This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A063440 #68 Mar 14 2023 04:37:53
%S A063440 1,2,4,4,4,4,6,9,6,4,8,8,4,8,16,8,6,6,8,16,8,4,12,18,6,8,16,8,8,8,10,
%T A063440 20,8,8,24,12,4,8,24,12,8,8,8,24,12,4,16,24,9,12,16,8,8,16,24,24,8,4,
%U A063440 16,16,4,12,36,24,16,8,8,16,16,8,18,18,4,12,24,16,16,8,16,40,10,4,16
%N A063440 Number of divisors of n-th triangular number.
%C A063440 a(n) = 4 iff either n is in A005383 or n/2 is in A005384.
%C A063440 a(n) is odd iff n is in A001108.
%C A063440 a(n) = 6 if either n = 18 or n = q^2 where q is in A048161 or n = 2 q^2 - 1 where q is in A106483. - _Robert Israel_, Oct 26 2015
%C A063440 From _Bernard Schott_, Aug 29 2020: (Start)
%C A063440 a(n-1) is the number of solutions in positive integers (x, y, z) to the simultaneous equations (x + y - z = n, x^2 + y^2 - z^2 = n) for n > 1. See the British Mathematical Olympiad link. In this case, one always has z > x and z > y.
%C A063440 For n = 12 as in the Olympiad problem, the a(11) = 8 solutions are (13,78,79), (14,45,47), (15,34,37), (18,23,29), (23,18,29), (34,15,37), (45,14,47), (78,13,79). (End)
%D A063440 Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 2 of the British Mathematical Olympiad 2007, page 28.
%H A063440 Ray Chandler, <a href="/A063440/b063440.txt">Table of n, a(n) for n = 1..10000</a>
%H A063440 British Mathematical Olympiad 2007/2008, Round 1, <a href="https://bmos.ukmt.org.uk/home/bmo1-2008.pdf">Problem 2</a>.
%H A063440 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%F A063440 a(n) = A000005(A000217(n)).
%F A063440 From _Robert Israel_, Oct 26 2015: (Start)
%F A063440 a(2k) = A000005(k)*A000005(2k+1).
%F A063440 a(2k+1) = A000005(2k+1)*A000005(k+1).
%F A063440 gcd(a(2k), a(2k+1)) = A000005(2k+1) * A060778(k). (End)
%e A063440 a(6) = 4 since 1+2+3+4+5+6 = 21 has four divisors {1,3,7,21}.
%p A063440 seq(numtheory:-tau(n*(n+1)/2), n=1..100); # _Robert Israel_, Oct 26 2015
%t A063440 DivisorSigma[0,#]&/@Accumulate[Range[90]] (* _Harvey P. Dale_, Apr 15 2019 *)
%o A063440 (PARI) for (n=1, 10000, write("b063440.txt", n, " ", numdiv(n*(n + 1)/2)) ) \\ _Harry J. Smith_, Aug 21 2009
%o A063440 (PARI) a(n)=factorback(apply(numdiv,if(n%2,[n,(n+1)/2],[n/2,n+1]))) \\ _Charles R Greathouse IV_, Dec 27 2014
%o A063440 (PARI) vector(100, n, numdiv(n*(n+1)/2)) \\ _Altug Alkan_, Oct 26 2015
%Y A063440 Cf. A000005, A000217.
%Y A063440 Cf. A001108, A005383, A005384, A048161, A060778, A081978 (greedy inverse), A106483, A101755 (indices of records), A101756 (records).
%K A063440 nonn,easy
%O A063440 1,2
%A A063440 _Henry Bottomley_, Jul 24 2001