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A302300
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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.
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2
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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
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OFFSET
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0,4
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COMMENTS
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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.
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LINKS
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FORMULA
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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.
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EXAMPLE
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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
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MAPLE
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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):
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MATHEMATICA
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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];
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PROG
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(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))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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