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a(0) = 0, a(1) = 1; for n > 1, a(n) = a(n-1) - A001065(n) if nonnegative and not already in the sequence, otherwise a(n) = a(n-1) + A001065(n), where A001065(n) is the sum of the proper divisors of n.
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%I #33 Jan 09 2021 21:04:35

%S 0,1,2,3,6,5,11,10,17,13,21,20,4,5,15,24,9,8,29,28,50,39,25,26,62,56,

%T 40,27,55,54,12,13,44,59,79,66,121,120,98,81,31,30,84,83,43,76,102,

%U 101,177,169,126,105,151,150,216,199,135,112,80,81,189,188,154,113,176,157,235,234,292

%N a(0) = 0, a(1) = 1; for n > 1, a(n) = a(n-1) - A001065(n) if nonnegative and not already in the sequence, otherwise a(n) = a(n-1) + A001065(n), where A001065(n) is the sum of the proper divisors of n.

%C This sequences uses the same rules as Recamán's sequence A005132 except that, instead of adding or subtracting n each term, the sum of the proper divisors of n is used. See A001065.

%C For the first 10 million terms the smallest value not appearing is 46. As there are infinite abundant numbers it is likely that eventually all values are visited, although this is unknown.

%C In the same range the maximum value is a(9529524) = 65424507, and 633238 terms repeat a previously visited value, the first time this occurs is a(13) = a(5) = 5. The longest run of consecutive increasing terms is 8, starting at a(7069152) = 1040443, while the longest run of consecutive decreasing terms is 11, starting at a(106380) = 621777.

%e a(2) = 2. As 2 is prime A001065(2) = 1, and 1 has previously appeared, a(2) = a(1) + 1 = 2.

%e a(4) = 6. As A001065(4) = 1+2 = 3, and 0 has previously appeared, a(4) = a(3) + 3 = 6.

%e a(5) = 5. As 5 is prime A001065(5) = 1, and 5 has not previously appeared, a(5) = a(4) - 1 = 5.

%Y Cf. A005132, A001065, A335382, A336760, A336761, A000040.

%K nonn

%O 0,3

%A _Scott R. Shannon_, Aug 16 2020