cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

A125122 First differences of A034888.

Original entry on oeis.org

0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0
Offset: 0

Views

Author

Keywords

Comments

This sequence is not periodic because log(3)/log(10) is an irrational number. - T. D. Noe, Jan 10 2007

Examples

			a(1)=0 because 3^(1+1)=9 (one digit) 3^1=3 (one digit) and the difference is 0
a(4)=1 because 3^(4+1)=243 (three digits) 3^(4)=81 (two digits) and the difference is 1
		

Programs

  • Maple
    P:=proc(n) local i,j,k,w,old; k:=3; for i from 1 by 1 to n do j:=k^i; w:=0; while j>0 do w:=w+1; j:=trunc(j/10); od; if i>1 then print(w-old); old:=w; else old:=w; fi; od; end: P(1000);
  • Mathematica
    Differences[IntegerLength[3^Range[0,110]]] (* Harvey P. Dale, Jan 28 2015 *)

Formula

a(n)=Number_of_digits{3^(n+1)}-Number_of digits{3^(n)} with n>=0.

A008564 Digits of powers of 3.

Original entry on oeis.org

1, 3, 9, 2, 7, 8, 1, 2, 4, 3, 7, 2, 9, 2, 1, 8, 7, 6, 5, 6, 1, 1, 9, 6, 8, 3, 5, 9, 0, 4, 9, 1, 7, 7, 1, 4, 7, 5, 3, 1, 4, 4, 1, 1, 5, 9, 4, 3, 2, 3, 4, 7, 8, 2, 9, 6, 9, 1, 4, 3, 4, 8, 9, 0, 7, 4, 3, 0, 4, 6, 7, 2, 1, 1, 2, 9, 1, 4, 0, 1, 6, 3, 3, 8, 7, 4, 2, 0, 4, 8, 9, 1, 1, 6, 2, 2, 6, 1, 4, 6, 7, 3
Offset: 0

Views

Author

Keywords

Comments

Irregular table with row length sequence A034888. - Jason Kimberley, Nov 26 2012
The constant whose decimal expansion is this sequence is irrational (Mahler, 1981). - Amiram Eldar, Mar 23 2025

Crossrefs

Programs

  • GAP
    Flat(List([0..30],n->ListOfDigits(3^n))); # Muniru A Asiru, Sep 29 2018
  • Mathematica
    Flatten[Table[IntegerDigits[3^n], {n, 0, 26}]] (* Alonso del Arte, Sep 29 2018 *)

A210434 Number of digits in 4^n.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 8, 8, 9, 10, 10, 11, 11, 12, 13, 13, 14, 14, 15, 16, 16, 17, 17, 18, 19, 19, 20, 20, 21, 22, 22, 23, 23, 24, 25, 25, 26, 26, 27, 28, 28, 29, 29, 30, 31, 31, 32, 32, 33, 34, 34, 35, 35, 36, 37, 37, 38, 38, 39, 40, 40, 41, 41
Offset: 0

Views

Author

Luc Comeau-Montasse, Mar 21 2012

Keywords

Comments

Since log10(4) = A114493 ~ 0.60205 (= twice log10(2) = 0.30102999566...), the first 98 terms are equal to floor(n*3/5)+1. - M. F. Hasler, Mar 31 2025

Examples

			a(4) = 3 because 4^4 = 256, which has 3 digits.
a(5) = 4 because 4^5 = 1024, which has 4 digits.
		

Crossrefs

Programs

  • Magma
    [#Intseq(4^n): n in [0..68]]; // Bruno Berselli, Mar 22 2012
    
  • Maple
    a:= n-> length(4^n): seq(a(n), n=0..100); # Alois P. Heinz, Mar 22 2012
  • Mathematica
    Table[Length[IntegerDigits[4^n]], {n, 0, 68}] (* Bruno Berselli, Mar 22 2012 *)
  • PARI
    apply( {A210434(n)=logint(4^n,10)+1}, [0..66]) \\ M. F. Hasler, Mar 31 2025
    
  • PARI
    a(n)=log(4)*n\log(10)+1 \\ correct up to n ~ 10^precision, with default precision = 38. - M. F. Hasler, Mar 31 2025
    
  • Python
    from math import log
    def A210434(n): return int(n*log(4,10))+1 if n<1e16 else "not enough precision" # M. F. Hasler, Mar 31 2025

Formula

a(n) = A055642(A000302(n)) = A055642(4^n) = floor(log_10(10*(4^n))). - Jonathan Vos Post, Mar 22 2012

A210435 Number of digits in 5^n.

