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A355176
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a(n) is the smallest index k such that prime(n) divides both A090252(k) and A090252(2*k+1).
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2
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2, 3, 14, 32, 60, 96, 120, 128, 132, 244, 264, 388, 480, 484, 488, 2064, 1056, 571, 776, 960, 968, 976, 980, 2112, 2128, 1143, 1536, 1552, 1556, 1920, 3872, 1937, 3904, 3920, 1961, 4128, 4256, 3104, 6224, 3113, 3844, 3848, 7808, 7824, 7840, 8256, 8448, 8452
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OFFSET
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1,1
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COMMENTS
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For n > 2, a(n) is not the smallest k such that prime(n) divides A090252(k), but it is the smallest k such that prime(n) divides both A090252(k) and A090252(2*k+1). If k_(0) = a(n) we may find either an infinite or finite range of indices where prime(n) divides A090252 using the recurrence k_(n) = 2*k_(n-1)+1, but there is a caveat: in very rare cases, some k values of this recurrence may be wrong by +-1, and the next iteration will then fit again. This uncertainty is caused by the fact that two terms of A090252 will be governed by the same floor(n/2) history. For yet unknown reasons, there may be an upper limit where such a recurrence may break.
This works because in A090252 the number of primes which do not divide the last floor(n/2) terms is growing faster than they are used up by this sequence. For each prime p then there exists an index k into A090252 where the supply of unused factors is so large that, when p becomes coprime to the last floor(n/2) terms, we can always immediately find a matching second prime to build a yet-unused semiprime or use p as a yet-unused power of itself.
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LINKS
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FORMULA
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A090252(f^m(a(n))) mod A000040(n) = 0, with f(x) = 2*x+1. The range of m is yet unknown.
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EXAMPLE
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2*2+1 = 5; 2*5+1 = 11; 2*11+1 = 23; 2*23+1 = 47.
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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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