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

Author

Amarnath Murthy, Nov 24 2002

Keywords

• base
• nonn
• frac

Comments

Conjecture: sequence is not equal to the sequence of denominators presented in A172495 and A172506. - Jaroslav Krizek, Feb 05 2010

The conjecture is false for both other sequences; see A172495 and A172506 for proofs. - Michael S. Branicky, Nov 30 2022

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

Crossrefs

Cf. A058183, A078258 (numerators), A172495, A172506.

Programs

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