cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-9 of 9 results.

A173982 a(n) = numerator of (Zeta(0,2,1/3) - Zeta(0,2,n+1/3)), where Zeta is the Hurwitz Zeta function.

Original entry on oeis.org

0, 9, 153, 7641, 192789, 32757741, 525987081, 190358321841, 23076404893161, 577743530648769, 578407918658769, 556370890030917009, 160916328686946575601, 220439117509451225357769
Offset: 0

Views

Author

Artur Jasinski, Mar 04 2010

Keywords

Comments

All numbers in this sequence are divisible by 9.

Crossrefs

Programs

  • Magma
    [0] cat [Numerator((&+[9/(3*k+1)^2: k in [0..n-1]])): n in [1..20]]; // G. C. Greubel, Aug 23 2018
  • Maple
    A173982 := proc(n) add( 1/(1/3+i)^2,i=0..n-1) ; numer(%) ; end proc: seq(A173982(n),n=0..20) ; # R. J. Mathar, Apr 22 2010
  • Mathematica
    Table[FunctionExpand[-Zeta[2, (3*n + 1)/3] + Zeta[2, 1/3]], {n, 0, 20}] // Numerator (* Vaclav Kotesovec, Nov 13 2017 *)
    Numerator[Table[Sum[9/(3*k + 1)^2, {k, 0, n - 1}], {n, 0, 20}]] (* G. C. Greubel, Aug 23 2018 *)
  • PARI
    for(n=0,20, print1(numerator(sum(k=0,n-1, 9/(3*k+1)^2)), ", ")) \\ G. C. Greubel, Aug 23 2018
    

Formula

a(n) = numerator of 2*(Pi^2)/3 + J - Zeta(2,(3*n+1)/3), where Zeta is the Hurwitz Zeta function and the constant J is A173973.
A173982(n)/A173984(n) = sum_{i=0..n} 1/(1/3+i)^2 = 9*sum_{i=0..n} 1/(1+3i)^2 = psi'(1/3) - psi'(n+1/3). - R. J. Mathar, Apr 22 2010
a(n) = numerator of Sum_{k=0..(n-1)} 9/(3*k+1)^2. - G. C. Greubel, Aug 23 2018

Extensions

Name simplified by Peter Luschny, Nov 14 2017

A173983 a(n) = numerator((Zeta(2, 1/3) - Zeta(2, n + 1/3))/9), where Zeta(n, z) is the Hurwitz Zeta function.

Original entry on oeis.org

0, 1, 17, 849, 21421, 3639749, 58443009, 21150924649, 2564044988129, 64193725627641, 64267546517641, 61818987781213001, 17879592076327397289, 24493235278827913928641, 24506988360923903264741
Offset: 0

Views

Author

Artur Jasinski, Mar 04 2010

Keywords

Comments

From Wolfdieter Lang, Nov 12 2017: (Start)
a(n+1)/A173984(n+1) gives, for n >= 0, the partial sum Sum_{k=0..n} 1/(1+3*k)^2.
The limit n -> infinity is given in A214550 as the Hurwitz Zeta function or the Polygamma function (1/9)*Zeta(2, 1/3) = (1/9)*Psi(1, 1/3) = 1.121733... (End)

Examples

			The rationals a(n)/A173984(n) begin 0/1, 1/1, 17/16, 849/784, 21421/19600, 3639749/3312400, 58443009/52998400, 21150924649/19132422400, ... - _Wolfdieter Lang_, Nov 12 2017
		

Crossrefs

Programs

  • Magma
    [0] cat [Numerator((&+[1/(3*k+1)^2: k in [0..n-1]])): n in [1..20]]; // G. C. Greubel, Aug 23 2018
  • Maple
    a := n -> numer((Zeta(0,2,1/3) - Zeta(0,2,n+1/3))/9):
    seq(a(n), n=0..14); # Peter Luschny, Nov 12 2017
  • Mathematica
    Table[FunctionExpand[-Zeta[2, (3*n + 1)/3] + Zeta[2, 1/3]]/9, {n, 0, 20}] // Numerator (* Vaclav Kotesovec, Nov 13 2017 *)
    Numerator[Table[Sum[1/(3*k + 1)^2, {k, 0, n - 1}], {n, 0, 20}]] (* G. C. Greubel, Aug 23 2018 *)
  • PARI
    for(n=0,20, print1(numerator(sum(k=0,n-1, 1/(3*k+1)^2)), ", ")) \\ G. C. Greubel, Aug 23 2018
    

