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a(n+1) = n*(a(n) + 1) for n >= 1, a(1) = 1.
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%I #55 May 11 2024 21:54:59

%S 1,2,6,21,88,445,2676,18739,149920,1349289,13492900,148421911,

%T 1781062944,23153818285,324153456004,4862301840075,77796829441216,

%U 1322546100500689,23805829809012420,452310766371235999,9046215327424720000,189970521875919120021

%N a(n+1) = n*(a(n) + 1) for n >= 1, a(1) = 1.

%H Vincenzo Librandi, <a href="/A033540/b033540.txt">Table of n, a(n) for n = 1..200</a>

%H David Carlson, <a href="https://dl.acm.org/doi/10.5555/3144687.3144716">Sophomores meet the traveling salesperson problem</a>, Journal of Computing Sciences in Colleges, 33(3), 126-133, 2018. See proof of the formula by Benoit Cloitre.

%H N. J. A. Sloane and Brady Haran, <a href="https://www.youtube.com/watch?v=0zT16q3p24M">A Sequence with a Mistake</a>, Numberphile video (2021).

%F a(n) = n!*(1 +1/0! +1/1! +...+ 1/(n-1)!). - Jon Bentley (jlb(AT)research.bell-labs.com)

%F For n>=1, a(n+1) = floor((1+e)*n!) - 1. - _Benoit Cloitre_, Sep 07 2002

%F From _Vladeta Jovovic_, Feb 02 2003: (Start)

%F a(n) = n! + A007526(n).

%F E.g.f.: (1+x*exp(x))/(1-x). (End)

%F a(n) = (n+1)*a(n-1) - (2*n-3)*a(n-2) + (n-3)*a(n-3) for n>=4. - _Jaume Oliver Lafont_, Sep 11 2009

%F a(n) = n! + floor(e*n!) - 1, n>0. - _Gary Detlefs_, Jun 06 2010

%p seq(coeff(series( (1+x*exp(x))/(1-x), x, n+1)*n!, x, n), n = 0..30); # _G. C. Greubel_, Oct 13 2019

%p # second Maple program:

%p a:= proc(n) option remember;

%p `if`(n=1, 1, (n-1)*(a(n-1)+1))

%p end:

%p seq(a(n), n=1..23); # _Alois P. Heinz_, May 12 2021

%t FoldList[#1*#2 + #2 &, 1, Range[19]] (* _Robert G. Wilson v_, Jul 07 2012 *)

%t nxt[{a_,n_}]:={n(a+1),n+1}; Transpose[NestList[nxt,{1,1},20]][[1]] (* _Harvey P. Dale_, Jun 20 2014 *)

%o (Magma) I:=[1,2,6]; [n le 3 select I[n] else (n+1)*Self(n-1)-(2*n-3)*Self(n-2)+(n-3)*Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Jun 21 2014

%o (PARI) my(x='x+O('x^30)); Vec(serlaplace( (1+x*exp(x))/(1-x) )) \\ _G. C. Greubel_, Oct 13 2019

%o (Sage) [factorial(n)*( (1+x*exp(x))/(1-x) ).series(x,n+1).list()[n] for n in (0..30)] # _G. C. Greubel_, Oct 13 2019

%o (GAP) a:=[1,2,6];; for n in [4..30] do a[n]:=(n+1)*a[n-1]-(2*n-3)*a[n-2] +(n-3)*a[n-3]; od; a; # _G. C. Greubel_, Oct 13 2019

%Y Cf. A007526, A090805.

%K nonn,easy

%O 1,2

%A _Antti Karttunen_