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A280356
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Number of ways to write n as x^4 + y^3 + z^2 + 2^k, where x,y,z are nonnegative integers and k is a positive integer.
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5
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0, 1, 3, 4, 4, 4, 3, 3, 5, 5, 4, 5, 6, 5, 2, 3, 7, 8, 7, 7, 8, 5, 1, 4, 9, 8, 5, 7, 8, 6, 3, 8, 14, 11, 7, 8, 7, 4, 4, 8, 13, 9, 4, 8, 8, 5, 4, 8, 11, 5, 5, 8, 8, 6, 4, 6, 9, 6, 6, 10, 6, 2, 3, 4, 10, 10, 9, 13, 12, 7, 2, 7, 11, 9, 7, 9, 6, 2, 3, 7
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OFFSET
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1,3
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 23, 1135, 6415, 6471.
(ii) If P(x,y) is one of the polynomials 3*x^4 + y^3 and x^6 + 3*y^2, then any positive integer n can be written as P(x,y) + z^2 + 2^k with x,y,z and k nonnegative integers.
We have verified that a(n) > 0 for all n = 2..2*10^7, and that part (ii) of the conjecture holds for all n = 1..10^7.
We also find finitely many polynomials of the form a*x^m + b*y^2 (including x^4 + y^2 and 10*x^5 + y^2) with a and b positive integers and m <= 5, for which it seems that any positive integer can be written as P(x,y) + z^2 + 2^k with x,y,z,k nonnegative integers.
See also A280153 for a similar conjecture involving powers of 4 or 8.
Qing-Hu Hou at Tianjin Univ. has verified that a(n) > 0 for all n = 2..10^9. In 2017, the author announced to offer US $234 as the prize for the first correct solution to his conjecture that a(n) > 0 for all n > 1. - Zhi-Wei Sun, Dec 30 2017
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LINKS
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EXAMPLE
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a(2) = 1 since 2 = 0^4 + 0^3 + 0^2 + 2^1.
a(23) = 1 since 23 = 2^4 + 1^3 + 2^2 + 2^1.
a(1135) = 1 since 1135 = 0^4 + 7^3 + 28^2 + 2^3.
a(6415) = 1 since 6415 = 1^4 + 13^3 + 11^2 + 2^12.
a(6471) = 1 since 6471 = 1^4 + 13^3 + 57^2 + 2^10.
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MATHEMATICA
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SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
In[2]:= Do[r=0; Do[If[SQ[n-2^k-x^4-y^3], r=r+1], {k, 1, Log[2, n]}, {x, 0, (n-2^k)^(1/4)}, {y, 0, (n-2^k-x^4)^(1/3)}]; Print[n, " ", r]; Continue, {n, 1, 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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