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A001705 Generalized Stirling numbers: a(n) = n! * Sum_{k=0..n-1} (k+1)/(n-k).
(Formerly M3944 N1625)
42
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

Partial sum of first n harmonic numbers multiplied by n!: a(n) = n!*Sum[Sum[1/k,{k,1,m}],{m,1,n}] = n!*Sum[H(m),{m,1,n}], whrere H(m) = Sum[1/k,{k,1,m}] = A001008(m)/A002805(m) is m-th Harmonic number.

Contribution from Emeric Deutsch, Sep 22 2008: (Start)

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.

a(n) = sum(k*A143947(n,k), k = n..n*(n+1)/2). (End)

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

G. C. Greubel, Table of n, a(n) for n = 0..445 (terms 0 through 100 were provided by T. D. Noe)

J.-L. Baril, S. Kirgizov, The pure descent statistic on permutations, Preprint, 2016.

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 406

D. S. Mitrinovic, M. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling, Univ. Beograd. Pubi. Elektrotehn. Fak. Ser. Mat. Fiz. 77 (1962).

Robert E. Moritz, On the sum of products of n consecutive integers, Univ. Washington Publications in Math., 1 (No. 3, 1926), 44-49 [Annotated scanned copy]

FORMULA

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) = A112486(n, 1).

a(n) = a(n-1)*(n+1)+n! = A000254(n+1)-A000142(n+1) = A067176(n+1, 1). [Henry Bottomley, Jan 09 2002]

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!*Sum_{k=1..n}(((-1)^(k+1)*binomial(n+1,k+1))/k). - Vladimir Kruchinin, Oct 10 2016

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

From Olivier Gérard, Dec 31 2012: (Start)

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

a := n-> add((n+1)!/k, k=2..n+1): seq(a(n), n=0..21); # Zerinvary Lajos, Jan 22 2008; edited Johannes W. Meijer, Nov 28 2012

a := n -> ((n+1)!*(h(n+1)-1)): h := n-> harmonic(n): seq(a(n), n=0..21); # Gary Detlefs, Dec 18 2009; corrected by Johannes W. Meijer, Nov 28 2012

MATHEMATICA

Table[n!*Sum[Sum[1/k, {k, 1, m}], {m, 1, n}], {n, 0, 20}] (* Alexander Adamchuk, Apr 14 2006 *)

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

CROSSREFS

Cf. A000254 (total number of cycles in permutations, including fixed points), A006675.

Cf. A001008, A002805.

Cf. A143947.

Related to n!*the k-th successive summation of the harmonic numbers:

  (k=0) A000254, (k=1) A001705, (k=2) A001711, (k=3) A001716,

  (k=4) A001721, (k=5) A051524, (k=6) A051545, (k=7) A051560,

  (k=8) A051562, (k=9) A051564.

Cf. A002104 (number of different cycles in permutations, without fixed points).

Cf. A006231 (number of different cycles in permutations, including fixed points).

Sequence in context: A263134 A082029 A081047 * A185108 A209672 A057793

Adjacent sequences:  A001702 A001703 A001704 * A001706 A001707 A001708

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane

EXTENSIONS

More terms from Sascha Kurz, Mar 22 2002

2 cross-refs added by Olivier Gérard, Dec 31 2012

STATUS

approved

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Last modified November 21 22:32 EST 2017. Contains 295054 sequences.