A210062 -id:A210062 - cp's OEIS frontend

A008568 Digits of powers of 7.

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

Offset: 0

N. J. A. Sloane

Irregular table with row length sequence A210062. - Jason Kimberley, Nov 26 2012

The constant whose decimal expansion is this sequence is irrational (Mahler, 1981). - Amiram Eldar, Mar 23 2025

			Triangle begins:
  1;
  7;
  4, 9;
  3, 4, 3;
  2, 4, 0, 1;
  1, 6, 8, 0, 7;
  1, 1, 7, 6, 4, 9;
  8, 2, 3, 5, 4, 3;
  ...

Seiichi Manyama, Rows n = 0..100, flattened
Kurt Mahler, On some irrational decimal fractions, Journal of Number Theory, Vol. 13, No. 2 (1981), pp. 268-269.

Cf. A000420, A210062.

Cf. A000455, A008564, A008565, A008566, A008567, A008569, A008570, A008571, A008572, A008573, A010054.

Flatten[IntegerDigits[7^Range[0,20]]]  (* Harvey P. Dale, Mar 03 2011 *)

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

Luc Comeau-Montasse, Mar 21 2012

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

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

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A000302, A034887, A034888, A210062, A210435, A210436.

[#Intseq(4^n): n in [0..68]]; // Bruno Berselli, Mar 22 2012

a:= n-> length(4^n): seq(a(n), n=0..100); # Alois P. Heinz, Mar 22 2012

Table[Length[IntegerDigits[4^n]], {n, 0, 68}] (* Bruno Berselli, Mar 22 2012 *)

apply( {A210434(n)=logint(4^n,10)+1}, [0..66]) \\ M. F. Hasler, Mar 31 2025

a(n)=log(4)*n\log(10)+1 \\ correct up to n ~ 10^precision, with default precision = 38. - M. F. Hasler, Mar 31 2025

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

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

Luc Comeau-Montasse, Mar 21 2012

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

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Ana Rechtman, Octobre 2015, 3e Défi, Images des Mathématiques, CNRS, 2015 (in French).

Cf. A000351, A055642.

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

[#Intseq(5^n): n in [0..67]]; // Bruno Berselli, Mar 22 2012

a:= n-> length(5^n): seq(a(n), n=0..100); # Alois P. Heinz, Mar 22 2012

Table[Length[IntegerDigits[5^n]], {n, 0, 67}] (* Bruno Berselli, Mar 22 2012 *)
IntegerLength[5^Range[0,70]] (* Harvey P. Dale, Mar 26 2013 *)

a(n) = #Str(5^n); \\ Michel Marcus, Oct 27 2015

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

Luc Comeau-Montasse, Mar 21 2012

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

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A000400, A034887, A034888, A210062, A210434, A210435.

[#Intseq(6^n): n in [0..67]]; // Bruno Berselli, Mar 22 2012

a:= n-> length(6^n): seq (a(n), n=0..100); # Alois P. Heinz, Mar 22 2012

Table[Length[IntegerDigits[6^n]], {n, 0, 99}] (* Alonso del Arte, Mar 22 2012 *)

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

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

Seiichi Manyama, Oct 22 2021

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

Seiichi Manyama, Table of n, a(n) for n = 0..10000

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).

Cf. A001020, A055642.

a[n_] := IntegerLength[11^n]; Array[a, 100, 0] (* Amiram Eldar, Oct 22 2021 *)

PARI
```
a(n) = #Str(11^n);
```

def a(n): return len(str(11**n))
print([a(n) for n in range(98)]) # Michael S. Branicky, Oct 22 2021

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

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A008568 Digits of powers of 7.

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A210434 Number of digits in 4^n.

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A210435 Number of digits in 5^n.

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A210436 Number of digits in 6^n.

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A348553 Number of digits in 11^n.

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