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A338026
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a(1) = 8; for n > 1, a(n) is the largest integer m such that m = ((2x * a(n-1)) /(x+1)) - x, with x a positive nontrivial divisor of m.
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0
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8, 9, 10, 12, 15, 20, 28, 42, 66, 110, 190, 342, 630, 1190, 2278, 4422, 8646, 17030, 33670, 66822, 132900, 264758, 528034, 1053990, 2105077, 4205820, 8405840, 16803405, 33595212, 67173930, 134324628, 268616475, 537185908, 1074305622, 2148516546
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OFFSET
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1,1
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COMMENTS
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There are no primes in the sequence and, excepting for a(1), no powers of 2.
For each n > 1 there are two integers f, g such that f*g = a(n) and f + g = 2*a(n-1) - a(n) - 1. (Empirical observation)
Excluding the condition that each term should be the largest one, the terms which satisfies the remaining conditions performs an irregular infinite net.
If the condition "be the largest term" is replaced for "be the smallest one" with the rest of conditions remaining, A209724 sequence is obtained. (This is true, at least, from a’(1) to a’(100))
Similar sequences are obtained with a’(1) a nonprime integer larger than 6, either a power of two or not. However, if a’(1) is not a power of two, there exist at least, one integer: a’(0) = ( a’(1) +x)(1+x)) / (2x) with x a positive nontrivial divisor of a’(1). Example: a’(1) = 26, a’(0) = 21, a’(-1) = 16. (As a’(-1) is a power of 2 there is not an a’(-2) term). If a’(1) = 6, a’(0) = a’(1) = a’(2)…= a’(k) = 6.
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LINKS
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EXAMPLE
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a(5) = 15 = ( (2*3*12) / 4) - 3 or ( (2*5*12) / 6) - 5 = 15; Also 14 = ((2*2*12) / 3) - 2, but 15 is larger.
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MATHEMATICA
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w[n_] := Module[{x, p}, Max[p /. List@ToRules@Reduce[p == (2 n*x)/(x + 1) - x == x*y && x > 1 && y > 1, p, Integers]]]; n := 8; k := {n}; m = 1; While[m < 35, {AppendTo[k, w[n]], n = w[n]}; m++]; k
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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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