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 A157805 Numerator of Euler(n,3). 3

%I

%S 1,5,6,55,30,125,126,2015,510,2075,2046,15685,8190,38225,32766,118975,

%T 131070,3726575,524286,-217736285,2097150,4730505125,8388606,

%U -968249463115,33554430,14717801331875,134217726,-2093659805510855,536870910,86125674710684825

%N Numerator of Euler(n,3).

%H Vladimir Shevelev, <a href="https://arxiv.org/abs/1708.08096">On a Luschny question</a>, arXiv:1708.08096 [math.NT], 2017.

%H Vladimir Shevelev, <a href="http://list.seqfan.eu/pipermail/seqfan/2017-September/017929.html">A formula for numerator of Euler(n,k)</a>, Wed Sep 06 2017.

%F From _Vladimir Shevelev_, Sep 04 2017: (Start)

%F For even n, a(n) = 2^(n+1) - 2 + delta(n,0), where delta is the Kronecker symbol; for n == 1 (mod 4), a(n) = (2^(n+1)-2)*A006519(n+1) + A002425((n+1)/2); for n == 3 (mod 4), a(n) = (2^(n+1)-2)*A006519(n+1) - A002425((n+1)/2).

%F A generalization: Let N(n,k) denote numerator of Euler(n,k), k >= 1 is integer. Set u(n,k) = 2*Sum_{1 <= i <= k-1}(-1)^(i-1)*(k-i)^n. Then, for even n, N(n,k) = u(n,k) + (-1)^(k-1)^delta(n,0); for n == 1 (mod 4), N(n,k) = u(n,k)*A006519(n+1) + (-1)^(k-1)*A002425((n+1)/2); for n == 3 (mod 4), N(n,k) = u(n,k)* A006519 (n+1) - (-1)^(k-1)*A002425((n+1)/2). (End)

%e By the formula, we have a(1) = 2*2 + 1 = 5, a(3) = 14*4 - 1 = 55, a(5) = 62*2 + 1 = 125, a(7) = 254*8 - 17 = 2015, a(9) = 1022*2 + 31 = 2075, etc. - _Vladimir Shevelev_, Sep 04 2017

%t a2425[n_] := (-1)^n/n*(1 - 4^n)*BernoulliB[2*n]*2^IntegerExponent[2*n, 2];

%t a6519[n_] := 2^IntegerExponent[n, 2];

%t a[n_] := Switch[Mod[n, 4], 0 | 2, 2^(n + 1) - 2 + KroneckerDelta[n, 0], 1, (2^(n + 1) - 2)*a6519[n + 1] + a2425[(n + 1)/2], 3, (2^(n + 1) - 2)*a6519[n + 1] - a2425[(n + 1)/2]];

%t Table[a[n], {n, 0, 30}]

%t (* or *)

%t Table[EulerE[n, 3] // Numerator, {n, 0, 30}] (* _Jean-François Alcover_, Jul 14 2018 *)

%Y For denominators see A006519.

%Y Cf. A002425.

%K sign,frac

%O 0,2

%A _N. J. A. Sloane_, Nov 10 2009

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Last modified July 15 22:48 EDT 2019. Contains 325061 sequences. (Running on oeis4.)