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.

Previous Showing 41-50 of 74 results. Next

A090998 Decimal expansion of lim_{k -> +-oo} k^2*(1 - Gamma(1+i/k)) where i^2 = -1 and Gamma is the Gamma function.

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Feb 29 2004

Keywords

Comments

Limit_{k->oo} k*(1-Gamma(1+1/k)) = -Gamma'(1) = gamma = 0.577....
Decimal expansion of the higher-order exponential integral constant gamma(2,1). The higher-order exponential integrals, see A163931, are defined by E(x,m,n) = x^(n-1)*Integral_{t=x..oo} (E(t,m-1,n)/t^n) dt for m >= 1 and n >= 1, with E(x,m=0,n) = exp(-x). The series expansions of the higher-order exponential integrals are dominated by the gamma(k,n) and the alpha(k,n) constants, see A163927. - Johannes W. Meijer and Nico Baken, Aug 13 2009

Examples

			G(2,1) = 0.9890559953279725553953956515...
		

Crossrefs

Cf. A163931 (E(x,m,n)), A163927 (alpha(k,n)), A001620 (gamma).
The structure of the G(k,n=1) formulas lead (replace gamma with G and Zeta with Z) to A036039. - Johannes W. Meijer and Nico Baken, Aug 13 2009
Cf. A081855.

Programs

  • Magma
    SetDefaultRealField(RealField(100)); R:= RealField(); (6*EulerGamma(R)^2 + Pi(R)^2)/12; // G. C. Greubel, Feb 01 2019
    
  • Maple
    ncol:=1; nmax:=5; kmax:=nmax; for n from 1 to nmax do G(0,n):=1 od: for n from 1 to nmax do for k from 1 to kmax do G(k,n):= expand((1/k)*((gamma-sum(p^(-1),p=1..n-1))* G(k-1,n)+sum((Zeta(k-i)-sum(p^(-(k-i)),p=1..n-1))*G(i,n),i=0..k-2))) od; od: for k from 0 to kmax do G(k,ncol):=G(k,ncol) od; # Johannes W. Meijer and Nico Baken, Aug 13 2009
  • Mathematica
    RealDigits[(6*EulerGamma^2 + Pi^2)/12, 10, 104][[1]] (* Jean-François Alcover, Mar 04 2013 *)
  • PARI
    default(realprecision, 100); (6*Euler^2 +Pi^2)/12 \\ G. C. Greubel, Feb 01 2019
    
  • Sage
    numerical_approx((6*euler_gamma^2 + pi^2)/12, digits=100) # G. C. Greubel, Feb 01 2019

Formula

From Johannes W. Meijer and Nico Baken, Aug 13 2009: (Start)
G(2,1) = gamma(2,1) = gamma^2/2 + Pi^2/12.
G(k,n) = (1/k)*(gamma*G(k-1,n)) - (1/k)*Sum_{p=1..n-1} (p^(-1))* G(k-1,n) + (1/k) * Sum_{i=0..k-2} (Zeta(k-i) * G(i,n)) - (1/k)*Sum_{i=0..k-2}(Sum_{p=1..n-1} (p^(i-k)) * G(i,n)) with G(0,n) = 1 for k >= 0 and n >= 1.
G(k,n+1) = G(k,n) - G(k-1,n)/n.
GF(z,n) = GAMMA(n-z)/GAMMA(n).
(gamma - G(1,n)) = A001008(n-1)/A002805(n-1) for n >= 2. (End)
Equals A081855/2. - Hugo Pfoertner, Mar 12 2024

A163934 Triangle related to the asymptotic expansion of E(x,m=4,n).

