

A108750


Even numbers n such that there exists a solution to lcm(r+s,t) = n1, lcm(s,r+t) = n2, r,s>0, t>1, r+s+t <= n.


0



58, 146, 156, 206, 288, 466, 478, 496, 498, 562, 596, 610, 640, 716, 738, 782, 834, 838, 870, 982, 1028, 1068, 1162, 1234, 1276, 1314, 1336, 1366, 1636, 1706, 1718, 1856, 1888, 1982, 2110, 2148, 2186, 2206, 2228, 2416, 2452, 2612, 2626, 2642, 2666, 2668
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OFFSET

1,1


COMMENTS

The alternating groups A_n with n even can be generated by a 3cycle x=(1,2,3) and an (n1)cycle y=(2,3,...n) whose product has cycle type 2 x (n2). These are called standard generators of A_n. If we assume only that o(y)=n1 and o(xy)=n2, then for the values of n in the sequence, y could be the product of 2 cycles (cycle type (r+s).t) and xy can have cycle type s.(r+t). Thus the elements of the sequence give those alternating groups A_n which contain pairs of elements which look like standard generators if one only looks at their orders.


LINKS



EXAMPLE

a(1)=58 because we can take r=11, s=8, t=3 giving lcm(19,3)=57, lcm(8,14)=56


MATHEMATICA

maxTerm = 3000; f[n_] := Module[{cond = False}, Do[r = rss; If[r == rtt && r > 0 && r+s+t <= n, If[LCM[r+s, t] == n1 && LCM[s, r+t] == n2, cond = True; Break[]]], {t, Rest[Divisors[n1]]}, {s, Divisors[n2]}, {rs, Divisors[n1]}, {rt, Divisors[n2]}]; cond]; Reap[For[n = 0; k = 2, k <= maxTerm, k = k+2, If[f[k], n++; Print["a(", n, ") = ", k]; Sow[k]]]][[2, 1]] (* JeanFrançois Alcover, May 06 2017 *)


CROSSREFS



KEYWORD

nonn


AUTHOR

Simon Nickerson (simonn(AT)maths.bham.ac.uk), Jun 23 2005


STATUS

approved



