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 1-10 of 21 results. Next

A120298 Largest prime factor of the n-th central factorial number A001819(n).

Original entry on oeis.org

5, 7, 41, 479, 59, 266681, 63397, 514639, 178939, 242101, 18500393, 40799043101, 190297, 1844659, 78206663, 616798327, 38512247, 86364397717734821, 421950627598601, 84786899, 52030193, 31411862913089, 1422157053067, 47961582392203, 8408339, 6269019298981
Offset: 2

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Author

Alexander Adamchuk, Jul 11 2006

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Max[FactorInteger[Sum[1/i^2,{i,1,n}]/Product[1/i^2,{i,1,n}]]],{n,2,30}]
    Table[FactorInteger[n!^2*HarmonicNumber[n,2]][[-1,1]],{n,2,30}] (* Harvey P. Dale, Jul 29 2023 *)
  • PARI
    a(n) = vecmax(factor(n!^2 * sum(k=1, n, 1/k^2))[,1]); \\ Daniel Suteu, Jun 10 2022

Formula

a(n) = A006530(A001819(n)). - Daniel Suteu, Jun 10 2022

Extensions

More terms from Daniel Suteu, Jun 10 2022

A007407 a(n) = denominator of Sum_{k=1..n} 1/k^2.

Original entry on oeis.org

1, 4, 36, 144, 3600, 3600, 176400, 705600, 6350400, 1270080, 153679680, 153679680, 25971865920, 25971865920, 129859329600, 519437318400, 150117385017600, 150117385017600, 54192375991353600, 10838475198270720, 221193371393280
Offset: 1

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Author

Keywords

Comments

Denominators of the Eulerian numbers T(-2,k) for k = 0,1..., if T(n,k) is extended to negative n by the recurrence T(n,k) = (k+1)*T(n-1,k) + (n-k)*T(n-1,k-1) (indexed as in A173018). - Michael J. Collins, Oct 10 2024

Examples

			1/1^2 + 1/2^2 + 1/3^2 = 1/1 + 1/4 + 1/9 = 49/36, so a(3) = 36. - _Jon E. Schoenfield_, Dec 26 2014
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A007406 (numerators), A000290, A035166.

Programs

  • Haskell
    import Data.Ratio ((%), denominator)
    a007407 n = a007407_list !! (n-1)
    a007407_list = map denominator $
                       scanl1 (+) $ map (1 %) $ tail a000290_list
    -- Reinhard Zumkeller, Jul 06 2012
    
  • Maple
    ZL:=n->sum(1/i^2, i=2..n): a:=n->floor(denom(ZL(n))): seq(a(n), n=1..21); # Zerinvary Lajos, Mar 28 2007
  • Mathematica
    s=0;lst={};Do[s+=n^2/n^4;AppendTo[lst,Denominator[s]],{n,3*4!}];lst (* Vladimir Joseph Stephan Orlovsky, Jan 24 2009 *)
    Table[Denominator[Pi^2/6 - Zeta[2, x]], {x, 1, 22}] (* Artur Jasinski, Mar 03 2010 *)
    Denominator[Accumulate[1/Range[30]^2]] (* Harvey P. Dale, Nov 08 2012 *)
  • PARI
    a(n)=denominator(sum(k=1,n,1/k^2)) \\ Charles R Greathouse IV, Nov 20 2012
    
  • Python
    from fractions import Fraction
    def A007407(n): return sum(Fraction(1,k**2) for k in range(1,n+1)).denominator # Chai Wah Wu, Apr 03 2021

Formula

a(n) = denominator of (Pi^2)/6 - zeta(2, x). - Artur Jasinski, Mar 03 2010
a(n) = A001044(n) / gcd(A001819(n), A001044(n)). - Daniel Suteu, Dec 25 2016