Formula

a(n) = numerator of (1/9)(2(Pi^2)/3 + J - Zeta(2,(3n+1)/3)) where J is the constant A173973.
a(n) = numerator of Sum_{k=0..(n-1)} 1/(3*k+1)^2. - G. C. Greubel, Aug 23 2018

Extensions

Name simplified by Peter Luschny, Nov 12 2017

A173984 a(n) is the denominator of (Zeta(0,2,1/3) - Zeta(0,2,n+1/3)) where Zeta is the Hurwitz Zeta function.

Original entry on oeis.org

1, 1, 16, 784, 19600, 3312400, 52998400, 19132422400, 2315023110400, 57875577760000, 57875577760000, 55618430227360000, 16073726335707040000, 22004931353582937760000, 22004931353582937760000
Offset: 0

Views

Author

Artur Jasinski, Mar 04 2010

Keywords

Crossrefs

Programs

  • Magma
    [1,1] cat [Denominator((&+[9/(3*k+1)^2: k in [1..n-1]])): n in [2..20]]; // G. C. Greubel, Aug 24 2018
  • Maple
    a := n -> Zeta(0,2,1/3) - Zeta(0,2,n+1/3):
    seq(denom(a(n)), n=0..14); # Peter Luschny, Nov 14 2017
  • Mathematica
    Table[FunctionExpand[-Zeta[2, (3*n + 1)/3] + Zeta[2, 1/3]], {n, 0, 20}] // Denominator (* Vaclav Kotesovec, Nov 13 2017 *)
    Denominator[Table[Sum[9/(3*k + 1)^2, {k, 1, n - 1}], {n, 0, 30}]] (* G. C. Greubel, Aug 24 2018 *)
  • PARI
    for(n=0,20, print1(denominator(sum(k=1,n-1, 9/(3*k+1)^2)), ", ")) \\ G. C. Greubel, Aug 24 2018
    

Formula

a(n) = denominator of 2*(Pi^2)/3 + J - Zeta(2,(3*n+1)/3), where Zeta is the Hurwitz Zeta function and the constant J is A173973.
a(n) = denominator of Sum_{k=1..(n-1)} 9/(3*k+1)^2. - G. C. Greubel, Aug 24 2018

Extensions

Name simplified by Peter Luschny, Nov 14 2017

A173986 a(n) = numerator((Psi(1, 2/3) - Psi(1, n+2/3))/9), where Psi(1, z) is the Trigamma function.

Original entry on oeis.org

0, 1, 29, 489, 60769, 3026081, 884023809, 890877733, 474015890357, 80471258049933, 67921427083803253, 1089963588226225073, 1092655876391630769, 395273284628034202009, 665644988593672027490729
Offset: 0

Views

Author

Artur Jasinski, Mar 04 2010

Keywords

Comments

a(n+1)/A173987(n+1) gives, for n >= 0, the partial sum Sum_{k=0..n} 1/(3*k+2)^2. The limit n -> infinity is given in A294967 as the Hurwitz Zeta function or the Trigamma function (1/9)*Zeta(2, 2/3) = (1/9)*Psi(1, 2/3) = 0.3404306010 ... - Wolfdieter Lang, Nov 12 2017

Examples

			The rationals a(n)/A173987(n) begin 0/1, 1/4, 29/100, 489/1600, 60769/193600, 3026081/9486400, 884023809/2741569600, 890877733/2741569600, ... - _Wolfdieter Lang_, Nov 12 2017
		

