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Search: a105516 -id:a105516
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Number of times 1 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.
+10
11
0, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 7, 7, 7, 7, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 12, 13, 13, 13, 13, 14, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 18, 18, 18, 18, 19, 20, 20, 20, 20, 21, 22, 22, 22, 22, 23, 23, 23, 23, 23, 24
OFFSET
0,3
LINKS
FORMULA
a(n) = #{k: A008963(k) = 1 and 0<=k<=n};
a(A105501(n)) = a(A105501(n) - 1) + 1;
n = a(n) + A105512(n) + A105513(n) + A105514(n) + A105515(n) + A105516(n) + A105517(n) + A105518(n) + A105519(n).
a(n) ~ log_10(2) * n. - Amiram Eldar, Jan 12 2023
MATHEMATICA
Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]] == 1, 1, 0], {n, 0, 100}]] (* Amiram Eldar, Jan 12 2023 *)
PROG
(PARI)
(leadingdigit(n, b=10) = n \ 10^logint(n, b));
(isok(n) = leadingdigit(fibonacci(n))==1);
(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
Number of times 9 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.
+10
11
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 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
OFFSET
0,36
LINKS
FORMULA
a(n) = #{k: A008963(k) = 9 and 0<=k<=n};
a(A105509(n)) = a(A105509(n) - 1) + 1;
n = A105511(n) + A105512(n) + A105513(n) + A105514(n) + A105515(n) + A105516(n) + A105517(n) + A105518(n) + a(n).
a(n) ~ (1 - log_10(9)) * n. - Amiram Eldar, Jan 12 2023
MATHEMATICA
Table[If[First[IntegerDigits[Fibonacci[n]]]==9, 1, 0], {n, 0, 110}]// Accumulate (* Harvey P. Dale, Nov 27 2018 *)
PROG
(PARI)
(leadingdigit(n, b=10) = n \ 10^logint(n, b));
(isok(n) = leadingdigit(fibonacci(n))==9);
(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 18 2023
KEYWORD
nonn,base
AUTHOR
Reinhard Zumkeller, Apr 11 2005
STATUS
approved
Number of times 2 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.
+10
10
0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16
OFFSET
0,9
LINKS
FORMULA
a(n) = #{k: A008963(k) = 2 and 0<=k<=n};
a(A105502(n)) = a(A105502(n) - 1) + 1;
n = A105511(n) + a(n) + A105513(n) + A105514(n) + A105515(n) + A105516(n) + A105517(n) + A105518(n) + A105519(n).
a(n) ~ log_10(3/2) * n. - Amiram Eldar, Jan 12 2023
MATHEMATICA
Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]] == 2, 1, 0], {n, 0, 100}]] (* Amiram Eldar, Jan 12 2023 *)
PROG
(PARI)
(leadingdigit(n, b=10) = n \ 10^logint(n, b));
(isok(n) = leadingdigit(fibonacci(n))==2);
(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
Number of times 3 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.
+10
10
0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11
OFFSET
0,10
LINKS
FORMULA
a(n) = #{k: A008963(k) = 3 and 0<=k<=n};
a(A105503(n)) = a(A105503(n) - 1) + 1;
n = A105511(n) + A105512(n) + a(n) + A105514(n) + A105515(n) + A105516(n) + A105517(n) + A105518(n) + A105519(n).
a(n) ~ log_10(4/3) * n. - Amiram Eldar, Jan 12 2023
MATHEMATICA
Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]] == 3, 1, 0], {n, 0, 100}]] (* Amiram Eldar, Jan 12 2023 *)
PROG
(PARI)
(leadingdigit(n, b=10) = n \ 10^logint(n, b));
(isok(n) = leadingdigit(fibonacci(n))==3);
(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
Number of times 4 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.
+10
10
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9
OFFSET
0,25
LINKS
FORMULA
a(n) = #{k: A008963(k) = 4 and 0<=k<=n};
a(A105504(n)) = a(A105504(n) - 1) + 1;
n = A105511(n) + A105512(n) + A105513(n) + a(n) + A105515(n) + A105516(n) + A105517(n) + A105518(n) + A105519(n).
a(n) ~ log_10(5/4) * n. - Amiram Eldar, Jan 12 2023
MATHEMATICA
Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]] == 4, 1, 0], {n, 0, 100}]] (* Amiram Eldar, Jan 12 2023 *)
PROG
(PARI)
(leadingdigit(n, b=10) = n \ 10^logint(n, b));
(isok(n) = leadingdigit(fibonacci(n))==4);
(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
Number of times 5 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.
+10
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
Number of times 7 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.
+10
10
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
OFFSET
0,45
LINKS
FORMULA
a(n) = #{k: A008963(k) = 7 and 0<=k<=n};
a(A105507(n)) = a(A105507(n) - 1) + 1;
n = A105511(n) + A105512(n) + A105513(n) + A105514(n) + A105515(n) + A105516(n) + a(n) + A105518(n) + A105519(n).
a(n) ~ log_10(8/7) * n. - Amiram Eldar, Jan 12 2023
MATHEMATICA
Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]]==7, 1, 0], {n, 0, 120}]] (* Harvey P. Dale, Apr 29 2018 *)
PROG
(PARI)
(leadingdigit(n, b=10) = n \ 10^logint(n, b));
(isok(n) = leadingdigit(fibonacci(n))==7);
(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
Number of times 8 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.
+10
10
0, 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, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 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
OFFSET
0,12
LINKS
FORMULA
a(n) = #{k: A008963(k) = 8 and 0<=k<=n};
a(A105508(n)) = a(A105508(n) - 1) + 1;
n = A105511(n) + A105512(n) + A105513(n) + A105514(n) + A105515(n) + A105516(n) + A105517(n) + a(n) + A105519(n).
a(n) ~ log_10(9/8) * n. - Amiram Eldar, Jan 12 2023
MATHEMATICA
Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]] == 8, 1, 0], {n, 0, 100}]] (* Amiram Eldar, Jan 12 2023 *)
PROG
(PARI)
(leadingdigit(n, b=10) = n \ 10^logint(n, b));
(isok(n) = leadingdigit(fibonacci(n))==8);
(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
Numbers m such that 6 is the leading digit of the n-th Fibonacci number in decimal representation.
+10
9
15, 20, 39, 63, 82, 87, 106, 130, 149, 154, 173, 197, 216, 221, 240, 259, 264, 283, 288, 307, 326, 331, 350, 355, 374, 393, 398, 417, 422, 441, 460, 465, 484, 508, 527, 532, 551, 575, 594, 599, 618, 642, 661, 666, 685, 709, 728, 733, 752, 771, 776, 795, 800
OFFSET
1,1
COMMENTS
A008963(a(n)) = 6; A105516(a(n)) = A105516(a(n) - 1) + 1.
FORMULA
a(n) ~ kn by the equidistribution theorem, where k = log(10)/(log(7) - log(6)) = 14.9372.... - Charles R Greathouse IV, Oct 07 2016
EXAMPLE
a(10)=154: A008963(154) = A000030(A000045(154)) =
A000030(68330027629092351019822533679447) = 6.
PROG
(PARI) is(n)=digits(fibonacci(n))[1]==6 \\ Charles R Greathouse IV, Oct 07 2016
KEYWORD
nonn,base
AUTHOR
Reinhard Zumkeller, Apr 11 2005
STATUS
approved

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