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A271518
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Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with x + 3*y + 5*z a square, where w, x, y and z are nonnegative integers.
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141
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1, 2, 2, 2, 2, 1, 1, 1, 1, 3, 3, 2, 2, 2, 4, 2, 2, 5, 5, 3, 2, 2, 2, 3, 1, 5, 5, 2, 2, 5, 8, 1, 2, 6, 3, 3, 2, 3, 7, 5, 2, 8, 6, 1, 4, 6, 6, 2, 2, 6, 9, 5, 4, 3, 7, 6, 2, 6, 7, 5, 2, 1, 6, 6, 2, 10, 9, 6, 3, 3, 6, 2, 3, 8, 12, 5, 5, 7, 11, 5, 1
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OFFSET
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0,2
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 4^k*6 (k = 0,1,2,...), 16^k*m (k = 0,1,2,... and m = 5, 7, 8, 31, 43, 61, 116).
(ii) Any integer n > 15 can be written as w^2 + x^2 + y^2 + z^2 with w, x, y, z nonnegative integers and 6*x + 10*y + 12*z a square.
(iii) Each nonnegative integer n not among 7, 15, 23, 71, 97 can be written as w^2 + x^2 + y^2 + z^2 with w, x, y, z nonnegative integers and 2*x + 6*y + 10*z a square. Also, any nonnegative integer n not among 7, 43, 79 can be written as w^2 + x^2 + y^2 + z^2 with w, x, y, z nonnegative integers and 3*x + 5*y + 6*z a square.
a(n) > 0 verified for all n <= 3*10^7. - Zhi-Wei Sun, Nov 28 2016
Qing-Hu Hou at Tianjin Univ. has verified a(n) > 0 and parts (ii) and (iii) of the above conjecture for n up to 10^9. - Zhi-Wei Sun, Dec 04 2016
The conjecture that a(n) > 0 for all n = 0,1,2,... is called the 1-3-5-Conjecture and the author has announced a prize of 1350 US dollars for its solution. - Zhi-Wei Sun, Jan 17 2017
Qing-Hu Hou has finished his verification of a(n) > 0 for n up to 10^10. - Zhi-Wei Sun, Feb 17 2017
The 1-3-5 conjecture was finally proved by António Machiavelo and Nikolaos Tsopanidis in a JNT paper published in 2021. This is a great achivement! - Zhi-Wei Sun, Mar 31 2021
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LINKS
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EXAMPLE
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a(5) = 1 since 5 = 2^2 + 1^2 + 0^2 + 0^2 with 1 + 3*0 + 5*0 = 1^2.
a(6) = 1 since 6 = 2^2 + 1^2 + 1^2 + 0^2 with 1 + 3*1 + 5*0 = 2^2.
a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^2 with 1 + 3*1 + 5*1 = 3^2.
a(8) = 1 since 8 = 0^2 + 0^2 + 2^2 + 2^2 with 0 + 3*2 + 5*2 = 4^2.
a(24) = 1 since 24 = 4^2 + 0^2 + 2^2 + 2^2 with 0 + 3*2 + 5*2 = 4^2.
a(31) = 1 since 31 = 1^2 + 5^2 + 2^2 + 1^2 with 5 + 3*2 + 5*1 = 4^2.
a(43) = 1 since 43 = 1^2 + 1^2 + 5^2 + 4^2 with 1 + 3*5 + 5*4 = 6^2.
a(61) = 1 since 61 = 6^2 + 0^2 + 0^2 + 5^2 with 0 + 3*0 + 5*5 = 5^2.
a(116) = 1 since 116 = 10^2 + 4^2 + 0^2 + 0^2 with 4 + 3*0 + 5*0 = 2^2.
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MATHEMATICA
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SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[x+3y+5z], r=r+1], {x, 0, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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