A228453 Numbers k such that tau(k+1) - tau(k) = 5, where tau(k) = the number of divisors of k (A000005).

35, 169, 289, 529, 961, 1369, 2809, 3135, 4489, 7921, 9409, 10609, 10815, 11881, 12769, 16129, 18495, 18769, 22201, 22801, 26569, 27889, 32041, 33855, 38809, 44521, 49729, 51529, 52441, 53823, 58081, 61503, 69169, 72361, 76729, 78961, 80089, 96721

Offset: 1

Author

Jaroslav Krizek, Nov 03 2013

Keywords

• nonn

Comments

Numbers k such that A051950(k+1) = 5.

Numbers k such that A049820(k) - A049820(k+1) = 4.

Either k or k+1 is a square. - Amiram Eldar, Apr 17 2024

Examples

			35 is in sequence because tau(36) - tau(35) = 9 - 4 = 5.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A049820, A051950.

Numbers k such that tau(k+1) - tau(k) = m: A055927 (m = 1), A230115 (m = 2), A230653 (m = 3), A230654 (m = 4), this sequence (m = 5).

Programs

Mathematica

Select[ Range[ 50000], DivisorSigma[0, # ] + 5 == DivisorSigma[0, # + 1] &]

PARI

lista(kmax) = {my(d); for(k = 2, kmax, d = numdiv(k^2); if(d == numdiv(k^2-1) + 5, print1(k^2-1, ", ")); if(d == numdiv(k^2+1) - 5, print1(k^2, ", ")));} \\ Amiram Eldar, Apr 17 2024