Mia Boudreau - cp's OEIS frontend

User: Mia Boudreau

Mia Boudreau has authored 2 sequences.

A386253 a(n) = (smallest digit of n)^(largest digit of n) + n.

1, 2, 6, 30, 260, 3130, 46662, 823550, 16777224, 387420498, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 20, 22, 26, 31, 40, 57, 90, 155, 284, 541, 30, 32, 40, 60, 115, 278, 765, 2224, 6599, 19722, 40, 42, 58, 124, 300, 1069, 4142, 16431, 65584, 262193, 50

Offset: 0

Mia Boudreau, Jul 16 2025

0^0 is an indeterminate form, but for the purpose of a(0) it is taken to be 1. - Robert Israel, Jul 22 2025

			a(563) = 1292 because 3^6 + 563 = 1292.

Mia Boudreau, Table of n, a(n) for n = 0..10000

Cf. A054054, A054055, A097385.

f:= proc(n) local L; L:= convert(n,base,10); n + min(L)^max(L) end proc:
map(f, [$0..100]); # Robert Israel, Jul 22 2025

Unprotect[Power]; 0^0:=1; Protect[Power]; a[n_]:= (Min[IntegerDigits[n]])^(Max[IntegerDigits[n]]) + n; Array[a,51,0] (* Stefano Spezia, Jul 17 2025 *)

def a(n): return int(min(s:=str(n)))**int(max(s)) + n
print([a(n) for n in range(51)]) # Michael S. Branicky, Jul 21 2025

a(n) = A054054(n)^A054055(n) + n.

A384184 Order of the permutation of {0,...,n-1} formed by successively swapping elements at i and 2*i mod n, for i = 0,...,n-1.

Original entry on oeis.org

1, 2, 1, 4, 2, 2, 2, 8, 3, 4, 5, 4, 6, 4, 6, 16, 4, 6, 9, 8, 4, 10, 28, 8, 10, 12, 9, 8, 14, 12, 12, 32, 5, 8, 70, 12, 18, 18, 24, 16, 10, 8, 7, 20, 210, 56, 126, 16, 110, 20, 60, 24, 26, 18, 120, 16, 9, 28, 29, 24, 30, 24, 60, 64, 6, 10, 33, 16

Offset: 1

Mia Boudreau, May 29 2025

nonn

a(2*n) = 2*a(n) since the cycle lengths of the permutation with size 2*n is effectively that of size n twice, doubled. Thus, the LCM/order is doubled.

			For n = 11, the permutation is {0,3,4,7,8,1,2,9,10,5,6} and it has order a(11) = 5.

Mia Boudreau, Table of n, a(n) for n = 1..10000
Mia Boudreau, Java program, GitHub

Cf. A003418, A000027, A051732, A004626, A065119, A001122, A155072, A001133, A001134, A001135, A225759.

from sympy.combinatorics import Permutation
def a(n):
   L = list(range(n))
   for i in range(n):
       if (j:= (i << 1) % n) != i:
           L[i],L[j] = L[j],L[i]
   return Permutation(L).order() # Darío Clavijo, Jun 05 2025

a(2*n) = 2*a(n).
a(2^n) = 2^n.
Conjecture: a(2^n + 2^x) = 2^n * (x-n) if x > n.
a(2^n - 1) = A003418(n-1).
s(2^n + 1) = A000027(n).
a(2*n - 1) = A051732(n).
a(A004626(n)) % 2 = 1.
a(A065119(n)) = n/3.
a(A001122(n)) = (n-1) / 2.
a(A155072(n)) = (n-1) / 4.
a(A001133(n)) = (n-1) / 6.
a(A001134(n)) = (n-1) / 8.
a(A001135(n)) = (n-1) / 10.
a(A225759(n)) = (n-1) / 16.

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User: Mia Boudreau

A386253 a(n) = (smallest digit of n)^(largest digit of n) + n.

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