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A023037
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a(n) = n^0 + n^1 + ... + n^(n-1), or a(n) = (n^n-1)/(n-1) with a(0)=0; a(1)=1.
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36
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0, 1, 3, 13, 85, 781, 9331, 137257, 2396745, 48427561, 1111111111, 28531167061, 810554586205, 25239592216021, 854769755812155, 31278135027204241, 1229782938247303441, 51702516367896047761, 2314494592664502210319, 109912203092239643840221
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OFFSET
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0,3
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COMMENTS
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For prime n, a(n) is conjectured to be the period of Bell numbers (mod n). See A054767. - T. D. Noe, Oct 12 2007
For prime n, a(n) is a multiple of the period of Bell numbers mod n (and conjectured to be exactly the period, as mentioned above). - Charles R Greathouse IV, Jul 31 2012
For n >= 1, a(n) is the number whose base n representation is a string of n ones. For example, 11111 in base 5 is a(5) = 781. - Melvin Peralta, May 23 2016
For n > 0, n^(a(n)-1) == 1 (mod a(n)), so for n > 1, a(n) is a prime or a Fermat pseudoprime to base n. - Thomas Ordowski, Mar 15 2021
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LINKS
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FORMULA
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EXAMPLE
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a(3) = 3^0 + 3^1 + 3^2 = 1+3+9 = 13.
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MAPLE
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MATHEMATICA
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Join[{0, 1}, Table[(n^n-1)/(n-1), {n, 2, 20}]] (* Harvey P. Dale, Aug 01 2014 *)
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PROG
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(Sage) [lucas_number1(n, n+1, n) for n in range(0, 19)] # Zerinvary Lajos, May 16 2009
(PARI) a(n) = if(n==1, 1, (n^n-1)/(n-1)); \\ Altug Alkan, Oct 04 2017
(Python)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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Entry improved by Tobias Nipkow (nipkow(AT)in.tum.de).
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STATUS
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approved
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