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A272072
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Nonnegative integers n such that 10^n is not of the form x^3 + y^3 + z^3 where x > y > z > 0.
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0
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OFFSET
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1,3
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COMMENTS
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This sequence is finite because 10^n can be written as the sum of three distinct positive cubes for all n > 10.
Proof: If 10^n = x^3 + y^3 + z^3, then 10^(n+3) = 10^3 * 10^n = 10^3 * (x^3 + y^3 + z^3) = (10*x)^3 + (10*y)^3 + (10*z)^3. It is clear that if x, y and z are distinct, then k*x, k*y and k*z are also distinct for all nonzero values of k. So with 10^n also 10^(n+3) is not in the sequence. Therefore, if we can find an integer n such that 10^n, 10^(n+1) and 10^(n+2) are not in the sequence then all 10^m with m >= n will also not be in the sequence. In the example section is shown that 10^n is not in the sequence for n = 11, 12 and 13. Thus this sequence is finite, and all its terms can be found in limited range 0 <= n <= 10.
Note that 10^n, for positive n == 1 (mod 3), can be written as a sum of three nondistinct positive cubes because 10^(1+3*k) = (10^k)^3 + (10^k)^3 + (2*10^k)^3 for all k >= 0. This applies to 10^n with n = 1, 4, 7, 10.
Also note that 10^0, 10^2, 10^3 and 10^5 cannot be the sum of three positive cubes (see the table for A003072) which applies especially to the nondistinct case. Nondistinct representations for 10^1 and 10^4 have been considered above. 10^n can be written as the sum of three positive cubes (not necessarily distinct) for all n > 5. For n >= 11 the existence of distinct representations has been proved above. For n = 7 and 10 see the nondistinct case given in the 1 (mod 3) comment. For n = 6, 8 and 9 see the example section for distinct cases.
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LINKS
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EXAMPLE
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10^6 = 35^3 + 70^3 + 85^3.
10^8 = 196^3 + 312^3 + 396^3.
10^9 = 84^3 + 658^3 + 894^3.
10^11 = 1960^3 + 3120^3 + 3960^3.
10^12 = 9^3 + 2991^3 + 9910^3.
10^13 = 6150^3 + 14575^3 + 18825^3.
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MATHEMATICA
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Select[Range[0, 10], NoneTrue[PowersRepresentations[10^#, 3, 3], #[[3]] > #[[2]] > #[[1]] > 0 &] &] (* Michael De Vlieger, Apr 20 2016, Version 10 *)
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CROSSREFS
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KEYWORD
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nonn,fini,full
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AUTHOR
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STATUS
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approved
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