Original entry on oeis.org

1, 6, 4, 35, 40, 10, 225, 340, 150, 20, 1624, 2940, 1750, 420, 35, 13132, 27076, 19600, 6440, 980, 56, 118124, 269136, 224490, 90720, 19110, 2016, 84, 1172700, 2894720, 2693250, 1265460, 330750, 48720, 3780, 120
Offset: 1

Views

Author

Johannes W. Meijer, Aug 13 2009

Keywords

Comments

The higher order exponential integrals E(x,m,n) are defined in A163931 while the general formula for their asymptotic expansion can be found in A163932.
We used the latter formula and the asymptotic expansion of E(x,m=3,n), see A163932, to determine that E(x,m=4,n) ~ (exp(-x)/x^4)*(1 - (6+4*n)/x + (35+40*n+ 10*n^2)/x^2 - (225+340*n+ 150*n^2+20*n^3)/x^3 + ... ). This formula leads to the triangle coefficients given above.
The asymptotic expansion leads for the values of n from one to five to known sequences, see the cross-references.
The numerators of the o.g.f.s. of the right hand columns of this triangle lead for z=1 to A000457, see A163939 for more information.
The first Maple program generates the sequence given above and the second program generates the asymptotic expansion of E(x,m=4,n).

Examples

			The first few rows of the triangle are:
1;
6, 4;
35, 40, 10;
225, 340, 150, 20;
		

Crossrefs

Cf. A163931 (E(x,m,n)), A163932 and A163939.
Cf. A048994 (Stirling1), A000454 (row sums).
A000399, 4*A000454, 10*A000482, 20*A001233, 35*A001234 equal the first five left hand columns.
A000292, A027777 and A163935 equal the first three right hand columns.
The asymptotic expansion leads to A000454 (n=1), A001707 (n=2), A001713 (n=3), A001718 (n=4) and A001723 (n=5).
Cf. A130534 (m=1), A028421 (m=2), A163932 (m=3).

Programs

  • Maple
    with(combinat): A163934 := proc(n,m): (-1)^(n+m)* binomial(m+2, 3) *stirling1(n+2, m+2) end: seq(seq(A163934(n,m), m=1..n), n=1..8);
    with(combinat): imax:=6; EA:=proc(x,m,n) local E, i; E:=0: for i from m-1 to imax+2 do E:=E + sum((-1)^(m+k+1)*binomial(k,m-1)*n^(k-m+1)* stirling1(i, k), k=m-1..i)/x^(i-m+1) od: E:= exp(-x)/x^(m)*E: return(E); end: EA(x,4,n);
    # Maple programs revised by Johannes W. Meijer, Sep 11 2012
  • Mathematica
    a[n_, m_] /; n >= 1 && 1 <= m <= n = (-1)^(n+m)*Binomial[m+2, 3] * StirlingS1[n+2, m+2]; Flatten[Table[a[n, m], {n, 1, 8}, {m, 1, n}]][[1 ;; 36]] (* Jean-François Alcover, Jun 01 2011, after formula *)

Formula

a(n,m) = (-1)^(n+m)*C(m+2,3)*stirling1(n+2,m+2) for n >= 1 and 1<= m <= n.

A001706 Generalized Stirling numbers.

Original entry on oeis.org

1, 9, 71, 580, 5104, 48860, 509004, 5753736, 70290936, 924118272, 13020978816, 195869441664, 3134328981120, 53180752331520, 953884282141440, 18037635241029120, 358689683932346880, 7483713725055744000, 163478034254755584000, 3731670622213083648000
Offset: 0

Views

Author

Keywords

Comments

The asymptotic expansion of the higher order exponential integral E(x,m=3,n=2) ~ exp(-x)/x^3*(1 - 9/x + 71/x^2 - 580/x^3 + 5104/x^4 - 48860/x^5 + the sequence given above). See A163931 and A163932 for more information. - Johannes W. Meijer, Oct 20 2009
a(n-1) is equal to -1 times the coefficient of x of the characteristic polynomial of the n X n matrix whose (i,j)-entry is equal to i+3 if i=j and is equal to 1 otherwise. - John M. Campbell, May 24 2011

References

  • 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).

