# User:KlokkoVanDenBerg/My own sequences

a(0) = 0

a(1) = 1, because 0 is < 1 we add 1 (we did not have a 1 earlier as a term in the sequence)

a(2) = 3, because 1 is < 2 we add 2 (we did not have a 2 earlier as a term in the sequence)

a(3) = 7, because 3 < 4. the first time we skip adding or subtracting number (we did have a 3 earlier in the sequence. We add 4 instead because 3 was a earlier term in the sequence (the previous terrm)

a(4) = 2, because 7 > 5 we can either add or subtract 5, and by Recamán rules, we subtract because that does not give a earlier result ) (we skipped a number (3), so a(n) = adding or subtracting n+1 from now)

a(5) = 8 , because 2 < 6 we add 6

a(6) = 17, because 8 < 9 we add 9, (we’ve skipped 7 and 8, because they appeared as terms in the sequence, like we did for 3 earlier , and from now on. a(n) = adding or subtracting n+3 from now._

a(7) = 27, because 17 > 10, we can either add or subtract 10, and by Recamán rules, we add because subtraction does give a earlier result )

a(8) = 16, because 27 > 11. we can either add or subtract 11, and by Recamán rules, we subtract because subtraction does not give a earlier result)

a(9) = 4, because 16 > 12 we can either add or subtract 12, and by Recamán rules, we subtract because subtraction does not give a earlier result)

a(10) = 17, because 4 < 13. we add 13

a(11) = 31, because 17 > 14 we can either add or subtract 14, and by Recamán rules, we add because subtraction does give a earlier result )

a(12) = 46, because 31 > 15 we can either add or subtract 15, and by Recamán rules, we add because subtraction does give a earlier result )

a(13) = 28 because 46 > 18 we can either add or subtract 18, and by Recamán rules, we subtract because subtraction does not give a earlier result ) (we’ve skipped adding/subtracting 16 and 17, because they appeared as terms in the sequence, like we did for 3, 7 and 8 earlier , and from now on. a(n) = adding or subtracting n+5 from now.)