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.

A003983 Array read by antidiagonals with T(n,k) = min(n,k).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 3, 2, 1, 1, 2, 3, 4, 3, 2, 1, 1, 2, 3, 4, 4, 3, 2, 1, 1, 2, 3, 4, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 7, 6, 5, 4, 3, 2, 1
Offset: 1

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Author

Keywords

Comments

Also, "correlation triangle" for the constant sequence 1. - Paul Barry, Jan 16 2006
Antidiagonal sums are in A002620.
As a triangle, row sums are A002620. T(2n,n)=n+1. Diagonal sums are A001399. Construction: Take antidiagonal triangle of MM^T where M is the sequence array for the constant sequence 1 (lower triangular matrix with all 1's). - Paul Barry, Jan 16 2006
From Franklin T. Adams-Watters, Sep 25 2011: (Start)
As a triangle, count up to ceiling(n/2) and back down again (repeating the central term when n is even).
When the first two instances of each number are removed from the sequence, the original sequence is recovered.
(End)

Examples

			Triangle version begins
  1;
  1, 1;
  1, 2, 1;
  1, 2, 2, 1;
  1, 2, 3, 2, 1;
  1, 2, 3, 3, 2, 1;
  1, 2, 3, 4, 3, 2, 1;
  1, 2, 3, 4, 4, 3, 2, 1;
  1, 2, 3, 4, 5, 4, 3, 2, 1;
  ...
		

Crossrefs

Programs

  • Haskell
    a003983 n k = a003983_tabl !! (n-1) !! (k-1)
    a003983_tabl = map a003983_row [1..]
    a003983_row n = hs ++ drop m (reverse hs)
       where hs = [1..n' + m]
             (n',m) = divMod n 2
    -- Reinhard Zumkeller, Aug 14 2011
    
  • Maple
    a(n) = min(floor(1/2 + sqrt(2*n)) - (2*n + round(sqrt(2*n)) - round(sqrt(2*n))^2)/2+1, (2*n + round(sqrt(2*n)) - round(sqrt(2*n))^2)/2) # Leonid Bedratyuk, Dec 13 2009
  • Mathematica
    Flatten[Table[Min[n-k+1, k], {n, 1, 14}, {k, 1, n}]] (* Jean-François Alcover, Feb 23 2012 *)
  • PARI
    T(n,k) = min(n,k) \\ Charles R Greathouse IV, Feb 06 2017
    
  • Python
    from math import isqrt
    def A003983(n):
        a = (m:=isqrt(k:=n<<1))+(k>m*(m+1))
        x = n-(a*(a-1)>>1)
        return min(x,a-x+1) # Chai Wah Wu, Jun 14 2025

Formula

Number triangle T(n, k) = Sum_{j=0..n} [j<=k][j<=n-k]. - Paul Barry, Jan 16 2006
G.f.: 1/((1-x)*(1-x*y)*(1-x^2*y)). - Christian G. Bower, Jan 17 2006
a(n) = min(floor( 1/2 + sqrt(2*n)) - (2*n + round(sqrt(2*n)) - round(sqrt(2*n))^2)/2+1, (2*n + round(sqrt(2*n)) - round(sqrt(2*n))^2)/2). - Leonid Bedratyuk, Dec 13 2009

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Nov 08 2000
Entry revised by N. J. A. Sloane, Dec 05 2006

A173953 a(n) = numerator of (Zeta(2, 3/4) - Zeta(2, n-1/4)), where Zeta is the Hurwitz Zeta function.

Original entry on oeis.org

0, 16, 928, 119344, 3078464, 1132669904, 606887707616, 49610806397296, 48006150564413056, 48265162121607952, 8192066749392160288, 15200753287254377716912, 33677610844789597790454208
Offset: 1

Views

Author

Artur Jasinski, Mar 03 2010

Keywords

Comments

All numbers in this sequence are divisible by 16. For A173953/16 see A173955.
a(n+2)/A173954(n+2) is, for n >= 0, the partial sum Sum_{k=0..n} 1/(k + 3/4)^2 = 16*Sum_{k=0..n} 1/(4*k + 3)^2. The limit n -> infinity is given in A282824 as Zeta(2, 3/4) = Psi(1, 3/4) = Pi^2 - 8*Catalan, with the trigamma function Psi(1, z) and the Catalan constant A006752.