Programs

  • Mathematica
    Table[-Coefficient[CharacteristicPolynomial[Array[KroneckerDelta[#1,#2]((((#1+3)))-1)+1&,{n,n}],x],x,1],{n,1,10}] (* John M. Campbell, May 24 2011 *)

Formula

E.g.f. (with offset 2): log(1 - x)^2 / (2 * (1 - x)^2).
a(n) = Sum_{k=0..n}(-1)^(n+k)*binomial(k+2, 2)*2^k*stirling1(n+2, k+2). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
a(n-1) = (1/2)*Sum_{i=0..n} binomial(n, i)*A000254(i)*A000254(n-i). - Benoit Cloitre, Mar 09 2004
If we define f(n,i,a)=sum(binomial(n,k)*stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then a(n-1) = |f(n,2,2)|, for n>=2. - Milan Janjic, Dec 21 2008
a(n) = (n+3)!*((gamma-1)*Psi(n+4)+2+gamma^2-17*gamma/6+sum(Psi(i+4)/(i+4),i = 0 .. n-1)). - Mark van Hoeij, Oct 26 2011

Extensions

More terms from Christian G. Bower

A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

Original entry on oeis.org

0, 1, 1, 9, 4, 25, 9, 49, 16, 81, 25, 121, 36, 169, 49, 225, 64, 289, 81, 361, 100, 441, 121, 529, 144, 625, 169, 729, 196, 841, 225, 961, 256, 1089, 289, 1225, 324, 1369, 361, 1521, 400, 1681, 441, 1849, 484, 2025, 529, 2209, 576, 2401, 625, 2601
Offset: 0

Views

Author

Paul Curtz, Nov 18 2009

Keywords

Comments

From Paul Curtz, Mar 26 2011: (Start)
Successive A026741(n) * A026741(n+p):
p=0: 0, 1, 1, 9, 4, 25, 9, a(n),
p=1: 0, 1, 3, 6, 10, 15, 21, A000217,
p=2: 0, 3, 2, 15, 6, 35, 12, A142705,
p=3: 0, 2, 5, 9, 14, 20, 27, A000096,
p=4: 0, 5, 3, 21, 8, 45, 15, A171621,
p=5: 0, 3, 7, 12, 18, 25, 33, A055998,
p=6: 0, 7, 4, 27, 10, 55, 18,
p=7: 0, 4, 9, 15, 22, 30, 39, A055999,
p=8: 0, 9, 5, 33, 12, 65, 21, (see A061041),
p=9: 0, 5, 11, 18, 26, 35, 45, A056000. (End)
The moment generating function of p(x, m=2, n=1, mu=2) = 4*x*E(x, 2, 1), see A163931 and A274181, is given by M(a) = (-4 * log(1-a) - 4 * polylog(2, a))/a^2. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 03 2016
Multiplicative because both A129194 and A040001 are. - Andrew Howroyd, Jul 26 2018

Crossrefs

Programs

  • Magma
    I:=[0,1,1,9,4,25]; [n le 6 select I[n] else 3*Self(n-2)-3*Self(n-4)+Self(n-6): n in [1..60]]; // Vincenzo Librandi, Jul 10 2016
    
  • Maple
    a := proc(n): n^2*(5-3*(-1)^n)/8 end: seq(a(n), n=0..46); # Johannes W. Meijer, Jul 03 2016
  • Mathematica
    LinearRecurrence[{0,3,0,-3,0,1},{0,1,1,9,4,25},60] (* Harvey P. Dale, May 14 2011 *)
    f[n_] := Numerator[(n/2)^2]; Array[f, 60, 0] (* Robert G. Wilson v, Dec 18 2012 *)
    CoefficientList[Series[x(1+x+6x^2+x^3+x^4)/((1-x)^3(1+x)^3), {x,0,60}], x] (* Vincenzo Librandi, Jul 10 2016 *)
  • PARI
    concat(0, Vec(x*(1+x+6*x^2+x^3+x^4)/((1-x)^3*(1+x)^3) + O(x^60))) \\ Altug Alkan, Jul 04 2016
    
  • PARI
    a(n) = lcm(4, n^2)/4; \\ Andrew Howroyd, Jul 26 2018
    
  • Sage
    (x*(1+x+6*x^2+x^3+x^4)/(1-x^2)^3).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 20 2019

