%I #11 Dec 06 2018 03:44:13
%S 6,10,15,66,153,406,435,561,861,903,1378,1540,1770,2211,2346,2556,
%T 2926,3655,3916,4186,4371,5151,5778,6555,7626,9453,10011,10296,11175,
%U 11325,12720,14535,14878,16110,16836,17205,17391,17766,18336,19306,19503,20301,20706
%N Triangular numbers T(k) that have the same number of divisors as their successors T(k+1).
%C Not every term T(k) has the same prime signature as its successor triangular number T(k+1); the first counterexample is the pair (T(52), T(53)) = (1378, 1431) = (2 * 13 * 53, 3^3 * 53), each of which has 8 divisors. The first counterexample in which the two triangular numbers have the same number of distinct prime factors is (T(45630), T(45631)) = (1041071265, 1041116896) = (3^3 * 5 * 13^2 * 45631, 2^5 * 23 * 31 * 45631), each of which has 48 divisors.
%H Muniru A Asiru, <a href="/A319035/b319035.txt">Table of n, a(n) for n = 1..10000</a>
%e T(2) = 6 is a term because 6 = 2 * 3 has 4 divisors (1, 2, 3, 6) and T(3) = 10 = 2 * 5 also has 4 divisors (1, 2, 5, 10).
%e T(17) = 153 is a term because 153 = 3^2 * 17 has 6 divisors (1, 3, 9, 17, 51, 153) and T(18) = 171 = 3^2 * 19 also has 6 divisors (1, 3, 9, 19, 57, 171).
%t t[n_] := n(n+1)/2; aQ[n_] := DivisorSigma[0, t[n]] == DivisorSigma[0, t[n+1]]; t[Select[Range[100], aQ]] (* _Amiram Eldar_, Dec 06 2018 *)
%o (PARI) lista(nn) = {for (n=1, nn, if (numdiv(t=n*(n+1)/2) == numdiv((n+1)*(n+2)/2), print1(t, ", ")););} \\ _Michel Marcus_, Dec 06 2018
%o (GAP) T:=List([1..210],n->n*(n+1)/2);; a:=List(Filtered([1..Length(T)-1],i->Tau(T[i])=Tau(T[i+1])),k->T[k]); # _Muniru A Asiru_, Dec 06 2018
%Y Cf. A000005, A000217, A081978.
%Y Cf. A276542 (indices of these triangular numbers).
%K nonn
%O 1,1
%A _Jon E. Schoenfield_, Dec 05 2018
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