A006752 -id:A006752 - cp's OEIS frontend

A050469 a(n) = Sum_{ d divides n, n/d=1 mod 4} d - Sum_{ d divides n, n/d=3 mod 4} d.

1, 2, 2, 4, 6, 4, 6, 8, 7, 12, 10, 8, 14, 12, 12, 16, 18, 14, 18, 24, 12, 20, 22, 16, 31, 28, 20, 24, 30, 24, 30, 32, 20, 36, 36, 28, 38, 36, 28, 48, 42, 24, 42, 40, 42, 44, 46, 32, 43, 62, 36, 56, 54, 40, 60, 48, 36, 60, 58, 48, 62, 60, 42, 64, 84, 40

Offset: 1

N. J. A. Sloane, Dec 23 1999

Multiplicative with a(p^e)=p^e if p=2, (p^(e+1)-1)/(p-1) if p==1 (mod 4), else (p^(e+1)+(-1)^e)/(p+1). - Michael Somos, May 02 2005

Multiplicative because it is the Dirichlet convolution of A000027 = n and A101455 = [1 0 -1 0 1 0 -1 ...], which are both multiplicative. - Christian G. Bower, May 17 2005

Seiichi Manyama, Table of n, a(n) for n = 1..10000
J. W. L. Glaisher, On the representations of a number as the sum of two, four, six, eight, ten, and twelve squares, Quart. J. Math. 38 (1907), 1-62 (see p. 4 and p. 8).
Index entries for sequences mentioned by Glaisher.

Cf. A006752, A050470, A050471, A050468, A175647, A243381.

max = 70; s = Sum[n*x^(n-1)/(1+x^(2*n)), {n, 1, max}] + O[x]^max; CoefficientList[s, x] (* Jean-François Alcover, Dec 02 2015 *)
f[p_, e_] := Which[p == 2, p^e, Mod[p, 4] == 1, (p^(e + 1) - 1)/(p - 1), Mod[p, 4] == 3, (p^(e + 1) + (-1)^e)/(p + 1)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Nov 06 2022 *)

a(n)=if(n<1,0,sumdiv(n,d,d*((n/d%4==1)-(n/d%4==3))))

{a(n)=local(A,p,e); if(n<2, n==1, A=factor(n); prod(k=1,matsize(A)[1], if(p=A[k,1], e=A[k,2]; if(p==2, p^e, if(p%4==1, (p^(e+1)-1)/(p-1), (p^(e+1)+(-1)^e)/(p+1)))))) } /* Michael Somos, May 02 2005 */

a(n)=if(n<1,0,polcoeff(sum(k=1,n,k*x^k/(1+x^(2*k)),x*O(x^n)),n))

G.f.: Sum_{n>=1} n*x^n/(1+x^(2*n)). - Vladeta Jovovic, Oct 16 2002
L.g.f.: Sum_{k>=1} arctan(x^k). - Ilya Gutkovskiy, Dec 16 2019
O.g.f.: Sum_{n >= 1} (-1)^(n+1) * x^(2*n-1)/(1 - x^(2*n-1))^2. - Peter Bala, Jan 04 2021
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{primes p == 1 (mod 4)} 1/(1-1/p^2) * Product_{primes p == 3 (mod 4)} 1/(1+1/p^2) = (1/2) * A175647 / A243381 = A006752/2 = 0.4579827970... . - Amiram Eldar, Nov 06 2022, Nov 05 2023
a(n) = Sum_{d|n} (n/d)*sin(d*Pi/2). - Ridouane Oudra, Sep 26 2024

A060968 Number of solutions to x^2 + y^2 == 1 (mod n).

