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A122768
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Number of combinations which can be taken from the integer partitions of n. Total number of cases in the (n,m)-fragmentation process.
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56
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0, 1, 3, 7, 15, 29, 54, 95, 163, 270, 439, 696, 1088, 1669, 2530, 3780, 5591, 8173, 11845, 17000, 24215, 34210, 48008, 66895, 92660, 127554, 174651, 237830, 322297, 434625, 583524, 779972, 1038356, 1376787, 1818755, 2393775, 3139812, 4104433, 5348375, 6947545, 8998201, 11620313, 14965126, 19220569
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OFFSET
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0,3
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COMMENTS
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Consider a fragmentation process of an n-object which consists of n unlabeled elements (= 1-parts). By definition the n-object can scatter into up to n m-parts where an m-part consists of 1 up to n elements. A 4-object can split up for example into 4 1-parts which corresponds to the integer partition [1,1,1,1], or it can, for example, rest unfragmented which corresponds to [4]. Since the number of integer partitions of n=4 equals 5, there are 5 n=4-fragmentation processes.
Now we ask for the probability of getting an m-part after an n-fragmentation. Think of a Greek statue which had been broken into n parts and covered by earth. We could find several m-parts, in the most lucky case we would find all m-parts which add up to m_1+m_2+...+m_n=n. Then the statue could be restored.
For example for n=4 we could ask for the probability prob(n=4,m=2) of just a single 2-part. We have 2 cases for a 2-part and we have 15 cases in total, thus prob(n=4,m=2)=2/15 (the 2 cases come from [1,1,2] and [2,2]). The chances to find the two 2-parts from the [2,2]-fragmentation are 1/15 only. The chances to find the n=4-object unsplitted are also 1/15 only.
This sequence is generated over the unordered partitions; for example, when n = 4 there are 1+3+2+5+4 = 15 cases. If we allow a null case for each of the five partitions then we have 15+5 = 20 which is A000712(4). - Alford Arnold, Dec 12 2006
Number of partitions into two kinds of parts with the first kind of parts used in each partition. - Joerg Arndt, Jun 21 2011
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LINKS
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FORMULA
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G.f.: 1/P(x)^2 - 1/P(x) where P(x)=prod(k>=1, 1-x^k ). - Joerg Arndt, Jun 21 2011
With sum_i^P(n) = the sum over all P(n) integer partitions of n, sum_j^p(i) = the sum over all p(i) parts of the i-th integer partition, prttn(i) = the i-th partition whereat prttn(i) is a list, choose(L,k) = construct the list LC of combinations of a list L (see Maple), |LC| = number of elements of list LC (=Maple's nops command) we have a(n) = sum_i^P(n) sum_j^p(i) |choose(prttn,j)|
a(n) ~ exp(2*Pi*sqrt(n/3)) / (4*3^(3/4)*n^(5/4)) * (1 - (Pi/12 + 45/(16*Pi))/sqrt(3*n)). - Vaclav Kotesovec, Mar 31 2017
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EXAMPLE
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a(n=4) = 15 because the possible combinations of all five integer partitions of n=4 are: [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1], [2], [1, 1], [1, 2], [1, 1, 2], [2], [2, 2], [1], [3], [1, 3], [4].
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MAPLE
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A122768 := proc(n::integer) local i, j, prttnlst, prttn, ZahlTeile, H; prttnlst:=partition(n); H := NULL; for i from 1 to nops(prttnlst) do prttn := prttnlst[i]; ZahlTeile := nops(prttn); for j from 1 to ZahlTeile do H := H, op(choose(prttn, j)); od; od; print(n, H, nops([H])); end proc;
A000712 := proc(n) option remember ; add(combinat[numbpart](k)*combinat[numbpart](n-k), k=0..n) ; end: A000041 := proc(n) combinat[numbpart](n) ; end: A122768 := proc(n::integer) RETURN( A000712(n)-A000041(n)) ; end: for n from 0 to 80 do printf("%d, ", A122768(n)) ; od: # R. J. Mathar, Aug 25 2008
# third Maple program:
b:= proc(n, k) option remember; `if`(n=0, 1, add(
k*numtheory[sigma](j)*b(n-j, k), j=1..n)/n)
end:
a:= n-> b(n, 2)-b(n, 1):
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MATHEMATICA
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PROG
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(PARI) x='x+O('x^66); /* that many terms */
Vec(1/eta(x)^2-1/eta(x)) /* show terms (omitting initial zero) */
(Haskell)
a122768 n = a122768_list !! n
a122768_list = 0 : f (tail a000041_list) [1] where
f (p:ps) rs = (sum $ zipWith (*) rs $ tail a000041_list) : f ps (p : rs)
(Python)
from sympy import npartitions
def A122768(n): return (sum(npartitions(k)*npartitions(n-k) for k in range(1, n+1>>1))<<1) + (0 if n&1 else npartitions(n>>1)**2) + npartitions(n) if n else 0 # Chai Wah Wu, Sep 25 2023
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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