A008955 Triangle of central factorial numbers |t(2n,2n-2k)| read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 4, 1, 14, 49, 36, 1, 30, 273, 820, 576, 1, 55, 1023, 7645, 21076, 14400, 1, 91, 3003, 44473, 296296, 773136, 518400, 1, 140, 7462, 191620, 2475473, 15291640, 38402064, 25401600, 1, 204, 16422, 669188, 14739153, 173721912, 1017067024, 2483133696, 1625702400
Offset: 0

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Author

Keywords

Comments

Discussion of Central Factorial Numbers by N. J. A. Sloane, Feb 01 2011: (Start)
Here is Riordan's definition of the central factorial numbers t(n,k) given in Combinatorial Identities, Section 6.5:
For n >= 0, expand the polynomial
x^[n] = x*Product{i=1..n-1} (x+n/2-i) = Sum_{k=0..n} t(n,k)*x^k.
The t(n,k) are not always integers. The cases n even and n odd are best handled separately.
For n=2m, we have:
x^[2m] = Product_{i=0..m-1} (x^2-i^2) = Sum_{k=1..m} t(2m,2k)*x^(2k).
E.g. x^[8] = x^2(x^2-1^2)(x^2-2^2)(x^2-3^2) = x^8-14x^6+49x^4-36x^2,
which corresponds to row 4 of the present triangle.
So the m-th row of the present triangle gives the absolute values of the coefficients in the expansion of Product_{i=0..m-1} (x^2-i^2).
Equivalently, and simpler, the n-th row gives the coefficients in the expansion of Product_{i=1..n-1}(x+i^2), highest powers first.
For n odd, n=2m+1, we have:
x^[2m+1] = x*Product_{i=0..m-1}(x^2-((2i+1)/2)^2) = Sum_{k=0..m} t(2m+1,2k+1)*x^(2k+1).
E.g. x^[5] = x(x^2-(1/2)^2)(x^2-(3/2)^2) = x^5-10x^3/4+9x/16,
which corresponds to row 2 of the triangle in A008956.
We now rescale to get integers by replacing x by x/2 and multiplying by 2^(2m+1) (getting 1, -10, 9 from the example).
The result is that row m of triangle A008956 gives the coefficients in the expansion of x*Product_{i=0..m} (x^2-(2i+1)^2).
Equivalently, and simpler, the n-th row of A008956 gives the coefficients in the expansion of Product_{i=0..n-1} (x+(2i+1)^2), highest powers first.
Note that the n-th row of A182867 gives the coefficients in the expansion of Product_{i=1..n} (x+(2i)^2), highest powers first.
(End)
Contribution from Johannes W. Meijer, Jun 18 2009: (Start)
We define Beta(n-z,n+z)/Beta(n,n) = Gamma(n-z)*Gamma(n+z)/Gamma(n)^2 = sum(EG2[2m,n]*z^(2m), m = 0..infinity) with Beta(z,w) the Beta function. The EG2[2m,n] coefficients are quite interesting, see A161739. Our definition leads to EG2[2m,1] = 2*eta(2m) and the recurrence relation EG2[2m,n] = EG2[2m,n-1] - EG2[2m-2,n-1]/(n-1)^2 for m = -2, -1, 0, 1, 2, ... and n = 2, 3, ... , with eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. We found for the matrix coefficients EG2[2m,n] = sum((-1)^(k+n)*t1(n-1,k-1)*2*eta(2*m-2*n+2*k)/((n-1)!)^2,k=1..n) with the central factorial numbers t1(n,m) as defined above, see also the Maple program.
From the EG2 matrix we arrive at the ZG2 matrix, see A161739 for its odd counterpart, which is defined by ZG2[2m,1] = 2*zeta(2m) and the recurrence relation ZG2[2m,n] = ZG2[2m-2,n-1]/(n*(n-1))-(n-1)*ZG2[2m,n-1]/n for m = -2, -1, 0, 1, 2, ... and n = 2, 3, ... . We found for the ZG2[2m,n] = Sum_{k=1..n} (-1)^(k+1)*t1(n-1,k-1)* 2* zeta(2*m-2*n+2*k)/((n-1)!*(n)!), and we see that the central factorial numbers t1(n,m) once again play a crucial role.
(End)

Examples

			Triangle begins:
  1;
  1,   1;
  1,   5,   4;
  1,  14,  49,  36;
  1,  30, 273, 820, 576;
  ...
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part 1, Springer-Verlag 1985.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.

