Revision History for A254629
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#7 by Bruno Berselli at Tue Feb 03 11:52:01 EST 2015
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#6 by Zhi-Wei Sun at Tue Feb 03 10:34:22 EST 2015
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#5 by Zhi-Wei Sun at Tue Feb 03 10:34:00 EST 2015
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#4 by Zhi-Wei Sun at Tue Feb 03 10:32:42 EST 2015
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| COMMENTS
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(ii) If f(x) is one amongof the polynomials 3x^2+x, (7x^2+3x)/2, (9x^2-x)/2, (11x^2-7x)/2, (15x^2-7x)/2, (15x^2-11x)/2, then any nonnegative integer n can be written as x^2+ + y*(y+1)+) + f(z) with x,y,z nonnegative integers.
(iii) If g(x) is among the polynomials (5x^2-x)/2, (7x^2-3x)/2, (13x^2-9x)/2, (15x^2-11x)/2, then any nonnegative integer n can be written as the sum of x*(x+1)/2 + y^2 + f(z) with x,y,z nonnegative integers.
We have proved that for each n = 0,1,... there are integers x,y,z such that n = x^2 + y*(y+1) + z*(4z+1).
It is known that {x^2+y(*(y+1): x,y=0,1,...} = {x(*(x+1)/2+y(*(y+1)/2: x,y=0,1,...}.
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| LINKS
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Zhi-Wei Sun, <a href="http://arxiv.org/abs/0905.0635">On universal sums of polygonal numbers</a>, arXiv:0905.0635 [math.NT], 2009-2015.
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| EXAMPLE
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a(103) = 1 since 103 = 8^2 + 0*1 + 3*(4*3+1).
a(122) = 1 since 122 = 9^2 + 1*2 +3*(4*3+1).
a(143) = 1 since 143 = 6^2 + 1*2 + 5*(4*5+1).
a(167) = 1 since 167 = 3^2 + 9*10 + 4*(4*4+1).
a(248) = 1 since 248 = 5^2 + 4*5 + 7*(4*7+1).
a(338) = 1 since 338 = 5^2 + 10*11 + 7*(4*7+1).
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| MATHEMATICA
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SQ[n_]:=IntegerQ[Sqrt[n]]
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| CROSSREFS
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Cf. A000217, A000290, A007742, A254623.
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#3 by Zhi-Wei Sun at Tue Feb 03 09:46:17 EST 2015
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| NAME
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Number of ways to write n as x^2 + y*(y+1) + z*(4*z+1) with x,y,z nonnegative integers.
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| COMMENTS
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Conjecture: (i) a(n) > 0 for all n, and a(n) > 1 for all n > 338.
(iii) If g(x) is among the polynomials (5x^2-x)/2, (7x^2-3x)/2, (13x^2-9x)/2, (15x^2-11x)/2, then any nonnegative integer n can be written as the sum of x*(x+1)/2 + y^2 + f(z) with x,y,z a nonnegative integerintegers.
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| LINKS
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Zhi-Wei Sun, <a href="/A254629/b254629.txt">Table of n, a(n) for n = 0..10000</a>
Zhi-Wei Sun, <a href="http://arxiv.org/abs/0905.0635">On universal sums of polygonal numbers</a>, arXiv:0905.0635 [math.NT], 2009-2015.
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| MATHEMATICA
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SQ[n_]:=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-y(y+1)-z*(4z+1)], r=r+1], {y, 0, (Sqrt[4n+1]-1)/2}, {z, 0, (Sqrt[16(n-y(y+1))+1]-1)/8}];
Print[n, " ", r]; Continue, {n, 0, 100}]
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| CROSSREFS
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Cf. A000217, A000290, A254623.
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#2 by Zhi-Wei Sun at Tue Feb 03 09:40:50 EST 2015
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| NAME
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allocated for Zhi-Wei Sun
Number of ways to write n as x^2 + y*(y+1) + z*(4*z+1) with x,y,z nonnegative integers.
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| DATA
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1, 1, 1, 1, 1, 1, 3, 2, 1, 2, 1, 3, 2, 1, 1, 2, 3, 1, 3, 1, 3, 5, 2, 1, 3, 3, 2, 3, 2, 3, 3, 3, 1, 2, 4, 1, 5, 1, 2, 5, 2, 3, 5, 4, 1, 4, 4, 3, 4, 4, 2, 5, 2, 1, 4, 5, 5, 3, 1, 1, 7, 5, 1, 3, 4, 2, 5, 3, 2, 6, 5, 3, 4, 4, 5, 5, 4, 4, 5, 3, 1, 8, 2, 4, 7, 3, 4, 3, 5, 3, 6, 3, 3, 3, 6, 4, 5, 5, 2, 9, 2
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| OFFSET
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0,7
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| COMMENTS
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Conjecture: (i) a(n) > 0 for all n, and a(n) > 1 for all n > 338.
(ii) If f(x) is among the polynomials 3x^2+x, (7x^2+3x)/2, (9x^2-x)/2, (15x^2-7x)/2, then any nonnegative integer n can be written as x^2+y*(y+1)+f(z) with x,y,z nonnegative integers.
(iii) If g(x) is among the polynomials (5x^2-x)/2, (7x^2-3x)/2, (13x^2-9x)/2, (15x^2-11x)/2, then any nonnegative integer n can be written as the sum of x*(x+1)/2 + y^2 + f(z) with x,y,z a nonnegative integer.
It is known that {x^2+y(y+1): x,y=0,1,...} = {x(x+1)/2+y(y+1)/2: x,y=0,1,...}.
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| CROSSREFS
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Cf. A000217, A000290, A254623.
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| KEYWORD
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allocated
nonn
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| AUTHOR
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Zhi-Wei Sun, Feb 03 2015
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| STATUS
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approved
editing
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#1 by Zhi-Wei Sun at Tue Feb 03 09:40:50 EST 2015
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| NAME
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allocated for Zhi-Wei Sun
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| KEYWORD
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allocated
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| STATUS
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approved
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