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%I #8 Aug 13 2014 18:54:14
%S 1,1,2,4,8,17,38,87,205,493,1203,2969,7389,18504,46561,117596,297883,
%T 756388,1924484,4904830,12519121,31995286,81864992,209681349,
%U 537562018,1379332297,3542013533,9102191107,23406301490,60226845008,155059899921
%N Antidiagonal sums of triangle A107105: a(n) = Sum_{k=0..n} A107105(n-k,k), where A107105(n,k) = C(n,k)*(C(n,k) + 1)/2.
%C Limit a(n+1)/a(n) = (sqrt(5)+3)/2.
%F a(n) = (A051286(n) + A000045(n+1))/2, where A000045(n+1) = Fibonacci(n+1) and A051286(n) = Whitney number of level n.
%F G.f.: ( 1/(1-x-x^2) + 1/sqrt( (1+x+x^2)*(1-3*x+x^2) ) )/2. - _Michael Somos_, Jul 27 2007
%F G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} x^k * C(n,k)*(C(n,k) + 1)/2. - _Paul D. Hanna_, Aug 13 2014
%o (PARI) a(n)=(sum(k=0,n,binomial(n-k,k)^2)+fibonacci(n+1))/2
%o (PARI) {a(n)= if(n<0, 0, polcoeff( (1/(1-x-x^2) +1/sqrt((1+x+x^2)* (1-3*x+x^2)+ x*O(x^n)))/2, n))} /* _Michael Somos_, Jul 27 2007 */
%Y Cf. A107105, A051286, A000045.
%K nonn
%O 0,3
%A _Paul D. Hanna_, May 22 2005
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