Crossrefs

Cf. A036969.
Columns include A000330, A000596, A000597. Right-hand columns include A001044, A001819, A001820, A001821. Row sums are in A101686.
Appears in A160464 (Eta triangle), A160474 (Zeta triangle), A160479 (ZL(n)), A161739 (RSEG2 triangle), A161742, A161743, A002195, A002196, A162440 (EG1 matrix), A162446 (ZG1 matrix) and A163927. - Johannes W. Meijer, Jun 18 2009, Jul 06 2009 and Aug 17 2009
Cf. A234324 (central terms).

Programs

  • GAP
    T:= function(n,k)
        if k=0 then return 1;
        elif k=n then return (Factorial(n))^2;
        else return n^2*T(n-1,k-1) + T(n-1,k);
        fi;
      end;
    Flat(List([0..8], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Sep 14 2019
  • Haskell
    a008955 n k = a008955_tabl !! n !! k
    a008955_row n = a008955_tabl !! n
    a008955_tabl = [1] : f [1] 1 1 where
       f xs u t = ys : f ys v (t * v) where
         ys = zipWith (+) (xs ++ [t^2]) ([0] ++ map (* u^2) (init xs) ++ [0])
         v = u + 1
    -- Reinhard Zumkeller, Dec 24 2013
    
  • Magma
    T:= func< n,k | Factorial(2*(n+1))*(&+[(-1)^j*Binomial(n,k-j)*(&+[2^(m-2*k)*StirlingFirst(2*(n-k+1)+m, 2*(n-k+1))*Binomial(2*(n-k+1)+2*j-1, 2*(n-k+1)+m-1)/Factorial(2*(n-k+1)+m): m in [0..2*j]]): j in [0..k]]) >;
    [T(n,k): k in [0..n], n in [0..8]]; // G. C. Greubel, Sep 14 2019
    
  • Maple
    nmax:=7: for n from 0 to nmax do t1(n, 0):=1: t1(n, n):=(n!)^2 end do: for n from 1 to nmax do for k from 1 to n-1 do t1(n, k) := t1(n-1, k-1)*n^2 + t1(n-1, k) end do: end do: seq(seq(t1(n, k), k=0..n), n=0..nmax); # Johannes W. Meijer, Jun 18 2009, Revised Sep 16 2012
    t1 := proc(n,k)
            sum((-1)^j*stirling1(n+1,n+1-k+j)*stirling1(n+1,n+1-k-j),j=-k..k) ;
    end proc: # Mircea Merca, Apr 02 2012
    # third Maple program:
    T:= proc(n, k) option remember; `if`(k=0, 1,
          add(T(j-1, k-1)*j^2, j=1..n))
        end:
    seq(seq(T(n, k), k=0..n), n=0..8);  # Alois P. Heinz, Feb 19 2022
  • Mathematica
    t[n_, 0]=1; t[n_, n_]=(n!)^2; t[n_ , k_ ]:=t[n, k] = n^2*t[n-1, k-1] + t[n-1, k]; Flatten[Table[t[n, k], {n,0,8}, {k,0,n}] ][[1 ;; 42]]
    (* Jean-François Alcover, May 30 2011, after recurrence formula *)
  • Maxima
    T(n,m):=(2*(n+1))!*sum((-1)^k*binomial(n,m-k)*sum((2^(i-2*m)*stirling1(2*(n-m+1)+i,2*(n-m+1))*binomial(2*(n-m+1)+2*k-1,2*(n-m+1)+i-1))/(2*(n-m+1)+i)!,i,0,2*k),k,0,m); /* Vladimir Kruchinin, Oct 05 2013 */
    
