[seqfan] I need help with defining these 3 sequences

Ali Sada pemd70 at yahoo.com
Sun Mar 1 07:20:48 CET 2020


Hi Everyone,
As usual, I am finding difficulties defining the three sequences below. I would really appreciate your help.
Sequence 1: We distribute the natural numbers on “cycles.” The first cycle has only one element (1), the second cycle has two elements (2,3), the third cycle has three elements (4,5,6), etc.
1
2,3
4,5,6
7,8,9,10
11,12,13,14,15

We start with a(1)=1. Then a(2) is the first element in the second cycle. a(2) = 2.
Since a(2)=2, then a(3) is the second element in the third cycle. a(3) = 5.
a(4) is the fifth element in the fourth cycle, and that would be 7 (since we have only 4 elements then the cycle goes back to the first element.) a(4) = 7
a(5) is the seventh element in the fifth cycle and that would be 12.

This is the sequence we get. 
1, 2, 5, 7, 12, 21, 28, 32, 45, 50, 61, 67, 80, 101, 116, 124, 141, 168, 187, 197, 218, 251, 274, 286, 311, 350, 377, 391, 420, 465, 496

Sequence 2:    Finding terms in this sequence doesn’t go lower indices to higher indices. We find terms in a way similar to the movement of a ping pong table. I don’t know if this is an acceptable way to creating sequence.

We start with a(1) = 1 and a(2) = 2. 
a(3) = a(1) + a(2) = 3. We don’t have any gap here, so we continue. 
We add a(2) and a(3) and we get 5. So, a(5) = 5. To find a(4), we add a(5) + a(3) = 8.
We filled the gap between 3 and 5, so we continue.
We add a(4) + a(5) = 5+8= 13, so a(13) =1 3. Now, we need to fill the gaps between 6 and 12.
a(6) = a(5) + a(13) = 5+13 = 18
a(12) = a(6) + a(13) = 18 + 13 = 31
a(7) = a(6) + a(12) = 18 + 31 =49
a(11) = a(7) + a(12) = 49 +31 = 80
a (8) = a(7) + a(11) = 49 + 80= 129
a(10)= a(8) + a(11) = 129 +80 = 209 
a(9)= a(8) + a(10) = 209 + 129 = 388.
Now, we filled all the gaps, so we continue.
a(12) + a(13)= 44, so, a(44) = 44. And we try to fill the gaps the same way.
The sequence we get is below. 


1, 2, 3, 8, 5, 18, 49, 129, 338, 209, 80, 31, 13, 57, 158, 417, 1093, 2808, 7331, 19185, 50224, 131487, 344237, 901224, 2359435, 6177081, 16171808, 42338343, 68504878, 26166535, 9994727, 3817646, 1458211, 556987, 212750, 81263, 31039, 11854, 4523, 1715, 676, 259, 101, 44, 189, 523, 1380, 3617, 9471, 24796, 64917, 169955, 274993, 655024, 1690079, 4415213, 7140347, 17005828, 43877137

Sequence 3:    I have a definition for this sequence, but I am sure it could be improved.
The definition is “a(1) =1; a(n) is the lexicography earliest positive integer such that n+1 and the summation of terms up to a(n) are co-primes.”
We start with a(1) = 1. 
To find a(2), we need the earliest positive integer such that a(1) + a(2) is co-prime with 3. So, a(2) = 3. To find a(3), we cannot put neither 2 nor 4, because 6 and 8 have a common factor with 4, so we put a(3) = 5. And so on. 
The sequence we get is: 

1, 3, 5, 2, 6, 7, 9, 4, 10, 11, 13, 8, 14, 16, 12, 17, 19, 15, 21, 18, 22, 23, 25, 20, 26, 28, 24, 29, 31, 27, 33, 30, 34, 35, 37
Best,
Ali 




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