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User: Peter J. Marko

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Peter J. Marko has authored 1 sequences.

A318490 Irregular triangle read by rows T(n,k): T(1,1) = 0; for n > 1, row n lists distinct prime factors of the n-th highly composite number (A002182(n)), where column k = 1, 2, 3, ..., omega(A002182(n)) = A108602(n).

Original entry on oeis.org

0, 2, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 5, 2, 3, 5, 2, 3, 5, 2, 3, 5, 2, 3, 5, 2, 3, 5, 2, 3, 5, 7, 2, 3, 5, 7, 2, 3, 5, 7, 2, 3, 5, 7, 2, 3, 5, 7, 2, 3, 5, 7, 2, 3, 5, 7, 2, 3, 5, 7, 2, 3, 5, 7, 2, 3, 5, 7, 2, 3, 5, 7, 11, 2, 3, 5, 7
Offset: 1

Author

Peter J. Marko, Aug 27 2018

Keywords

Comments

The exponents of factors in row n are given by A212182(n).

Examples

			Triangle begins:
  0;
  2;
  2;
  2, 3;
  2, 3;
  2, 3;
  2, 3;
  2, 3;
  2, 3, 5;
  2, 3, 5;
  2, 3, 5;
  2, 3, 5;
  2, 3, 5;
  2, 3, 5;
  2, 3, 5, 7;
  ...
1st row: A002182(1) = 1 so T(1,1) = 0;
2nd row: A002182(2) = 2 so T(2,1) = 2;
3rd row: A002182(3) = 4 = 2^2 so T(3,1) = 2;
4th row: A002182(4) = 6 = 2 * 3 so T(4,1) = 2 and T(4,2) = 3;
5th row: A002182(5) = 12 = 2^2 * 3 so T(5,1) = 2 and T(5,2) = 3;
6th row: A002182(6) = 24 = 2^3 * 3 so T(6,1) = 2 and T(6,2) = 3.
		

Crossrefs

Row n has length A108602(n), n >= 2.