  • PARI
    T(n,k)=if(k==0,1, if(k==n, (n!)^2, n^2*T(n-1, k-1) + T(n-1, k)));
    for(n=0,8, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Sep 14 2019
    
  • Sage
    # This triangle is (0,0)-based.
    def A008955(n, k) :
        if k==0 : return 1
        if k==n : return factorial(n)^2
        return n^2*A008955(n-1, k-1) + A008955(n-1, k)
    for n in (0..7) : print([A008955(n, k) for k in (0..n)]) # Peter Luschny, Feb 04 2012
    

Formula

The n-th row gives the coefficients in the expansion of Product_{i=1..n-1}(x+i^2), highest powers first (see Comments section).
The triangle can be obtained from the recurrence t1(n,k) = n^2*t1(n-1,k-1) + t1(n-1,k) with t1(n,0) = 1 and t1(n,n) = (n!)^2.
t1(n,k) = Sum_{j=-k..k} (-1)^j*s(n+1,n+1-k+j)*s(n+1,n+1-k-j) = Sum_{j=0..2*(n+1-k)} (-1)^(n+1-k+j)*s(n+1,j)*s(n+1,2*(n+1-k)-j), where s(n,k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 02 2012
E.g.f.: cosh(2/sqrt(t)*asin(sqrt(t)*z/2)) = 1 + z^2/2! + (1 + t)*z^4/4! + (1 + 5*t + 4*t^2)*z^6/6! + ... (see Berndt, p.263 and p.306). - Peter Bala, Aug 29 2012
T(n,m) = (2*(n+1))!*Sum_{k=0..m} ((-1)^k*binomial(n,m-k)*Sum_{i=0..2*k} ((2^(i-2*m)*stirling1(2*(n-m+1)+i,2*(n-m+1))*binomial(2*(n-m+1)+2*k-1, 2*(n-m+1)+i-1))/(2*(n-m+1)+i)!)). - Vladimir Kruchinin, Oct 05 2013

Extensions

There's an error in the last column of Riordan's table (change 46076 to 21076).
More terms from Vladeta Jovovic, Apr 16 2000
Link added and cross-references edited by Johannes W. Meijer, Aug 17 2009
Discussion of Riordan's definition of central factorial numbers added by N. J. A. Sloane, Feb 01 2011

A066989 a(n) = (n!)^3 * Sum_{i=1..n} 1/i^3.

Original entry on oeis.org

1, 9, 251, 16280, 2048824, 444273984, 152759224512, 78340747014144, 57175952894078976, 57223737619918848000, 76212579497951858688000, 131758938842553681444864000, 289584291977410916858462208000, 794860754824699647616459210752000
Offset: 1

Views

Author

Benoit Cloitre, Jan 27 2002

Keywords

Comments

p^2 divides a(p-1) for prime p>5. - Alexander Adamchuk, Jul 11 2006

Crossrefs

Cf. A007408.
Column k=3 of A291556.

Programs

  • Mathematica
    f[k_] := k^3; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 22}] (* A066989 *)
    (* Clark Kimberling, Dec 29 2011 *)
    Table[(n!)^3 * Sum[1/i^3, {i, 1, n}], {n, 1, 20}] (* Vaclav Kotesovec, Aug 27 2017 *)

Formula

Recurrence: a(1) = 1, a(2) = 9, a(n+2) = (2*n+3)*(n^2+3*n+3)*a(n+1) - (n+1)^6*a(n). b(n) = n!^3 satisfies the same recurrence with the initial conditions b(1) = 1, b(2) = 8. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(1-1^6/(9-2^6/(35-3^6/(91-...-(n-1)^6/((2n-1)*(n^2-n+1)))))) for n >= 2, leading to the infinite continued fraction expansion zeta(3) = 1/(1-1^6/(9-2^6/(35-3^6/(91-...-(n-1)^6/((2n-1)*(n^2-n+1)-...))))). Compare with A001819. - Peter Bala, Jul 19 2008
a(n) ~ Zeta(3) * (2*Pi)^(3/2) * n^(3*n+3/2) / exp(3*n). - Vaclav Kotesovec, Aug 27 2017
Sum_{n>=1} a(n) * x^n / (n!)^3 = polylog(3,x) / (1 - x). - Ilya Gutkovskiy, Jul 14 2020

A162990 Triangle of polynomial coefficients related to 3F2([1,n+1,n+1],[n+2,n+2],z).