Original entry on oeis.org

1, 2, 4, 8, 4, 8, 8, 16, 12, 8, 12, 32, 12, 16, 16, 32, 16, 24, 20, 32, 32, 24, 24, 64, 20, 24, 36, 64, 28, 32, 32, 64, 48, 32, 32, 96, 36, 40, 48, 64, 40, 64, 44, 96, 48, 48, 48, 128, 56, 40, 64, 96, 52, 72, 48, 128, 80, 56, 60, 128, 60, 64, 96, 128, 48, 96, 68, 128, 96, 64, 72

Offset: 1

Ahmed Fares (ahmedfares(AT)my-deja.com), May 09 2001

From Jianing Song, Nov 05 2019: (Start)
a(n) is also the order of the group SO(2,Z_n), i.e., the group of 2 X 2 matrices A over Z_n such that A*A^T = E = [1,0;0,1] and det(A) = 1. Elements in SO(2,Z_n) are of the form [x,y;-y,x] where x^2+y^2 == 1 (mod n). For example, SO(2,Z_4) = {[1,0;0,1], [0,1;3,0], [1,2;2,1], [2,1;3,2], [3,0;0,3], [0,3;1,0], [3,2;2,3], [2,3;1,2]}. Note that SO(2,Z_n) is abelian, and it is isomorphic to the multiplicative group G_n := {x+yi: x^2 + y^2 = 1, x,y in Z_n} where i = sqrt(-1), by the mapping [x,y;-y,x] <-> x+yi. See my link below for the group structure of SO(2,Z_n).
The exponent of SO(2,Z_n) (i.e., least e > 0 such that x^e = E for every x in SO(2,Z_n)) is given by A235863(n).
The rank of SO(2,Z_n) (i.e., the minimum number of generators) is omega(n) if n is not divisible by 4, omega(n)+1 if n is divisible by 4 but not by 8 and omega(n)+2 if n is divisible by 8, omega = A001221. (End)
In general, let R be any commutative ring with unity, O(m,R) be the group of m X m matrices A over R such that A*A^T = E and SO(m,R) be the group of m X m matrices A over R such that A*A^T = E and det(A) = 1, then O(m,R)/SO(m,R) = {square roots of unity in R*}, where R* is the multiplicative group of R. This is because if we define f(M) = det(M) for M in O(m,R), then f is a surjective homomorphism from O(m,R) to {square roots of unity in R*}, and SO(m,R) is its kernel. See also A182039. - Jianing Song, Nov 08 2019

			a(3) = 4 because the 4 solutions are (0,1), (0,2), (1,0), (2,0).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Jianing Song, Structure of the group SO(2,Z_n).
László Tóth, Counting solutions of quadratic congruences in several variables revisited, arXiv preprint arXiv:1404.4214 [math.NT], 2014.
László Tóth, Counting Solutions of Quadratic Congruences in Several Variables Revisited, J. Int. Seq. 17 (2014), Article 14.11.6.

Cf. A006752, A060594, A087784.
Cf. A027748, A124010.
Cf. A235863, A182039.

a060968 1 = 1
a060968 n = (if p == 2 then (if e == 1 then 2 else 2^(e+1)) else 1) *
   (product $ zipWith (*) (map (\q -> q - 2 + mod q 4) ps'')
                          (zipWith (^) ps'' (map (subtract 1) es'')))
   where (ps'', es'') = if p == 2 then (ps, es) else (ps', es')
         ps'@(p:ps) = a027748_row n; es'@(e:es) = a124010_row n
-- Reinhard Zumkeller, Aug 05 2014

fa=FactorInteger; phi[p_,s_] := Which[Mod[p,4] == 1, p^(s-1)*(p-1), Mod[p,4]==3, p^(s-1)*(p+1), s==1, 2, True, 2^(s+1)]; phi[1]=1; phi[n_] := Product[phi[fa[n][[i,1]], fa[n][[i,2]]], {i, Length[fa[n]]}]; Table[phi[n], {n,1,100}]

a(n)=my(f=factor(n)[,1]);n*prod(i=if(n%2,1,2),#f,if(f[i]%4==1, 1-1/f[i], 1+1/f[i]))*if(n%4,1,2) \\ Charles R Greathouse IV, Apr 16 2012

