A276542 Numbers k such that the k-th and (k+1)st triangular numbers have the same number of divisors.

3, 4, 5, 11, 17, 28, 29, 33, 41, 42, 52, 55, 59, 66, 68, 71, 76, 85, 88, 91, 93, 101, 107, 114, 123, 137, 141, 143, 149, 150, 159, 170, 172, 179, 183, 185, 186, 188, 191, 196, 197, 201, 203, 208, 213, 215, 217, 219, 227, 232, 235, 236, 239, 243, 244, 247, 265

Offset: 1

Author

K. D. Bajpai, Apr 09 2017

Keywords

• nonn

Comments

The k-th triangular number T(k) = k*(k+1)/2.

The lesser member of each twin-prime pair appears in this sequence. Hence, A001359 is a subset of this sequence.

Examples

			a(3) = 5; T(5) = 5*(5+1)/2 = 15; T(5+1) = 6*(6+1)/2 = 21; 15 and 21 have 4 divisors each.
a(6) = 28; T(28) = 28*(28+1)/2 = 406; T(28+1) = 29*(29+1)/2 = 435; 406 and 435 have 8 divisors each
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000040, A000217, A001359, A062832.

Cf. A319035 (the corresponding triangular numbers).

Programs

GAP

T:=List([1..270],n->n*(n+1)/2);;  a:=Filtered([1..Length(T)-1],i->Tau(T[i])=Tau(T[i+1])); # Muniru A Asiru, Dec 06 2018

Magma

[n: n in [1..300] | DivisorSigma(0, n*(n + 1) div 2) eq DivisorSigma(0, (n + 1)*(n + 1 + 1) div 2)]; // Vincenzo Librandi, Dec 06 2018

Maple

T:= seq(numtheory:-tau(n*(n+1)/2), n=1..1000):
select(t -> T[t]=T[t+1], [$1..999]); # Robert Israel, Apr 09 2017

Mathematica

Select[Range[1000], DivisorSigma[0, #*(# + 1)/2] == DivisorSigma[0, (# + 1)*(# + 1 + 1)/2] &]
SequencePosition[DivisorSigma[0,#]&/@Accumulate[Range[300]],{x_,x_}][[All, 1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 02 2018 *)

PARI

k=[]; for(n=1, 1000, a=numdiv(n*(n+1)/2); b=numdiv((n+1)*(n+1+1)/2); if(a==b, k=concat(k, n))); k

Python

from sympy import divisor_count
for n in range(1,20):
    if divisor_count(n*(n+1)/2)==divisor_count((n+1)*(n+2)/2):
        print(n, end=', ') # Stefano Spezia, Dec 06 2018