Original entry on oeis.org

4, 36, 9, 576, 144, 64, 14400, 3600, 1600, 900, 518400, 129600, 57600, 32400, 20736, 25401600, 6350400, 2822400, 1587600, 1016064, 705600, 1625702400, 406425600, 180633600, 101606400, 65028096, 45158400, 33177600, 131681894400
Offset: 1

Views

Author

Johannes W. Meijer, Jul 21 2009

Keywords

Comments

The hypergeometric function 3F2([1,n+1,n+1],[n+2,n+2],z) = (n+1)^2*Li2(z)/z^(n+1) - MN(z;n)/(n!^2*z^n) for n >= 1, with Li2(z) the dilogarithm. The polynomial coefficients of MN(z;n) lead to the triangle given above.
We observe that 3F2([1,1,1],[2,2],z) = Li2(z)/z and that 3F2([1,0,0],[1,1],z) = 1.
The generating function for the EG1[3,n] coefficients of the EG1 matrix, see A162005, is GFEG1(z;m=2) = 1/(1-z)*(3*zeta(3)/2-2*z*log(2)* 3F2([1,1,1],[2,2],z) + sum((2^(1-2*n)* factorial(2*n-1)*z^(n+1)*3F2([1,n+1,n+1],[n+2,n+2],z))/(factorial(n+1)^2), n=1..infinity)).
The zeros of the MN(z;n) polynomials for larger values of n get ever closer to the unit circle and resemble the full moon, hence we propose to call the MN(z;n) the moon polynomials.

Examples

			The first few rows of the triangle are:
  [4]
  [36, 9]
  [576, 144, 64]
  [14400, 3600, 1600, 900]
The first few MN(z;n) polynomials are:
  MN(z;n=1) = 4
  MN(z;n=2) = 36 + 9*z
  MN(z;n=3) = 576 + 144*z + 64*z^2
  MN(z;n=4) = 14400 + 3600*z + 1600*z^2 + 900*z^3
		

References

  • Lewin, L., Polylogarithms and Associated Functions. New York, North-Holland, 1981.

Crossrefs

A162995 is a scaled version of this triangle.
A001819(n)*(n+1)^2 equals the row sums for n>=1.
A162991 and A162992 equal the first and second right hand columns.
A001048, A052747, A052759, A052778, A052794 are related to the square root of the first five right hand columns.
A001044, A162993 and A162994 equal the first, second and third left hand columns.
A000142, A001710, A002301, A133799, A129923, A001715 are related to the square root of the first six left hand columns.
A027451(n+1) equals the denominators of M(z, n)/(n!)^2.
A129202(n)/A129203(n) = (n+1)^2*Li2(z=1)/(Pi^2) = (n+1)^2/6.
Cf. A002378 and A035287.

Programs

  • Maple
    a := proc(n, m): ((n+1)!/m)^2 end: seq(seq(a(n, m), m=1..n), n=1..7); # Johannes W. Meijer, revised Nov 29 2012
  • Mathematica
    Table[((n+1)!/m)^2, {n, 10}, {m, n}] (* Paolo Xausa, Mar 30 2024 *)

Formula

a(n,m) = ((n+1)!/m)^2 for n >= 1 and 1 <= m <= n.

A142995 a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 3)*a(n) - n^4*a(n-1), n >= 1.

