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Numbers k such that tau(k+1) - tau(k) = 5, where tau(k) = the number of divisors of k (A000005).
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%I #21 Apr 17 2024 02:31:46

%S 35,169,289,529,961,1369,2809,3135,4489,7921,9409,10609,10815,11881,

%T 12769,16129,18495,18769,22201,22801,26569,27889,32041,33855,38809,

%U 44521,49729,51529,52441,53823,58081,61503,69169,72361,76729,78961,80089,96721

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

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

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

%C Either k or k+1 is a square. - _Amiram Eldar_, Apr 17 2024

%H Amiram Eldar, <a href="/A228453/b228453.txt">Table of n, a(n) for n = 1..10000</a>

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

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

%o (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

%Y Cf. A000005, A049820, A051950.

%Y 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).

%K nonn

%O 1,1

%A _Jaroslav Krizek_, Nov 03 2013