A372534 Expansion of g.f. A(x) satisfying A(x)^2 = A( x*A(x) + 3*x*A(x)^2 ).
1, 3, 12, 63, 372, 2322, 15102, 101439, 698340, 4900914, 34931808, 252185238, 1840242546, 13551558336, 100579610790, 751610709279, 5650352546628, 42702935642082, 324256445598816, 2472613511240754, 18926918200655928, 145379893260849876, 1120198916414984148, 8656357557290045382
Offset: 1
Keywords
Examples
G.f.: A(x) = x + 3*x^2 + 12*x^3 + 63*x^4 + 372*x^5 + 2322*x^6 + 15102*x^7 + 101439*x^8 + 698340*x^9 + 4900914*x^10 + 34931808*x^11 + 252185238*x^12 + ... where A(x)^2 = A( x*A(x) + 3*x*A(x)^2 ). Also, A(x) = x * (1 + 3*A(x)) * (1 + 3*A(x)^2) * (1 + 3*A(x)^4) * (1 + 3*A(x)^8) * (1 + 3*A(x)^16) * ... * (1 + 3*A(x)^(2^n)) * ... RELATED SERIES. A(x)^2 = x^2 + 6*x^3 + 33*x^4 + 198*x^5 + 1266*x^6 + 8388*x^7 + 57033*x^8 + 396090*x^9 + 2798718*x^10 + 20056824*x^11 + 145438146*x^12 + ... x*A(x) + 3*x*A(x)^2 = x^2 + 6*x^3 + 30*x^4 + 162*x^5 + 966*x^6 + 6120*x^7 + 40266*x^8 + 272538*x^9 + 1886610*x^10 + 13297068*x^11 + ... SPECIFIC VALUES. A(1/9) = 0.20017482594200170883488591841314367600913783... A(1/10) = 0.15939222988059047986391116283589184626082823... A(1/11) = 0.13474373940944085584086064879196682498369755... A(1/12) = 0.11741441277153705906655653078308588616286400...
Links
- Paul D. Hanna, Table of n, a(n) for n = 1..600
Programs
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PARI
{a(n) = my(A = serreverse( x/prod(k=0, #binary(n), 1 + 3*x^(2^k) +x*O(x^n)) )); polcoeff(A, n)} for(n=1, 30, print1(a(n), ", "))
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PARI
{a(n) = my(A=[0, 1]); for(i=1, n, A = concat(A, 0); F=Ser(A); A[#A] = polcoeff( subst(F, x, x*F*(1 + 3*F) ) - F^2, #A) ); A[n+1]} for(n=1, 30, print1(a(n), ", "))
Formula
G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following formulas.
(1) A(x)^2 = A( x*A(x) + 3*x*A(x)^2 ).
(2) A(x)^4 = A( x*A(x)^3*(1 + 3*A(x))*(1 + 3*A(x)^2) ).
(3) A(x)^8 = A( x*A(x)^7*(1 + 3*A(x))*(1 + 3*A(x)^2)*(1 + 3*A(x)^4) ).
(4) A(x)^(2^n) = A( x*A(x)^(2^n-1)*Product_{k=0..n-1} (1 + 3*A(x)^(2^k)) ) for n > 0.
(5) A(x) = x * Product_{n>=0} (1 + 3*A(x)^(2^n)).
(6) A(x) = x * Sum_{n>=0} 3^A000120(n) * A(x)^n, where A000120(n) = number of 1's in binary expansion of n.
(7) A(x) = Series_Reversion( x / Product_{n>=0} (1 + 3*x^(2^n)) ).
The radius of convergence r of g.f. A(x) and A(r) satisfy 1 = Sum_{n>=0} 3*2^n * A(r)^(2^n) / (1 + 3*A(r)^(2^n)) and r = A(r) / Product_{n>=0} (1 + 3*A(r)^(2^n)), where r = 0.121354219013536538658862726712953201279180864478537... and A(r) = 0.301069983372147236415588688159692129761365234627514...
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