Original entry on oeis.org

0, 1, 7, 89, 1836, 56164, 2390832, 135213840, 9809203968, 888117094656, 98167241088000, 13010123816064000, 2036436482119680000, 371699564417796096000, 78251077775510986752000
Offset: 0

Views

Author

Peter Bala, Jul 18 2008

Keywords

Comments

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + m^2 + m + 1)*a(n) - n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series Sum_{k>=1} 1/k^2 for the constant zeta(2). For other cases see A001819 (m=0), A142996 (m=2), A142997 (m=3) and A142998 (m=4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!^2*p_m(n)*Sum_{k = 1..n} 1/(k^2*p_m(k-1)*p_m(k)), where p_m(x) := Sum_{k = 0..m} C(m,k)^2*C(x+k,m) = Sum_{k = 0..m} C(m,k)*C(m+k,k)*C(x,k) is the Ehrhart polynomial of the polytope formed from the convex hull of a root system of type A_m (equivalently, the polynomial that generates the crystal ball sequence for the A_m lattice [Bacher et al.]).
The first few are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 3*x^2 + 3*x + 1 and p_3(x) = (10*x^3 + 15*x^2 + 11*x + 3)/3. The o.g.f. for the p_m(x) is ((1-t^2)^x/(1-t)^(2x+1))*Legendre_P(x,(1+t^2)/(1-t^2)) = 1 + (2*x+1)*t + (3*x^2+3*x+1)*t^2 + ... [Gogin & Hirvensalo, Theorem 1 with N = -1].
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x+1)^2*f(x+1) + x^2*f(x-1) = (2*x^2 + 2*x + m^2 + m + 1)*f(x), normalized so that f(0) = 1. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_m(x-1), m = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p. 4 of [BUMP et al.]).
The general recurrence in the first paragraph above has a second solution b(n) = n!^2*p_m(n) with initial conditions b(0) = 1, b(1) = m^2 + m + 1. Hence the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = Sum_{k>=1} 1/(k^2*p_m(k-1)*p_m(k)) = 1/((m^2 + m + 1) - 1^4/((m^2 + m + 5) - 2^4/((m^2 + m + 13) - ... - n^4/((2*n^2 + 2*n + m^2 + m + 1) - ...)))) = 2*Sum_{k>=1} (-1)^(k+1)/(m+k)^2. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).
For related results see A142999. For corresponding results for the constants e, log(2) and zeta(3) see A000522, A142979 and A143003 respectively.

References

  • Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Crossrefs

Cf. A000522, A001819, A003215 (A_2 lattice), A005902 (A_3 lattice), A008384 (A_4 lattice), A008386 (A_5 lattice), A108625, A142979, A142996, A142997, A142998, A143003.

Programs

  • Maple
    p := n -> 2*n+1: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);
  • Mathematica
    a[n_] := -1/6*n!^2*(2*n*(Pi^2-12) + Pi^2 - 6*(2*n+1)*PolyGamma[1, n+1]) // Simplify; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Mar 06 2013 *)

Formula

a(n) = n!^2*p(n)*Sum_{k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = 2*n+1. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 3)*a(n) - n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 3. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 - 1^4/(7 - 2^4/(15 - 3^4/(27 - ... - (n-1)^4/(2*n^2 - 2*n + 3))))), for n >= 2. Lim_{n -> infinity} a(n)/b(n) = 1/(3 - 1^4/(7 - 2^4/(15 - 3^4/(27 - ... - n^4/((2*n^2 + 2*n + 3) - ...))))) = Sum_{k>=1} 1/(k^2*(4*k^2 - 1)) = 2 - zeta(2).

A099827 Generalized harmonic number H(n,5) = Sum_{k=1..n} 1/k^5 multiplied by (n!)^5.

Original entry on oeis.org

0, 1, 33, 8051, 8252000, 25795462624, 200610400564224, 3371852494046112768, 110492114540967125581824, 6524555433591956305924325376, 652461835742417609568446054400000, 105080260346474296336209157187174400000
Offset: 0

Views

Author

Alexander Adamchuk, Oct 27 2004

Keywords

Comments

Note that a(n) is divisible by n, except when n is prime. Also, a(n+1) is divisible by n, except when n is prime or n = 0.

