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a(n) = number of terms of A003052 that are <= n.
1

%I #7 Feb 27 2017 17:22:07

%S 1,1,2,2,3,3,4,4,5,5,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,7,7,7,7,

%T 7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,9,10,10,10,

%U 10,10,10,10,10,10,10,10,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12

%N a(n) = number of terms of A003052 that are <= n.

%H U. Zannier, <a href="http://www.ams.org/journals/proc/1982-085-01/S0002-9939-1982-0647887-4/">On the distribution of self-numbers</a>, Proc. Amer. Math. Soc. 85 (1982), 10-14.

%F Zannier shows that a(n) = L*n + O(*log x)^2), where L is approximately 10.227...

%p # Assumes the array b52 contains a list of the terms in A003052.

%p p:=[]; t:=1; m:=b52[t]; c:=1;

%p for n from 1 to 1000 do

%p if n=m then c:=c+1; t:=t+1; m:=b52[t]; fi;

%p p:=[op(p),c];

%p od:

%p p;

%Y Cf. A003052.

%K nonn

%O 1,3

%A _N. J. A. Sloane_, Feb 27 2017