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Number of numbers k <= n such that tau(k) = tau(k+1) where tau(x) = A000005(x) is the number of divisors of x.
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%I #13 Jun 25 2022 00:48:32

%S 0,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,4,4,5,6,

%T 6,6,6,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,9,9,

%U 9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,11,12,12,12,12,12,12,12,13,14

%N Number of numbers k <= n such that tau(k) = tau(k+1) where tau(x) = A000005(x) is the number of divisors of x.

%F Is a(n) asymptotic to c*n with c=0.1......?

%t Accumulate[If[#[[1]]==#[[2]],1,0]&/@Partition[DivisorSigma[ 0,Range[ 100]],2,1]] (* _Harvey P. Dale_, Jan 27 2021 *)

%o (PARI) a(n)=sum(i=1,n,if(numdiv(i)-numdiv(i+1),0,1))

%Y Cf. A000005 (tau), A005237.

%Y Partial sums of A130638.

%K easy,nonn

%O 1,14

%A _Benoit Cloitre_, Sep 08 2002