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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). 1

%I #17 Nov 30 2022 11:20:32

%S 11,303,123123,6170617,246902469,1929001929,12345671234567,

%T 617283906172839,123456789123456789,123456789101234567891,

%U 12345678910111234567891011,15432098637639015432098637639,1234567891011121312345678910111213,6172839455055606570617283945505560657

%N 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).

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

%C From _Michael S. Branicky_, Nov 30 2022: (Start)

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

%C 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 A078527(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 A078527(n) beyond those allowed by the unreduced numerator of a(n). (End)

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

%o (Python)

%o from itertools import count, islice

%o def agen(): # generator of terms

%o k, den, pow = 0, 1, 0

%o for n in count(1):

%o sn = str(n)

%o k = k*10**len(sn) + n

%o den *= 10**len(sn)

%o pow += len(sn)

%o nr, c2, c5 = k*(den+1), pow, pow

%o while nr%2 == 0 and c2 > 0: nr //= 2; c2 -= 1

%o while nr%5 == 0 and c5 > 0: nr //= 5; c5 -= 1

%o yield nr

%o print(list(islice(agen(), 19))) # _Michael S. Branicky_, Nov 30 2022

%Y Cf. A078257, A078258.

%K nonn,base

%O 1,1

%A _Jaroslav Krizek_, Feb 05 2010

%E a(9) and beyond from _Michael S. Branicky_, Nov 30 2022

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Last modified February 21 14:42 EST 2024. Contains 370236 sequences. (Running on oeis4.)