A357498 - OEIS

%I #42 May 10 2023 07:28:21

%S 1,1,3,1,2,5,1,2,4,9,1,2,3,5,11,1,2,3,5,8,17,1,2,3,4,6,10,21,1,2,3,4,

%T 6,9,14,29,1,2,3,4,5,7,10,16,33,1,2,3,4,5,7,9,13,20,41,1,2,3,4,5,6,8,

%U 11,15,23,47,1,2,3,4,5,6,8,10,13,18,28,57

%N Triangle read by rows where each term in row n is the next greater multiple of n..1 divided by n..1.

%C This triangle is related to A357431. Terms there are divisible by n..1 and here that division is performed, leaving the respective multiple of each.

%C Row n has length n and columns are numbered k = 1..n corresponding to multiples n..1.

%C Row n begins with n/n = 1. The end-most terms of the rows are A007952.

%H Neal Gersh Tolunsky, <a href="/A357498/b357498.txt">Table of n, a(n) for n = 1..10011</a> (141 rows, flattened)

%F T(n,k) = A357431(n,k) / (n-k+1).

%F T(n,1) = 1.

%F T(n,k) = (T(n,k-1)*(n-k+2) + (n-k+1) - (T(n,k-1)*(n-k+2)) mod (n-k+1))/(n-k+1), for k >= 2.

%F T(n,n) = A007952(n).

%e Triangle begins:

%e   n/k|  1   2   3   4   5   6   7

%e   --------------------------------

%e   1  |  1;

%e   2  |  1,  3;

%e   3  |  1,  2,  5;

%e   4  |  1,  2,  4,  9;

%e   5  |  1,  2,  3,  5, 11;

%e   6  |  1,  2,  3,  5,  8, 17;

%e   7  |  1,  2,  3,  4,  6, 10, 21;

%e   ...

%e For row n=6, we have:

%e   A357431 row  6 10 12 15 16 17

%e   divided by   6  5  4  3  2  1

%e   results in   1  2  3  5  8 17

%t row[n_] := Module[{k = n, s = Table[0, n], r}, s[[1]] = 1; Do[k++; k += If[(r = Mod[k, i]) == 0, 0, i - Mod[k, i]]; s[[n + 1 - i]] = k/i, {i, n - 1, 1, -1}]; s]; Array[row, 12] // Flatten (* _Amiram Eldar_, Oct 01 2022 *)

%o (PARI) row(n) = my(v=vector(n)); v[1] = n; for (k=2, n, v[k] = v[k-1] + (n-k+1) - (v[k-1] % (n-k+1));); vector(n, k, v[k]/(n-k+1)); \\ _Michel Marcus_, Nov 16 2022

%Y Cf. A358435 (row sums), A357431, A007952 (right diagonal).

%K nonn,tabl,easy

%O 1,3

%A _Tamas Sandor Nagy_, Oct 01 2022