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A336761
a(0) = 0, a(1) = 1; for n > 1, a(n) = a(n-1) - lpf(n) if nonnegative and not already in the sequence, otherwise a(n) = a(n-1) + lpf(n), where lpf(n) is the least prime dividing n.
7
0, 1, 3, 6, 4, 9, 7, 14, 12, 15, 13, 2, 4, 17, 19, 16, 18, 35, 33, 52, 50, 47, 45, 22, 20, 25, 23, 26, 24, 53, 51, 82, 80, 77, 75, 70, 68, 31, 29, 32, 30, 71, 69, 112, 110, 107, 105, 58, 56, 49, 51, 48, 46, 99, 97, 92, 90, 87, 85, 144, 142, 81, 79, 76, 74, 79, 81, 148, 146, 143, 141, 212, 210
OFFSET
0,3
COMMENTS
This sequences uses the same rules as Recamán's sequence A005132 except that, instead of adding or subtracting n each term, the least prime dividing n is used. See A020639.
For the first 100 million terms the smallest value not appearing is 5. As any term for prime n can be the previous term minus n there is no apparent lower bound for the terms as n increases. For example a(16367081) = 601, the previous term being a(16367080) = 16367682. Thus it is possible 5, and eventually all values, are visited, although this is unknown.
In the same range the maximum value is a(98782561) = 602622357, and 7627043 terms repeat a previously visited value, the first time this occurs is a(12) = a(4) = 4. The longest run of consecutive increasing terms is 47, starting at a(96135288) = 26062, while the longest run of consecutive decreasing terms is 238, starting at a(32357989) = 160443385.
EXAMPLE
a(2) = 3. As 2 is prime lpf(2) = 2 thus a(2) = a(1) + 2 = 1 + 2 = 3.
a(6) = 7. As lpf(6) = 2 and as 7 has not been previously visited and is nonnegative, a(6) = a(5) - 2 = 9 - 2 = 7.
CROSSREFS
KEYWORD
nonn
AUTHOR
Scott R. Shannon, Aug 03 2020
STATUS
approved