A048861 -id:A048861 - cp's OEIS frontend

A385302 Least common multiple of {n^n-1, n^(n-1)-1, ..., n-1}.

1, 0, 3, 104, 5355, 15107664, 2684295425, 2261529015616800, 97901171532649325295, 671549787473885210310580160, 113480471243172592617657936025689, 67423693602852027222491606156048516615143200, 52627558956534012662100374980910169826998422190695

Offset: 0

Avery Diep, Jun 24 2025

nonn

For prime powers q (i.e. A000961), A351657(q) divides q*a(q). Follows from Niven's Fermat's theorem for matrices when considering Fibonacci q-step matrices over GF(q).

			a(1) = LCM{1-1} = 0, a(2) = LCM{2^2-1,2-1} = 3, a(3) = LCM{3^3-1,3^2-1,3-1} = 104.

Alois P. Heinz, Table of n, a(n) for n = 0..44
Ivan Niven, Fermat’s theorem for matrices, Duke Math. J., Vol. 15, No. 3 (1948), pp. 823-826.

Cf. A000217, A000961, A003418, A048861, A156291, A351657.

a:= n-> ilcm(n^j-1$j=1..n):
seq(a(n), n=0..12);  # Alois P. Heinz, Jun 25 2025

a[n_] := LCM @@ (n^Range[n] - 1); a[0] = 1; Array[a, 14, 0] (* Amiram Eldar, Jun 25 2025 *)

a(n) = lcm(vector(n, k, n^k-1)); \\ Michel Marcus, Jun 25 2025

a(n) = LCM{n^n-1, n^(n-1)-1, ..., n-1}.

Conjecture: a(n) = Product_{k=1..n} Cyclotomic[k, n], i.e., a(n) = A156291(A000217(n)-1).

A109675 Numbers k such that the sum of the digits of (k^k - 1) is divisible by k.

Original entry on oeis.org

1, 4, 5, 10, 25, 50, 100, 446, 1000, 9775, 10000, 100000, 995138, 996544, 998866, 1000000

Offset: 1

Ryan Propper, Aug 06 2005

k = 10^m is a term of the sequence for all m >= 0. Proof: Let k = 10^m for some nonnegative integer m. Then k^k - 1 has m*10^m 9's and no other digits, so its digits sum to 9*m*10^m = 9*m*k, a multiple of k.

			The digits of 9775^9775 - 1 sum to 175950 and 175950 is divisible by 9775, so 9775 is in the sequence.

Cf. A007953, A048861.

sumdigs:= n -> convert(convert(n,base,10),`+`);
select(n -> sumdigs(n^n-1) mod n = 0, [$1..10^5]); # Robert Israel, Dec 03 2014

Do[k = n^n - 1; s = Plus @@ IntegerDigits[k]; If[Mod[s, n] == 0, Print[n]], {n, 1, 10^5}]

A109675_list = [n for n in range(1,10**4) if not sum([int(d) for d in str(n**n-1)]) % n]
# Chai Wah Wu, Dec 03 2014

a(13)-a(16) from Michael S. Branicky, Jun 25 2023

A345360 a(n) = n^n*n - n.

Original entry on oeis.org

0, 0, 6, 78, 1020, 15620, 279930, 5764794, 134217720, 3486784392, 99999999990, 3138428376710, 106993205379060, 3937376385699276, 155568095557812210, 6568408355712890610, 295147905179352825840, 14063084452067724990992, 708235345355337676357614, 37589973457545958193355582

Offset: 0

Matt Donahoe, Jun 28 2021

Braingle, Impossible Series

Cf. A007778, A048861.

a(n) = n^(n+1) - n.
a(n) = A007778(n) - n.
E.g.f.: -LambertW(-x)/(1+LambertW(-x))^3-x*exp(x). - Alois P. Heinz, Jul 12 2021
a(n) = n*A048861(n). - Kevin Ryde, Jul 12 2021

cp's OEIS Frontend

A385302 Least common multiple of {n^n-1, n^(n-1)-1, ..., n-1}.

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A109675 Numbers k such that the sum of the digits of (k^k - 1) is divisible by k.

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Author

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Maple

Mathematica

Python

Extensions

A345360 a(n) = n^n*n - n.

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