Multiplicative, with a(2^e) = 2 if e = 1 or 2^(e+1) if e > 1, a(p^e) = (p-1)*p^(e-1) if p == 1 (mod 4), a(p^e) = (p+1)*p^(e-1) if p == 3 (mod 4). - David W. Wilson, Jun 19 2001
a(n) = n * (Product_{prime p|n, p == 1 (mod 4)} (1 - 1/p)) * (Product_{prime p|n, p == 3 (mod 4)} (1 + 1/p)) * (1 + [4|n]) where "[ ]" is the Iverson bracket. - Ola Veshta (olaveshta(AT)my-deja.com), May 18 2001
a(n) = A182039(n)/A060594(n). - Jianing Song, Nov 08 2019
Sum_{k=1..n} a(k) ~ c * n^2 + O(n*log(n)), where c = 5/(8*G) = 0.682340..., where G is Catalan's constant (A006752) (Tóth, 2014). - Amiram Eldar, Oct 18 2022

A077591 Maximum number of regions into which the plane can be divided using n (concave) quadrilaterals.

Original entry on oeis.org

1, 2, 18, 50, 98, 162, 242, 338, 450, 578, 722, 882, 1058, 1250, 1458, 1682, 1922, 2178, 2450, 2738, 3042, 3362, 3698, 4050, 4418, 4802, 5202, 5618, 6050, 6498, 6962, 7442, 7938, 8450, 8978, 9522, 10082, 10658, 11250, 11858, 12482, 13122, 13778

Offset: 0

Joshua Zucker and the Castilleja School MathCounts club, Nov 07 2002

Sequence found by reading the segment (1, 2) together with the line from 2, in the direction 2, 18, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 05 2011
For a(n) > 1, a(n) are the numbers such that phi(sum of the odd divisors of a(n)) = phi(sum of even divisors of a(n)). - Michel Lagneau, Sep 14 2011
Apart from first term, subsequence of A195605. - Bruno Berselli, Sep 21 2011
Engel expansion of 1F2(1; 1/2, 1/2; 1/8). - Benedict W. J. Irwin, Jun 21 2018
Let f(n) = 4*n^2 - 5, then (x, y, z) = (a(n+1), -f(n), -f(n + 1)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 512. - XU Pingya, Apr 25 2022

			a(2) = 18 if you draw two concave quadrilaterals such that all four sides of one cross all four sides of the other.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Keyang Li, when n=1, number of regions is 2; when n=2, maximum number of regions is 18
Keyang Li, when n=3, maximum number of regions is 50
Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).

Cf. A006752, A077588, A239186.
Cf. A071974, A071975.
Cf. A016754, A069129.

Concatenation([1], List([1..2000], n->8*n^2 - 8*n + 2)); # Muniru A Asiru, Jan 29 2018

A077591:=n->`if`(n=0, 1, 8*n^2 - 8*n + 2); seq(A077591(n), n=0..50); # Wesley Ivan Hurt, Mar 12 2014

Table[2*(4*n^2 - 4*n + 1), {n,0,50}] (* G. C. Greubel, Jul 15 2017 *)

isok(n) = (sod = sumdiv(n, d, (d%2)*d)) && (sed = sumdiv(n, d, (1 - d%2)*d)) && (eulerphi(sod) == eulerphi(sed)); \\ from Michel Lagneau comment; Michel Marcus, Mar 15 2014

