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A338026 revision #23

A338026
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.
0
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
OFFSET
1,1
COMMENTS
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.
LINKS
EXAMPLE
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.
MATHEMATICA
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
CROSSREFS
Cf. A209724.
Sequence in context: A114842 A169928 A067683 * A359881 A164276 A154967
KEYWORD
nonn
AUTHOR
STATUS
editing