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Search: id:A101194
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| A101194 |
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G.f. defined as the limit: A(x) = limit_{n->oo} F(n)^(1/5^(n-1)) where F(n) is the n-th iteration of: F(0) = 1, F(n) = F(n-1)^5 + (5x)^((5^n-1)/4) for n>=1. |
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3
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| 1, 5, 0, 0, 0, 0, 3125, -62500, 781250, -7812500, 68359375, -546875000, 4082031250, -28417968750, 179443359375, -939941406250, 2685546875000, 23010253906250, -569122314453125, 7669982910156250, -84739685058593750, 836715698242187500, -7611751556396484375
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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The Euler transform of the power series A(x) at x=1/5 converges to the constant: c = Sum_{n=0..infty} Sum_{k=0..n} C(n,k)*a(k)/5^k)/2^(n+1)) = 2.012346619142363112612326559... which is the limit of S(n)^(1/5^(n-1)) where S(0)=1, S(n+1) = S(n)^5 +1.
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FORMULA
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G.f. begins: A(x) = (1+m*x) + m^m*x^(m+1)/(1+m*x)^(m-1) +... at m=5.
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EXAMPLE
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The iteration begins:
F(0) = 1,
F(1) = 1 +5*x
F(2) = 1 +25*x +250*x^2 +1250*x^3 +3125*x^4 +3125*x^5 +15625*x^6
F(3) = 1 +125*x +7500*x^2 +287500*x^3 +... + 5^31*x^31.
The 5^(n-1)-th root of F(n) tend to the limit of A(x):
F(1)^(1/5^0) = 1 +5*x
F(2)^(1/5^1) = 1 +5*x +3125*x^6 -62500*x^7 +781250*x^8 +...
F(3)^(1/5^2) = 1 +5*x +3125*x^6 -62500*x^7 +781250*x^8 +...
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PROGRAM
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(PARI) {a(n)=local(F=1, A, L); if(n==0, A=1, L=ceil(log(n+1)/log(5)); for(k=1, L, F=F^5+(5*x)^((5^k-1)/4)); A=polcoeff((F+x*O(x^n))^(1/5^(L-1)), n)); A}
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CROSSREFS
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Cf. A101189, A101192, A101193.
Sequence in context: A048894 A047766 A005078 this_sequence A106222 A090750 A036297
Adjacent sequences: A101191 A101192 A101193 this_sequence A101195 A101196 A101197
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KEYWORD
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sign
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Dec 07 2004
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