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

2,2

Comments

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

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

Links

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

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

Example

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.

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 *)

Prog

(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

Crossrefs

Cf. A000217, A005383, A077065.

Sequence in context: A158284 A090941 A090917 * A309572 A366679 A296107

Adjacent sequences: A226537 A226538 A226539 * A226541 A226542 A226543

Keyword

nonn,easy

Author

Paolo P. Lava, Jun 10 2013

Status

approved