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Sum(sum(binomial(i,j),i=n..2*n),j=0..n).
4

%I #42 Oct 01 2013 14:49:50

%S 1,5,22,91,366,1454,5748,22691,89590,354010,1400268,5544334,21973420,

%T 87158972,345977832,1374249251,5461704870,21717305762,86391846492,

%U 343800647066,1368639516420,5450093895812,21708897213912,86492537630606

%N Sum(sum(binomial(i,j),i=n..2*n),j=0..n).

%C As the definition indicates, each term is the sum of numbers from Pascal's Triangle in an (n+1) X (n+1) square arrangement.

%C Example for a(2):

%C 1

%C 1 1

%C |1 2 1 |

%C |1 3 3 |1

%C |1 4 6 |4 1

%C 1 5 10 10 5 1

%C 1 6 15 20 15 6 1

%C ...

%C and

%C (1+2+1) + (1+3+3) + (1+4+6) = 22 = a(2). Similarly, a(1) = (1+1)+(1+2) = 5. - _John Molokach_, Sep 17 2013

%H Vincenzo Librandi, <a href="/A085812/b085812.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = 4^n - 2^n + C(2*n+2, n+1)/2. - _Vaclav Kotesovec_, Oct 28 2012

%F (n+1)*a(n) +2*(-6*n-1)*a(n-1) +4*(13*n-9)*a(n-2) +8*(-12*n+19)*a(n-3) +32*(2*n-5)*a(n-4)=0. - _R. J. Mathar_, Oct 01 2013

%e a(1)=binomial(1,0)+binomial(2,0)+binomial(1,1)+binomial(2,1)=1+1+1+2=5

%p a := n->add(add(binomial(i,j),i=n..2*n),j=0..n); seq(a(n),n=0..25);

%t Table[4^n-2^n+Binomial[2*n+2,n+1]/2,{n,0,20}] (* _Vaclav Kotesovec_, Oct 28 2012 *)

%o (PARI) a(n)=4^n-2^n+binomial(2*n+2,n+1)/2; \\ _Joerg Arndt_, May 10 2013

%K nonn,easy

%O 0,2

%A Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jul 25 2003