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A242090
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Number of triples (a,b,c) with 0 < a < b < c < p and a + b + c == 0 mod p, where 2*b < p = prime(n).
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3
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0, 0, 0, 1, 5, 8, 16, 21, 33, 56, 65, 96, 120, 133, 161, 208, 261, 280, 341, 385, 408, 481, 533, 616, 736, 800, 833, 901, 936, 1008, 1281, 1365, 1496, 1541, 1776, 1825, 1976, 2133, 2241, 2408, 2581, 2640, 2945, 3008, 3136, 3201, 3605, 4033, 4181, 4256
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OFFSET
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1,5
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COMMENTS
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Given a + b < 2*b < p and c < p it follows that a + b + c < 2*p then the condition reduces to a + b + c = p. - Fausto A. C. Cariboni, Sep 30 2018
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LINKS
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FORMULA
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EXAMPLE
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For prime(4) = 7 there is 1 triple (a,b,c) with 0 < a < b < c < 7 and a+b+c == 0 mod 7, namely, 1+2+4 = 7, so a(4) = 1.
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MATHEMATICA
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Table[(1/2) Length[ Reduce[ Mod[a + b + c, Prime[n]] == 0 && 0 < a < b < c < Prime[n], {a, b, c}, Integers]], {n, 40}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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