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The number of digits of the greatest product from addends that sum up to 10^n.
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%I #24 Sep 14 2019 22:59:48

%S 1,2,16,160,1591,15905,159041,1590405,15904042,159040419,1590404183,

%T 15904041824,159040418240,1590404182399,15904041823989,

%U 159040418239888,1590404182398875,15904041823988748,159040418239887480,1590404182398874791,15904041823988747910,159040418239887479099

%N The number of digits of the greatest product from addends that sum up to 10^n.

%F a(n) = 1 + floor(log_10(36) + 10*log_10(27)*R_(n-1)), R_k being the k-th repunit, i.e., 111...111 with only digit 1 appearing k times.

%e The greatest product of numbers that sum up to 10 is 2*2*3*3 = 36 which has 2 digits. Thus a(1) = 2.

%e The greatest product of numbers that sum up to 100 is 2*2*3^(32) ~ 7.4*10^15 which has 16 digits. Hence a(2) = 16.

%e The greatest product of numbers that sum up to 1000 is 2*2*3^(332) ~ 1.0*10^159 which has 160 digits. Therefore a(3) = 160.

%o (PARI) a(n) = 1 + floor(log(4)/log(10) + ((10^n-1)/3-1)*log(3)/log(10)); \\ _Jinyuan Wang_, Aug 03 2019

%Y Cf. A000792.

%K nonn,base

%O 0,2

%A _Lekraj Beedassy_, Aug 03 2019