A211168
Exponent of alternating group An.
Original entry on oeis.org
1, 1, 3, 6, 30, 60, 420, 420, 1260, 2520, 27720, 27720, 360360, 360360, 360360, 360360, 6126120, 12252240, 232792560, 232792560, 232792560, 232792560, 5354228880, 5354228880, 26771144400, 26771144400, 80313433200, 80313433200, 2329089562800, 2329089562800
Offset: 1
For n = 7, lcm{1,...,5,7} = 420.
Even entries given by the sequence
A076100, or the odd entries in the sequence
A003418.
The records of this sequence are a subsequence of
A002809 and
A126098.
-
for n in [1..40] do
Exponent(AlternatingGroup(n));
end for;
-
for n in [1..40] do
if n mod 2 eq 0 then
L := [1..n-1];
else
L := Append([1..n-2],n);
end if;
LCM(L);
end for;
-
Table[If[Mod[n, 2] == 0, LCM @@ Range[n - 1],
LCM @@ Join[Range[n - 2], {n}]], {n, 1, 100}] (* or *)
a[1] = 1; a[2] = 1; a[3] = 3; a[n_] := a[n] =
If[Mod[n, 2] == 0, LCM[a[n - 1], n - 2], LCM[a[n - 2], n - 3, n]]; Table[a[n], {n, 1, 40}]
-
a(n)=lcm(if(n%2,concat([2..n-2],n),[2..n-1])) \\ Charles R Greathouse IV, Mar 02 2014
A352881
a(n) is the minimal number z having the largest number of solutions to the Diophantine equation 1/z = 1/x + 1/y such that 1 <= x <= y <= 10^n.
Original entry on oeis.org
2, 12, 60, 840, 9240, 55440, 720720, 6126120, 116396280, 232792560, 5354228880, 26771144400, 465817912560, 4813451763120, 24067258815600, 144403552893600, 2671465728531600, 36510031623265200, 219060189739591200, 4709794079401210800, 18839176317604843200, 221360321731856907600
Offset: 1
For n=1, we have the following, where r = (x*y) mod (x+y). (In the last four columns, each number marked by an asterisk is a square.)
.
r z x y x*y x+y x*y*z x^2+y^2+z^2
- - - - --- --- ----- -----------
0 1 2 2 4* 4* 4* 9* (solution)
2 1 2 4 8 6 8 21
4 1 2 6 12 8 12 41
6 1 2 8 16* 10 16* 69
3 1 3 3 9 6 9* 19
0 2 3 6 18* 9* 36* 49* (solution)
3 2 3 9 27 12 54 94
0 2 4 4 16* 8 32 36* (solution)
8 2 4 8 32 12 64* 84
5 2 5 5 25* 10 50 54
0 3 6 6 36* 12 108 81* (solution)
7 3 7 7 49* 14 147 107
0 4 8 8 64* 16* 256* 144* (solution)
9 4 9 9 81* 18 324* 178
.
z = 2 has the largest number of solutions, so a(1) = 2.
The number of solutions for the resulting z cannot exceed A018892(z).
-
a(n)=my(bc=0,bk=0,L=10^n);for(k=1,L-1,my(c=0,k2=k^2);for(d=max(1,k2\(L-k)+1),k,if(k2%d==0,c++););if(c>bc,bc=c;bk=k););return(bk); \\ Darío Clavijo, Mar 03 2025
-
def a(n):
# k=x*y and d=x+y
bc, bk, L = 0, None, 10**n
for k in range(1, L):
c, k2 = 0, k * k
for d in range(max(1, k2 // (L - k) + 1), k + 1):
if k2 % d == 0: c += 1
if c > bc:
bc, bk = c, k
return bk
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