A357498 Triangle read by rows where each term in row n is the next greater multiple of n..1 divided by n..1.

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

Offset

1,3

Comments

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.

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

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

Links

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10011 (141 rows, flattened)

Formula

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

T(n,1) = 1.

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.

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

Example

Triangle begins:
  n/k|  1   2   3   4   5   6   7
  --------------------------------
  1  |  1;
  2  |  1,  3;
  3  |  1,  2,  5;
  4  |  1,  2,  4,  9;
  5  |  1,  2,  3,  5, 11;
  6  |  1,  2,  3,  5,  8, 17;
  7  |  1,  2,  3,  4,  6, 10, 21;
  ...
For row n=6, we have:
  A357431 row  6 10 12 15 16 17
  divided by   6  5  4  3  2  1
  results in   1  2  3  5  8 17

Mathematica

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 *)

Prog

(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

Crossrefs

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

Sequence in context: A354348 A211025 A125704 * A131127 A113141 A134225

Adjacent sequences: A357495 A357496 A357497 * A357499 A357500 A357501

Keyword

nonn,tabl,easy

Author

Tamas Sandor Nagy, Oct 01 2022

Status

approved