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%I #58 Jan 20 2019 12:03:10
%S 2,2,3,4,7,11,21,36,71,127,253,463,925,1717,3433,6436,12871,24311,
%T 48621,92379,184757,352717,705433,1352079,2704157,5200301,10400601,
%U 20058301,40116601,77558761,155117521,300540196,601080391,1166803111
%N a(n) = binomial(n, floor(n/2)) + 1.
%C With the exception of the initial 2s, these are numbers such that if Pascal's triangle is written in base a(n) - 1, the first n - 2 rows give the digits of the powers of a(n) written in that base. This is most often noticed for the powers of 11 since of course we use decimal. - _Alonso del Arte_, Jul 10 2011
%H Alois P. Heinz, <a href="/A051920/b051920.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: -((2*x-1)*(3*x-1) +(x-1)*sqrt(1 - 4*x^2))/(2*x*(x-1)*(2*x-1)). - _Thomas Baruchel_, Jun 26 2018
%F 0 = 1 +a(n)*(-2 +4*a(n+1) -2*a(n+2)) +a(n+1)*(-1 -2*a(n+1) +a(n+2)) +a(n+2) for all n>=0. - _Michael Somos_, Jun 30 2018
%p a:= proc(n) option remember; `if`(n<3, [2, 2, 3][n+1],
%p ((n^2+n-4)*a(n-1) +2*(n-1)*(2*n-5)*a(n-2)
%p -4*(n-1)*(n-2)*a(n-3)) / ((n+1)*(n-2)))
%p end:
%p seq(a(n), n=0..50); # _Alois P. Heinz_, Mar 03 2014
%t a[n_] := a[n] = (4(n-1) a[n-2] + 2a[n-1] - 3n + 3)/(n+1); a[0] = a[1] = 2; Array[a, 50, 0] (* _Jean-François Alcover_, Jan 19 2017 *)
%t Table[Binomial[n,Floor[n/2]],{n,0,40}]+1 (* _Harvey P. Dale_, Jan 20 2019 *)
%o (PARI) a(n)=binomial(n,n\2)+1 \\ _Charles R Greathouse IV_, Feb 05 2013
%Y Cf. A001405, A006481, A007318, A193242.
%K nonn
%O 0,1
%A _N. J. A. Sloane_, Dec 18 1999
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