Crossrefs

Programs

  • Magma
    [0] cat [Numerator((&+[2/(3*k+2)^2: k in [0..n-2]])): n in [2..20]]; // G. C. Greubel, Aug 23 2018
  • Maple
    r := n -> (Psi(1, 2/3) - Psi(1, n+2/3))/9:
    seq(numer(simplify(r(n))), n=0..14); # Peter Luschny, Nov 13 2017
  • Mathematica
    Table[Numerator[FunctionExpand[(4*Pi^2/3 - Zeta[2, 1/3] - Zeta[2, (3*n + 2)/3])/9]], {n, 0, 20}] (* Vaclav Kotesovec, Nov 14 2017 *)
    Numerator[Table[Sum[2/(3*k + 2)^2, {k, 0, n - 2}], {n, 1, 20}]] (* G. C. Greubel, Aug 23 2018 *)
  • PARI
    for(n=1,20, print1(numerator(sum(k=0,n-2, 2/(3*k+2)^2)), ", ")) \\ G. C. Greubel, Aug 23 2018
    

Formula

a(n) = numerator(r(n)) with r(n) = (1/9)*(4*(Pi^2)/3 - Zeta(2, 1/3) - Zeta(2, (3*n+2)/3)) = (1/9)*(Zeta(2, 2/3) - Zeta(2, (3*n+2)/3)) with the Hurwitz Zeta function Zeta(2, q). This becomes the formula given in the name. - Wolfdieter Lang, Nov 13 2017
a(n) = numerator of (1/9)*(2(Pi^2)/3 - J - Zeta(2, (3n+2)/3)) where J is the constant A173973 [which becomes the preceding formula].
a(n) = numerator of Sum_{k=0..(n-2)} 2/(3*k+2)^2. - G. C. Greubel, Aug 23 2018

Extensions

Name simplified by Peter Luschny, Nov 13 2017

A173985 a(n) = numerator of (Zeta(0,2,2/3) - Zeta(0,2,n+2/3)), where Zeta is the Hurwitz Zeta function.

Original entry on oeis.org

0, 9, 261, 4401, 546921, 27234729, 7956214281, 8017899597, 4266143013213, 724241322449397, 611292843754229277, 9809672294036025657, 9833902887524676921, 3557459561652307818081, 5990804897343048247416561
Offset: 0

Views

Author

Artur Jasinski, Mar 04 2010

Keywords

Comments

All numbers in this sequence are divisible by 9.

Crossrefs

Programs

  • Magma
    [0] cat [Numerator((&+[9/(3*k+1)^2: k in [0..n-1]])): n in [1..20]]; // G. C. Greubel, Aug 23 2018
  • Maple
    A173985 := proc(n) add( 1/(2/3+i)^2,i=0..n-1) ; numer(%) ; end proc: seq(A173985(n),n=0..20) ; # R. J. Mathar, Apr 22 2010
  • Mathematica
    Table[FunctionExpand[4*(Pi^2)/3 - Zeta[2, 1/3] - Zeta[2, (3*n + 2)/3]], {n, 0, 20}] // Numerator (* Vaclav Kotesovec, Nov 13 2017 *)
    Numerator[Table[Sum[9/(3*k + 1)^2, {k, 0, n - 1}], {n, 0, 20}]] (* G. C. Greubel, Aug 23 2018 *)
  • PARI
    for(n=0,20, print1(numerator(9*sum(k=0,n-1, 1/(3*k+1)^2)), ", ")) \\ G. C. Greubel, Aug 23 2018
    

Formula

a(n) = numerator of 2*(Pi^2)/3 - J - Zeta(2, (3*n+2)/3), where Zeta is the Hurwitz Zeta function and the constant J is A173973.
a(n)/A173987(n) = sum_{i=0..n-1} 1/(i+2/3)^2 = psi'(2/3)-psi'(2/3+n). - R. J. Mathar, Apr 22 2010
a(n) = numerator of Sum_{k=0..(n-1)} 9/(3*k+1)^2. - G. C. Greubel, Aug 23 2018

Extensions

Name simplified by Peter Luschny, Nov 14 2017

A294967 Decimal expansion of (1/9)*Hurwitz Zeta(2, 2/3) = (1/9)*Psi(1, 2/3), with the Polygamma function Psi.

