A172506 a(n) = numerator of fraction a/b, where gcd(a, b) = 1, whose decimal representation has the form (1)(2)(3)...(n-1)(n).(1)(2)(3)...(n-1)(n).

11, 303, 123123, 6170617, 246902469, 1929001929, 12345671234567, 617283906172839, 123456789123456789, 123456789101234567891, 12345678910111234567891011, 15432098637639015432098637639, 1234567891011121312345678910111213, 6172839455055606570617283945505560657

Offset: 1

Author

Jaroslav Krizek, Feb 05 2010

Keywords

• nonn
• base

Comments

Sequence of denominators: 10, 25, 1000, 5000, 20000, 15625, 10000000, 50000000, ... Conjecture: this sequence is not equal to the sequence A078257.

From Michael S. Branicky, Nov 30 2022: (Start)

The conjecture is false: the denominators here are the same as in A078257.

Proof. Let Cn denote the concatenation (1)(2)(3)...(n-1)(n) and en its number of decimal digits. The unreduced numerator and denominator for a(n) are Cn and 10^en, respectively. For A078257(n), they are Cn*(10^en + 1) and 10^en. Since (10^en + 1) is never divisible by 2 or 5, no reductions can be made in the denominator of A078257(n) beyond those allowed by the unreduced numerator of a(n). (End)

Examples

			a(6) = 1929001929; 1929001929/15625 = 123456.123456.
		

Crossrefs

Cf. A078257, A078258.

Programs

Python

from itertools import count, islice
def agen(): # generator of terms
    k, den, pow = 0, 1, 0
    for n in count(1):
        sn = str(n)
        k = k*10**len(sn) + n
        den *= 10**len(sn)
        pow += len(sn)
        nr, c2, c5 = k*(den+1), pow, pow
        while nr%2 == 0 and c2 > 0: nr //= 2; c2 -= 1
        while nr%5 == 0 and c5 > 0: nr //= 5; c5 -= 1
        yield nr
print(list(islice(agen(), 19))) # Michael S. Branicky, Nov 30 2022

Extensions

a(9) and beyond from Michael S. Branicky, Nov 30 2022