A340530 Irregular triangle read by rows T(n,k) in which row n has length is A000070(n-1) and every column k is A006218, (n >= 1, k >= 1),

1, 3, 1, 5, 3, 1, 1, 8, 5, 3, 3, 1, 1, 1, 10, 8, 5, 5, 3, 3, 3, 1, 1, 1, 1, 1, 14, 10, 8, 8, 5, 5, 5, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 16, 14, 10, 10, 8, 8, 8, 5, 5, 5, 5, 5, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 20, 16, 14, 14, 10, 10, 10, 8, 8, 8, 8, 8, 5, 5, 5, 5, 5, 5, 5

Offset

1,2

Comments

The sum of row n equals A284870(n), the total number of parts in all partitions of all positive integers <= n. It is conjectured that this property is due to the correspondence between divisors and partitions. For more information see A336811.

Links

Table of n, a(n) for n=1..94.

Formula

a(m) = A006218(A176206(m)), assuming A176206 has offset 1.

T(n,k) = A006218(A176206(n,k)), assuming A176206 has offset 1.

Example

Triangle begins:
   1;
   3,  1;
   5,  3,  1,  1;
   8,  5,  3,  3, 1, 1, 1;
  10,  8,  5,  5, 3, 3, 3, 1, 1, 1, 1, 1;
  14, 10,  8,  8, 5, 5, 5, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1;
...
For n = 5 the length of row 5 is A000070(4) = 12.
The sum of row 5 is 10 + 8 + 5 + 5 + 3 + 3 + 3 + 1 + 1 + 1 + 1 + 1 = 42, equaling A284870(5).

Crossrefs

Row sums give A284870.

Cf. A340526 (a regular version).

Members of the same family are: A176206, A337209, A339258, A340531.

Cf. A339278, A339304, A340423, A340529.

Cf. A000070, A006128, A336811.

Sequence in context: A134743 A099547 A137758 * A147754 A360325 A084305

Adjacent sequences: A340527 A340528 A340529 * A340531 A340532 A340533

Keyword

nonn,tabf

Author

Omar E. Pol, Jan 10 2021

Status

approved