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Numbers n such that tau(n+1) - tau(n) = 4, where tau(n) = the number of divisors of n (A000005).
5

%I #9 Oct 06 2017 01:05:40

%S 11,17,19,31,39,43,55,65,67,69,77,87,97,129,134,163,175,183,185,194,

%T 207,211,221,237,241,247,249,254,265,283,295,309,321,327,331,337,343,

%U 351,365,398,404,417,437,454,458,459,469,471,473,482,493,494,497,505,517

%N Numbers n such that tau(n+1) - tau(n) = 4, where tau(n) = the number of divisors of n (A000005).

%C Numbers n such that A051950(n+1) = 4. Numbers n such that A049820(n) - A049820(n+1) = 3. Sequence of starts of first run of n (n>=2) consecutive integers m_1, m_2, ..., m_n such that tau(m_k) - tau(m_k-1) = 4, for all k=n...2: 11, 458, 3013, ... (a(5) > 100000); example for n=4: tau(3013) = 4, tau(3014) = 8, tau(3015) = 12, tau(3016) = 16.

%H Jaroslav Krizek, <a href="/A230654/b230654.txt">Table of n, a(n) for n = 1..4000</a>

%e 19 is in sequence because tau(20) - tau(19) = 6 - 2 = 4.

%t Select[ Range[ 50000], DivisorSigma[0, # ] + 4 == DivisorSigma[0, # + 1] &]

%Y Cf. A055927 (numbers n such that tau(n+1) - tau(n) = 1), A230115 (numbers n such that tau(n+1) - tau(n) = 2), A230653 (numbers n such that tau(n+1) - tau(n) = 3), A000005.

%K nonn

%O 1,1

%A _Jaroslav Krizek_, Nov 03 2013