A078257 a(n) = denominator(N) where N = 0.123...n (concatenation of 1 to n after decimal point).
10, 25, 1000, 5000, 20000, 15625, 10000000, 50000000, 1000000000, 10000000000, 10000000000000, 125000000000000, 100000000000000000, 5000000000000000000, 200000000000000000000, 25000000000000000000000, 10000000000000000000000000, 500000000000000000000000000, 100000000000000000000000000000
Offset: 1
Examples
a(1) = 10 as 10*0.1 = 1, a(2) = 25 as 25*0.12 = 3.
Links
- Michael S. Branicky, Table of n, a(n) for n = 1..369
Programs
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PARI
a(n) = {my(s = ""); for (k=1, n, s = concat(s, Str(k))); denominator(eval(s)/10^(#s));} \\ Michel Marcus, Jan 15 2019 -
Python
from itertools import count, islice def agen(): # generator of terms num, den, pow = 0, 1, 0 for n in count(1): sn = str(n) num = num*10**len(sn) + n den *= 10**len(sn) pow += len(sn) nr, dr, c2, c5 = num, den, pow, pow while nr%2 == 0 and c2 > 0: nr //= 2; dr //= 2; c2 -= 1 while nr%5 == 0 and c5 > 0: nr //= 5; dr //= 5; c5 -= 1 yield dr print(list(islice(agen(), 19))) # Michael S. Branicky, Nov 30 2022
Formula
a(n) = denominator(Sum_{k=1..n} k/10^A058183(k)). - Stefano Spezia, Nov 30 2022
Extensions
More terms from Sascha Kurz, Jan 04 2003
More terms from Michel Marcus, Jan 15 2019
Comments