|
%I #16 May 19 2015 20:42:48
%S 2,6,4,20,12,42,8,18,40,110,24,156,84,60,16,272,36,342,80,126,220,506,
%T 48,100,312,54,168,812,120,930,32,330,544,210,72,1332,684,468,160,
%U 1640,252,1806,440,180,1012,2162,96,294,200,816,624,2756,108,550,336,1026
%N a(n) = n*(p-1) where p is the largest prime factor of n.
%C It is conjectured that sequence gives period length of the periodic sequence {A088957(k) mod n}_{k>n}.
%C The records of this sequence are given by A036689 (product of a prime and the previous number). - _Michel Marcus_, May 19 2015
%H Ivan Neretin, <a href="/A088659/b088659.txt">Table of n, a(n) for n = 2..1000</a>
%F For p the k-th prime, a(p) = A036689(k). - _Michel Marcus_, May 19 2015
%F a(n) = n*A070777(n). - _Michel Marcus_, May 19 2015
%p seq(n*(max(numtheory:-factorset(n))-1), n=2..100); # _Robert Israel_, May 19 2015
%t Table[n*(FactorInteger[n][[-1, 1]] - 1), {n, 2, 57}] (* _Ivan Neretin_, May 19 2015 *)
%o (PARI) a(n)=n*(component(factor(n),1)-1)
%K nonn
%O 2,1
%A _Benoit Cloitre_, Nov 21 2003
|
