A226540 Maximum of the proper divisors of the triangular numbers.

1, 3, 5, 5, 7, 14, 18, 15, 11, 33, 39, 13, 35, 60, 68, 51, 57, 95, 105, 77, 23, 138, 150, 65, 117, 189, 203, 145, 155, 248, 264, 187, 119, 315, 333, 37, 247, 390, 410, 287, 301, 473, 495, 345, 47, 564, 588, 245, 425, 663, 689, 477, 495, 770, 798, 551, 59, 885

Offset: 2

Author

Paolo P. Lava, Jun 10 2013

Keywords

• nonn
• easy

Comments

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

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

Examples

			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.
		

Links

• Paolo P. Lava, Table of n, a(n) for n = 2..1000

Crossrefs

Cf. A000217, A005383, A077065.

Programs

Maple

with(numtheory); A226540:=proc(q) local a,n;
for n from 2 to q do a:=sort([op(divisors(n*(n+1)/2))]);
print(a[nops(a)-1]); od; end: A226540(10^6);

Mathematica

Table[Divisors[(n(n+1))/2][[-2]],{n,2,60}] (* Harvey P. Dale, Apr 09 2021 *)

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

Formula

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