a(n) = 8*n^2 - 8*n + 2 = 2*(2*n-1)^2, n > 0, a(0)=1.
Proof from Keyang Li, Jun 18 2022: (Start)
Represent the configuration of n concave quadrilaterals by a planar graph with a node for each vertex of the quadrilaterals and for each intersection point. Let there be v_n nodes and e_n edges. By Euler's formula for planar graphs, a(n) = e_n - v_n + 2. When we go from n to n+1 quadrilaterals, each side of the new quadrilateral can meet each side of the existing quadrilaterals at most 4 times, so Dv_n := v_{n+1} - v_n <= 4*4n = 16n.
Each of these intersection points increases the number of edges in the graph by 2, so De_n := e_{n+1} - e_n = 4 + 2*Dv_n, Da_n := a(n+1) - a(n) = 4 + Dv_n <= 4+16*n.
These upper bounds can be achieved by taking n interwoven concave quadrilaterals (for n=1,2,3 see the attached Keyang Li links), and we achieve a(n) = 8n^2 - 8n + 2 (and v_n = 8n^2 - 4n, e_n = 4n*(4n-3)) for n > 0. QED (End)
For n > 0: A071974(a(n)) = 2*n+1, A071975(a(n)) = 2. - Reinhard Zumkeller, Jul 10 2011
a(n) = 1 + A069129(n), if n >= 1. - Omar E. Pol, Sep 05 2011
a(n) = 2*A016754(n-1), if n >= 1. - Omar E. Pol, Sep 05 2011
G.f.: (1 - x + 15*x^2 + x^3)/(1-x)^3. - Colin Barker, Feb 23 2012
E.g.f.: (8*x^2 + 2)*exp(x) - 1. - G. C. Greubel, Jul 15 2017
From Amiram Eldar, Jan 29 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/16.
Sum_{n>=1} (-1)^(n+1)/a(n) = G/2, where G is Catalan constant (A006752).
Product_{n>=1} (1 + 1/a(n)) = cosh(Pi/sqrt(8)).
Product_{n>=1} (1 - 1/a(n)) = cos(Pi/sqrt(8)). (End)

A016946 a(n) = (6*n+3)^2.

Original entry on oeis.org

9, 81, 225, 441, 729, 1089, 1521, 2025, 2601, 3249, 3969, 4761, 5625, 6561, 7569, 8649, 9801, 11025, 12321, 13689, 15129, 16641, 18225, 19881, 21609, 23409, 25281, 27225, 29241, 31329, 33489, 35721, 38025, 40401, 42849, 45369, 47961, 50625, 53361, 56169

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A002378, A006752, A016754, A016945, A086729.

[(6*n+3)^2: n in [0..60]]; // Vincenzo Librandi, May 05 2011

A016946:=n->(6*n+3)^2: seq(A016946(n), n=0..50); # Wesley Ivan Hurt, Oct 13 2014

(6 Range[0, 50] + 3)^2 (* or *)
CoefficientList[Series[9 (1 + 6 x + x^2)/(1 - x)^3, {x, 0, 50}], x] (* Wesley Ivan Hurt, Oct 13 2014 *)
LinearRecurrence[{3,-3,1},{9,81,225},40] (* Harvey P. Dale, Jul 13 2015 *)

a(n)=(6*n+3)^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 36*A002378(n)+9. - Jean-Bernard François, Oct 12 2014
From Wesley Ivan Hurt, Oct 13 2014: (Start)
G.f.: 9*(1+6*x+x^2)/(1-x)^3.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3).
a(n) = A016945(n)^2 = A000290(A016945(n)). (End)
Sum_{n>=0} 1/a(n) = A086729. - Amiram Eldar, Nov 16 2020
a(n) = 9*A016754(n). - R. J. Mathar, Dec 11 2020
Sum_{n>=0} (-1)^n/a(n) = G/9, where G is Catalan's constant (A006752). - Amiram Eldar, Mar 30 2022
E.g.f.: 9*exp(x)*(1 + 8*x + 4*x^2). - Stefano Spezia, Aug 19 2022

A173947 a(n) = numerator of (Zeta(2, 1/4) - Zeta(2, n+1/4)), where Zeta is the Hurwitz Zeta function.

Original entry on oeis.org

0, 16, 416, 34096, 5794624, 1680121936, 82501802464, 2065646660464, 1739147340740224, 210617970218777104, 288533264855755545376, 485294472126860897387056, 485518650207447822251456

Offset: 0

Artur Jasinski, Mar 03 2010

For A173947/16 see A173949.

a(n+1)/A173948(n+1) =: r(n) = (Zeta(2, 1/4) - Zeta(2, n + 5/4)), the partial sum Sum_{k=0..n} 1/(k + 1/4)^2, n >= 0. The limit is Zeta(2, 1/4) = A282823 = 16*A222183. - Wolfdieter Lang, Nov 14 2017

G. C. Greubel, Table of n, a(n) for n = 0..250
Eric Weisstein's World of Mathematics, Hurwitz Zeta Function
Eric Weisstein's World of Mathematics, Trigamma Function

Cf. A006752, A120268, A173945, A173948 (denominators), A173949.

