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A038545
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a(n) = Sum_{i=0..10^n} i^10.
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0
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1, 14914341925, 959924142434241924250, 91409924241424243424241924242500, 9095909924242414242424342424241924242425000, 909140909924242424142424242434242424241924242424250000, 90909590909924242424241424242424243424242424241924242424242500000
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listen;
history;
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internal format)
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OFFSET
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0,2
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COMMENTS
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Jacob Bernoulli gives a(3) in Ars Conjectandi. - Charles R Greathouse IV, Dec 18 2019
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LINKS
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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)
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FORMULA
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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
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MAPLE
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a:= n-> sum(i^10, i=0..10^n):
seq(a(n), n=0..10); # Alois P. Heinz, Jan 19 2021
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MATHEMATICA
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Table[Sum[i^10, {i, 10^n}], {n, 0, 5}] (* Harvey P. Dale, Jul 02 2016 *)
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PROG
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(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
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CROSSREFS
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Cf. A023002.
Sequence in context: A172617 A216015 A246109 * A186094 A023050 A233704
Adjacent sequences: A038542 A038543 A038544 * A038546 A038547 A038548
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KEYWORD
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easy,nonn
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AUTHOR
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Marvin Ray Burns
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STATUS
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approved
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