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 9, 10, 10, 11, 12, 12, 13, 14, 14, 15, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 23, 24, 24, 25, 26, 26, 27, 28, 28, 29, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 44, 45, 45, 46, 47, 47
Offset: 0

Views

Author

Luc Comeau-Montasse, Mar 21 2012

Keywords

Examples

			a(4) = 3 because 5^4 = 625, which has 3 digits.
a(5) = 4 because 5^5 = 3125, which has 4 digits.
		

Crossrefs

Number of digits in b^n: A034887 (b=2), A034888 (b=3), A210434 (b=4), A210435 (b=5), A210436 (b=6), A210062 (b=7).

Programs

  • Magma
    [#Intseq(5^n): n in [0..67]]; // Bruno Berselli, Mar 22 2012
    
  • Maple
    a:= n-> length(5^n): seq(a(n), n=0..100); # Alois P. Heinz, Mar 22 2012
  • Mathematica
    Table[Length[IntegerDigits[5^n]], {n, 0, 67}] (* Bruno Berselli, Mar 22 2012 *)
    IntegerLength[5^Range[0,70]] (* Harvey P. Dale, Mar 26 2013 *)
  • PARI
    a(n) = #Str(5^n); \\ Michel Marcus, Oct 27 2015

Formula

a(n) = A055642(A000351(n)) = A055642(5^n) = floor(log_10(10*(5^n))). [Jonathan Vos Post, Mar 22 2012]
a(n) + A034887(n) = n+1. - Michel Marcus, Oct 27 2015

A210436 Number of digits in 6^n.

Original entry on oeis.org

1, 1, 2, 3, 4, 4, 5, 6, 7, 8, 8, 9, 10, 11, 11, 12, 13, 14, 15, 15, 16, 17, 18, 18, 19, 20, 21, 22, 22, 23, 24, 25, 25, 26, 27, 28, 29, 29, 30, 31, 32, 32, 33, 34, 35, 36, 36, 37, 38, 39, 39, 40, 41, 42, 43, 43, 44, 45, 46, 46, 47, 48, 49, 50, 50, 51, 52, 53
Offset: 0

Views

Author

Luc Comeau-Montasse, Mar 21 2012

Keywords

Examples

			a(4) = 4 because 6^4 = 1296, which has 4 digits.
a(5) = 4 because 6^5 = 7776, which has 4 digits.
		

Crossrefs

Programs

  • Magma
    [#Intseq(6^n): n in [0..67]]; // Bruno Berselli, Mar 22 2012
  • Maple
    a:= n-> length(6^n): seq (a(n), n=0..100); # Alois P. Heinz, Mar 22 2012
  • Mathematica
    Table[Length[IntegerDigits[6^n]], {n, 0, 99}] (* Alonso del Arte, Mar 22 2012 *)

Formula

a(n) = A055642(A000400(n)) = A055642(6^n) = floor(log_10(10*(6^n))). - Jonathan Vos Post, Mar 23 2012

A210062 Number of digits in 7^n.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 6, 6, 7, 8, 9, 10, 11, 11, 12, 13, 14, 15, 16, 17, 17, 18, 19, 20, 21, 22, 22, 23, 24, 25, 26, 27, 28, 28, 29, 30, 31, 32, 33, 33, 34, 35, 36, 37, 38, 39, 39, 40, 41, 42, 43, 44, 44, 45, 46, 47, 48, 49, 50, 50, 51, 52, 53, 54, 55, 55, 56, 57
Offset: 0

Views

Author

Luc Comeau-Montasse, Mar 16 2012

Keywords

Crossrefs

Number of digits in b^n: A034887 (b=2), A034888 (b=3), A210434 (b=4), A210435 (b=5), A210436 (b=6), this sequence (b=7).

