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%I #27 Nov 12 2017 03:10:08
%S 1,3,17,211,5793,339491,41326513,10282961907,5181436229441,
%T 5258784071302723,10717167529963833681,43779339268428732008723,
%U 358114286723184561034838497,5862685570087914880854259126371,192026370558313054275618817346778353
%N a(0) = 1 and, for n >= 1, (BM)a(n) = 2*a(n-1), where BM is the BinomialMean transform.
%C The BinomialMean transform BM is defined by (BM)a(n) = (M^n)a(0) where (M)a(n) is the mean (a(n) + a(n+1))/2, or, alternatively, by (BM)a(n) = (Sum_{k=0..n} binomial(n,k)*a(k))/(2^n).
%C The BinomialMean transform of this sequence is given in A075272.
%H Alois P. Heinz, <a href="/A075271/b075271.txt">Table of n, a(n) for n = 0..50</a>
%H Michael Z. Spivey and Laura L. Steil, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Spivey/spivey7.html">The k-Binomial Transforms and the Hankel Transform</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
%F O.g.f. as a continued fraction: A(x) = 1/(1 + x - 2^2*x/(1 - 2*(2 - 1)^2*x/(1 + x - 2^4*x/(1 - 2*(2^2 - 1)^2*x/(1 + x - 2^6*x/(1 - 2*(2^3 - 1)^2*x/(1 + x - 2^8*x/(1 - 2*(2^4 - 1)^2*x/(1 + x - ... ))))))))). Cf. A075272. - _Peter Bala_, Nov 10 2017
%e Given that a(0)=1 and a(1)=3. Then (BM)a(2) = (1 + 2*3 + a(2))/4 = 2a(1) = 6, hence a(2)=17.
%p iBM:= proc(p) proc(n) option remember; add(2^(k)*p(k)*(-1)^(n-k) *binomial(n, k), k=0..n) end end: a:= iBM(aa): aa:= n-> `if`(n=0, 1, 2*a(n-1)): seq(a(n), n=0..16); # _Alois P. Heinz_, Sep 09 2008
%t iBM[p_] := Module[{proc}, proc[n_] := proc[n] = Sum[2^k*p[k]*(-1)^(n-k) * Binomial[n, k], {k, 0, n}]; proc]; a = iBM[aa]; aa[n_] := If[n == 0, 1, 2*a[n-1]]; Table[a[n], {n, 0, 16}] (* _Jean-François Alcover_, Nov 08 2015, after _Alois P. Heinz_ *)
%t Table[Sum[QFactorial[k, 2] Binomial[n + 1, k]/2, {k, 0, n + 1}], {n, 0, 15}] (* _Vladimir Reshetnikov_, Oct 16 2016 *)
%Y Cf. A075272.
%K eigen,nonn
%O 0,2
%A _John W. Layman_, Sep 11 2002
%E More terms from _Alois P. Heinz_, Sep 09 2008
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