

A003149


a(n) = Sum_{k=0..n} k!(nk)!.
(Formerly M1496)


18



1, 2, 5, 16, 64, 312, 1812, 12288, 95616, 840960, 8254080, 89441280, 1060369920, 13649610240, 189550368000, 2824077312000, 44927447040000, 760034451456000, 13622700994560000, 257872110354432000, 5140559166898176000
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OFFSET

0,2


COMMENTS

The sequence (origin 1) is the resistance between opposite corners of an ndimensional hypercube of unit resistors, multiplied by n!.
The resistances for n = 1,2,3,... are 1 1 5/6 2/3 8/15 13/30 151/420 32/105 83/315 73/315 1433/6930 ...
Number of {12,21*,2*1}avoiding signed permutations in the hyperoctahedral group.
a(n) is the sum of the reciprocals of the binomial coefficients C(n,k), multiplied by n!; example: a(4) = 4!*(1/1 + 1/4 + 1/6 + 1/4 + 1/1) = 64.  Philippe Deléham, May 12 2005
a(n) = number of permutations on [n+1] that avoid the pattern 132. The absence of a dash between 1 and 3 means the "1" and "3" must be consecutive in the permutation; the vertical bar means the "2" must occur at the end of the permutation. For example, 24153 fails to avoid this pattern: 243 is an offending subpermutation.  David Callan, Nov 02 2005
n!/A003149(n) is the probability that a random walk on an (n+1)dimensional hypercube will visit the diagonally opposite vertex before it returns to its starting point. 2^n*A003149(n)/n! is the expected length of a random walk from one vertex of an (n+1)dimensional hypercube to the diagonally opposite vertex (a walk which may include one or more passes through the starting point). These "random walk" examples are solutions to IBM's "Ponder This" puzzle for April, 2006.  Graeme McRae, Apr 02 2006
a(n) = number of strong fixed points in all permutations of {1,2,...,n+1} (a permutation p of {1,2,...,n} is said to have j as a strong fixed point (splitter) if p(k)<j for k<j and p(k)>j for k>j). Example: a(2)=5 because the permutations of {1,2,3}, with marked strong fixed points, are: 1'2'3', 1'32, 312, 213', 231 and 321.  Emeric Deutsch, Oct 28 2008
It appears that a(n)=2^(n)*(A103213(n)+n!H(n)) with H(n) harmonic number of order n.  Groux Roland, Dec 18 2010


REFERENCES

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Stanley, R. P., Enumerative Combinatorics, Volume 1 (1986), p. 49. [From Emeric Deutsch, Oct 28 2008]


LINKS

T. D. Noe, Table of n, a(n) for n=0..100
Joerg Arndt, Generating Random Permutations, PhD thesis, Australian National University, Camberra, Australia, (2010).
Fred Curtis, Resistancenetwork Problems.
IBM, "Ponder This" puzzle for April, 2006
Todd Feil, Gary Kennedy and David Callan, Problem E3467, Amer. Math. Monthly, 100 (1993), 800801. [From Emeric Deutsch, Oct 28 2008]
T. Mansour and J. West, Avoiding 2letter signed patterns, arXiv:math/0207204 [math.CO], 2002.
F. Nedemeyer and Y. Smorodinsky, Resistance in the multidimensional cube, Quantum 7:1 (1996) 1215 and 63.
R. Sprugnoli, Moments of Reciprocals of Binomial Coefficients, Journal of Integer Sequences, 14 (2011), #11.7.8.
V. Strehl, The average number of splitters in a random permutation [Unpublished; included here with the author's permission.]
B. Sury, Sum of the reciprocals of the binomial coefficients, Europ. J. Comb., 14 (1993), 351353.
Eric Weisstein's World of Mathematics, Incomplete Beta Function.
Eric Weisstein's World of Mathematics, Lerch Transcendent.


FORMULA

a(n) = n!+((n+1)/2)a(n1), n >= 1.  Leroy Quet, Sep 06 2002
a(n) = ((3n+1)/2)a(n1)(n^2/2)a(n2), n >= 2.  David W. Wilson, Sep 06, 2002
G.f.: (Sum_{k>=0} k!*x^k)^2.  Vladeta Jovovic, Aug 30 2002
E.g.f: log(1x)/(x/21) if offset 1.
Convolution of A000142 [factorial numbers] with itself.  Ross La Haye, Oct 29 2004
a(n) = Sum(k*A145878(n+1,k),k=0..n+1).  Emeric Deutsch, Oct 28 2008
a(n) = A084938(n+2,2).  Philippe Deléham, Dec 17 2008
a(n) = 2*int(Ei(t)*exp(2*t)*t^(n+1),t=0..infinity) where Ei is the exponential integral function.
O.g.f.: 1/(1I(x))^2 where I(x) is o.g.f. for A003319.  Geoffrey Critzer, Apr 27 2012
a(n) ~ 2*n!.  Vaclav Kotesovec, Oct 04 2012
a(n) = (n+1)!/2^n * Sum_{k=0..n} 2^k/(k+1).  Vaclav Kotesovec, Oct 27 2012
E.g.f.: 2/((x1)*(x2)) + 2*x/(x2)^2*G(0) where G(k) = 1 + x*(2*k+1)/(2*(k+1)  4*x*(k+1)^2/(2*x*(k+1) + (2*k+3)/G(k+1) )); (recursively defined continued fraction).  Sergei N. Gladkovskii, Dec 14 2012
a(n) = 2 * n! * (1 + Sum_{k>=1} A005649(k1)/n^k).  Vaclav Kotesovec, Aug 01 2015
From Vladimir Reshetnikov, Nov 12 2015: (Start)
a(n) = (n+1)!*Re(Beta(2; n+2, 0))/2^(n+1), where Beta(z; a, b) is the incomplete Beta function.
a(n) = 2*(n+1)!*Re(LerchPhi(2, 1, n+2)), where LerchPhi(z, s, a) is the Lerch transcendent.
(End)


MATHEMATICA

Table[Sum[k!(nk)!, {k, 0, n}], {n, 0, 20}] (* Harvey P. Dale, Mar 28 2012 *)
Table[(n+1)!/2^n*Sum[2^k/(k+1), {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 27 2012 *)
Round@Table[2 (n+1)! Re[LerchPhi[2, 1, n+2]], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 12 2015 *)


PROG

(PARI) a(n)=sum(k=0, n, k!*(nk)!)
(PARI) a(n)=if(n<0, 0, (n+1)!*polcoeff(log(1x+x^2*O(x^n))/(x/21), n+1))


CROSSREFS

Cf. A046825, A046878, A046879.
Cf. A052186, A006932, A145878.  Emeric Deutsch, Oct 28 2008
Sequence in context: A185998 A127083 A131178 * A027046 A000522 A182290
Adjacent sequences: A003146 A003147 A003148 * A003150 A003151 A003152


KEYWORD

nonn,easy,nice,changed


AUTHOR

N. J. A. Sloane, Henry Gould


EXTENSIONS

More terms from Michel ten Voorde (seqfan(AT)tenvoorde.org) Apr 11 2001
Additional comments from Michael Somos, Feb 14 2002
Second formula corrected by N. Sato, Jan 27 2010


STATUS

approved



