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A105515
Number of times 5 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.
10
0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9
OFFSET
0,11
LINKS
FORMULA
a(n) = #{k: A008963(k) = 5 and 0<=k<=n};
a(A105505(n)) = a(A105505(n) - 1) + 1;
n = A105511(n) + A105512(n) + A105513(n) + A105514(n) + a(n) + A105516(n) + A105517(n) + A105518(n) + A105519(n).
a(n) ~ log_10(6/5) * n. - Amiram Eldar, Jan 12 2023
MATHEMATICA
Accumulate[If[First[IntegerDigits[#]]==5, 1, 0]&/@Fibonacci[Range[0, 110]]] (* Harvey P. Dale, Nov 02 2014 *)
PROG
(PARI)
(leadingdigit(n, b=10) = n \ 10^logint(n, b));
(isok(n) = leadingdigit(fibonacci(n))==5);
(lista(n)=my(a=vector(1+n), r=0); for (i=1, n, r+=isok(i); a[1+i]=r); a) \\ Winston de Greef, Mar 17 2023
KEYWORD
nonn,base
AUTHOR
Reinhard Zumkeller, Apr 11 2005
STATUS
approved