A107597 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.
1, 1, 2, 4, 8, 17, 38, 87, 205, 493, 1203, 2969, 7389, 18504, 46561, 117596, 297883, 756388, 1924484, 4904830, 12519121, 31995286, 81864992, 209681349, 537562018, 1379332297, 3542013533, 9102191107, 23406301490, 60226845008, 155059899921
Offset: 0
Keywords
Programs
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PARI
a(n)=(sum(k=0,n,binomial(n-k,k)^2)+fibonacci(n+1))/2
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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 */
Formula
a(n) = (A051286(n) + A000045(n+1))/2, where A000045(n+1) = Fibonacci(n+1) and A051286(n) = Whitney number of level n.
G.f.: ( 1/(1-x-x^2) + 1/sqrt( (1+x+x^2)*(1-3*x+x^2) ) )/2. - Michael Somos, Jul 27 2007
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
Comments