

A354424


Numbers k for which the ratio A008475(k)/k reaches a record low.


0



2, 6, 10, 12, 15, 20, 28, 30, 40, 42, 56, 60, 84, 105, 120, 140, 168, 180, 210, 252, 280, 315, 330, 360, 385, 390, 420, 616, 630, 660, 770, 780, 840, 924, 1092, 1155, 1260, 1540, 1820, 1848, 1980, 2184, 2310, 2520, 2730, 3080, 3465, 3640, 3960, 4095, 4290, 4620, 5460, 6552, 6930
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OFFSET

1,1


COMMENTS

Sequence gives the numbers k for which m/k reaches a record low, where m is minimal so that the symmetric group S_m has an element of order k.


LINKS



EXAMPLE

First, an element of order 2 shows up in S_2, so the smallest ratio we've seen so far is 1. This is the smallest ratio we see until we reach 6, since there's an element of order 6 in S_5. Next is 10, since there's an element of order 10 in S_7, and 7/10 is the next ratio smaller than 5/6. Then comes 12, since S_7 also has an element of order 12, and 7/12 is the next ratio less than 7/10, etc.


MATHEMATICA

s = {}; fm = 2; Do[If[(f = Plus @@ Power @@@ FactorInteger[n]/n) < fm, fm = f; AppendTo[s, n]], {n, 2, 7000}]; s (* Amiram Eldar, Jul 12 2022 *)


PROG

(Sage)
memo = {1: (2, 1)}
def a(n):
if n in memo.keys(): return memo[n]
_ = a(n1)
prev, prevRatio = memo[n1]
ratio = 1
N = prev
while ratio >= prevRatio:
N += 1
# compute m so that S_m has an element of order N
principalDivisors = list(factor(N))
m = sum([a^b for (a, b) in principalDivisors])
ratio = m/N
memo[n] = (N, ratio)
return N
(PARI) b(n) = my(f=factor(n)); vecsum(vector(#f~, i, f[i, 1]^f[i, 2])); \\ A008475
lista(nn) = my(m=oo, list=List(), x); for (n=2, nn, if ((x=b(n)/n) < m, m = x; listput(list, n); ); ); Vec(list); \\ Michel Marcus, Jul 12 2022


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