[1] cat [Numerator((&+[1/(4*k+1)^2: k in [0..n-1]])): n in [1..20]]; // G. C. Greubel, Aug 22 2018

r := n -> Psi(1, 1/4) - Zeta(0, 2, n+1/4):
seq(numer(simplify(r(n))), n=0..13); # Peter Luschny, Nov 14 2017

Table[Numerator[FunctionExpand[8*Catalan + Pi^2 - Zeta[2, (4*n + 1)/4]]], {n, 0, 20}] (* Vaclav Kotesovec, Nov 14 2017 *)
Numerator[Table[128*n*Sum[(1 + 4*k + 2*n) / ((1 + 4*k)^2*(1 + 4*k + 4*n)^2), {k, 0, Infinity}], {n, 0, 20}]] (* Vaclav Kotesovec, Nov 14 2017 *)
Numerator[Table[16*Sum[1/(4*k + 1)^2, {k, 0, n - 1} ], {n, 0, 20}]] (* Vaclav Kotesovec, Nov 14 2017 *)

for(n=0,20, print1(numerator(sum(k=0,n-1, 1/(4*k+1)^2)), ", ")) \\ G. C. Greubel, Aug 22 2018

a(n) = numerator of 8*Catalan + Pi^2 - Zeta(2, (4 n + 1)/4), with the Catalan constant given in A006752.

a(n) = numerator(r(n)) with r(n) = Zeta(2, 1/4) - Zeta(2, n + 1/4), with the Hurwitz Zeta function (see the name). With Zeta(2, 1/4) = Psi(1, 1/4) = 8*Catalan + Pi^2 this is the preceding formula, where Psi(1, z) is the Trigamma function. - Wolfdieter Lang, Nov 14 2017

Name simplified and offset set to 0 by Peter Luschny, Nov 14 2017

A173948 a(n) = denominator of (Zeta(2, 1/4) - Zeta(2, n+1/4)), where Zeta is the Hurwitz Zeta function.

Original entry on oeis.org

1, 1, 25, 2025, 342225, 98903025, 4846248225, 121156205625, 101892368930625, 12328976640605625, 16878369020989100625, 28372538324282678150625, 28372538324282678150625, 1390254377889851229380625, 3905224547492592103330175625, 1409786061644825749302193400625, 5245813935380396613153461643725625

Offset: 0

Artur Jasinski, Mar 03 2010

Presumably conjectures:
For n>=2 numbers in this sequence are divisible by 25.
For n>=7 numbers in this sequence are divisible by 25^2.

G. C. Greubel, Table of n, a(n) for n = 0..250

Cf. A006752, A120268, A173945, A173947 (numerators).

[1] cat [Denominator((&+[1/(4*k+1)^2: k in [0..n-1]])): n in [1..20]]; // G. C. Greubel, Aug 22 2018

r := n -> Psi(1, 1/4) - Zeta(0, 2, n+1/4):
seq(denom(simplify(r(n))), n=0..16); # Peter Luschny, Nov 14 2017

Table[Denominator[FunctionExpand[8*Catalan + Pi^2 - Zeta[2, (4*n + 1)/4]]], {n, 0, 20}] (* Vaclav Kotesovec, Nov 14 2017 *)
Denominator[Table[Sum[1/(4*k + 1)^2, {k, 0, n-1} ], {n, 0, 20}]] (* Vaclav Kotesovec, Nov 14 2017 *)

for(n=0,20, print1(denominator(sum(k=0,n-1, 1/(4*k+1)^2)), ", ")) \\ G. C. Greubel, Aug 22 2018

a(n) = denominator of 8*Catalan + Pi^2 - Zeta(2, (4*n + 1)/4), with the Hurwitz Zeta function, and Catalan is given in A006752. [See the name with Zeta(2, 1/4) = Psi(1, 1/4) = 8*Catalan + Pi^2, and the Trigamma function Psi(1, z).]

