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%I #19 Sep 02 2014 06:39:52
%S 1,1,2,6,15,39,116,340,1001,3041,9322,28718,89363,279987,881376,
%T 2788464,8860677,28256709,90407666,290124182,933482527,3010689527,
%U 9731366060,31516942060,102259648701,332347297141,1081810639970,3526399820374,11510355762339,37616896717155
%N G.f.: Sum_{n>=0} x^n / (1-x^2)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * x^(2*k)]^2.
%C Bisections form A246570 and A246571.
%C Compare formula for a(n) to a formula for tribonacci numbers:
%C A000073(n+2) = Sum_{k=0..[n/2]} Sum_{j=0..k} C(n-k-j,k) * C(k,j).
%H Paul D. Hanna, <a href="/A246563/b246563.txt">Table of n, a(n) for n = 0..500</a>
%F G.f.: Sum_{n>=0} x^(2*n) / (1-x)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * x^k]^2.
%F G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * x^k * Sum_{j=0..k} C(k,j)^2 * x^j.
%F G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * x^(n-k) * Sum_{j=0..k} C(k,j)^2 * x^(2*j).
%F a(n) = Sum_{k=0..[n/2]} Sum_{j=0..k} C(n-k-j,k)^2 * C(k,j)^2.
%F Recurrence: n^2*(2*n - 7)*(2*n - 5)*a(n) = (2*n - 7)*(4*n^3 - 14*n^2 + 10*n - 3)*a(n-1) + (8*n^4 - 64*n^3 + 174*n^2 - 198*n + 87)*a(n-2) + (2*n - 3)*(20*n^3 - 150*n^2 + 362*n - 281)*a(n-3) + 3*(2*n - 5)*a(n-4) - (2*n - 7)*(20*n^3 - 150*n^2 + 362*n - 279)*a(n-5) - (8*n^4 - 96*n^3 + 414*n^2 - 742*n + 447)*a(n-6) - (2*n - 3)*(4*n^3 - 46*n^2 + 170*n - 197)*a(n-7) + (n-5)^2*(2*n - 5)*(2*n - 3)*a(n-8). - _Vaclav Kotesovec_, Sep 02 2014
%F a(n) ~ c * d^n / (Pi*n), where d = ((54+6*sqrt(33))^(2/3) + 12 + 3*(54+6*sqrt(33))^(1/3)) / (3*(54+6*sqrt(33))^(1/3)) = 3.3829757679062374941227... is the root of the equation -1 - d - 3*d^2 + d^3 = 0, c = 1/12*(199+3*sqrt(33))^(1/3) + 17/(6*(199+3*sqrt(33))^(1/3)) + 7/12 = 1.55556563078009965666864... . - _Vaclav Kotesovec_, Sep 02 2014
%e G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 15*x^4 + 39*x^5 + 116*x^6 + 340*x^7 +...
%e where the g.f. is given by the binomial series identity:
%e A(x) = 1/(1-x^2) + x/(1-x^2)^3 * (1 + x^2)^2
%e + x^2/(1-x^2)^5 * (1 + 2^2*x^2 + x^4)^2
%e + x^3/(1-x^2)^7 * (1 + 3^2*x^2 + 3^2*x^4 + x^6)^2
%e + x^4/(1-x^2)^9 * (1 + 4^2*x^2 + 6^2*x^4 + 4^2*x^6 + x^8)^2
%e + x^5/(1-x^2)^11 * (1 + 5^2*x^2 + 10^2*x^4 + 10^2*x^6 + 5^2*x^8 + x^10)^2 +...
%e equals the series
%e A(x) = 1/(1-x) + x^2/(1-x)^3 * (1 + x)^2
%e + x^4/(1-x)^5 * (1 + 2^2*x + x^2)^2
%e + x^6/(1-x)^7 * (1 + 3^2*x + 3^2*x^2 + x^3)^2
%e + x^8/(1-x)^9 * (1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4)^2
%e + x^10/(1-x)^11 * (1 + 5^2*x + 10^2*x^2 + 10^2*x^3 + 5^2*x^4 + x^5)^2 +...
%e Curiously, the BISECTIONS of the g.f. are squares of integer series:
%e let A(x) = B0(x^2) + x*B1(x^2), then
%e B0(x) = 1 + 2*x + 15*x^2 + 116*x^3 + 1001*x^4 + 9322*x^5 + 89363*x^6 +...+ A246570(n)*x^n +...
%e sqrt(B0(x)) = 1 + x + 7*x^2 + 51*x^3 + 425*x^4 + 3879*x^5 + 36527*x^6 +...+ A246572(n)*x^n +...
%e B1(x) = 1 + 6*x + 39*x^2 + 340*x^3 + 3041*x^4 + 28718*x^5 + 279987*x^6 +...+ A246571(n)*x^n +...
%e sqrt(B1(x)) = 1 + 3*x + 15*x^2 + 125*x^3 + 1033*x^4 + 9385*x^5 + 88531*x^6 +...+ A246573(n)*x^n +...
%t Table[Sum[Sum[Binomial[n-k-j,k]^2 * Binomial[k,j]^2,{j,0,k}],{k,0,Floor[n/2]}],{n,0,20}] (* _Vaclav Kotesovec_, Sep 02 2014 *)
%o (PARI) /* By definition: */
%o {a(n)=local(A=1); A=sum(m=0, n, x^m/(1-x^2)^(2*m+1) * sum(k=0, m, binomial(m, k)^2 * x^(2*k))^2 +x*O(x^n)); polcoeff(A, n)}
%o for(n=0, 35, print1(a(n), ", "))
%o (PARI) /* By a binomial identity: */
%o {a(n)=local(A=1); A=sum(m=0, n, x^(2*m)/(1-x)^(2*m+1) * sum(k=0, m, binomial(m, k)^2 * x^k)^2 +x*O(x^n)); polcoeff(A, n)}
%o for(n=0, 35, print1(a(n), ", "))
%o (PARI) /* By a binomial identity: */
%o {a(n)=polcoeff(sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2 * x^k * sum(j=0, k, binomial(k, j)^2 * x^j )+x*O(x^n))), n)}
%o for(n=0, 35, print1(a(n), ", "))
%o (PARI) /* By a binomial identity: */
%o {a(n)=polcoeff(sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2 * x^(m-k) * sum(j=0, k, binomial(k, j)^2 * x^(2*j) )+x*O(x^n))), n)}
%o for(n=0, 35, print1(a(n), ", "))
%o (PARI) /* From a formula for a(n): */
%o {a(n)=sum(k=0, n\2, sum(j=0, min(k,n-2*k), binomial(n-k-j, k)^2 * binomial(k, j)^2 ))}
%o for(n=0, 35, print1(a(n), ", "))
%Y Cf. A246570, A246571, A246572, A246573.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Aug 29 2014
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