cp's OEIS Frontend

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.

A364060 Triangle read by rows where T(n,k) is the number of integer partitions of n with rounded mean k.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 2, 2, 0, 1, 0, 2, 4, 0, 0, 1, 0, 2, 5, 3, 0, 0, 1, 0, 4, 7, 0, 3, 0, 0, 1, 0, 4, 8, 5, 4, 0, 0, 0, 1, 0, 4, 14, 7, 4, 0, 0, 0, 0, 1, 0, 7, 21, 8, 0, 5, 0, 0, 0, 0, 1, 0, 7, 22, 11, 10, 0, 5, 0, 0, 0, 0, 1
Offset: 0

Views

Author

Gus Wiseman, Jul 07 2023

Keywords

Comments

We use the "rounding half to even" rule, see link.

Examples

			Triangle begins:
  1
  0  1
  0  1  1
  0  1  1  1
  0  2  2  0  1
  0  2  4  0  0  1
  0  2  5  3  0  0  1
  0  4  7  0  3  0  0  1
  0  4  8  5  4  0  0  0  1
  0  4 14  7  4  0  0  0  0  1
  0  7 21  8  0  5  0  0  0  0  1
  0  7 22 11 10  0  5  0  0  0  0  1
  0  7 36 15 12  0  6  0  0  0  0  0  1
  0 12 32 36 14  0  6  0  0  0  0  0  0  1
  0 12 53 23 23 16  0  7  0  0  0  0  0  0  1
  0 12 80 30 27 19  0  0  7  0  0  0  0  0  0  1
Row n = 7 counts the following partitions:
  .  (31111)    (511)   .  (61)  .  .  (7)
     (22111)    (421)      (52)
     (211111)   (4111)     (43)
     (1111111)  (331)
                (322)
                (3211)
                (2221)
		

Crossrefs

Row sums are A000041.
The rank statistic for this triangle is A363489.
The version for low mean is A363945, rank statistic A363943.
The version for high mean is A363946, rank statistic A363944.
Column k = 1 is A363947 (A026905 tripled).
A008284 counts partitions by length, A058398 by mean.
A026905 redoubled counts partitions with high mean 2, ranks A363950.
A051293 counts subsets with integer mean, median A000975.
A067538 counts partitions with integer mean, strict A102627, ranks A316413.
More triangles: A124943, A124944, A363952, A363953.

Programs

  • Mathematica
    Table[If[n==k==0,1,Length[Select[IntegerPartitions[n], Round[Mean[#]]==k&]]],{n,0,15},{k,0,n}]