Programs

  • Magma
    [#Intseq(7^n): n in [0..67]]; // Bruno Berselli, Mar 22 2012
  • Mathematica
    Table[Length[IntegerDigits[7^n]], {n, 0, 100}] (* T. D. Noe, Mar 20 2012 *)

Formula

a(n) = A055642(A000420(n)) = A055642(7^n) = floor(log_10(10*(7^n))). [Jonathan Vos Post, Mar 23 2012]

A055255 Number of even digits in 3^n.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 1, 2, 2, 2, 2, 1, 2, 2, 4, 4, 5, 4, 6, 6, 7, 6, 4, 5, 8, 9, 8, 6, 8, 8, 8, 6, 9, 6, 8, 6, 7, 8, 7, 8, 10, 9, 7, 9, 11, 11, 10, 11, 12, 10, 10, 8, 16, 13, 12, 11, 9, 16, 18, 14, 14, 17, 19, 16, 19, 13, 17, 14, 14, 14, 14, 14, 17, 16, 15, 16, 16, 15, 25, 23, 22
Offset: 0

Views

Author

Asher Auel, May 05 2000

Keywords

Crossrefs

Programs

  • PARI
    a(n) = #select(x->((x%2) == 0), digits(3^n)); \\ Michel Marcus, Nov 22 2019

A055256 Number of odd digits in 3^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 5, 4, 5, 3, 4, 3, 5, 3, 4, 3, 5, 7, 6, 4, 3, 5, 7, 6, 6, 7, 9, 7, 10, 9, 11, 11, 10, 12, 11, 10, 11, 14, 12, 10, 11, 12, 12, 11, 14, 14, 17, 9, 13, 14, 16, 18, 12, 10, 15, 15, 13, 11, 15, 12, 19, 15, 18, 19, 19, 20, 20, 18, 19, 21, 20, 21, 22
Offset: 0

Views

Author

Asher Auel, May 05 2000

Keywords

Crossrefs

Programs

  • PARI
    a(n) = #select(x->((x%2) == 1), digits(3^n)); \\ Michel Marcus, Nov 22 2019

A348553 Number of digits in 11^n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75
Offset: 0

Views

Author

Seiichi Manyama, Oct 22 2021

Keywords

Examples

			a(24) = 25 because 11^24 = 9849732675807611094711841, which has 25 digits.
a(25) = 27 because 11^25 = 108347059433883722041830251, which has 27 digits.
		

Crossrefs

Number of digits in b^n: A034887 (b=2), A034888 (b=3), A210434 (b=4), A210435 (b=5), A210436 (b=6), A210062 (b=7), this sequence (b=11).

Programs

  • Mathematica
    a[n_] := IntegerLength[11^n]; Array[a, 100, 0] (* Amiram Eldar, Oct 22 2021 *)
  • PARI
    a(n) = #Str(11^n);
    
  • Python
    def a(n): return len(str(11**n))
    print([a(n) for n in range(98)]) # Michael S. Branicky, Oct 22 2021

Formula

a(n) = A055642(A001020(n)) = A055642(11^n).

A385950 Number of digits in 3^(n!).

Original entry on oeis.org

1, 1, 1, 3, 12, 58, 344, 2405, 19238, 173138, 1731378, 19045154, 228541845, 2971043978, 41594615682, 623919235225, 9982707763599, 169706031981174, 3054708575661119, 58039462937561246, 1160789258751224920, 24376574433775723304, 536284637543065912676
Offset: 0

Views

Author

Vincenzo Librandi, Jul 27 2025

Keywords

Crossrefs

Subsequence of A034888.

Programs

  • Mathematica
    Array[Floor[#! Log10@3+1]&,22]

Formula

a(n) = A055642(A100731(n)).
Showing 1-10 of 10 results.