|
%I #10 Nov 08 2014 17:56:37
%S 1,5,35,285,2519,23545,228715,2284365,23294551,241366025,2532599675,
%T 26845322925,286946021495,3088651368025,33443864305675,
%U 363983410742925,3979005075583255,43667580737050025,480884378835323675,5311978947724802925,58839469859529979319,653372409529941364345
%N G.f.: Sum_{n>=0} x^n / (1-4*x)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * 2^k * x^k] * [Sum_{k=0..n} C(n,k)^2 * 4^k * x^k].
%C Compare this sequence to its dual, A249921.
%F G.f.: Sum_{n>=0} x^n / (1-x)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * x^k] * [Sum_{k=0..n} C(n,k)^2 * 4^(n-k) * 2^k * x^k].
%F G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * Sum_{j=0..k} C(k,j)^2 * 4^(k-j) * 2^j * x^j.
%F G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * 4^(n-k) * Sum_{j=0..k} C(k,j)^2 * 2^j * x^j.
%F a(n) = Sum_{k=0..[n/2]} 2^k * Sum_{j=0..n-2*k} C(n-k, k+j)^2 * C(k+j, j)^2 * 4^j.
%F a(n) ~ sqrt(24+17*sqrt(2)) * (6+4*sqrt(2))^n / (4*Pi*n). - _Vaclav Kotesovec_, Oct 04 2014
%e G.f.: A(x) = 1 + 5*x + 35*x^2 + 285*x^3 + 2519*x^4 + 23545*x^5 +...
%e where the g.f. is given by the binomial series identity:
%e A(x) = 1/(1-4*x) + x/(1-4*x)^3 * (1 + 2*x) * (1 + 4*x)
%e + x^2/(1-4*x)^5 * (1 + 2^2*2*x + 4*x^2) * (1 + 2^2*4*x + 16*x^2)
%e + x^3/(1-4*x)^7 * (1 + 3^2*2*x + 3^2*4*x^2 + 8*x^3) * (1 + 3^2*4*x + 3^2*16*x^2 + 64*x^3)
%e + x^4/(1-4*x)^9 * (1 + 4^2*2*x + 6^2*4*x^2 + 4^2*8*x^3 + 16*x^4) * (1 + 4^2*4*x + 6^2*16*x^2 + 4^2*64*x^3 + 2561*x^4)
%e + x^5/(1-4*x)^11 * (1 + 5^2*2*x + 10^2*4*x^2 + 10^2*8*x^3 + 5^2*16*x^4 + 32*x^5) * (1 + 5^2*4*x + 10^2*16*x^2 + 10^2*64*x^3 + 5^2*256*x^4 + 1024*x^5) +...
%e equals the series
%e A(x) = 1/(1-x) + x/(1-x)^3 * (1 + x) * (4 + 2*x)
%e + x^2/(1-x)^5 * (1 + 2^2*x + x^2) * (16 + 2^2*4*2*x + 4*x^2)
%e + x^3/(1-x)^7 * (1 + 3^2*x + 3^2*x^2 + x^3) * (64 + 3^2*16*2*x + 3^2*4*4*x^2 + 8*x^3)
%e + x^4/(1-x)^9 * (1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4) * (256 + 4^2*64*2*x + 6^2*16*4*x^2 + 4^2*4*8*x^3 + 16*x^4)
%e + x^5/(1-x)^11 * (1 + 5^2*x + 10^2*x^2 + 10^2*x^3 + 5^2*x^4 + x^5) * (1024 + 5^2*256*2*x + 10^2*64*4*x^2 + 10^2*16*8*x^3 + 5^2*4*16*x^4 + 32*x^5) +...
