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A167000 G.f. A(x) satisfies: Sum_{n>=0} log( A(2^n*x) )^n / n! = 1 + Sum_{n>=0} 2^(2^n-n) * x^(2^n). 4

%I #13 May 12 2018 06:55:58

%S 1,1,-1,0,-16,-1767,-493164,-422963721,-1130568823448,

%T -9811523398109059,-287512372919585565730,-29365896347484186250530846,

%U -10704256920972727382240940549099,-14165930844739651162632827455464483815,-68918096446337603401330634164181238008617534

%N G.f. A(x) satisfies: Sum_{n>=0} log( A(2^n*x) )^n / n! = 1 + Sum_{n>=0} 2^(2^n-n) * x^(2^n).

%H Paul D. Hanna, <a href="/A167000/b167000.txt">Table of n, a(n) for n = 0..50</a>

%F The coefficient of x^(2^n) in A(x)^(2^(2^n)) equals 2^(2^n-n):

%F [x^(2^n)] A(x)^(2^(2^n)) = 2^(2^n-n); while

%F [x^n] A(x)^(2^n) = 0 when n>0 is not a power of 2, with A(0)=1.

%e G.f.: A(x) = 1 + x - x^2 - 16*x^4 - 1767*x^5 - 493164*x^6 -...

%e log(A(x)) = x - 3*x^2/2 + 4*x^3/3 - 71*x^4/4 - 8744*x^5/5 - 2948592*x^5/5 -...

%e ILLUSTRATE THE SERIES DEFINITION:

%e 1 + log(A(2x)) + log(A(4x))^2/2! + log(A(8x))^3/3! + log(A(16x))^4/4! +...

%e = 1 + 2*x + 2*x^2 + 4*x^4 + 32*x^8 + 4096*x^16 + 134217728*x^32 +...

%e = 1 + 2^(1-0)*x + 2^(2-1)*x^2 + 2^(4-2)*x^4 + 2^(8-3)*x^8 + 2^(16-4)*x^16 +...

%e ILLUSTRATE (2^n)-th POWERS OF G.F. A(x).

%e The coefficients in the expansion of A(x)^(2^n) for n>=0 begin:

%e [(1),1,-1,0,-16,-1767,-493164,-422963721,-1130568823448,...];

%e [1,(2),-1,-2,-31,-3566,-989830,-846910236,-2261982587754,...];

%e [1,4,(2),-8,-69,-7252,-1993858,-1697772536,-4527350821567,...];

%e [1,8,20,(0),-198,-15088,-4045944,-3411523840,-9068291678061,...];

%e [1,16,104,320,(4),-33344,-8341216,-6888386304,-18191329536118,...];

%e [1,32,464,3968,21064,(0),-17646208,-14050624512,-36604843747036];

%e [1,64,1952,37632,511376,5030400,(0),-29063442432,-74124859451768];

%e [1,128,8000,325120,9649952,222432256,4056470528,(0),...];

%e [1,256,32384,2698240,166530624,8117172224,325157844992,10872157339648, (32),...]; ...

%e where the coefficients along the diagonal (in parenthesis) begin:

%e [1,2,2,0,4,0,0,0,32,0,0,0,0,0,0,0,4096,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 134217728,...]

%e and equal 2^(2^m-m) at positions n=2^m for m>=0, with zeros elsewhere (except for the initial '1').

%t max = 25; f[x_] := Sum[a[k]*x^k, {k, 0, max}]; init (* to speed-up computation *) = {1, 1, -1, 0}; skip = Length[init]; a[n_ /; n < skip] := init[[n+1]]; coes = CoefficientList[Series[Sum[Log[f[2^n*x]]^n/n!, {n, 0, max}] - 1 - Sum[2^(2^n-n)*x^2^n, {n, 0, Log[2, max]//Floor}], {x, 0, max} ], x]; Do[coes = coes /. (sol[k-1] = Solve[coes[[k]] == 0][[1, 1]]), {k, skip+1, Length[coes]}]; Join[init, Table[a[k] /. sol[k], {k, skip, max}]] (* _Jean-François Alcover_, Mar 06 2013, updated Sep 04 2017 *)

%o (PARI) {a(n) = my(A=[1,1],B=[1,2]); for(i=1,n, A=concat(A,0); B=Vec( sum(m=0,#A, log( subst(Ser(A),x,2^m*x) )^m/m!) ); A[#A] = -floor( B[#A]/2^(#A-1) )); A[n+1]}

%o for(n=0,20, print1(a(n),", "))

%Y Cf. A167001, A167002, A167003.

%K nice,sign

%O 0,5

%A _Paul D. Hanna_, Nov 13 2009

%E Typos in examples fixed by _Paul D. Hanna_, Nov 15 2009

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Last modified June 8 00:48 EDT 2024. Contains 373206 sequences. (Running on oeis4.)