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Sum_{k = 1..10^n} tau_4(k), where tau_4 is the number of ordered factorizations into 4 factors (A007426)
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%I #7 Mar 31 2012 10:24:14

%S 1,89,3575,93237,1951526,35270969,578262093,8840109380,128217432396,

%T 1784942188189,24045237260214,315312623543840,4042957241191810,

%U 50862246063060180,629513636928477232,7681900592647818929

%N Sum_{k = 1..10^n} tau_4(k), where tau_4 is the number of ordered factorizations into 4 factors (A007426)

%C Using that tau_4 = tau_2 ** tau_2, where ** means Dirichlet convolution and tau_2 is (A000005), one can calculate a(n) faster than in O(10^n) operations - namely in O(10^(3n/4)) or even in O(10^(2n/3)). See links for details.

%H A. V. Lelechenko <a href="http://taac.org.ua/files/a2011/proceedings/UA-1-Andrew%20Vladimirovich%20Lelechenko-83.pdf">The summation of the multiplicative functions</a> (in Russian)

%F a(n) = A061202(10^n) = sum(k = 1..10^n, A007426(n))

%Y Cf. A057494 - partial sums up to 10^n of the divisors function tau_2 (A000005), A180361 - of the unitary divisors function tau_2* (A034444), A180365 - of the 3-divisors function tau_3 (A007425).

%Y Also see A072692 for such sums of the sum of divisors function (A000203), A084237 for sums of Moebius function (A008683), A064018 for sums of Euler totient function (A000010).

%K nonn

%O 0,2

%A _Andrew Lelechenko_, Apr 15 2011