Examples

			a(2) = (2!)^5 * (1 + 1/2^5) = 2^5 + 1 = 33,
a(3) = (3!)^5 * (1 + 1/2^5 + 1/3^5) = 6^5 + 3^5 + 1 = 8051.
		

Crossrefs

Column k = 5 of A291556.

Programs

  • Mathematica
    Table[(n!)^5*Sum[1/k^5, {k, 1, n}], {n, 0, 13}] or Table[(n!)^5*HarmonicNumber[n, 5], {n, 0, 13}]

Formula

a(n) = (n!)^5 * Sum_{k=1..n} 1/k^5 = (n!)^5 * HarmonicNumber[n, 5] = (n!)^5 * A099828(n)/A069052(n).
a(0) = 0, a(1) = 1, a(n+1) = (n^5 + (n+1)^5)*a(n) - n^10*a(n-1) for n > 0. - Seiichi Manyama, Aug 24 2017
a(n) ~ Zeta(5) * (2*Pi)^(5/2) * n^(5*n+5/2) / exp(5*n). - Vaclav Kotesovec, Aug 27 2017
Sum_{n>=0} a(n) * x^n / (n!)^5 = polylog(5,x) / (1 - x). - Ilya Gutkovskiy, Jul 14 2020

Extensions

a(0) = 0 prepended by Seiichi Manyama, Aug 23 2017
Name edited by Petros Hadjicostas, May 10 2020

A291556 Square array A(n,k), n>=0, k>=0, read by antidiagonals: A(n,k) = (n!)^k * Sum_{i=1..n} 1/i^k.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 3, 3, 0, 1, 5, 11, 4, 0, 1, 9, 49, 50, 5, 0, 1, 17, 251, 820, 274, 6, 0, 1, 33, 1393, 16280, 21076, 1764, 7, 0, 1, 65, 8051, 357904, 2048824, 773136, 13068, 8, 0, 1, 129, 47449, 8252000, 224021776, 444273984, 38402064, 109584, 9
Offset: 0

Views

Author

Seiichi Manyama, Aug 26 2017

Keywords

Examples

			Square array begins:
   0,  0,   0,     0,      0, ...
   1,  1,   1,     1,      1, ...
   2,  3,   5,     9,     17, ...
   3, 11,  49,   251,   1393, ...
   4, 50, 820, 16280, 357904, ...
		

Crossrefs

Rows n=0-3 give: A000004, A000012, A000051, A074528.
Main diagonal gives A060943.

Programs

  • Maple
    A:= (n, k)-> n!^k * add(1/i^k, i=1..n):
    seq(seq(A(n, d-n), n=0..d), d=0..10);  # Alois P. Heinz, Aug 26 2017
  • Mathematica
    A[0, ] = 0; A[1, ] = 1; A[n_, k_] := A[n, k] = ((n-1)^k + n^k) A[n-1, k] - (n-1)^(2k) A[n-2, k];
    Table[A[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, May 11 2019 *)

Formula

A(0, k) = 0, A(1, k) = 1, A(n+1, k) = (n^k+(n+1)^k)*A(n, k) - n^(2*k)*A(n-1, k).

A291456 a(n) = (n!)^6 * Sum_{i=1..n} 1/i^6.

Original entry on oeis.org

0, 1, 65, 47449, 194397760, 3037656102976, 141727869124448256, 16674281388691716870144, 4371079210518164503303028736, 2322975003299339366419974718488576, 2322977286679362958150790503464960000000
Offset: 0

Views

Author

Seiichi Manyama, Aug 24 2017

Keywords

Crossrefs

Column k=6 of A291556.
Cf. A000254 (k=1), A001819 (k=2), A066989 (k=3), A203229 (k=4), A099827 (k=5).

