|
| |
|
|
A001705
|
|
Generalized Stirling numbers: a(n) = n! * Sum_{k=0..n-1} (k+1)/(n-k).
(Formerly M3944 N1625)
|
|
49
|
|
|
|
0, 1, 5, 26, 154, 1044, 8028, 69264, 663696, 6999840, 80627040, 1007441280, 13575738240, 196287356160, 3031488633600, 49811492505600, 867718162483200, 15974614352793600, 309920046408806400, 6320046028584960000, 135153868608460800000, 3024476051557847040000
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
a(n) is also the sum of the positions of the right-to-left minima in all permutations of [n]. Example: a(3)=26 because the positions of the right-to-left minima in the permutations 123,132,213,231,312 and 321 are 123, 13, 23, 3, 23 and 3, respectively and 1 + 2 + 3 + 1 + 3 + 2 + 3 + 3 + 2 + 3 + 3 = 26. - Emeric Deutsch, Sep 22 2008
The asymptotic expansion of the higher order exponential integral E(x,m=2,n=2) ~ exp(-x)/x^2*(1 - 5/x + 26/x^2 - 154/x^3 + 1044/x^4 - 8028/x^5 + 69264/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009
a(n) is the total number of cycles (excluding fixed points) in all permutations of [n+1]. - Olivier Gérard, Oct 23 2012; Dec 31 2012
A length n sequence is formed by randomly selecting (one-by-one) n real numbers in (0,1). a(n)/(n+1)! is the expected value of the sum of the new maximums in such a sequence. For example for n=3: If we select (in this order): 0.591996, 0.646474, 0.163659 we would add 0.591996 + 0.646474 which would be a bit above the average of a(3)/4! = 26/24. - Geoffrey Critzer, Oct 17 2013
|
|
|
REFERENCES
|
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
|
|
|
LINKS
|
|
|
|
FORMULA
|
Partial sum of first n harmonic numbers multiplied by n!.
a(n) = n!*Sum_{m=1..n} Sum_{k=1..m} 1/k = n!*Sum_{m=1..n} H(m), where H(m) = Sum_{k=1..m} 1/k = A001008(m)/A002805(m) is m-th Harmonic number.
E.g.f.: - log (1 - x) / (1 - x)^2.
a(n) = (n+1)! * H(n) - n*n!, H(n) = Sum_{k=1..n} (1/k).
a(n) = Sum_{k=0..n-1} ((-1)^(n-1+k) * (k+1) * 2^k * Stirling1(n, k+1)). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
With alternating signs: Ramanujan polynomials psi_2(n, x) evaluated at 0. - Ralf Stephan, Apr 16 2004
a(n) = Sum_{k=1..n} (k*StirlingCycle(n+1,k+1)). - David Callan, Sep 25 2006
For n >= 1, a(n) = Sum_{j=0..n-1} ((-1)^(n-j-1) * 2^j * (j+1) * Stirling1(n,j+1)). - Milan Janjic, Dec 14 2008
a(n) = (2*n+1)*a(n-1) - n^2*a(n-2). - Gary Detlefs, Nov 27 2009
a(n) = (n+1)!*(H(n+1) - 1) where H(n) is the n-th harmonic number. - Gary Detlefs, Dec 18 2009
a(n) = (n+1)!*Sum_{k = 1..n} (-1)^(k+1)*binomial(n+1,k+1)*k/(k+1). - Peter Bala, Feb 15 2022
a(n) = Gamma(n + 2) * (Digamma(n + 2) + EulerGamma - 1). - Peter Luschny, Feb 19 2022
a(n) = -Sum_{k=0..n} k!*A066667(n, k+1).
a(n) = Sum_{k=0..n} k!*A132159(n, k+1). (End)
a(n) = n*(n + 1)!*hypergeom([1, 1, 1 - n], [2, 3], 1)/2. - Peter Luschny, Jun 22 2022
|
|
|
EXAMPLE
|
(1-x)^-2 * (-log(1-x)) = x + 5/2*x^2 + 13/3*x^3 + 77/12*x^4 + ...
Examples: a(6) = 6!*(1/6 + 2/5 + 3/4 + 4/3 + 5/2 + 6/1) = 8028; a(20) = 20!*(1/20 + 2/19 + 3/18 + 4/17 + 5/16 + ... + 16/5 + 17/4 + 18/3 + 19/2 + 20/1) = 135153868608460800000. - Alexander Adamchuk, Oct 09 2004
The cycle decomposition of all permutations of 4 elements gives the following list: {{{1},{2},{3},{4}}, {{1},{2},{3,4}}, {{1},{2,3},{4}}, {{1},{2,4,3}}, {{1},{2,3,4}}, {{1},{2,4},{3}}, {{1,2},{3},{4}}, {{1,2},{3,4}}, {{1,3,2},{4}},{{1,4,3,2}}, {{1,3,4,2}}, {{1,4,2},{3}}, {{1,2,3},{4}}, {{1,2,4,3}},{{1,3},{2},{4}}, {{1,4,3},{2}}, {{1,3},{2,4}}, {{1,4,2,3}}, {{1,2,3,4}}, {{1,2,4},{3}}, {{1,3,4},{2}}, {{1,4},{2},{3}}, {{1,3,2,4}}, {{1,4},{2,3}}}.
Deleting the fixed points gives the following 26 items: {{3,4}, {2,3}, {2,4,3}, {2,3,4}, {2,4}, {1,2}, {1,2}, {3,4}, {1,3,2}, {1,4,3,2}, {1,3,4,2}, {1,4,2}, {1,2,3}, {1,2,4,3}, {1,3}, {1,4,3}, {1,3}, {2,4}, {1,4,2,3}, {1,2,3,4}, {1,2,4}, {1,3,4}, {1,4}, {1,3,2,4}, {1,4}, {2,3}}. (End)
|
|
|
MAPLE
|
|
|
|
MATHEMATICA
|
Table[n!*Sum[Sum[1/k, {k, 1, m}], {m, 1, n}], {n, 0, 20}] (* Alexander Adamchuk, Apr 14 2006 *)
a[n_] := (n + 1)! (EulerGamma - 1 + PolyGamma[n + 2]);
|
|
|
PROG
|
(Maxima)
a(n):=n!*sum(((-1)^(k+1)*binomial(n+1, k+1))/k, k, 1, n); /* Vladimir Kruchinin, Oct 10 2016 */
(PARI) for(n=0, 25, print1(n!*sum(k=0, n-1, (k+1)/(n-k)), ", ")) \\ G. C. Greubel, Jan 20 2017
(Python)
from math import factorial
f = factorial(n)
return sum(f*(k+1)//(n-k) for k in range(n)) # Chai Wah Wu, Jun 23 2022
|
|
|
CROSSREFS
|
Cf. A000254 (total number of cycles in permutations, including fixed points).
Cf. A002104 (number of different cycles in permutations, without fixed points).
Cf. A006231 (number of different cycles in permutations, including fixed points).
Related to n!*the k-th successive summation of the harmonic numbers:
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