Name simplified by Peter Luschny, Nov 14 2017

Formula reformulated. - Wolfdieter Lang, Nov 14 2017.

A065072 Number of ways to tile a square of side 2n by dominoes (rectangles of size 2 X 1 or 1 X 2) is 2^n * a(n)^2 (see A004003).

Original entry on oeis.org

1, 1, 3, 29, 901, 89893, 28793575, 29607089625, 97725875584681, 1035449388414303593, 35216739783694029601963, 3844747107219467355553841461, 1347358497824862447450096142795629, 1515633798331963142551890627742773295309

Offset: 0

Nicolau C. Saldanha (nicolau(AT)mat.puc-rio.br), Nov 08 2001

nonn

A099390 is the main entry for this problem. - N. J. A. Sloane, Mar 15 2015

			G.f. = 1 + x + 3*x^2 + 29*x^3 + 901*x^4 + 89893*x^5 + 28793575*x^6 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..50 (terms n=1..25 from T. D. Noe)
N. Allegra, Exact solution of the 2d dimer model: Corner free energy, correlation functions and combinatorics, arXiv:1410.4131 [cond-mat.stat-mech], 2014.
H. Cohn, 2-adic behavior of numbers of domino tilings, arXiv:math/0008222 [math.CO], 2000.
Philippe Di Francesco, Twenty Vertex model and domino tilings of the Aztec triangle, arXiv:2102.02920 [math.CO], 2021. Mentions this sequence.
Laura Florescu, Daniela Morar, David Perkinson, Nicholas Salter, and Tianyuan Xu, Sandpiles and Dominos, Electronic Journal of Combinatorics, Volume 22(1), 2015. [Mentions this sequence together with a different sequence (A256043) with the same initial terms]
Peter E. John and Horst Sachs, On a strange observation in the theory of the dimer problem, Discrete Math. 216 (2000), no. 1-3, 211-219. [_N. J. A. Sloane_, Feb 06 2012]
James Propp, Some 2-Adic Conjectures Concerning Polyomino Tilings of Aztec Diamonds, Integers (2023) Vol. 23, Art. A30.

Cf. A099390, A004003, A256043.

a[n_] := With[{L = 2n}, Sqrt[Product[4 Cos[p Pi/(L+1)]^2 + 4 Cos[q Pi/(L+1)]^2, {p, 1, L/2}, {q, 1, L/2}]/2^(L/2)] // Round];
Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Nov 11 2018 *)
Table[Resultant[ChebyshevU[2*n, x/2], ChebyshevU[2*n, I*x/2], x]^(1/4) / 2^(n/2), {n, 0, 15}] (* Vaclav Kotesovec, Dec 30 2020 *)

a(n) ~ exp(G*(2*n + 1)^2/(2*Pi)) / (2^((n-1)/2) * (1 + sqrt(2))^(n + 1/2)), where G is Catalan's constant A006752. - Vaclav Kotesovec, Apr 14 2020, updated Dec 30 2020

a(0)=1 prepended by Alois P. Heinz, Mar 25 2015

A173949 a(n) = numerator of (Zeta(2, 1/4) - Zeta(2, n+1/4))/16, where Zeta is the Hurwitz Zeta function.

