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A308314 Decimal expansion of Sum_{k>=1} (1/A055642(k)^A055642(k)) where A055642(k) is the number of digits of the integer k.

%I A308314 #44 Jun 30 2019 04:47:52
%S A308314 1,6,8,0,5,2,4,5,3,7,5,2,6,2,1,6,8,9,4,9,0,8,5,6,7,3,3,2,0,5,5,6,7,2,
%T A308314 4,5,2,1,9,6,5,2,6,7,9,9,7,1,9,8,4,9,5,0,4,9,1,5,5,7,0,3,5,9,8,1,4,3,
%U A308314 7,9,8,3,4,8,1,7,5,7,0,8,8,9,4,8,3,4,6,1,6,4,4,4,5,0,7,8,4,8,6,4
%N A308314 Decimal expansion of Sum_{k>=1} (1/A055642(k)^A055642(k)) where A055642(k) is the number of digits of the integer k.
%C A308314 With summation by parts to obtain 1st formula:
%C A308314 Sum_{k>=1} (1/length(k)^length(k)) =
%C A308314 Sum_{m=1..9} (1/1^1) + Sum_{m=10..99} (1/2^2) + Sum_{m=100...999} (1/3^3) + Sum_{m=1000...9999} (1/4^4) + ... =
%C A308314 9*(1/1^1) + 90*(1/2^2) + 900*(1/3^3) + 9000*(1/4^4) + 90000*(1/5^5) + ... =
%C A308314 9 ( 1/1^1 + 10^1/2^2 + 10^2/3^3 + 10^3/4^4 + 10^4/5^5 + ... =
%C A308314 (9/10) * (10^1/1^1 + 10^2/2^2 + 10^3/3^3 + 10^4/4^4 + 10^5/5^5 + ... =
%C A308314 (9/10) * ( (10/1)^1 + (10/2)^2 + (10/3)^3 + (10/4)^4 + (10/5)^5 + ... =
%C A308314 (9/10) * Sum_{m>=1} (10/m)^m.
%D A308314 Xavier Merlin, Methodix Analyse, Ellipses, 1997, Exercice 22 p. 120.
%D A308314 J.-M. Monier, Analyse, Tome 3, 2ème année, MP.PSI.PC.PT, Dunod, 1997, Exercice 3.2.1.h" p. 248.
%F A308314 Equals (9/10) * Sum_{k>=1} (10/k)^k.
%F A308314 Equals Sum_{n>=1} (1/A138908(n)).
%e A308314 168.05245375262168949085673320556724...
%p A308314 evalf((9/10) * Sum((10/n)^n, n=1..infinity), 100);
%o A308314 (PARI) (9/10) * suminf(k=1, (10/k)^k) \\ _Michel Marcus_, Jun 08 2019
%Y A308314 Cf. A055642, A138908.
%K A308314 nonn,base,cons
%O A308314 3,2
%A A308314 _Bernard Schott_, May 19 2019