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A038546
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Numbers n such that n-th Fibonacci number has initial digits n.
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7
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0, 1, 5, 43, 48, 53, 3301, 48515, 348422, 406665, 1200207, 6698641, 190821326, 2292141445, 257125021372, 5843866639660, 45173327533483, 46312809996150, 59358981837795, 129408997210988, 1450344802530203, 5710154240910003
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OFFSET
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1,3
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COMMENTS
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LINKS
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FORMULA
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n>5 is in the sequence if a=(1+sqrt(5))/2 b=1/sqrt(5) and n==floor(b*(a^n)/10^(floor((log(b) +n*log(a))/log(10))-floor(log(n)/log(10))) ). - Benoit Cloitre, Feb 27 2002
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EXAMPLE
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a(3)=43 since 43rd Fibonacci number starts with 43 -> {43}3494437.
Fibonacci(53) is 53316291173, which begins with 53, so 53 is a term in the sequence.
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MATHEMATICA
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a = N[ Log[10, Sqrt[5]/5], 24]; b = N [Log[10, GoldenRatio], 24]; Do[ If[ IntegerPart[10^FractionalPart[a + n*b]*10^Floor[ Log[10, n]]] == n, Print[n]], {n, 225000000}] (* Robert G. Wilson v, May 09 2005 *)
(* confirmed with: *) fQ[n_] := (FromDigits[ Take[ IntegerDigits[ Fibonacci[n]], Floor[ Log[10, n] + 1]]] == n)
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PROG
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(PARI) /* To obtain terms > 5: */ a=(1+sqrt(5))/2; b=1/sqrt(5); for(n=1, 3500, if(n==floor(b*(a^n)/10^( floor(log(b *(a^n))/log(10))-floor(log(n)/log(10)))), print1(n, ", "))) \\ Benoit Cloitre, Feb 27 2002
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CROSSREFS
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KEYWORD
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nonn,base,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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