Examples

			The rationals r(n) = Zeta(2, 3/4) - Zeta(2, n-1/4) begin:  0/1, 16/9, 928/441, 119344/53361, 3078464/1334025, 1132669904/481583025, 606887707616/254757420225, 49610806397296/20635351038225, ... - _Wolfdieter Lang_, Nov 14 2017
		

Crossrefs

Denominators are in A173954.

Programs

  • Magma
    [0] cat [Numerator((&+[16/(4*k+3)^2: k in [0..n-2]])): n in [2..20]]; // G. C. Greubel, Aug 23 2018
  • Maple
    r := n -> Zeta(0, 2, 3/4) - Zeta(0, 2, n-1/4):
    seq(numer(simplify(r(n))), n=1..13); # Peter Luschny, Nov 14 2017
  • Mathematica
    Table[Numerator[FunctionExpand[Pi^2 - 8*Catalan - Zeta[2, (4*n - 1)/4]]], {n, 1, 20}] (* Vaclav Kotesovec, Nov 14 2017 *)
    Numerator[Table[128*n*Sum[(4*k - 1 + 2*n) / ((4*k - 1)^2 * (4*k - 1 + 4*n)^2), {k, 1, Infinity}], {n, 0, 20}]] (* Vaclav Kotesovec, Nov 14 2017 *)
    Numerator[Table[16*Sum[1/(4*k + 3)^2, {k, 0, n-1}], {n, 1, 20}]] (* Vaclav Kotesovec, Nov 15 2017 *)
  • PARI
    for(n=1,20, print1(numerator(16*sum(k=0,n-2, 1/(4*k+3)^2)), ", ")) \\ G. C. Greubel, Aug 23 2018
    

Formula

a(n) = Numerator of (Pi^2 - 8*Catalan - Zeta(2, (4 n - 1)/4)).
Numerator of 128*n*Sum_{k>=1} (4*k - 1 + 2*n) / ((4*k - 1)^2 * (4*k - 1 + 4*n)^2). - Vaclav Kotesovec, Nov 14 2017
Numerator of 16*Sum_{k=0..n-2} 1/(4*k + 3)^2, n >= 2, with a(1) = 0. See a comment above. - Wolfdieter Lang, Nov 14 2017

Extensions

Name simplified by Peter Luschny, Nov 14 2017

A173955 a(n) = numerator of (Zeta(2, 3/4) - Zeta(2, n-1/4))/16 where Zeta(n, a) is the Hurwitz Zeta function.

Original entry on oeis.org

0, 1, 58, 7459, 192404, 70791869, 37930481726, 3100675399831, 3000384410275816, 3016572632600497, 512004171837010018, 950047080453398607307, 2104850677799349861903388, 609822785846772474028096357, 611130542819711220012487366
Offset: 1

Views

Author

Artur Jasinski, Mar 03 2010

Keywords

Comments

The denominators are given in A173954.
a(n+2)/A173954(n+2) = (Zeta(2, 3/4) - Zeta(2, n + 7/4))/16 gives, for n >= 0, the partial sum Sum_{k=0..n} 1/(4*n + 3). In the limit n -> infinity the series value is Zeta(2,3/4)/16, with the Hurwitz Zeta function, and it is given in A247037. - Wolfdieter Lang, Nov 15 2017

Crossrefs

Programs

  • Magma
    [0] cat [Numerator((&+[1/(4*k+3)^2: k in [0..n-2]])): n in [2..20]]; // G. C. Greubel, Aug 23 2018
  • Maple
    r := n -> (Zeta(0, 2, 3/4) - Zeta(0, 2, n-1/4))/16:
    seq(numer(simplify(r(n))), n=1..15); # Peter Luschny, Nov 14 2017
  • Mathematica
    Table[Numerator[FunctionExpand[(Pi^2 - 8*Catalan - Zeta[2, (4*n - 1)/4])/16]], {n, 1, 20}] (* Vaclav Kotesovec, Nov 14 2017 *)
    Numerator[Table[8*n*Sum[(4*k - 1 + 2*n) / ((4*k - 1)^2 * (4*k - 1 + 4*n)^2), {k, 1, Infinity}], {n, 0, 20}]] (* Vaclav Kotesovec, Nov 14 2017 *)
    Numerator[Table[Sum[1/(4*k + 3)^2, {k, 0, n-2}], {n, 1, 20}]] (* Vaclav Kotesovec, Nov 15 2017 *)
  • PARI
    for(n=1,20, print1(numerator(sum(k=0,n-2, 1/(4*k+3)^2)), ", ")) \\ G. C. Greubel, Aug 23 2018
    

