OFFSET
0,2
COMMENTS
Jacob Bernoulli gives a(3) in Ars Conjectandi. - Charles R Greathouse IV, Dec 18 2019
LINKS
Jacob Bernoulli, Ars Conjectandi (1713). (German translation?, see p. 99)
Burkard Polster, Power sum MASTER CLASS: How to sum quadrillions of powers ... by hand! (Euler-Maclaurin formula), Mathologer video (2019)
FORMULA
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.
a(n) = A023002(10^n). - Charles R Greathouse IV, Dec 18 2019
MAPLE
a:= n-> sum(i^10, i=0..10^n):
seq(a(n), n=0..10); # Alois P. Heinz, Jan 19 2021
MATHEMATICA
Table[Sum[i^10, {i, 10^n}], {n, 0, 5}] (* Harvey P. Dale, Jul 02 2016 *)
PROG
(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
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
STATUS
approved