A286885 -id:A286885 - cp's OEIS frontend

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.

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}]

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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.

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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.

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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.

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cp's OEIS Frontend

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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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.