Original entry on oeis.org

0, 1, 26, 2131, 362164, 105007621, 5156362654, 129102916279, 108696708796264, 13163623138673569, 18033329053484721586, 30330904507928806086691, 30344915637965488890716, 1487479897654682071525709

Offset: 0

Artur Jasinski, Mar 03 2010

For the Catalan constant see A006752.
The denominators are given in A173948.
a(n+1)/A173948(n+1), for n>= 0, gives the partial sum Sum_{k=0..n} 1/(4*k + 1)^2. For {(4*k + 1)^2}A016814.%20The%20limit%20n%20-%3E%20infinity%20is%20given%20in%20A222183%20as%201.074833072...%20.%20-%20_Wolfdieter%20Lang">{k>=0} see A016814. The limit n -> infinity is given in A222183 as 1.074833072... . - _Wolfdieter Lang, Nov 14 2017

			The rationals r(n) begin: 0/1, 1/1, 26/25, 2131/2025, 362164/342225, 105007621/98903025, 5156362654/4846248225, 129102916279/121156205625, 108696708796264/101892368930625, 13163623138673569/12328976640605625, ... - _Wolfdieter Lang_, Nov 14 2017

G. C. Greubel, Table of n, a(n) for n = 0..250
Eric Weisstein's World of Mathematics, Hurwitz Zeta Function
Eric Weisstein's World of Mathematics, Polygamma Function

Cf. A016814, A120268, A173945, A173947, A173948, A222183.

[0] cat [Numerator((&+[1/(4*k+1)^2: k in [0..n-1]])): n in [1..20]]; // G. C. Greubel, Aug 23 2018

r := n -> (Psi(1, 1/4) - Zeta(0, 2, n+1/4))/16:
seq(numer(simplify(r(n))), n=0..13); # Peter Luschny, Nov 14 2017

Table[Numerator[FunctionExpand[(8*Catalan + Pi^2 - Zeta[2, (4*n + 1)/4])/16]], {n, 0, 20}] (* Vaclav Kotesovec, Nov 14 2017 *)
Numerator[Table[Sum[1/(4*k + 1)^2, {k, 0, n-1}], {n, 0, 20}]] (* Vaclav Kotesovec, Nov 14 2017 *)

for(n=0,20, print1(numerator(sum(k=0,n-1, 1/(4*k+1)^2)), ", ")) \\ G. C. Greubel, Aug 23 2018

a(n) = numerator of expression (8*Catalan + Pi^2 - Zeta(2, (4*n + 1)/4))/16.

a(n) = numerator(r(n)) with r(n) = (Zeta(2,1/4) - Zeta(2, n + 1/4))/16, with the Hurwitz Zeta function Z(2, k). With Zeta(2,1/4) = 8 Catalan + Pi^2 this is the preceding formula, and Zeta(2, n + 1/4) = Psi(1, n + 1/4) with the polygamma (trigamma) function Psi(1, k). - Wolfdieter Lang, Nov 14 2017

Edited by Wolfdieter Lang, Nov 14 2017

Name changed according to a formula of Lang by Peter Luschny, Nov 14 2017

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

Artur Jasinski, Mar 03 2010

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.

			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

G. C. Greubel, Table of n, a(n) for n = 1..250
Eric Weisstein's World of Mathematics, Hurwitz Zeta Function
Eric Weisstein's World of Mathematics, Trigamma Function

Denominators are in A173954.

Cf. A006752, A120268, A173945, A173947, A173948, A173949, A173955.

[0] cat [Numerator((&+[16/(4*k+3)^2: k in [0..n-2]])): n in [2..20]]; // G. C. Greubel, Aug 23 2018

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

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 *)

for(n=1,20, print1(numerator(16*sum(k=0,n-2, 1/(4*k+3)^2)), ", ")) \\ G. C. Greubel, Aug 23 2018

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

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

Artur Jasinski, Mar 03 2010

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

G. C. Greubel, Table of n, a(n) for n = 1..250

Cf. A006752, A120268, A173945, A173947, A173948, A173949, A173953, A173954, A247037.

[0] cat [Numerator((&+[1/(4*k+3)^2: k in [0..n-2]])): n in [2..20]]; // G. C. Greubel, Aug 23 2018

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

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 *)

for(n=1,20, print1(numerator(sum(k=0,n-2, 1/(4*k+3)^2)), ", ")) \\ G. C. Greubel, Aug 23 2018

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

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

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