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A247890
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Number of digits in (R_n)^n.
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1
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1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, 241, 273, 307, 343, 381, 421, 464, 508, 554, 602, 652, 704, 758, 814, 872, 932, 994, 1058, 1124, 1192, 1262, 1334, 1408, 1484, 1562, 1642, 1724, 1808, 1895, 1983, 2073, 2165, 2259, 2355, 2453, 2553, 2655, 2759, 2865
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OFFSET
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1,2
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COMMENTS
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R_n is the n-th repunit (i.e., R_n = 11...111 with n 1's).
The number of digits of m is floor(log(m)/log(10)) + 1 for m > 0.
R_n = (10^n - 1) / 9 = (10 - 10^(1-n))/9 * 10^(n-1). Its number of digits is floor(log((10 - 10^(1-n))/9)) / log(10)) + n * (n - 1) + 1. (End)
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LINKS
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FORMULA
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a(n) = n^2 - n + 1 = A002061(n), for 1 <= n <= 21.
a(n) = n^2 - n + 2 = A014206(n-1), for 22 <= n <= 43.
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MATHEMATICA
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Table[IntegerLength[((10^n - 1)/9)^n], {n, 54}] (* or *)
Table[IntegerLength[FromDigits[Table[1, {n}]]^n], {n, 54}] (* Michael De Vlieger, Jun 27 2016 *)
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PROG
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(PARI) vector(100, n, #Str(((10^n-1)/9)^n))
(PARI) a(n) = logint(((10 - 10^(1-n))/9)^n\1, 10)+n^2-n+1 \\ David A. Corneth, Jun 27 2016
(Magma) [#Intseq(Floor((10^n-1)/9)^n): n in [1..50]]; // Marius A. Burtea, May 20 2019
(Python)
def a(n): return len(str(int("1"*n)**n))
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CROSSREFS
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KEYWORD
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nonn,easy,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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