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A007781
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a(n) = (n+1)^(n+1) - n^n for n>0, a(0) = 1.
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12
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1, 3, 23, 229, 2869, 43531, 776887, 15953673, 370643273, 9612579511, 275311670611, 8630788777645, 293959006143997, 10809131718965763, 426781883555301359, 18008850183328692241, 808793517812627212561
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OFFSET
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0,2
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COMMENTS
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(12n^2 + 6n + 1)^2 divides a(6n+1), where (12n^2 + 6n + 1) = (2n+1)^3 - (2n)^3 = A127854(n) = A003215(2n) are the hex (or centered hexagonal) numbers. The prime numbers of the form 12n^2 + 6n + 1 belong to A002407. - Alexander Adamchuk, Apr 09 2007
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REFERENCES
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Richard P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see equation (6.7).
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LINKS
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Andrew Cusumano, Problem H-656, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 45, No. 2 (2007), p. 187; A Sequence Tending To e, Solution to Problem H-656, ibid., Vol. 46-47, No. 3 (2008/2009), pp. 285-287.
Ronald K. Hoeflin, Mega Test. [Wayback Machine link]
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FORMULA
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a(n) = abs(discriminant(x^(n+1)-x+1)).
E.g.f.: W(-x)/(1+W(-x)) - W(-x)/((1+W(-x))^3*x) where W is the Lambert W function. - Robert Israel, Aug 19 2015
Limit_{n->oo} (a(n+2)/a(n+1) - a(n+1)/a(n)) = e (Cusumano, 2007). - Amiram Eldar, Jan 03 2022
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EXAMPLE
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a(14) = 10809131718965763 = 3 * 61^2 * 968299894201.
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MAPLE
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seq( `if`(n=0, 1, (n+1)^(n+1) -n^n), n=0..20); # G. C. Greubel, Mar 05 2020
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MATHEMATICA
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Join[{1}, Table[(n+1)^(n+1)-n^n, {n, 20}]] (* Harvey P. Dale, Feb. 09 2011 *)
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PROG
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(PARI) first(m)=vector(m, i, i--; (i+1)^(i+1) - i^i) /* Anders Hellström, Aug 18 2015 */
(Sage) [1]+[(n+1)^(n+1) -n^n for n in (1..20)] # G. C. Greubel, Mar 05 2020
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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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Peter McCormack (peter.mccormack(AT)its.csiro.au)
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STATUS
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approved
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