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%I #22 Jan 19 2021 21:13:32
%S 1,14914341925,959924142434241924250,91409924241424243424241924242500,
%T 9095909924242414242424342424241924242425000,
%U 909140909924242424142424242434242424241924242424250000,90909590909924242424241424242424243424242424241924242424242500000
%N a(n) = Sum_{i=0..10^n} i^10.
%C Jacob Bernoulli gives a(3) in Ars Conjectandi. - _Charles R Greathouse IV_, Dec 18 2019
%H Jacob Bernoulli, Ars Conjectandi (1713). (<a href="https://quod.lib.umich.edu/u/umhistmath/abz9501.0001.001/282?rgn=full+text;view=pdf">German translation?, see p. 99</a>)
%H Burkard Polster, <a href="https://www.youtube.com/watch?v=fw1kRz83Fj0">Power sum MASTER CLASS: How to sum quadrillions of powers ... by hand! (Euler-Maclaurin formula)</a>, Mathologer video (2019)
%F a(n) = 1/11*(10^n+1)^11 - 1/2*(10^n+1)^10 + 5/6*(10^n+1)^9 - (10^n+1)^7 + (10^n+1)^5 - 1/2*(10^n + 1)^3 + 5/66*10^n + 5/66.
%F a(n) = A023002(10^n). - _Charles R Greathouse IV_, Dec 18 2019
%p a:= n-> sum(i^10, i=0..10^n):
%p seq(a(n), n=0..10); # _Alois P. Heinz_, Jan 19 2021
%t Table[Sum[i^10,{i,10^n}],{n,0,5}] (* _Harvey P. Dale_, Jul 02 2016 *)
%o (PARI) a(n)=(6*(10^n)^11 + 33*(10^n)^10 + 55*(10^n)^9 - 66*(10^n)^7 + 66*(10^n)^5 - 33*(10^n)^3 + 5*(10^n))/66 \\ _Charles R Greathouse IV_, Dec 18 2019, modified by _Sean A. Irvine_, Jan 19 2021
%Y Cf. A023002.
%K easy,nonn
%O 0,2
%A _Marvin Ray Burns_
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