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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

Table of n, a(n) for n=0..6.

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

Marvin Ray Burns

STATUS

approved

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