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%I A007925 #55 Jul 04 2022 19:48:42
%S A007925 -1,-1,-1,17,399,7849,162287,3667649,91171007,2486784401,74062575399,
%T A007925 2395420006033,83695120256591,3143661612445145,126375169532421599,
%U A007925 5415486851106043649,246486713303685957375,11877172892329028459041,604107995057426434824791
%N A007925 a(n) = n^(n+1) - (n+1)^n.
%C A007925 From _Mathew Englander_, Jul 07 2020: (Start)
%C A007925 All a(n) are odd and for n even, a(n) == 3 (mod 4); for n odd and n != 1, a(n) == 1 (mod 4).
%C A007925 The correspondence between n and a(n) when considered mod 6 is as follows: for n == 0, 1, 2, or 3, a(n) == 5; for n == 4, a(n) == 3; for n == 5, a(n) == 1.
%C A007925 For all n, a(n)+1 is a multiple of n^2.
%C A007925 For n odd  and n >= 3, a(n)-1 is a multiple of (n+1)^2.
%C A007925 For n even and n >= 0, a(n)+1 is a multiple of (n+1)^2.
%C A007925 For proofs of the above, see the Englander link. (End)
%D A007925 G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
%H A007925 T. D. Noe, <a href="/A007925/b007925.txt">Table of n, a(n) for n = 0..100</a>
%H A007925 Mathew Englander, <a href="/A007925/a007925.pdf">Notes on OEIS A007925</a>
%H A007925 Sergio Silva, <a href="http://ginasiomental.no.sapo.pt/materialc/mteste/teste.htm">Teste Numerico</a>, Item 3.
%H A007925 H. J. Smith, <a href="http://harry-j-smith-memorial.com/Display/xyyx.html">Contour Plot of z = x^y - y^x</a>
%F A007925 Asymptotic expression for a(n) is a(n) ~ n^n * (n - e). - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 19 2001
%F A007925 From _Mathew Englander_, Jul 07 2020: (Start)
%F A007925 a(n) = A111454(n+4) - 1.
%F A007925 a(n) = A055651(n, n+1).
%F A007925 a(n) = A220417(n+1, n) for n >= 1.
%F A007925 a(n) = A007778(n) - A000169(n+1).
%F A007925 (End)
%F A007925 E.g.f.: LambertW(-x)/((1+LambertW(-x))*x)-LambertW(-x)/(1+LambertW(-x))^3. - _Alois P. Heinz_, Jul 04 2022
%e A007925 a(2) = 1^2 - 2^1 = -1,
%e A007925 a(4) = 3^4 - 4^3 = 17.
%p A007925 A007925:=n->n^(n+1)-(n+1)^n: seq(A007925(n), n=0..25); # _Wesley Ivan Hurt_, Jan 10 2017
%t A007925 lst={};Do[AppendTo[lst, (n^(n+1)-((n+1)^n))], {n, 0, 4!}];lst (* _Vladimir Joseph Stephan Orlovsky_, Sep 19 2008 *)
%t A007925 #^(#+1)-(#+1)^#&/@Range[0,20] (* _Harvey P. Dale_, Oct 22 2011 *)
%o A007925 (Maxima) A007925[n]:=n^(n+1)-(n+1)^n$ makelist(A007925[n],n,0,30); /* _Martin Ettl_, Oct 29 2012 */
%o A007925 (PARI) a(n)=n^(n+1)-(n+1)^n \\ _Charles R Greathouse IV_, Feb 06 2017
%Y A007925 Cf. A051442.
%Y A007925 Cf. A166326, A099498, A141074, A174379, A123206, A045575, A082754.
%K A007925 sign,easy,nice
%O A007925 0,4
%A A007925 Dennis S. Kluk (mathemagician(AT)ameritech.net)

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