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%I #20 Jul 07 2017 03:26:32
%S 1,2,2,2,0,-2,-4,-2,-6,-2,4,50,78,34,-248,-146,248,1478,992,-2570,
%T -13918,-19026,-12744,68034,193088,380810,86880,-1362006,-5050422,
%U -6966190,1464256,39349110,99348446,130130462,-62116516,-664338662,-1811216390,-2530859462,-458383728,9647822334,29564081352,51363216966,35437251200,-72752696138,-343482383882,-735851485042,-1167503794928,-1270647820502,-634163168566,3532238523874,18635012627076
%N G.f. A(x) satisfies: 1 = ...((((A(x) - 2*x)^2 - 4*x^2)^2 - 8*x^3)^2 - 16*x^4)^2 -... - 2^n*x^n)^2 -...., a series of infinite nested squares.
%H Paul D. Hanna, <a href="/A274850/b274850.txt">Table of n, a(n) for n = 0..300</a>
%F G.f.: S(0) where S(k) = (2*x)^(k+1) + S(k+1)^(1/2). - _Michael Somos_, Jul 06 2017
%e G.f.: A(x) = 1 + 2*x + 2*x^2 + 2*x^3 - 2*x^5 - 4*x^6 - 2*x^7 - 6*x^8 - 2*x^9 + 4*x^10 + 50*x^11 + 78*x^12 + 34*x^13 - 248*x^14 - 146*x^15 + 248*x^16 +...
%e such that A(x) satisfies the series of infinite nested squares given by:
%e 1 = (((((((((A - 2*x)^2 - 2^2*x^2)^2 - 2^3*x^3)^2 - 2^4*x^4)^2 - 2^5*x^5)^2 - 2^6*x^6)^2 - 2^7*x^7)^2 - 2^8*x^8)^2 - 2^9*x^9)^2 ...
%e ILLUSTRATION.
%e Equivalently, we may start with B0 = A(x) - 2*x, and then continue
%e B1 = ( B0^2 - 2^2*x^2 )^(1/2)
%e B2 = ( B1^4 - 2^3*x^3 )^(1/4)
%e B3 = ( B2^8 - 2^4*x^4 )^(1/8)
%e B4 = ( B3^16 - 2^5*x^5 )^(1/16)
%e B5 = ( B4^32 - 2^6*x^6 )^(1/32)
%e B6 = ( B5^64 - 2^7*x^7 )^(1/64)
%e B7 = ( B6^128 - 2^8*x^8 )^(1/128)
%e B8 = ( B7^256 - 2^9*x^9 )^(1/256)
%e B9 = ( B8^512 - 2^10*x^10 )^(1/512)
%e ...
%e The above series converge to 1, as can be seen by:
%e B0 = 1 + 2*x^2 + 2*x^3 - 2*x^5 - 4*x^6 - 2*x^7 - 6*x^8 - 2*x^9 + 4*x^10 +...
%e B1 = 1 + 2*x^3 + 2*x^4 + 2*x^5 - 4*x^6 - 10*x^7 - 24*x^8 - 10*x^9 + 16*x^10 +...
%e B2 = 1 + 2*x^4 + 2*x^5 + 2*x^6 + 2*x^7 - 12*x^8 - 26*x^9 - 56*x^10 - 86*x^11 +...
%e B3 = 1 + 2*x^5 + 2*x^6 + 2*x^7 + 2*x^8 + 2*x^9 - 28*x^10 - 58*x^11 - 120*x^12 +...
%e B4 = 1 + 2*x^6 + 2*x^7 + 2*x^8 + 2*x^9 + 2*x^10 + 2*x^11 - 60*x^12 - 122*x^13 +...
%e B5 = 1 + 2*x^7 + 2*x^8 + 2*x^9 + 2*x^10 + 2*x^11 + 2*x^12 + 2*x^13 - 124*x^14 +...
%e B6 = 1 + 2*x^8 + 2*x^9 + 2*x^10 + 2*x^11 + 2*x^12 + 2*x^13 + 2*x^14 + 2*x^15 - 252*x^16 - 506*x^17 - 1016*x^18 - 1526*x^19 - 2548*x^20 +...
%e B7 = 1 + 2*x^9 + 2*x^10 + 2*x^11 + 2*x^12 + 2*x^13 + 2*x^14 + 2*x^15 + 2*x^16 + 2*x^17 - 508*x^18 - 1018*x^19 - 2040*x^20 - 3062*x^21 +...
%e B8 = 1 + 2*x^10 + 2*x^11 + 2*x^12 + 2*x^13 + 2*x^14 + 2*x^15 + 2*x^16 + 2*x^17 + 2*x^18 + 2*x^19 - 1020*x^20 - 2042*x^21 - 4088*x^22 +...
%e B9 = 1 + 2*x^11 + 2*x^12 + 2*x^13 + 2*x^14 + 2*x^15 + 2*x^16 + 2*x^17 + 2*x^18 + 2*x^19 + 2*x^20 + 2*x^21 - 2044*x^22 - 4090*x^23 +...
%e ...
%t a[ n_] := If[ n < 0, 0, Module[{A = 1 + O[x]^(n + 1), B = (2 x)^(n + 1)}, Do[ A = Sqrt[A] + (B /= 2 x), n]; SeriesCoefficient[A, n]]]; (* _Michael Somos_, Jul 06 2017 *)
%o (PARI) {a(n) = my(A=[1],B=1,m); for(i=1,n, A=concat(A,0); B = Ser(A);
%o for(m=0,#A-1, B = (B^(2^m) - 2^(m+1)*x^(m+1))^(1/2^m); A[#A] = -Vec(B)[#A]) ); A[n+1]}
%o for(n=0,50,print1(a(n),", "))
%o (PARI) {a(n) = my(A); if( n<0, 0, A = 1 + x * O(x^n); for(k=1, n, A = sqrt(A) + (2*x)^(n+1-k)); polcoeff(A, n))}; /* _Michael Somos_, Jul 06 2017 */
%K sign
%O 0,2
%A _Paul D. Hanna_, Jul 09 2016
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