A197126 -id:A197126 - cp's OEIS frontend

A222710 Total number of parts of multiplicity 10 in all partitions of n.

1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 13, 15, 22, 26, 37, 45, 61, 74, 99, 120, 158, 192, 247, 300, 382, 463, 582, 705, 877, 1059, 1309, 1575, 1931, 2319, 2823, 3381, 4094, 4886, 5886, 7007, 8401, 9971, 11905, 14090, 16756, 19781, 23436, 27594, 32586, 38268, 45050

Offset: 10

Alois P. Heinz, Feb 28 2013

nonn

Alois P. Heinz, Table of n, a(n) for n = 10..1000

Column k=10 of A197126.

b:= proc(n, p) option remember; `if`(n=0, [1, 0], `if`(p<1, [0, 0],
      add((l->`if`(m=10, l+[0, l[1]], l))(b(n-p*m, p-1)), m=0..n/p)))
    end:
a:= n-> b(n, n)[2]:
seq(a(n), n=10..60);

b[n_, p_] := b[n, p] = If[n == 0, {1, 0}, If[p < 1, {0, 0}, Sum[Function[l, If[m == 10, l + {0, l[[1]]}, l]][b[n - p*m, p - 1]], {m, 0, n/p}]]];
a[n_] := b[n, n][[2]];
Table[a[n], {n, 10, 60}] (* Jean-François Alcover, Apr 30 2018, after Alois P. Heinz *)

G.f.: (x^10/(1-x^10)-x^11/(1-x^11))/Product_{j>0}(1-x^j).

a(n) ~ exp(Pi*sqrt(2*n/3)) / (220*Pi*sqrt(2*n)). - Vaclav Kotesovec, May 24 2018

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.

Original entry on oeis.org

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

Emily Anible, Apr 04 2018

nonn

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.

			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

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.

Cf. A024786, A197126.

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

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 *)

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))

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.

cp's OEIS Frontend

A222710 Total number of parts of multiplicity 10 in all partitions of n.

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