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A228453
Numbers k such that tau(k+1) - tau(k) = 5, where tau(k) = the number of divisors of k (A000005).
4
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
1,1
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
LINKS
EXAMPLE
35 is in sequence because tau(36) - tau(35) = 9 - 4 = 5.
MATHEMATICA
Select[ Range[ 50000], DivisorSigma[0, # ] + 5 == DivisorSigma[0, # + 1] &]
PROG
(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
CROSSREFS
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).
Sequence in context: A015219 A195545 A270860 * A327651 A319042 A033851
KEYWORD
nonn
AUTHOR
Jaroslav Krizek, Nov 03 2013
STATUS
approved