%e We can also express the g.f. by another binomial series identity:
%e A(x) = 1 + x*(4 + (1+2*x)) + x^2*(16 + 2^2*4*(1+2*x) + (1+2^2*2*x+4*x^2))
%e + x^3*(64 + 3^2*16*(1+2*x) + 3^2*4*(1+2^2*2*x+4*x^2) + (1+3^2*2*x+3^2*4*x^2+8*x^3))
%e + x^4*(256 + 4^2*64*(1+2*x) + 6^2*16*(1+2^2*2*x+4*x^2) + 4^2*4*(1+3^2*2*x+3^2*4*x^2+8*x^3) + (1+4^2*2*x+6^2*4*x^2+4^2*8*x^3+16*x^4))
%e + x^5*(1024 + 5^2*256*(1+2*x) + 10^2*64*(1+2^2*2*x+4*x^2) + 10^2*16*(1+3^2*2*x+3^2*4*x^2+8*x^3) + 5^2*4*(1+4^2*2*x+6^2*4*x^2+4^2*8*x^3+16*x^4) + (1+5^2*2*x+10^2*4*x^2+10^2*8*x^3+5^2*16*x^4+32*x^5)) +...
%e equals the series
%e A(x) = 1 + x*(1 + (4+2*x)) + x^2*(1 + 2^2*(4+2*x) + (16+2^2*4*2*x+4*x^2))
%e + x^3*(1 + 3^2*(4+2*x) + 3^2*(16+2^2*4*2*x+4*x^2) + (64+3^2*16*2*x+3^2*4*4*x^2+8*x^3))
%e + x^4*(1 + 4^2*(4+2*x) + 6^2*(16+2^2*4*2*x+4*x^2) + 4^2*(64+3^2*16*2*x+3^2*4*4*x^2+8*x^3) + (256+4^2*64*2*x+6^2*16*4*x^2+4^2*8*4*x^3+16*x^4))
%e + x^5*(1 + 5^2*(4+2*x) + 10^2*(16+2^2*4*2*x+4*x^2) + 10^2*(64+3^2*16*2*x+3^2*4*4*x^2+8*x^3) + 5^2*(256+4^2*64*2*x+6^2*16*4*x^2+4^2*8*4*x^3+16*x^4) + (1024+5^2*256*2*x+10^2*64*4*x^2+10^2*16*8*x^3+5^2*4*16*x^4+32*x^5)) +...
%t Table[Sum[2^k * Sum[Binomial[n-k, k+j]^2 * Binomial[k+j, j]^2 * 4^j,{j,0,n-2*k}],{k,0,Floor[n/2]}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 04 2014 *)
%o (PARI) /* By definition: */
%o {a(n,p=4,q=2)=local(A=1); A=sum(m=0, n, x^m/(1-p*x)^(2*m+1) * sum(k=0, m, binomial(m, k)^2 * q^k * x^k) * sum(k=0, m, binomial(m,k)^2 * p^k * x^k) +x*O(x^n)); polcoeff(A, n)}
%o for(n=0, 25, print1(a(n,4,2), ", "))
%o (PARI) /* By a binomial identity: */
%o {a(n,p=4,q=2)=local(A=1); A=sum(m=0, n, x^m/(1-x)^(2*m+1) * sum(k=0, m, binomial(m, k)^2 * p^(m-k) * q^k * x^k) * sum(k=0, m, binomial(m, k)^2 * x^k) +x*O(x^n)); polcoeff(A, n)}
%o for(n=0, 25, print1(a(n,4,2), ", "))
%o (PARI) /* By a binomial identity: */
%o {a(n,p=4,q=2)=polcoeff(sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2 * sum(j=0, k, binomial(k, j)^2 * p^(k-j) * q^j * x^j)+x*O(x^n))), n)}
%o for(n=0, 25, print1(a(n,4,2), ", "))
%o (PARI) /* By a binomial identity: */
%o {a(n,p=4,q=2)=polcoeff(sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2 * p^(m-k) * sum(j=0, k, binomial(k, j)^2 * q^j * x^j)+x*O(x^n))), n)}
%o for(n=0, 25, print1(a(n,4,2), ", "))
%o (PARI) /* Formula for a(n): */
%o {a(n,p=4,q=2)=sum(k=0, n\2, sum(j=0, n-2*k, q^k * binomial(n-k, k+j)^2 * binomial(k+j, j)^2 * p^j))}
%o for(n=0, 25, print1(a(n,4,2), ", "))
%Y Cf. A249921, A246510, A246423, A246455, A246056.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Sep 30 2014
| 