This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A357498 #42 May 10 2023 07:28:21
%S A357498 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 A357498 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 A357498 11,15,23,47,1,2,3,4,5,6,8,10,13,18,28,57
%N A357498 Triangle read by rows where each term in row n is the next greater multiple of n..1 divided by n..1.
%C A357498 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 A357498 Row n has length n and columns are numbered k = 1..n corresponding to multiples n..1.
%C A357498 Row n begins with n/n = 1. The end-most terms of the rows are A007952.
%H A357498 Neal Gersh Tolunsky, <a href="/A357498/b357498.txt">Table of n, a(n) for n = 1..10011</a> (141 rows, flattened)
%F A357498 T(n,k) = A357431(n,k) / (n-k+1).
%F A357498 T(n,1) = 1.
%F A357498 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 A357498 T(n,n) = A007952(n).
%e A357498 Triangle begins:
%e A357498 n/k| 1 2 3 4 5 6 7
%e A357498 --------------------------------
%e A357498 1 | 1;
%e A357498 2 | 1, 3;
%e A357498 3 | 1, 2, 5;
%e A357498 4 | 1, 2, 4, 9;
%e A357498 5 | 1, 2, 3, 5, 11;
%e A357498 6 | 1, 2, 3, 5, 8, 17;
%e A357498 7 | 1, 2, 3, 4, 6, 10, 21;
%e A357498 ...
%e A357498 For row n=6, we have:
%e A357498 A357431 row 6 10 12 15 16 17
%e A357498 divided by 6 5 4 3 2 1
%e A357498 results in 1 2 3 5 8 17
%t A357498 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 A357498 (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 A357498 Cf. A358435 (row sums), A357431, A007952 (right diagonal).
%K A357498 nonn,tabl,easy
%O A357498 1,3
%A A357498 _Tamas Sandor Nagy_, Oct 01 2022