cp's OEIS Frontend

A122768 Number of combinations which can be taken from the integer partitions of n. Total number of cases in the (n,m)-fragmentation process.

%I A122768 #38 Sep 25 2023 14:28:14
%S A122768 0,1,3,7,15,29,54,95,163,270,439,696,1088,1669,2530,3780,5591,8173,
%T A122768 11845,17000,24215,34210,48008,66895,92660,127554,174651,237830,
%U A122768 322297,434625,583524,779972,1038356,1376787,1818755,2393775,3139812,4104433,5348375,6947545,8998201,11620313,14965126,19220569
%N A122768 Number of combinations which can be taken from the integer partitions of n. Total number of cases in the (n,m)-fragmentation process.
%C A122768 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.
%C A122768 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.
%C A122768 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.
%C A122768 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
%C A122768 Number of partitions into two kinds of parts with the first kind of parts used in each partition. - _Joerg Arndt_, Jun 21 2011
%H A122768 Alois P. Heinz, <a href="/A122768/b122768.txt">Table of n, a(n) for n = 0..10000</a>
%F A122768 G.f.: 1/P(x)^2 - 1/P(x) where P(x)=prod(k>=1, 1-x^k ). - _Joerg Arndt_, Jun 21 2011
%F A122768 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)|
%F A122768 a(n) = A000712(n) - A000041(n). - _Alford Arnold_, Dec 12 2006
%F A122768 a(n) = A144064(n,2)-A144064(n,1). - _Alois P. Heinz_, Mar 31 2017
%F A122768 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
%e A122768 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].
%p A122768 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;
%p A122768 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
%p A122768 # third Maple program:
%p A122768 b:= proc(n, k) option remember; `if`(n=0, 1, add(
%p A122768       k*numtheory[sigma](j)*b(n-j, k), j=1..n)/n)
%p A122768     end:
%p A122768 a:= n-> b(n,2)-b(n,1):
%p A122768 seq(a(n), n=0..50);  # _Alois P. Heinz_, Mar 31 2017
%t A122768 1/QPochhammer[x]^2 - 1/QPochhammer[x] + O[x]^50 // CoefficientList[#, x]& (* _Jean-François Alcover_, Feb 05 2017, after _Joerg Arndt_ *)
%o A122768 (PARI)  x='x+O('x^66); /* that many terms */
%o A122768 Vec(1/eta(x)^2-1/eta(x)) /* show terms (omitting initial zero) */
%o A122768 /* _Joerg Arndt_, Jun 21 2011 */
%o A122768 (Haskell)
%o A122768 a122768 n = a122768_list !! n
%o A122768 a122768_list = 0 : f (tail a000041_list) [1] where
%o A122768    f (p:ps) rs = (sum $ zipWith (*) rs $ tail a000041_list) : f ps (p : rs)
%o A122768 -- _Reinhard Zumkeller_, Nov 09 2015
%o A122768 (Python)
%o A122768 from sympy import npartitions
%o A122768 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
%Y A122768 Cf. A000041, A000079, A000262.
%Y A122768 Cf. A000712, A074139, A074141.
%Y A122768 Cf. A048574, A144064.
%K A122768 nonn
%O A122768 0,3
%A A122768 _Thomas Wieder_, Sep 11 2006
%E A122768 Extended by _R. J. Mathar_, Aug 25 2008