A038546 Numbers n such that n-th Fibonacci number has initial digits n.

0, 1, 5, 43, 48, 53, 3301, 48515, 348422, 406665, 1200207, 6698641, 190821326, 2292141445, 257125021372, 5843866639660, 45173327533483, 46312809996150, 59358981837795, 129408997210988, 1450344802530203, 5710154240910003

Offset: 1

Author

Jeff Burch

Keywords

• nonn
• base
• nice

Comments

The Mathematica coding used by Robert G. Wilson v implements Binet's Fibonacci number formula as suggested by David W. Wilson and incorporates Benoit Cloitre's use of logarithms to achieve a further increase in speed.

Fixed points of A020344. - Alois P. Heinz, Jul 08 2022

Examples

			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.
		

Links

• Ron Knott, Fibonacci Numbers and the Golden Section
• Eric Weisstein's World of Mathematics, Fibonacci numbers

Crossrefs

Cf. A000045, A020344, A052000, A000350, A050816.

Programs

Mathematica

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)

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

Formula

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

Extensions

Term a(6) from Patrick De Geest, Oct 15 1999

a(7) from Benoit Cloitre, Feb 27 2002

a(8)-a(11) from Robert G. Wilson v, May 09 2005

a(12) from Robert G. Wilson v, May 11 2005

More terms from Robert Gerbicz, Aug 22 2006