Original entry on oeis.org

3, 4, 0, 4, 3, 0, 6, 0, 1, 0, 3, 9, 8, 5, 7, 4, 8, 9, 9, 9, 8, 5, 9, 0, 8, 0, 3, 6, 9, 7, 2, 9, 8, 3, 5, 0, 3, 5, 9, 1, 8, 8, 3, 4, 3, 3, 7, 4, 8, 2, 3, 2, 6, 2, 2, 1, 5, 8, 6, 4, 7, 3, 7, 1, 2, 5, 4, 4, 8, 7, 1, 6, 7, 4, 2, 2, 8, 0, 1, 6, 8, 2, 1, 2, 9, 5, 5, 8, 3, 7, 0, 8, 1, 5, 6, 5, 6, 0, 5, 1, 8, 0, 1, 4, 7, 4, 1, 1, 0, 7, 7, 2, 2, 8, 6, 7, 7, 9, 7, 3, 1, 7, 9, 8, 3, 1
Offset: 0

Views

Author

Wolfdieter Lang, Nov 12 2017

Keywords

Comments

This is the value of the series Sum_{n=0..infinity} 1/(3*n+2)^2. For (3*n+2)^2 see A016790.
For the partial sums see A173986(n+1)/A173987(n+1), n >= 0.

Examples

			0.340430601039857489998590803697298350359188343374823262215864737125448716...
		

Crossrefs

Programs

Formula

From Amiram Eldar, Oct 02 2020: (Start)
Equals Integral_{1..oo} log(x)/(x^3-1) dx.
Equals 4*Pi^2/27 - A214550. (End)

Extensions

Data corrected by Amiram Eldar, Oct 02 2020

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

Views

Author

Artur Jasinski, Feb 26 2023

Keywords

Comments

The function (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) is rational for every positive integer n.
For denominators see A360966.
(Zeta(2*n+1,1/4) + Zeta(2*n+1,3/4))/Zeta(2*n+1) = 4*16^n - 2*4^n; see A193475.
For numerators of the function (Zeta(2*n,1/4) + Zeta(2*n,3/4))/Pi^(2*n) see A361007.
For denominators of the function (Zeta(2*n,1/4) + Zeta(2*n,3/4))/Pi^(2*n) see A036279.
(Zeta(2*n,1/4) - Zeta(2*n,3/4))/beta(2*n) = 16^n (see A001025) where beta is the Dirichlet beta function.
From the above formulas we can express Zeta(k,1/4) and Zeta(k,3/4) for every positive integer k.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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 *)
  • PARI
    a(n) = numerator(abs(eulerfrac(2*n))*(2*n + 1)*2^(2*n)/(2*n + 1)!); \\ Michel Marcus, Apr 11 2023

Formula

a(n) = A046982(2*n).
(Zeta(2*n + 1, 1/4) - Zeta(2*n + 1, 3/4))/(Pi^(2*n + 1)) = A000364(n)*(2*n + 1)*2^(2*n)/(2*n + 1)!.

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

Views

Author

Artur Jasinski, Apr 09 2023

Keywords

Comments

The function (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) is rational for every positive integer n.
For numerators see A360945.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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 *)
  • PARI
    a(n) = denominator(abs(eulerfrac(2*n))*(2*n + 1)*2^(2*n)/(2*n + 1)!); \\ Michel Marcus, Apr 11 2023

Formula

a(n) = A046983(2*n).
(Zeta(2*n + 1, 1/4) - Zeta(2*n + 1, 3/4))/(Pi^(2*n + 1)) = A000364(n)*(2*n + 1)*2^(2*n)/(2*n + 1)!.

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

Views

Author

Artur Jasinski, Mar 15 2023

Keywords

Comments

The function (zeta(2*n,1/4) + zeta(2*n,3/4))/Pi^(2*n) is rational for every positive integer n.

Examples

			tan(2*x) = 2*x + (8/3)*x^3 + (64/15)*x^5 + (2176/315)*x^7 + (31744/2835)*x^9 + ...
		

Crossrefs

Programs

  • Mathematica
    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
  • PARI
    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

Formula

a(n) = numerator( [x^(2*n-1)] tan(2*x) ).
a(n) = numerator( (-1)^(n + 1)*4^(2*n)*(2^(2*n) - 1)*B(2*n)/(2*(2*n)!) ) where B(2*n) are Bernoulli numbers.
Showing 1-9 of 9 results.