, 3, 3, 2, 5, 2, 3, 3, 2, 6, 3, 4, 3, 5, 6, 3, 3, 5, 3, 5, 4, 2, 5, 4, 7, 3, 2, 7, 4, 6, 2, 2, 4, 3, 8, 4, 1, 2, 4, 8, 6, 2, 5, 2, 7, 4, 4, 3, 4, 5, 2, 4, 5, 6, 4, 3, 2, 5, 2, 7, 4, 5, 5, 2, 5, 3, 6, 5, 4, 7, 3, 4, 3, 5, 9, 3, 4, 3, 5, 11, 4, 5, 5

Offset: 0

Zhi-Wei Sun, Aug 04 2017

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 4, 54, 159, 289, 999, 1175, 1404, 16391, 39688.
(ii) All the numbers 24*n+4 (n = 0,1,2,...) can be written as 3*x^4+24*y^2+z^2, where x,y,z are integers with x > 0 and z > 0.
(iii) All the numbers 24*n+13 (n = 0,1,2,...) can be written as 3*x^4+9*y^2+z^2 with x,y,z positive integers.
(iv) All the numbers 24*n+r (n = 0,1,2,...) can be written as a*x^4+b*y^2+c*z^2 with x an integer and y and z positive integers, provided that (r,a,b,c) is among the following quadruples: (5,3,1,1), (5,12,4,1), (7,3,6,1), (10,5,1,1), (11,3,8,3), (11,6,3,2), (17,9,4,1).
See A290491 for a similar conjecture.

			a(4) = 1 since 12*4+5 = 4*0^4 + 4*1^2 + 7^2.
a(54) = 1 since 12*54+5 = 4*0^4 + 4*11^2 + 13^2.
a(159) = 1 since 12*159+5 = 4*0^4 + 4*4^2 + 43^2.
a(289) = 1 since 12*289+5 = 4*1^4 + 4*19^2 + 45^2.
a(999) = 1 since 12*999+5 = 4*7^4 + 4*21^2 + 25^2.
a(1175) = 1 since 12*1175+5 = 4*3^4 + 4*55^2 + 41^2.
a(1404) = 1 since 12*1404+5 = 4*3^4 + 4*10^2 + 127^2.
a(16391) = 1 since 12*16391+5 = 4*5^4 + 4*207^2 + 151^2.
a(39688) = 1 since 12*39688+5 = 4*5^4 + 4*50^2 + 681^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), 1367-1396.
Zhi-Wei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 2015-2017.
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Cf. A000290, A000583, A270566, A286885, A286944, A287616, A290342, A290472, A290491.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[12n+5-4x^4-4y^2],r=r+1],{x,0,((12n+5)/4)^(1/4)},{y,1,Sqrt[(12n+5-4x^4)/4]}];Print[n," ",r],{n,0,100}]

cp's OEIS Frontend

A286944 Number of ways to write n as x^2 + 15*y^2 + z*(3z+1)/2, where x and y are nonnegative integers and z is an integer.

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A290342 Number of ways to write n as x^2 + 2*y^2 + z*(z+1)/2, where x is a nonnegative integer, and y and z are positive integers.

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A286885 Number of ways to write 6*n+1 as x^2 + 3*y^2 + 54*z^2 with x,y,z nonnegative integers.

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A290472 Number of ways to write 6*n+1 as x^2 + 3*y^2 + 7*z^2, where x is a positive integer, and y and z are nonnegative integers.

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A290491 Number of ways to write 12*n+1 as x^2 + 4*y^2 + 8*z^4, where x and y are positive integers and z is a nonnegative integer.

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A260418 Number of ways to write 12*n+5 as 4*x^4 + 4*y^2 + z^2, where x is a nonnegative integer, and y and z are positive integers.

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A286944 Number of ways to write n as x^2 + 15y^2 + z(3z+1)/2, where x and y are nonnegative integers and z is an integer.

A290342 Number of ways to write n as x^2 + 2y^2 + z(z+1)/2, where x is a nonnegative integer, and y and z are positive integers.

A286885 Number of ways to write 6n+1 as x^2 + 3y^2 + 54*z^2 with x,y,z nonnegative integers.

A290472 Number of ways to write 6n+1 as x^2 + 3y^2 + 7*z^2, where x is a positive integer, and y and z are nonnegative integers.

A290491 Number of ways to write 12n+1 as x^2 + 4y^2 + 8*z^4, where x and y are positive integers and z is a nonnegative integer.

A260418 Number of ways to write 12n+5 as 4x^4 + 4*y^2 + z^2, where x is a nonnegative integer, and y and z are positive integers.