Programs

  • Mathematica
    Table[(n!)^6 * Sum[1/i^6, {i, 1, n}], {n, 0, 15}] (* Vaclav Kotesovec, Aug 27 2017 *)

Formula

a(0) = 0, a(1) = 1, a(n+1) = (n^6 + (n+1)^6)*a(n) - n^12*a(n-1) for n > 0.
a(n) ~ 8 * Pi^9 * n^(6*n+3) / (945 * exp(6*n)). - Vaclav Kotesovec, Aug 27 2017
a(n) = (n!)^6 * A103345(n)/A103346(n). - Petros Hadjicostas, May 10 2020
Sum_{n>=0} a(n) * x^n / (n!)^6 = polylog(6,x) / (1 - x). - Ilya Gutkovskiy, Jul 15 2020

A002455 Central factorial numbers: unsigned 1st subdiagonal of A182867.

Original entry on oeis.org

0, 1, 20, 784, 52480, 5395456, 791691264, 157294854144, 40683662475264, 13288048674471936, 5349739088314368000, 2603081566154391552000, 1506057980251484454912000, 1021944601582419125993472000
Offset: 0

Views

Author

Keywords

Examples

			(arcsin x)^4 = x^4 + 2/3*x^6 + 7/15*x^8 + 328/945*x^10 + ...
		

References

  • B. Berndt, Ramanujan's Notebooks, Part I, page 263.
  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 110.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • GAP
    List([0..20], n-> 4^(n-1)*(Factorial(n))^2*Sum([1..n], k-> 1/k^2)); # G. C. Greubel, Jul 04 2019
  • Magma
    [0] cat [4^(n-1)*(Factorial(n))^2*(&+[1/k^2: k in [1..n]]): n in [1..20]]; // G. C. Greubel, Jul 04 2019
    
  • Maple
    A002455 := proc(n)
        arcsin(x)^4;
        coeftayl(%,x=0,2*n+2)*(2*n+2)!/4! ;
    end proc:
    seq(A002455(n),n=0..20) ; # R. J. Mathar, Jan 20 2025
  • Mathematica
    nmax = 13; coes = CoefficientList[ Series[ ArcSin[x]^4, {x, 0, 2*nmax + 2}], x]* Range[0, 2*nmax + 2]!/24; a[n_] := coes[[2*n + 3]]; Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Dec 08 2011 *)
    Table[4^(n-1)*(n!)^2*HarmonicNumber[n,2], {n,0,20}] (* G. C. Greubel, Jul 04 2019 *)
  • PARI
    a(n)=if(n<0,0,(2*n+2)!*polcoeff(asin(x+O(x^(2*n+3)))^4/4!,2*n+2))
    
  • PARI
    a(n)=-(-1)^n*polcoeff(prod(k=0,2*n,x+2*k-2*n),3)
    
  • Sage
    [4^(n-1)*(factorial(n))^2*sum(1/k^2 for k in (1..n)) for n in (0..20)] # G. C. Greubel, Jul 04 2019
    

Formula

(-1)^(n-1)*a(n) is the coefficient of x^3 in Product_{k=0..2*n} (x+2*k-2*n). - Benoit Cloitre and Michael Somos, Nov 22 2002
E.g.f.: (arcsin x)^4; that is, a_k is the coefficient of x^(2*k+2) in (arcsin x)^4 multiplied by (2*k+2)! and divided by 4! Also a(n) = 2^(2*n-2)*(n!)^2 * Sum_{k=1..n} 1/k^2. - Joe Keane (jgk(AT)jgk.org)
a(n) = 4*(2*n^2 - 2*n + 1)*a(n-1) - 16*(n-1)^4*a(n-2). - Vaclav Kotesovec, Feb 23 2015
a(n) ~ Pi^3 * 2^(2*n-2) * n^(2*n+1) / (3 * exp(2*n)). - Vaclav Kotesovec, Feb 23 2015

Extensions

More terms from Joe Keane (jgk(AT)jgk.org)
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