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A264927 Terms satisfy: a(2*n) = b(n)^2 and a(2*n+1) = b(n)*b(n+1) for n>=0 with a(0)=1, where A(x)^2 = Sum_{n>=0} b(n)*x^n and g.f. A(x) = Sum_{n>=0} a(n)*x^n. 1

%I #7 Dec 15 2015 18:31:37

%S 1,1,4,18,81,396,1936,9416,45796,234972,1205604,6188328,31764496,

%T 162249168,828748944,4224034452,21529399441,113085801048,593996986944,

%U 3126735424128,16458794619136,86717178000000,456890625000000,2407551579000000,12686416153856064,66648950832253248,350144800010358336,1838080630013808960,9648980656951105600,50559912359694495680

%N Terms satisfy: a(2*n) = b(n)^2 and a(2*n+1) = b(n)*b(n+1) for n>=0 with a(0)=1, where A(x)^2 = Sum_{n>=0} b(n)*x^n and g.f. A(x) = Sum_{n>=0} a(n)*x^n.

%C What is the limit a(n)^(1/n) = ? The limit is at least 5.3...

%e G.f.: A(x) = 1 + x + 4*x^2 + 18*x^3 + 81*x^4 + 396*x^5 + 1936*x^6 + 9416*x^7 + 45796*x^8 + 234972*x^9 + 1205604*x^10 +...

%e where

%e A(x)^2 = 1 + 2*x + 9*x^2 + 44*x^3 + 214*x^4 + 1098*x^5 + 5636*x^6 + 28788*x^7 + 146729*x^8 + 770712*x^9 + 4056944*x^10 + 21375000*x^11 + 112633992*x^12 + 591730344*x^13 + 3106280840*x^14 + 16276671352*x^15 + 85215198522*x^16 +...

%e Illustration of initial terms, both a(n) of A(x) and b(n) of A(x)^2:

%e a(0) = 1; b(0) = 1;

%e a(1) = 1; b(1) = 2 = 1*1 + 1*1;

%e a(2) = b(1)*b(1) = 4; b(2) = 9 = 1*4 + 1*1 + 4*1;

%e a(3) = b(1)*b(2) = 18; b(3) = 44 = 1*18 + 1*4 + 4*1 + 18*1;

%e a(4) = b(2)*b(2) = 81; b(4) = 214 = 1*81 + 1*18 + 4*4 + 18*1 + 81*1;

%e a(5) = b(2)*b(3) = 396; b(5) = 1098;

%e a(6) = b(3)*b(3) = 1936; b(6) = 5636;

%e a(7) = b(3)*b(4) = 9416; b(7) = 28788;

%e a(8) = b(4)*b(4) = 45796; b(8) = 146729; ...

%o (PARI) {a(n) = my(A=1+x); for(k=2, n, B = A^2; A = A + polcoeff(B, k\2) * polcoeff(B, (k+1)\2) * x^k +x*O(x^n) ); polcoeff(A, n)}

%o for(n=0, 40, print1(a(n), ", "))

%o (PARI) {a(n) = my(A=[1,1]); for(k=2,n, B = Vec(Ser(A)^2); A = concat(A, B[k\2+1]*B[(k+1)\2+1]) ); A[n+1]}

%o for(n=0,40,print1(a(n),", "))

%o (PARI) /* Generates N terms rather quickly: */

%o N=500; A=[1, 1]; for(k=2, N, B = Vec(Ser(A)^2); A = concat(A, B[k\2+1]*B[(k+1)\2+1]) ); A

%Y Cf. A257889.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Dec 13 2015

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