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A006128
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Total number of parts in all partitions of n. Also, sum of largest parts of all partitions of n.
(Formerly M2552)
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237
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0, 1, 3, 6, 12, 20, 35, 54, 86, 128, 192, 275, 399, 556, 780, 1068, 1463, 1965, 2644, 3498, 4630, 6052, 7899, 10206, 13174, 16851, 21522, 27294, 34545, 43453, 54563, 68135, 84927, 105366, 130462, 160876, 198014, 242812, 297201, 362587, 441546, 536104, 649791, 785437, 947812, 1140945, 1371173, 1644136, 1968379, 2351597, 2805218, 3339869, 3970648, 4712040, 5584141, 6606438, 7805507, 9207637
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OFFSET
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0,3
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COMMENTS
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a(n) = degree of Kac determinant at level n as polynomial in the conformal weight (called h). (Cf. C. Itzykson and J.-M. Drouffe, Statistical Field Theory, Vol. 2, p. 533, eq.(98); reference p. 643, Cambridge University Press, (1989).) - Wolfdieter Lang
Also the number of one-element transitions from the integer partitions of n to the partitions of n-1 for labeled parts with the assumption that from any part z > 1 one can take an element of amount 1 in one way only. That means z is composed of z unlabeled parts of amount 1, i.e. z = 1 + 1 + ... + 1. E.g., for n=3 to n=2 we have a(3) = 6 and [111] --> [11], [111] --> [11], [111] --> [11], [12] --> [11], [12] --> [2], [3] --> [2]. For the case of z composed by labeled elements, z = 1_1 + 1_2 + ... + 1_z, see A066186. - Thomas Wieder, May 20 2004
Number of times a derivative of any order (not 0 of course) appears when expanding the n-th derivative of 1/f(x). For instance (1/f(x))'' = (2 f'(x)^2-f(x) f''(x)) / f(x)^3 which makes a(2) = 3 (by counting k times the k-th power of a derivative). - Thomas Baruchel, Nov 07 2005
Starting with offset 1 equals A000041: (1, 1, 2, 3, 5, 7, 11, ...) convolved with A000005: (1, 2, 2, 3, 2, 4, ...). - Gary W. Adamson, Jun 16 2009
More generally, the total number of parts >= k in all partitions of n equals the sum of k-th largest parts of all partitions of n. In this case k = 1. Apart from initial 0 the first column of A181187. - Omar E. Pol, Feb 14 2012
a(n) is also the total number of divisors of all positive integers in a sequence with n blocks where the m-th block consists of A000041(n-m) copies of m, with 1 <= m <= n. The mentioned divisors are also all parts of all partitions of n.
Apart from initial zero this is also as follows:
Number of ways to choose a part index of an integer partition of n, i.e., partitions of n with a selected position. Selecting a part value instead of index gives A000070. - Gus Wiseman, Apr 19 2021
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REFERENCES
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S. M. Luthra, On the average number of summands in partitions of n, Proc. Nat. Inst. Sci. India Part. A, 23 (1957), p. 483-498.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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G.f.: Sum_{n>=1} n*x^n / Product_{k=1..n} (1-x^k).
G.f.: Sum_{k>=1} x^k/(1-x^k) / Product_{m>=1} (1-x^m).
a(n) = Sum_{k=1..n} k*A008284(n, k).
a(n) = Sum_{m=1..n} of the number of divisors of m * number of partitions of n-m.
Note that the formula for the above comment is a(n) = Sum_{m=1..n} d(m)*p(n-m) = Sum_{m=1..n} A000005(m)*A000041(n-m), if n >= 1. - Omar E. Pol, Jan 21 2013
Erdős and Lehner show that if u(n) denotes the average largest part in a partition of n, then u(n) ~ constant*sqrt(n)*log n.
a(n) = O(sqrt(n)*log(n)*p(n)), where p(n) is the partition function A000041(n). - Peter Bala, Dec 23 2013
Asymptotics (Luthra, 1957): a(n) = p(n) * (C*N^(1/2) + C^2/2) * (log(C*N^(1/2)) + gamma) + (1+C^2)/4 + O(N^(-1/2)*log(N)), where N = n - 1/24, C = sqrt(6)/Pi, gamma is the Euler-Mascheroni constant A001620 and p(n) is the partition function A000041(n).
