login
Sum of sigma(j) for 1<=j<10^n, where sigma(j) = A048050(j) is the sum of the proper divisors >1 of j (excluding 1 and n).
1

%I #23 Feb 16 2020 03:19:46

%S 16,3034,320243,32226805,3224444759,322465138002,32246681892518,

%T 3224670122682648,322467031114802292,32246703322412473945,

%U 3224670334023621455211,322467033422357645316809,32246703342390510922780778,3224670334240928188556405242

%N Sum of sigma(j) for 1<=j<10^n, where sigma(j) = A048050(j) is the sum of the proper divisors >1 of j (excluding 1 and n).

%H Amiram Eldar, <a href="/A049030/b049030.txt">Table of n, a(n) for n = 1..36</a> (calculated from the b-file at A049000)

%H Carlos Rivera, <a href="http://www.primepuzzles.net/problems/prob_023.htm">Problem 23.- Divisors (V) Aliquot sequences, Sociable Numbers</a>, Prime Puzzles and Problems Connection.

%F At a(3) = 320243, for example, take a(3) from A049000: 820741 - 500498 = 320243. Compute 500498 from 999*1000/2 = 499500, split evenly and reverse to 500499 - 1 = 500498. Add a 9 and 0 for each successive term.

%F a(n) = A049000(n) - 10^n * (10^n + 1) / 2 + 2 ~ (Pi^2/12 - 1/2) * 10^(2*n). - _Amiram Eldar_, Feb 16 2020

%e For n = 1, the sum of sigma(j), for j < 10 is 0 + 0 + 0 + 2 + 0 + 5 + 0 + 6 + 3 = 16, so a(1) = 16.

%Y Cf. A001065, A046915, A048050, A048995, A049000.

%Y Cf. A072691 (Pi^2/12).

%K base,nonn

%O 1,1

%A _Enoch Haga_ and _Jud McCranie_

%E More terms from _Amiram Eldar_, Feb 16 2020