Search: a286944 -id:a286944
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A287616
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Number of ways to write n as x(x+1)/2 + y(3y+1)/2 + z(5z+1)/2 with x,y,z nonnegative integers.
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+10
7
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1, 1, 1, 3, 1, 2, 3, 1, 3, 1, 3, 3, 2, 4, 2, 3, 3, 3, 4, 3, 2, 5, 1, 2, 4, 3, 5, 4, 5, 4, 4, 3, 6, 3, 3, 2, 5, 2, 3, 7, 3, 7, 2, 6, 3, 5, 6, 7, 2, 4, 6, 3, 7, 2, 8, 4, 2, 6, 6, 3, 8, 3, 4, 6, 3, 7, 5, 6, 7, 4, 6, 9, 5, 6, 4, 4, 3, 4, 9, 5, 6
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OFFSET
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0,4
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COMMENTS
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Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 2, 4, 7, 9, 22.
It was proved in arXiv:1502.03056 that each n = 0,1,2,... can be written as x(x+1)/2 + y(3y+1)/2 + z(5z+1)/2 with x,y,z integers. The author would like to offer 135 US dollars as the prize for the first proof of the conjecture that a(n) is always positive.
See over 400 similar conjectures in the linked a-file.
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LINKS
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Zhi-Wei Sun, List of conjectural tuples (a,b,c,d,e,f) with {x*(ax+b)/2 + y*(cy+d)/2 + z*(ez+f)/2: x,y,z = 0,1,2,...} = {0,1,2,...}
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EXAMPLE
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a(4) = 1 since 4 = 1*(1+1)/2 + 0*(3*0+1)/2 + 1*(5*1+1)/2.
a(7) = 1 since 7 = 0*(0+1)/2 + 2*(3*2+1)/2 + 0*(5*0+1)/2.
a(9) = 1 since 9 = 3*(3+1)/2 + 0*(3*0+1)/2 + 1*(5*1+1)/2.
a(22) = 1 since 22 = 5*(5+1)/2 + 2*(3*2+1)/2 + 0*(5*0+1)/2.
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MATHEMATICA
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TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
Do[r=0; Do[If[TQ[n-x(3x+1)/2-y(5y+1)/2], r=r+1], {x, 0, (Sqrt[24n+1]-1)/6}, {y, 0, (Sqrt[40(n-x(3x+1)/2)+1]-1)/10}]; Print[n, " ", r], {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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A290342
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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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+10
5
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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
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OFFSET
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0,10
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COMMENTS
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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).
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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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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A286885
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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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+10
4
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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
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OFFSET
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0,9
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COMMENTS
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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.
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LINKS
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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.
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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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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A290472
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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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+10
4
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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
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OFFSET
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0,4
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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PROG
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(PARI) 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
(PARI) 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
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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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A290491
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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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+10
4
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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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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A260418
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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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+10
2
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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
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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, 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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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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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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