A226540 - OEIS

%I #16 Apr 09 2021 15:25:44

%S 1,3,5,5,7,14,18,15,11,33,39,13,35,60,68,51,57,95,105,77,23,138,150,

%T 65,117,189,203,145,155,248,264,187,119,315,333,37,247,390,410,287,

%U 301,473,495,345,47,564,588,245,425,663,689,477,495,770,798,551,59,885

%N Maximum of the proper divisors of the triangular numbers.

%C Solutions of A226540(n)=n are listed in A005383(n).

%C Solutions of A226540(n)=n+1 are listed in A005385(n).

%H Paolo P. Lava, <a href="/A226540/b226540.txt">Table of n, a(n) for n = 2..1000</a>

%F a(4n) = 4n^2 + n, 4n+1 <= a(4n+1) <= (8n^2 + 6n + 1)/3, 4n+3 <= a(4n+2) <= (8n^2 + 10n + 3)/3, a(4n+3) = 4n^2 + 7n + 3. - _Charles R Greathouse IV_, Jun 10 2013

%e For n = 28 we have n*(n+1)/2 = 406 and its proper divisors are 1, 2, 7, 14, 29, 58, 203. Hence a(28) = 203.

%p with(numtheory); A226540:=proc(q) local a,n;

%p for n from 2 to q do a:=sort([op(divisors(n*(n+1)/2))]);

%p print(a[nops(a)-1]); od; end: A226540(10^6);

%t Table[Divisors[(n(n+1))/2][[-2]],{n,2,60}] (* _Harvey P. Dale_, Apr 09 2021 *)

%o (PARI) a(n)=if(n==2,return(1));my(p=factor(n/gcd(n,2))[1,1],q=factor((n+1)/gcd(n+1,2))[1,1]); binomial(n+1,2)/min(p,q) \\ _Charles R Greathouse IV_, Jun 10 2013

%Y Cf. A000217, A005383, A077065.

%K nonn,easy

%O 2,2

%A _Paolo P. Lava_, Jun 10 2013