OFFSET
1,1
COMMENTS
All terms are divisible by 36, because the number k*(k+1)*(k+2) is always divisible by 6 so a product of at least 2 such factors is divisible by 36. The first 12 terms are even divisible by 720.
a(13) > 4.5*10^22 if it exists. - David A. Corneth, Jul 12 2024
b(F(2*k)^2-1) is a term for all k >= 2, where b(k) = k*(k+1)*(k+2) = A007531(k+2) and F = A000045 is the Fibonacci sequence, because b(F(2*k)^2-1) = b(F(2*k-1)-1)*b(F(2*k+1)-1). In particular, a(13) <= b(F(20)^2-1) = 95853241822852400000400. - Pontus von Brömssen, Jul 13 2024
LINKS
David A. Corneth, PARI program
EXAMPLE
With b(k) = k*(k+1)*(k+2) = A007531(k+2), we have the following factorizations of the first 12 terms:
720 = b(8) = 6*120 = b(1)*b(4);
262080 = b(63) = 120*2184 = b(4)*b(12);
43243200 = b(350) = 120*210*1716 = b(4)*b(5)*b(11);
85765680 = b(440) = 2184*39270 = b(12)*b(33);
14366626560 = b(2430) = 24*60*1716*5814 = b(2)*b(3)*b(11)*b(17);
27680637600 = b(3024) = 39270*704880 = b(33)*b(88);
8916100427520 = b(20735) = 704880*12649104 = b(88)*b(232);
2871098559070560 = b(142128) = 12649104*226980390 = b(232)*b(609);
5720836667515200 = b(178848) = 6*210*373176*12166770 = b(1)*b(5)*b(71)*b(229);
20123426048544000 = b(271998) = 6*210*328440*48626760 = b(1)*b(5)*b(68)*b(364);
924491486191094640 = b(974168) = 226980390*4073001576 = b(609)*b(1596);
297683700627082714560 = b(6677055) = 4073001576*73087057560 = b(1596)*b(4180).
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Pontus von Brömssen, Jul 07 2024
EXTENSIONS
a(8)-a(11) from Michael S. Branicky, Jul 07 2024
a(12) from David A. Corneth, Jul 12 2024
STATUS
approved