A050970
Numerator of S(n)/Pi^n, where S(n) = Sum_{k=-inf..+inf} (4k+1)^(-n).
Original entry on oeis.org
1, 1, 1, 1, 5, 1, 61, 17, 277, 31, 50521, 691, 540553, 5461, 199360981, 929569, 3878302429, 3202291, 2404879675441, 221930581, 14814847529501, 4722116521, 69348874393137901, 56963745931, 238685140977801337, 14717667114151
Offset: 1
The first few values of S(n)/Pi^n are 1/4, 1/8, 1/32, 1/96, 5/1536, 1/960, ...
- Vincenzo Librandi, Table of n, a(n) for n = 1..200
- N. D. Elkies, On the sums Sum((4k+1)^(-n),k,-inf,+inf), arXiv:math/0101168 [math.CA], 2001.
- N. D. Elkies, On the sums Sum_{k = -infinity .. infinity} (4k+1)^(-n), Amer. Math. Monthly, 110 (No. 7, 2003), 561-573.
- Z. K. Silagadze, Comment on the sums S(n) = sum(k=-inf..inf) 1/(4k+1)^n, (2012) arXiv:1207.2055
- Eric Weisstein's World of Mathematics, Favard Constants
-
S := proc(n, k) option remember; if k = 0 then `if`(n = 0, 1, 0) else
S(n, k - 1) + S(n - 1, n - k) fi end: EZ := n -> S(n, n)/(2^n * n!):
A050970 := n -> numer(EZ(n-1)): seq(A050970(n), n=1..26); # Peter Luschny, Aug 02 2017
# alternative
A050970 := proc(n)
if type(n,'even') then
(-1)^(n/2)*2^(n-2)/(n-1)!*euler(n-1,0) ;
else
(-1)^((n-1)/2)*2^(n-2)/(n-1)!*euler(n-1,1/2) ;
end if;
%/2^n ;
numer(%) ;
end proc:
seq(A050970(n),n=1..20) ; # R. J. Mathar, Jun 26 2024
-
s[n_] := Sum[(4*k + 1)^(-n), {k, -Infinity, Infinity}]; a[n_] := Numerator[FullSimplify[s[n]/Pi^n]]; a[1] = 1; Table[a[n], {n, 1, 26}] (* Jean-François Alcover, Oct 25 2012 *)
s[n_?EvenQ] := (-1)^(n/2-1)*(2^n-1)*BernoulliB[n]/(2*n!); s[n_?OddQ] := (-1)^((n-1)/2)*2^(-n-1)*EulerE[n-1]/(n-1)!; Table[s[n] // Numerator, {n, 1, 26}] (* Jean-François Alcover, May 13 2013 *)
a[n_] := 4*Sum[((-1)^k/(2*k+1))^n, {k, 0, Infinity}] /. Pi -> 1 // Numerator; Table[a[n], {n, 1, 26}] (* Jean-François Alcover, Jun 20 2014 *)
Table[4/(2 Pi)^n LerchPhi[(-1)^n, n, 1/2], {n, 21}] // Numerator (* Eric W. Weisstein, Aug 02 2017 *)
Table[4/Pi^n If[Mod[n, 2] == 0, DirichletLambda, DirichletBeta][n], {n, 21}] // Numerator (* Eric W. Weisstein, Aug 02 2017 *)
-
{a(n) = if( n<0, 0, numerator( polcoeff( 1 / (1 - tan(x/4 + x * O(x^n))), n)))}; /* Michael Somos, Nov 11 2014 */
A046983
Denominators of Taylor series for tan(x + Pi/4).
Original entry on oeis.org
1, 1, 1, 3, 3, 15, 45, 315, 63, 2835, 14175, 155925, 93555, 6081075, 42567525, 638512875, 127702575, 10854718875, 97692469875, 1856156927625, 371231385525, 194896477400625, 2143861251406875, 2900518163668125, 2275791174570375
Offset: 0
1 + 2*x + 2*x^2 + (8/3)*x^3 + (10/3)*x^4 + (64/15)*x^5 + (244/45)*x^6 + ...
- G. W. Caunt, Infinitesimal Calculus, Oxford Univ. Press, 1914, p. 477.
-
A046983 := proc(n)
coeftayl(tan(x+Pi/4),x=0,n) ;
denom(%) ;
end proc: # R. J. Mathar, Jan 22 2017
-
nmax = 24; t[0, 1] = 1; t[0, ] = 0; t[n, k_] := t[n, k] = (k-1)*t[n-1, k-1] + (k+1)*t[n-1, k+1]; Denominator[ Table[ Sum[ t[n, k]/n!, {k, 0, n+1}], {n, 0, nmax} ]] (* Jean-François Alcover, Nov 09 2011 *)
CoefficientList[Series[Tan[x+Pi/4],{x,0,30}],x]//Denominator (* Harvey P. Dale, Sep 05 2023 *)
A360945
a(n) = numerator of (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) where Zeta is the Hurwitz zeta function.
