A058183 Number of digits in concatenation of first n positive integers.
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125
Offset: 1
Examples
a(12) = 15 since 123456789101112 has 15 digits.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..10000
- GeeksforGeeks, Count total number of digits from 1 to n
- Eric Weisstein's World of Mathematics, Smarandache Number
Programs
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Maple
a:= proc(n) a(n):= `if`(n=0, 0, a(n-1) +length(n)) end: seq(a(n), n=1..100); # Alois P. Heinz, Nov 26 2013 a := proc(n) local d; d:=floor(log10(n))+1; (n+1)*d - (10^d-1)/9; end; # N. J. A. Sloane, Feb 20 2020
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Mathematica
Length/@ Flatten/@ IntegerDigits/@ Flatten/@ Rest[FoldList[List, {}, Range[70]]] (* Eric W. Weisstein, Nov 04 2015 *) Table[With[{d = IntegerLength[n]}, (n+1) d - (10^d -1)/9], {n, 70}] (* Eric W. Weisstein, Nov 06 2015 *) IntegerLength/@ FoldList[#2 + #1 10^IntegerLength[#2] &, Range[70]] (* Eric W. Weisstein, Nov 06 2015 *) Accumulate[ IntegerLength@ # & /@ Range @ 70] (* Robert G. Wilson v, Jul 31 2018 *) -
PARI
a(n)=my(t=log(10*n+.5)\log(10));n*t+t-10^t\9 \\ Charles R Greathouse IV, Sep 19 2012
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PARI
a(n) = sum(k=1, n, #digits(k)); \\ Michel Marcus, Jan 01 2017
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Python
def A058183(n): return (n+1)*(s:=len(str(n))) - (10**s-1)//9 # Chai Wah Wu, May 02 2023
Formula
a(n) = (n+1)*floor(log_10(10*n)) - (10^floor(log_10(10*n))-1)/(10-1).
a(n) = a(n-1) + floor(log_10(10*n)).
a(n) ~ n log_10 n + O(n). In particular lim inf (n log_10 n - a(n))/n = (1+log(10/9)+log(log(10)))/log(10) and the corresponding lim sup is 10/9. - Charles R Greathouse IV, Sep 19 2012
G.f.: (1-x)^(-2)*Sum_{j>=0} x^(10^j). - Robert Israel, Nov 04 2015
a(n) = b(n)*(n + 1) - (10^b(n) - 19)/9 - 2, where b(n) = A055642(n). - Lorenzo Sauras Altuzarra, May 09 2020
Comments