A302300 a(n) = Sum_{p in P} (Sum_{k_j = 1} 1)^2, where P is the set of partitions of n, and the k_j are the frequencies in p.

0, 1, 1, 5, 6, 12, 21, 33, 50, 79, 116, 169, 246, 346, 487, 675, 927, 1254, 1702, 2263, 3014, 3966, 5210, 6766, 8795, 11303, 14531, 18521, 23583, 29803, 37654, 47231, 59206, 73792, 91867, 113778, 140788, 173377, 213289, 261318, 319764, 389846, 474745, 576164

Offset

0,4

Comments

This sequence is part of the contribution to the b^2 term of C_{1-b,2}(q) for(1-b,2)-colored partitions - partitions in which we can label parts any of an indeterminate 1-b colors, but are restricted to using only 2 of the colors per part size. This formula is known to match the Han/Nekrasov-Okounkov hooklength formula truncated at hooks of size two up to the linear term in b.

It is of interest to enumerate and determine specific characteristics of partitions of n, considering each partition individually.

Links

Alois P. Heinz, Table of n, a(n) for n = 0..4000

Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, arXiv:0805.1398 [math.CO], 2008.

Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, Annales de l'institut Fourier, Tome 60 (2010) no. 1, pp. 1-29.

W. J. Keith, Restricted k-color partitions, Ramanujan Journal (2016) 40: 71.

Formula

a(n) = Sum_{p in P} (Sum_{k_j = 1} 1)^2, where P is the set of partitions of n, and k_j are the frequencies in p.

Example

For a(6), we sum over partitions of six. For each partition, we count 1 for each part which appears once, then square the total in each partition.
   6............1^2 = 1
   5,1..........2^2 = 4
   4,2..........2^2 = 4
   4,1,1........1^2 = 1
   3,3..........0^2 = 0
   3,2,1........3^2 = 9
   3,1,1,1......1^2 = 1
   2,2,2........0^2 = 0
   2,2,1,1......0^2 = 0
   2,1,1,1,1....1^2 = 1
   1,1,1,1,1,1..0^2 = 0
   --------------------
   Total.............21

Maple

b:= proc(n, i, p) option remember; `if`(n=0 or i=1, (
      `if`(n=1, 1, 0)+p)^2, add(b(n-i*j, i-1,
      `if`(j=1, 1, 0)+p), j=0..n/i))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..60);  # Alois P. Heinz, Apr 05 2018

Mathematica

Array[Total@ Map[Count[Split@ #, _?(Length@ # == 1 &)]^2 &, IntegerPartitions[#]] &, 43] (* Michael De Vlieger, Apr 05 2018 *)
b[n_, i_, p_] := b[n, i, p] = If[n == 0 || i == 1, (
     If[n == 1, 1, 0] + p)^2, Sum[b[n - i*j, i - 1,
     If[j == 1, 1, 0] + p], {j, 0, n/i}]];
a[n_] := b[n, n, 0];
a /@ Range[0, 60] (* Jean-François Alcover, Jun 06 2021, after Alois P. Heinz *)

Prog

(Python)
def frequencies(partition, n):
    tot = 0
    freq_list = []
    i = 0
    for p in partition:
        freq = [0 for i in range(n+1)]
        for i in p:
            freq[i] += 1
        for f in freq:
            if f == 0:
                tot += 1
        freq_list.append(freq)
    return freq_list
def sum_square_freqs_of_one(freq_part):
    tot = 0
    for f in freq_part:
        count = 0
        for i in f:
            if i == 1:
                count += 1
        tot += count*count
    return tot
import sympy.combinatorics
def A302300(n): # rewritten by _R. J. _Mathar_, 2023-03-24
    a =0
    if n ==0 :
        return 0
    part = sympy.combinatorics.IntegerPartition([n])
    partlist = []
    while True:
        part = part.next_lex()
        partlist.append(part.partition)
        if len(part.partition) <=1 :
            break
    freq_part = frequencies(partlist, n)
    return sum_square_freqs_of_one(freq_part)
for n in range(20): print(A302300(n))

Crossrefs

Cf. A024786, A197126.

Sequence in context: A276407 A022310 A126593 * A173074 A099473 A051572

Adjacent sequences: A302297 A302298 A302299 * A302301 A302302 A302303

Keyword

nonn

Author

Emily Anible, Apr 04 2018

Status

approved