A319035 Triangular numbers T(k) that have the same number of divisors as their successors T(k+1).

6, 10, 15, 66, 153, 406, 435, 561, 861, 903, 1378, 1540, 1770, 2211, 2346, 2556, 2926, 3655, 3916, 4186, 4371, 5151, 5778, 6555, 7626, 9453, 10011, 10296, 11175, 11325, 12720, 14535, 14878, 16110, 16836, 17205, 17391, 17766, 18336, 19306, 19503, 20301, 20706

Offset: 1

Author

Jon E. Schoenfield, Dec 05 2018

Keywords

• nonn

Comments

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.

Examples

			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).
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).
		

Links

• Muniru A Asiru, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A000217, A081978.

Cf. A276542 (indices of these triangular numbers).

Programs

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

Mathematica

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 *)

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