Search: id:A134084
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%I A134084
%S A134084 1,2,2,4,106,6948,1623788,1213437064,2912047916698,23264250235542100,
%T A134084 641982248042094828676,62929856484660987275500088,22331407793040258023249030997892,
%U A134084 29057717949243934527799656871001480808
%V A134084 1,2,2,-4,-106,-6948,-1623788,-1213437064,-2912047916698,-23264250235542100,
%W A134084 -641982248042094828676,-62929856484660987275500088,-22331407793040258023249030997892,
%X A134084 -29057717949243934527799656871001480808
%N A134084 G.f. A(x) = G(2x) where G(x) satisfies [x^(n+1)] G(x)^(2^n) = [x^n] G(x)^(2^n) 
               for n>=0.
%C A134084 G.f. A(x) satisfies: A(x/2)^2 is the g.f. of an integer sequence (A134085).
%F A134084 A134086(n) = [x^n] G(x)^(2^n) for n>=0. A134087(n) = [x^n] G(x)^(2^(n+1)) 
               for n>=0. G.f. A(x) satisfies: [x^(n+1)] A(x)^(2^n) = 2 * [x^n] A(x)^(2^n) 
               for n>=0.
%F A134084 G.f. A(x) satisfies: 1 = Sum_{n>=0} (1/2^n - x) * log( A(2^(n-1)*x) )^n 
               / n! = (1-x) + (1/2-x)log(A(x)) + (1/4-x)log(A(2x))^2/2! + (1/8-x)log(A(4x))^3/
               3! +... - Paul D. Hanna (pauldhanna(AT)juno.com), Jan 05 2008
%e A134084 G.f. A(x) = 1 + 2*x + 2*x^2 - 4*x^3 - 106*x^4 - 6948*x^5 - ...
%e A134084 Define G(x) = A(x/2); illustrate that
%e A134084 G(x) satisfies [x^(n+1)] G(x)^(2^n) = [x^n] G(x)^(2^n)
%e A134084 by listing powers G(x)^(2^n) as follows:
%e A134084 G(x)^1 = (1 + x) + 1/2*x^2 - 1/2*x^3 - 53/8*x^4 - 1737/8*x^5 -...;
%e A134084 G(x)^2 = 1+(2x + 2x^2) + 0x^3 - 14x^4 - 448x^5 - 51184x^6 -...;
%e A134084 G(x)^4 = 1 +4x +(8x^2 + 8x^3) - 24x^4 - 952x^5 - 104216x^6 -...;
%e A134084 G(x)^8 = 1 +8x +32x^2 +(80x^3 + 80x^4) - 1968x^5 - 216368x^6 -...;
%e A134084 G(x)^16 = 1 +16x +128x^2 +672x^3 +(2464x^4 + 2464x^5) -452704x^6 -...;
%e A134084 G(x)^32 = 1 +32x +512x^2+5440x^3 +42816x^4+(255808x^5 + 255808x^6) -...;
%e A134084 to show that the coefficients within the parenthesis are equal.
%e A134084 Note also that G(x)^2 consists entirely of integer coefficients.
%o A134084 (PARI) {a(n)=local(A=[1],B);for(i=1,n,A=concat(A,0); B=Vec(Ser(A)^(2^(#A-2)));
               A[ #A]=(B[ #B-1]-B[ #B])/2^(#A-2));2^n*A[n+1]}
%Y A134084 Cf. A134085, A134086, A134087, A134088, A134089.
%Y A134084 Sequence in context: A067700 A037010 A114695 this_sequence A100247 A011342 
               A084046
%Y A134084 Adjacent sequences: A134081 A134082 A134083 this_sequence A134085 A134086 
               A134087
%K A134084 sign
%O A134084 0,2
%A A134084 Paul D. Hanna (pauldhanna(AT)juno.com), Oct 25 2007

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