[seqfan] Re: I need help with defining these 3 sequences
Ali Sada
pemd70 at yahoo.com
Mon Mar 2 14:40:05 CET 2020
Thank you very much Dr. Hasler. I really appreciate it. And I am sorry about the blunder at a(9). It changed everything after.
And thank you for the explicit formula, but tbh, I don't understand the first definition.
Best,
Ali
On Sunday, March 1, 2020, 3:53:16 PM EST, M. F. Hasler <oeis at hasler.fr> wrote:
On Sun, Mar 1, 2020 at 2:30 AM Ali Sada wrote:
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
The n-th "cycle" is the interval [T(n)+1 .. T(n+1)] where T(n)=n(n+1)/2 are the triangular numbers A000217;the above triangle is also what is called T(k,j) = T(k-1)+j, 1 <= j <= k, in A000027, comment from Nov 19 2009.
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.
a(1) = 1; for n>1, a(n) = T(n, a(n-1) %% n) where x %% y := (x-1 mod y)+1 to get 1,...,y,1,...,y etc.or explicitly: a(n) = n(n-1)/2 + (a(n-1)-1 mod n) + 1.
(PARI) for(n=2,#a=Vec(1,100), a[n]=n*(n-1)/2 + (a[n-1]-1)%n + 1);a = [1, 2, 5, 7, 12, 21, 28, 32, 41, 46, 57, 75, 88, 95, 110, 134, 151, 160, 179, 209, 230, 241, 264, 300, 325, ...]
differs from your:
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
in a(9) = the 32-th element of "cycle 9" = 37..45 of length 9 = the (32 %% 9)-th element = the 31%9+1 = 5-th element, 36+5 = 42.
- Maximilian
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