Formula

a(n) = numerator of r(n) with r(n) = (Pi^2 - 8*Catalan - Zeta(2, n - 1/4))/16, with the Hurwitz Zeta function Z(2, z), and the Catalan constant is given in A006752. With Zeta(2, 3/4) = Pi^2 - 8*Catalan this is the formula given in the name.
Numerator of Sum_{k=0..n-2} 1/(4*k + 3)^2, n >= 2, with a(1) = 0. - G. C. Greubel, Aug 23 2018

Extensions

Numbers changed according to the old (or new) Mathematica program, and edited by Wolfdieter Lang, Nov 14 2017
Name simplified by Peter Luschny, Nov 14 2017

A282824 Decimal expansion of Pi^2 - 8*K, where K is Catalan's constant.

Original entry on oeis.org

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

Views

Author

Bruno Berselli, Mar 06 2017

Keywords

Examples

			2.5418796476716064983976628804170782491205044129874135522813644192459406...
		

Crossrefs

Programs

  • Magma
    SetDefaultRealField(RealField(100)); R:= RealField(); Pi(R)^2 - 8*Catalan(R); // G. C. Greubel, Aug 24 2018
  • Mathematica
    RealDigits[Pi^2 - 8 Catalan, 10, 100][[1]]
  • PARI
    Pi^2 - 8*Catalan \\ Charles R Greathouse IV, Jan 31 2018
    
  • PARI
    zetahurwitz(2,3/4) \\ Charles R Greathouse IV, Jan 31 2018
    

Formula

Equals 16*A247037.
Equals Psi(1, 3/4), where Psi(r, x) is the Polygamma function of order r.
Because this equals Zeta(2, 3/4), with the Hurwitz Zeta function, this is the value of the series Sum_{k>=0} 1/(k + 3/4)^2 = 16*Sum_{k>=0} 1/(4*k+3)^2 with partial sums {A173953/(n+2) / A173954(n+2)}{n>=0}. - _Wolfdieter Lang, Nov 14 2017
A282823 - this = 16*A006752. - R. J. Mathar, Jun 07 2024

A247037 Decimal expansion of Sum_{k >= 0} 1/(4*k+3)^2.

Original entry on oeis.org

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

Views

Author

Jean-François Alcover, Sep 10 2014

Keywords

Examples

			0.158867477979475406149853930026067390570031525811713347...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Sections 1.7 Catalan's constant p. 55.

Crossrefs

Programs

Formula

Equals Pi^2/16 - G/2, where G is Catalan's constant.
Equals A222068 - A006752/2.
Equals zeta(2, 3/4)/16 = Psi(1, 3/4)/16, with the Hurwitz zeta function and the Trigamma function Psi(1, z), and the partial sums of the series given in the name are {A173955(n+2) / A173954(n+2)}{n>=0}. - _Wolfdieter Lang, Nov 14 2017
Equals Integral_{x=1..oo} log(x)/(x^4 - 1) dx. - Amiram Eldar, Jul 17 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.

A287102 Numbers k such that (10^(4*k+1)*37 + 10^(2*k)*(-99) - 73)/99 is prime (k > 0).

Original entry on oeis.org

1, 9, 441, 2980
Offset: 1

Views

Author

Mikk Heidemaa, May 19 2017

Keywords

Comments

Or '37'||...'37'||'2'||'73'...||'73' in decimal form is prime (as a string, it consists of a middle '2' with the prefix composed of '37' concatenated k times, and the suffix composed of '73' concatenated k times).
a(5), if it exists, is greater than 58079. - Robert Price, May 03 2018

Examples

			9 is a term because (10^(4*9+1)*37 + 10^(2*9)*(-99) - 73)/99 = 3737373737373737372737373737373737373 (prime). As a string, it consists of a middle '2' with the prefix '373737373737373737' ('37' concatenated 9 times) and the suffix '737373737373737373' ('73' concatenated 9 times).
		

Crossrefs

Programs

  • Mathematica
    ParallelMap[ If[ PrimeQ[ (10^(4*#+1)*37+10^(2*#)*(-99)-73)/99], #, Nothing]&, Range[3000]]
Showing 1-9 of 9 results.