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A247940
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Least integer m > n such that m + n divides L(m) + L(n), where L(k) refers to the Lucas number A000032(k).
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11
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5, 5, 15, 5, 19, 30, 17, 19, 15, 13, 13, 24, 35, 236, 33, 34, 31, 90, 29, 23, 27, 25, 25, 84, 47, 80, 45, 190, 43, 54, 41, 35, 39, 1216, 37, 72, 59, 212, 57, 43, 55, 66, 53, 86, 51, 76, 49, 60, 71, 53, 69, 55, 67, 222, 65, 122, 63, 112, 61, 264
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OFFSET
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1,1
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COMMENTS
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Conjecture: Let A be any integer not congruent to 3 modulo 6. Define v(0) = 2, v(1) = A, and v(n+1) = A*v(n) + v(n-1) for n > 0. Then, for any integer n > 0, there are infinitely many positive integers m such that m + n divides v(m) + v(n).
This implies that a(n) exists for any n > 0.
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LINKS
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EXAMPLE
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a(3) = 15 since 15 + 3 = 18 divides L(15) + L(3) = 1364 + 4 = 18*76.
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MATHEMATICA
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Do[m=n+1; Label[aa]; If[Mod[LucasL[m]+LucasL[n], m+n]==0, Print[n, " ", m]; Goto[bb]]; m=m+1; Goto[aa]; Label[bb]; Continue, {n, 1, 60}]
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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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