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%I #32 Mar 24 2023 14:21:30
%S 0,1,1,5,6,12,21,33,50,79,116,169,246,346,487,675,927,1254,1702,2263,
%T 3014,3966,5210,6766,8795,11303,14531,18521,23583,29803,37654,47231,
%U 59206,73792,91867,113778,140788,173377,213289,261318,319764,389846,474745,576164
%N 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.
%C 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.
%C It is of interest to enumerate and determine specific characteristics of partitions of n, considering each partition individually.
%H Alois P. Heinz, <a href="/A302300/b302300.txt">Table of n, a(n) for n = 0..4000</a>
%H Guo-Niu Han, <a href="https://arxiv.org/abs/0805.1398">The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications</a>, arXiv:0805.1398 [math.CO], 2008.
%H Guo-Niu Han, <a href="https://doi.org/10.5802/aif.2515">The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications</a>, Annales de l'institut Fourier, Tome 60 (2010) no. 1, pp. 1-29.
%H W. J. Keith, <a href="https://doi.org/10.1007/s11139-015-9704-x">Restricted k-color partitions</a>, Ramanujan Journal (2016) 40: 71.
%F 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.
%e 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.
%e 6............1^2 = 1
%e 5,1..........2^2 = 4
%e 4,2..........2^2 = 4
%e 4,1,1........1^2 = 1
%e 3,3..........0^2 = 0
%e 3,2,1........3^2 = 9
%e 3,1,1,1......1^2 = 1
%e 2,2,2........0^2 = 0
%e 2,2,1,1......0^2 = 0
%e 2,1,1,1,1....1^2 = 1
%e 1,1,1,1,1,1..0^2 = 0
%e --------------------
%e Total.............21
%p b:= proc(n, i, p) option remember; `if`(n=0 or i=1, (
%p `if`(n=1, 1, 0)+p)^2, add(b(n-i*j, i-1,
%p `if`(j=1, 1, 0)+p), j=0..n/i))
%p end:
%p a:= n-> b(n$2, 0):
%p seq(a(n), n=0..60); # _Alois P. Heinz_, Apr 05 2018
%t Array[Total@ Map[Count[Split@ #, _?(Length@ # == 1 &)]^2 &, IntegerPartitions[#]] &, 43] (* _Michael De Vlieger_, Apr 05 2018 *)
%t b[n_, i_, p_] := b[n, i, p] = If[n == 0 || i == 1, (
%t If[n == 1, 1, 0] + p)^2, Sum[b[n - i*j, i - 1,
%t If[j == 1, 1, 0] + p], {j, 0, n/i}]];
%t a[n_] := b[n, n, 0];
%t a /@ Range[0, 60] (* _Jean-François Alcover_, Jun 06 2021, after _Alois P. Heinz_ *)
%o (Python)
%o def frequencies(partition, n):
%o tot = 0
%o freq_list = []
%o i = 0
%o for p in partition:
%o freq = [0 for i in range(n+1)]
%o for i in p:
%o freq[i] += 1
%o for f in freq:
%o if f == 0:
%o tot += 1
%o freq_list.append(freq)
%o return freq_list
%o def sum_square_freqs_of_one(freq_part):
%o tot = 0
%o for f in freq_part:
%o count = 0
%o for i in f:
%o if i == 1:
%o count += 1
%o tot += count*count
%o return tot
%o import sympy.combinatorics
%o def A302300(n): # rewritten by _R. J. _Mathar_, 2023-03-24
%o a =0
%o if n ==0 :
%o return 0
%o part = sympy.combinatorics.IntegerPartition([n])
%o partlist = []
%o while True:
%o part = part.next_lex()
%o partlist.append(part.partition)
%o if len(part.partition) <=1 :
%o break
%o freq_part = frequencies(partlist, n)
%o return sum_square_freqs_of_one(freq_part)
%o for n in range(20): print(A302300(n))
%Y Cf. A024786, A197126.
%K nonn
%O 0,4
%A _Emily Anible_, Apr 04 2018
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