A172495 -id:A172495 - cp's OEIS frontend

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

Amarnath Murthy, Nov 24 2002

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

			a(1) = 10 as 10*0.1 = 1, a(2) = 25 as 25*0.12 = 3.

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

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

a(n) = {my(s = ""); for (k=1, n, s = concat(s, Str(k))); denominator(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, 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

a(n) = denominator(Sum_{k=1..n} k/10^A058183(k)). - Stefano Spezia, Nov 30 2022

More terms from Sascha Kurz, Jan 04 2003

More terms from Michel Marcus, Jan 15 2019

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

A078257 a(n) = denominator(N) where N = 0.123...n (concatenation of 1 to n after decimal point).

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