Original entry on oeis.org
1, 2, 10, 244, 554, 202084, 2162212, 1594887848, 7756604858, 9619518701764, 59259390118004, 554790995145103208, 954740563911205348, 32696580074344991138888, 105453443486621462355224, 7064702291984369672858925136, 4176926860695042104392112698
Offset: 0
a(0) = 1 because lim_{n->0} (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) = 1.
a(3) = 244 because (Zeta(2*3+1,1/4) - Zeta(2*3+1,3/4))/Pi^(2*3+1) = 244/45.
Cf.
A000364,
A046982,
A173945,
A173947,
A173948,
A173949,
A173953,
A173954,
A173955,
A173982,
A173983,
A173984,
A173987,
A360966,
A361007,
A361007.
-
Table[(Zeta[2*n + 1, 1/4] - Zeta[2*n + 1, 3/4]) / Pi^(2*n + 1), {n, 1, 25}] // FunctionExpand // Numerator (* Vaclav Kotesovec, Feb 27 2023 *)
t[0, 1] = 1; t[0, _] = 0;
t[n_, k_] := t[n, k] = (k-1) t[n-1, k-1] + (k+1) t[n-1, k+1];
a[n_] := Sum[t[2n, k]/(2n)!, {k, 0, 2n+1}] // Numerator;
Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Mar 15 2023 *)
a[n_] := SeriesCoefficient[Tan[x+Pi/4], {x, 0, 2n}] // Numerator;
Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Apr 15 2023 *)
-
a(n) = numerator(abs(eulerfrac(2*n))*(2*n + 1)*2^(2*n)/(2*n + 1)!); \\ Michel Marcus, Apr 11 2023
A360966
a(n) = denominator of (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) where Zeta is the Hurwitz zeta function.
Original entry on oeis.org
1, 1, 3, 45, 63, 14175, 93555, 42567525, 127702575, 97692469875, 371231385525, 2143861251406875, 2275791174570375, 48076088562799171875, 95646113035463615625, 3952575621190533915703125, 1441527579493018251609375, 68739242628124575327993046875, 333120945043988326589504765625
Offset: 0
a(0) = 1 because lim_{n->0} (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) = 1.
a(3) = 45 because (Zeta(2*3+1,1/4) - Zeta(2*3+1,3/4))/Pi^(2*3+1) = 244/45.
Cf.
A000364,
A046982,
A173945,
A173947,
A173948,
A173949,
A173953,
A173954,
A173955,
A173982,
A173983,
A173984,
A173987,
A361007.
-
Table[(Zeta[2*n + 1, 1/4] - Zeta[2*n + 1, 3/4]) / Pi^(2*n + 1), {n, 0, 25}] // FunctionExpand // Denominator
(* Second program: *)
a[n_] := SeriesCoefficient[Tan[x + Pi/4], {x, 0, 2n}] // Denominator;
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Apr 16 2023 *)
-
a(n) = denominator(abs(eulerfrac(2*n))*(2*n + 1)*2^(2*n)/(2*n + 1)!); \\ Michel Marcus, Apr 11 2023
A361007
a(n) = numerator of (zeta(2*n,1/4) + zeta(2*n,3/4))/Pi^(2*n) where zeta is the Hurwitz zeta function.
Original entry on oeis.org
0, 2, 8, 64, 2176, 31744, 2830336, 178946048, 30460116992, 839461371904, 232711080902656, 39611984424992768, 955693069653508096, 1975371841521663868928, 1124025625663103358205952, 369906947004953565463576576, 278846808228005417477465964544
Offset: 0
tan(2*x) = 2*x + (8/3)*x^3 + (64/15)*x^5 + (2176/315)*x^7 + (31744/2835)*x^9 + ...
Cf.
A000364,
A046982,
A173945,
A173947,
A173948,
A173949,
A173953,
A173954,
A173955,
A173982,
A173983,
A173984,
A173987,
A360945,
A360966.
-
Table[(Zeta[2*n, 1/4] + Zeta[2*n, 3/4])/Pi^(2*n), {n, 0, 25}] //
FunctionExpand // Numerator
Table[4^(2 k) (2^(2 k) - 1) (-1)^(k + 1) BernoulliB[2 k]/(2 (2 k)!), {k, 0, 25}] // Numerator
-
my(x='x+O('x^50), v = Vec(tan(2*x)/x)); apply(numerator, vector(#v\2, k, v[2*k-1])) \\ Michel Marcus, Apr 09 2023
Showing 1-5 of 5 results.
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