A333068 - OEIS

A333068

a(1) = 1; for n > 1, a(n) = n*(n-1)/2 + ((a(n-1)-1) mod n) + 1, the a(n-1)-th term of the n-th row of the triangle of positive integers, indexed in cyclic manner.

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, 338, 365, 379, 408, 453, 484, 500, 533, 584, 619, 637, 674, 731, 770, 790, 831, 894, 937, 959, 1004, 1073, 1120, 1144, 1193, 1268, 1319, 1345, 1398

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OFFSET

1,2

COMMENTS

Write the positive integers A000027 in a triangle T(k,j) = k(k-1)/2 + j, k >= j >= 1. Then a(n) is the a(n-1)-th element of the n-th row, with indices taken in [1..n] modulo n. (If a row has fewer than a(n-1) elements, this is equivalent to repeating the elements in a cyclic/periodic manner up to the required length.)

LINKS

Table of n, a(n) for n=1..53.

EXAMPLE

Consider the positive integers written in a triangle, the n-th row going from n(n-1)/2 + 1 to n(n+1)/2 = A000217(n):

row k numbers T(k,j) = k(k-1)/2 + j, j = 1..k

----- ---------------------------------------

1 1;

2 2, 3;

3 4, 5, 6;

4 7, 8, 9, 10;

5 11, 12, 13, 14, 15;

etc.

After a(1) = 1, a(2) is the 1st term of the 2nd row of the triangle, a(2) = 2.

Then a(3) is the a(2)-th = 2nd term of the 3rd row, a(3) = 5.

Then a(4) is the a(3)-th = 5th term of row 4, with cyclic indexing: since the row has only 4 elements, we wrap around in a cyclic manner and come back to the 1st term, a(4) = 7.

Similarly, a(5) = 8 is the a(4)-th = 7th term of the 5th row, with cyclic indexing (so actually the 2nd term).

a(9) is the 32nd term of row 9 = 37..45 of length 9, i.e., taking modulo 9, the 5th term, 36+5 = 41.

MAPLE

a:= proc(n) option remember; `if`(n<2, n,

n*(n-1)/2 + ((a(n-1)-1) mod n) + 1)

end:

seq(a(n), n=1..60); # Alois P. Heinz, Apr 23 2020

MATHEMATICA

a[n_] := a[n] = If[n<2, n, n(n-1)/2 + Mod[a[n-1]-1, n] + 1];

Array[a, 60] (* Jean-François Alcover, Nov 30 2020, after Alois P. Heinz *)

PROG

(PARI) for(n=2, #a=Vec(1, 100), a[n]=n*(n-1)/2 + (a[n-1]-1)%n + 1); a