The formula a(n) = p(n) * (sqrt(3*n/(2*Pi)) * (log(n) + 2*gamma - log(Pi/6)) + O(log(n)^3)) in the abstract of the article by Kessler and Livingston (cited also in the book by Sandor, p. 495) is incorrect!
Right is: a(n) = p(n) * (sqrt(3*n/2)/Pi * (log(n) + 2*gamma - log(Pi^2/6)) + O(log(n)^3))
or a(n) ~ exp(Pi*sqrt(2*n/3)) * (log(6*n/Pi^2) + 2*gamma) / (4*Pi*sqrt(2*n)).
(End)
G.f.: (log(1-x) + psi_x(1))/(log(x) * (x)_inf), where psi_q(z) is the q-digamma function, and (q)_inf is the q-Pochhammer symbol (the Euler function). - Vladimir Reshetnikov, Nov 17 2016
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EXAMPLE
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For n = 4 the partitions of 4 are [4], [2, 2], [3, 1], [2, 1, 1], [1, 1, 1, 1]. The total number of parts is 12. On the other hand, the sum of the largest parts of all partitions is 4 + 2 + 3 + 2 + 1 = 12, equaling the total number of parts, so a(4) = 12. - Omar E. Pol, Oct 12 2018
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MAPLE
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g:= add(n*x^n*mul(1/(1-x^k), k=1..n), n=1..61):
a:= n-> coeff(series(g, x, 62), x, n):
seq(a(n), n=0..61);
# second Maple program:
a:= n-> add(combinat[numbpart](n-j)*numtheory[tau](j), j=1..n):
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MATHEMATICA
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a[n_] := Sum[DivisorSigma[0, m] PartitionsP[n - m], {m, 1, n}]; Table[ a[n], {n, 0, 41}]
CoefficientList[ Series[ Sum[n*x^n*Product[1/(1 - x^k), {k, n}], {n, 100}], {x, 0, 100}], x]
a[n_] := Plus @@ Max /@ IntegerPartitions@ n; Array[a, 45] (* Robert G. Wilson v, Apr 12 2011 *)
Join[{0}, ((Log[1 - x] + QPolyGamma[1, x])/(Log[x] QPochhammer[x]) + O[x]^60)[[3]]] (* Vladimir Reshetnikov, Nov 17 2016 *)
Length /@ Table[IntegerPartitions[n] // Flatten, {n, 50}] (* Shouvik Datta, Sep 12 2021 *)
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PROG
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(PARI) f(n)= {local(v, i, k, s, t); v=vector(n, k, 0); v[n]=2; t=0; while(v[1]<n, i=2; while(v[i]==0, i++); v[i]--; s=sum(k=i, n, k*v[k]); while(i>1, i--; s+=i*(v[i]=(n-s)\i)); t+=sum(k=1, n, v[k])); t } /* Thomas Baruchel, Nov 07 2005 */
(PARI) a(n) = sum(m=1, n, numdiv(m)*numbpart(n-m)) \\ Michel Marcus, Jul 13 2013
(Haskell)
a006128 = length . concat . ps 1 where
ps _ 0 = [[]]
ps i j = [t:ts | t <- [i..j], ts <- ps t (j - t)]
(Python)
from sympy import divisor_count, npartitions
def a(n): return sum([divisor_count(m)*npartitions(n - m) for m in range(1, n + 1)]) # Indranil Ghosh, Apr 25 2017
(GAP) List([0..60], n->Length(Flat(Partitions(n)))); # Muniru A Asiru, Oct 12 2018
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CROSSREFS
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The version for normal multisets is A001787.
The version for factorizations is A066637.
A000070 counts partitions with a selected part.
A336875 counts compositions with a selected part.
A339564 counts factorizations with a selected factor.
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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