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1, 1, 4, 4, 1, 1, 2, 4, 3, 1, 4, 4, 1, 1, 2, 8, 1, 3, 4, 4, 1, 1, 2, 4, 5, 1, 36, 4, 1, 1, 2, 16, 1, 1, 4, 12, 1, 1, 2, 4, 1, 1, 4, 4, 3, 1, 2, 8, 7, 5, 4, 4, 1, 9, 2, 4, 1, 1, 4, 4, 1, 1, 6, 32, 1, 1, 4, 4, 1, 1, 2, 12, 1, 1, 20, 4, 1, 1, 2, 8, 27, 1, 4, 4, 1, 1, 2, 4, 1, 3, 4, 4, 1, 1, 2
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OFFSET
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1,3
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COMMENTS
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Please look into A234001 for a more detailed description.
If n is squarefree and n == 1 (mod 4) or n == 2 (mod 4), then a(n) = 1.
If p^2 divides n for some prime p, a(n) is a multiple of p.
If n == 3 (mod 8), then a(n) is a multiple of 4 because numbers of the form x^2+n*y^2 cannot have any prime factors that are congruent to 2+n (mod 2n) raised to an odd power.
If n == 7 (mod 8), then a(n) is a multiple of 2 because numbers of the form x^2+n*y^2 can have prime factors that are congruent to 2+n (mod 2n) raised to an odd power, but they cannot be congruent to 2 (mod 4). So, we need to characterize the prime factor of 2 from the remaining prime factors that are congruent to 2+n (mod 2n) separately.
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KEYWORD
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nonn,uned
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AUTHOR
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STATUS
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approved
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