A078257 -id:A078257 - cp's OEIS frontend

A078258 a(n) = numerator(N), where N = 0.123...n (concatenation of 1 to n after decimal point).

1, 3, 123, 617, 2469, 1929, 1234567, 6172839, 123456789, 1234567891, 1234567891011, 15432098637639, 12345678910111213, 617283945505560657, 24691357820222426283, 3086419727527803285379, 1234567891011121314151617, 61728394550556065707580859, 12345678910111213141516171819

Offset: 1

Amarnath Murthy, Nov 24 2002

			a(4) = 617 = smallest integer multiple of 0.1234, 617 = 5000*(0.1234).

Michael S. Branicky, Table of n, a(n) for n = 1..369

Cf. A058183, A078257 (denominators).

a(n) = {my(s = ""); for (k=1, n, s = concat(s, Str(k))); numerator(eval(s)/10^(#s));} \\ Michel Marcus, Jan 15 2019

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, c2, c5 = num, 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

a(n) = numerator(Sum_{k=1..n} k/10^A058183(k)). - Stefano Spezia, Nov 30 2022. (This is simply a restatement of the definition.)

More terms from Sascha Kurz, Jan 04 2003

More terms from Michel Marcus, Jan 15 2019

A172495 a(n) = numerator of fraction whose decimal representation is (n).(1)(2)(3)...(n-1)(n).

Original entry on oeis.org

11, 53, 3123, 20617, 102469, 95679, 71234567, 406172839, 9123456789, 101234567891, 111234567891011, 1515432098637639, 1312345678910111213, 70617283945505560657, 3024691357820222426283, 403086419727527803285379, 171234567891011121314151617

Offset: 1

Jaroslav Krizek, Feb 05 2010

Sequence of denominators: 10, 25, 1000, 5000, 20000, 15625, 10000000, 50000000, 1000000000, 10000000000, ... Conjecture: this sequence is not equal to the sequence A078257.
I conjecture that it is the same as A078257. - Franklin T. Adams-Watters, Mar 29 2014
This sequence of denominators is the same as A078257 up to at least n=10000. - Jon E. Schoenfield, Mar 29 2014
From Michael S. Branicky, Nov 30 2022: (Start)
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 numerators and denominators of A078257(n) are Cn and 10^En; for a(n), they are (n*10^En + Cn) and 10^En. To find A078257(n), we continue to divide the unreduced numerator by 2 and 5 as long as that is possible. For a(n) to be smaller, we would have to "get past" all the decimal digits in Cn and divide n at least once. But if we could do that, it would be a contradiction to earlier terms of A078257. (End)

			a(6) = 95679; 95679/15625 = 6.123456.

Harvey P. Dale, Table of n, a(n) for n = 1..368

Numerator[#]GCD[Numerator[#],Denominator[#]]&/@Table[FromDigits[Join[{n},Flatten[ IntegerDigits/@Range[n]]]]/10^n,{n,20}] (* Harvey P. Dale, Dec 16 2019 *)

a(11)-a(17) from Jon E. Schoenfield, Dec 19 2017

A078260 a(n) = denominator(N), where N = 0.246...(2n) is the concatenation of the first n even numbers after decimal point.

Original entry on oeis.org

5, 25, 500, 2500, 100000, 25000000, 5000000000, 125000000000, 50000000000000, 500000000000000, 500000000000000000, 6250000000000000000, 5000000000000000000000, 250000000000000000000000, 10000000000000000000000000, 1250000000000000000000000000, 500000000000000000000000000000

Offset: 1

Amarnath Murthy, Nov 24 2002

Charlie Neder, Table of n, a(n) for n = 1..99

Cf. A078257 (similar, with concatenation of 1 to n), A078261 (numerators).

a:= n-> (t-> denom(t/10^length(t)))(parse(cat(2*i$i=1..n))):
seq(a(n), n=1..17);  # Alois P. Heinz, Jun 25 2025

a(n) = {my(s = ""); for (k=1, n, s = concat(s, Str(2*k))); denominator(eval(s)/10^(#s));} \\ Michel Marcus, Jan 15 2019

a(n) = c = concat(vector(n, i, Str(2*i))); denominator(eval(c) / 10^#Str(c)) \\ David A. Corneth, Jan 15 2019

More terms from Sascha Kurz, Jan 04 2003

a(5) and a(10) corrected, a(15)-a(17) from Charlie Neder, Jan 14 2019

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).

Original entry on oeis.org

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

Offset: 1

Jaroslav Krizek, Feb 05 2010

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)

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

Cf. A078257, A078258.

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

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

A357915 Concatenation of the decimal digits of {n, 1..n}.

Original entry on oeis.org

11, 212, 3123, 41234, 512345, 6123456, 71234567, 812345678, 9123456789, 1012345678910, 111234567891011, 12123456789101112, 1312345678910111213, 141234567891011121314, 15123456789101112131415, 1612345678910111213141516

Offset: 1

Mikk Heidemaa, Jan 18 2023

Concatenation of the consecutive integers 1..n, with n prepended (n > 0).
The terms a(1), a(7), a(31), and a(337) are primes (of the form n1...n).
a(3643) is a 13469-digit probable prime (the number of digits is also a prime).
These indices 7, 31, 337, 3643 are themselves primes of the form 6m+1.
For the known terms a(n) which are primes and for a(3643), a(n) == 2 (mod 3).
There is no other prime term for n < 15000 (and no prime term with prime index n < 25000).

			a(2) = 212 since it is the concatenation of the consecutive positive integers <= 2, with 2 prepended.

Winston de Greef, Table of n, a(n) for n = 1..367
G. L. Honaker, Jr. and Chris K. Caldwell, Prime Curios! 71234567

Cf. A007908, A078257, A172495.

aUpTo[n_] := Table[ FromDigits @ Flatten @ IntegerDigits @ {i, Range @ i}, {i,n}]; aUpTo[999]

a(n) = my(s=Str(n)); for(k=1, n, s=Str(s, k)); eval(s); \\ Michel Marcus, Jan 20 2023

def a(n): return int(str(n)+"".join(map(str, range(1, n+1))))
print([a(n) for n in range(1, 17)]) # Michael S. Branicky, Jan 20 2023

a(n) = concat(n, A007908(n)).

cp's OEIS Frontend

A078258 a(n) = numerator(N), where N = 0.123...n (concatenation of 1 to n after decimal point).

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A172495 a(n) = numerator of fraction whose decimal representation is (n).(1)(2)(3)...(n-1)(n).

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A078260 a(n) = denominator(N), where N = 0.246...(2n) is the concatenation of the first n even numbers after decimal point.

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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).

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Python

Extensions

A357915 Concatenation of the decimal digits of {n, 1..n}.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula