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%I #11 Jul 23 2015 02:40:55
%S 1,2,6,26,154,1182,11254,128522,1715802,26251118,453132214,8714516538,
%T 184817376154,4285657717694,107880712446966,2929921866938858,
%U 85399256991014746,2659096654189212558,88091786616208073398
%N Absolute row sums of triangle A107102, which is the matrix inverse of A100862.
%H W. Mlotkowski, A. Romanowicz, <a href="http://www.math.uni.wroc.pl/~pms/files/33.2/Article/33.2.19.pdf">A family of sequences of binomial type</a>, Probability and Mathematical Statistics, Vol. 33, Fasc. 2 (2013), pp. 401-408.
%F a(n) = 2*A043301(n-1) = 2^n*Sum_{k=0..n-1}(n+k-1)!/(n-k-1)!/k!/4^k for n>0, with a(0) = 1.
%o (PARI) a(n)=if(n==0,1,2^n*sum(k=0,n-1,(n+k-1)!/(n-k-1)!/k!/4^k))
%o (PARI) a(n)=local(X=x+x*O(x^n),Y=y+y*O(y^n)); sum(k=0,n,abs((matrix(n+1,n+1,m,j,if(m>=j, (m-1)!*polcoeff(polcoeff(exp(X+Y*X^2/2+X*Y),m-1,x),j-1,y)))^-1)[n+1,k+1]))
%Y Cf. A100862, A043301, A107102, A107103.
%K nonn
%O 0,2
%A _Paul D. Hanna_, May 21 2005
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