The OEIS is supported by the many generous donors to the OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A038545 a(n) = Sum_{i=0..10^n} i^10. 0
 1, 14914341925, 959924142434241924250, 91409924241424243424241924242500, 9095909924242414242424342424241924242425000, 909140909924242424142424242434242424241924242424250000, 90909590909924242424241424242424243424242424241924242424242500000 (list; graph; refs; listen; history; text; internal format)
 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 Cf. A023002. Sequence in context: A172617 A216015 A246109 * A186094 A023050 A233704 Adjacent sequences: A038542 A038543 A038544 * A038546 A038547 A038548 KEYWORD easy,nonn AUTHOR STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified January 29 20:08 EST 2023. Contains 359926 sequences. (Running on oeis4.)