A053062 -id:A053062 - cp's OEIS frontend

A061088 a(n) = A053062(n) / n.

1, 12, 123, 120304, 102030405, 10203040506, 1020304050607, 102030405060708, 10203040506070809, 10203040506070809010, 10203040506070809010011, 102030405060708009010011012, 1020304050607008009010011012013, 1020304050607008009010011012013014

Offset: 1

Amarnath Murthy, Apr 19 2001

			a(5) = 102030405 = 510152025/5, where 510152025 is obtained by concatenating 5,10,15,20 and finally 25 (which is 5^2).

Andrew Howroyd, Table of n, a(n) for n = 1..100

Cf. A053062.

a(n)={fromdigits(concat(vector(n, k, digits(k*n))))/n} \\ Andrew Howroyd, Nov 09 2019

Offset corrected and terms a(7) and beyond from Andrew Howroyd, Nov 09 2019

A326120 a(n) is the concatenation of n^1, n^2, ..., n^n.

Original entry on oeis.org

1, 24, 3927, 41664256, 5251256253125, 6362161296777646656, 749343240116807117649823543, 864512409632768262144209715216777216, 981729656159049531441478296943046721387420489, 10100100010000100000100000010000000100000000100000000010000000000

Offset: 1

Trevor Mulindi, Sep 10 2019

a(100) has 10200 digits.

a(n) is the concatenation of the n terms of the n-th row of A075363 triangle. - Michel Marcus, Sep 15 2019

Cf. A007908, A019521, A019522, A053062, A075363.

a:= n-> parse(cat(n^j$j=1..n)):
seq(a(n), n=1..14);  # Alois P. Heinz, Jan 24 2021

a[n_] := FromDigits@ Flatten@ IntegerDigits[n^Range[n]]; Array[a, 10] (* Giovanni Resta, Sep 12 2019 *)

def a(n): return int("".join(str(n**i) for i in range(1, n+1)))
print([a(n) for n in range(1, 11)]) # Michael S. Branicky, Jan 24 2021

More terms from Giovanni Resta, Sep 12 2019

cp's OEIS Frontend

A061088 a(n) = A053062(n) / n.

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A326120 a(n) is the concatenation of n^1, n^2, ..., n^n.

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