A038545 a(n) = Sum_{i=0..10^n} i^10.

1, 14914341925, 959924142434241924250, 91409924241424243424241924242500, 9095909924242414242424342424241924242425000, 909140909924242424142424242434242424241924242424250000, 90909590909924242424241424242424243424242424241924242424242500000

Offset: 0

Author

Marvin Ray Burns

Keywords

• easy
• nonn

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)

Crossrefs

Cf. A023002.

Programs

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 *)

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

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