A348599 Irregular triangle T(n,k) where row n is the partition of n with the least number of 3-smooth parts such that the product of parts is maximal.

1, 2, 3, 4, 3, 2, 6, 4, 3, 8, 9, 6, 4, 8, 3, 12, 9, 4, 8, 6, 9, 6, 16, 9, 8, 18, 16, 3, 12, 8, 12, 9, 16, 6, 9, 8, 6, 24, 16, 9, 18, 8, 27, 16, 12, 27, 2, 18, 12, 27, 4, 32, 24, 9, 18, 16, 27, 8, 36, 36, 1, 32, 6, 27, 12, 24, 16, 32, 9, 24, 18, 27, 16, 32, 12, 27, 18

Offset: 1

Author

Michael De Vlieger, Feb 23 2022

Keywords

• tabf
• nonn

Comments

Let k = 2^a * 3^b be a part such that n is the sum of at least one such k. Then we may represent k at (a,b) on a Cartesian grid. In Dimitrov, et al., these k are known as "digits" in a 2-dimensional number base system. Since these partitions have the least number of parts (digits) in order to represent n, in Dimitrov this is called a "canonic form" for n in base (2,3).

These numbers represent an extreme representation where the digits tend to be placed close together.

Examples

			Triangle begins:
   1;
   2;
   3;
   4;
   3, 2;    (product larger than (4,1))
   6;
   4, 3;    (product larger than (6,1))
   8;
   9;
   6, 4;    (product greatest of {(9,1), (8,2), (6,4)})
   8, 3;    (product larger than (9,2))
  12;
  ...
		

References

• Vassil Dimitrov, Graham Jullien, and Roberto Muscedere, Multiple Number Base System Theory and Applications, 2nd ed., CRC Press (2012), 35-39.

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10093 (rows n = 1..3600, flattened)
• Michael De Vlieger, Plot of parts in row n at (T(n,k), n), for n = 1..256.
• Michael De Vlieger, Plot T(n,k) at (T(n,k), n) for n = 1..10000.
• Michael De Vlieger, Annotated plot of m = A347860(n,k) and m = T(n,k) at (m, n) for n = 1..64, showing m in row n of this sequence in blue, m in row n of A347860 in red, but in black if these coincide.

Crossrefs

Cf. A003586, A237442, A276380.

Programs

Mathematica

nn = 45; ss = Union@ Flatten@ Table[2^a*3^b, {a, 0, Log2[nn]}, {b, 0, Log[3, nn/(2^a)]}]; Table[Block[{k = 1, w, t = TakeWhile[ss, # <= n &]}, While[{} == Set[w, IntegerPartitions[n, {k}, t]], k++]; MaximalBy[w, Times @@ # &][[1]]], {n, nn}] // Flatten

Formula

A237442(n) = length of row n.