Formula

From R. J. Mathar, Jan 22 2011: (Start)
G.f.: x*(1 + x + 6*x^2 + x^3 + x^4) / ((1-x)^3*(1+x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(n) = n^2*(5 - 3*(-1)^n)/8. (End)
a(n) = A026741(n)^2.
a(2*n) = A000290(n); a(2*n+1) = A016754(n).
a(n) - a(n-4) = 4*A064680(n+2). - Paul Curtz, Mar 27 2011
4*a(n) = A061038(n) * A010121(n+2) = A109043(n)^2, n >= 2. - Paul Curtz, Apr 07 2011
a(n) = A129194(n) / A040001(n). - Andrew Howroyd, Jul 26 2018
From Peter Bala, Feb 19 2019: (Start)
a(n) = numerator(n^2/(n^2 + 4)) = n^2/(gcd(n^2,4)) = (n/gcd(n,2))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 4, ...] is a purely periodic sequence of period 2. Thus a(n) is a quasi-polynomial in n.
O.g.f.: x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3.
Cf. A181318. (End)
From Werner Schulte, Aug 30 2020: (Start)
Multiplicative with a(2^e) = 2^(2*e-2) for e > 0, and a(p^e) = p^(2*e) for prime p > 2.
Dirichlet g.f.: zeta(s-2) * (1 - 3/2^s).
Dirichlet convolution with A259346 equals A000290.
Sum_{n>0} 1/a(n) = Pi^2 * 7 / 24. (End)
Sum_{k=1..n} a(k) ~ (5/24) * n^3. - Amiram Eldar, Nov 28 2022

A001712 Generalized Stirling numbers.

Original entry on oeis.org

1, 12, 119, 1175, 12154, 133938, 1580508, 19978308, 270074016, 3894932448, 59760168192, 972751628160, 16752851775360, 304473528961920, 5825460745532160, 117070467915075840, 2465958106403712000, 54336917746726272000, 1250216389189281024000
Offset: 0

Views

Author

Keywords

Comments

The asymptotic expansion of the higher order exponential integral E(x,m=3,n=3) ~ exp(-x)/x^3*(1 - 12/x + 119/x^2 - 1175/x^3 + 12154/x^4 - 133938/x^5 + ...) leads to the sequence given above. See A163931 and A163932 for more information. - Johannes W. Meijer, Oct 20 2009
From Petros Hadjicostas, Jun 11 2020: (Start)
For nonnegative integers n, m and complex numbers a, b (with b <> 0), the numbers R_n^m(a,b) were introduced by Mitrinovic (1961) using slightly different notation. They were further examined by Mitrinovic and Mitrinovic (1962).
These numbers are defined via the g.f. Product_{r=0..n-1} (x - (a + b*r)) = Sum_{m=0..n} R_n^m(a,b)*x^m for n >= 0.
As a result, R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b) for n >= m >= 1 with R_1^0(a,b) = a, R_1^1(a,b) = 1, and R_n^m(a,b) = 0 for n < m. (Because an empty product is by definition 1, we may let R_0^0(a,b) = 1.)
With a = 0 and b = 1, we get the Stirling numbers of the first kind S1(n,m) = R_n^m(a=0, b=1) = A048994(n,m). (Array A008275 is the same as array A048994 but with no zero row and no zero column.)
We have R_n^m(a,b) = Sum_{k=0}^{n-m} (-1)^k * a^k * b^(n-m-k) * binomial(m+k, k) * S1(n, m+k) for n >= m >= 0.
For the current sequence, a(n) = R_{n+2}^2(a=-3, b=-1) for n >= 0. (End)

References

  • 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

  • Maple
    A001712 := proc(n)
        add((-1)^(n+k)*binomial(k+2, 2)*3^k*Stirling1(n+2, k+2), k=0..n) ;
    end proc:
    seq(A001712(n), n=0..10) ; # R. J. Mathar, Jun 09 2018
  • Mathematica
    nn = 22; t = Range[0, nn]! CoefficientList[Series[Log[1 - x]^2/(2*(1 - x)^3), {x, 0, nn}], x]; Drop[t, 2] (* T. D. Noe, Aug 09 2012 *)
  • PARI
    a(n) = sum(k=0, n, (-1)^(n+k)*binomial(k+2, 2)*3^k*stirling(n+2, k+2, 1)) \\ Michel Marcus, Jan 20 2016
    
  • PARI
    b(n) = prod(r=0, n+1, r+3);
    c(n) = sum(i=0, n+1, sum(j=i+1, n+1, 1/((3+i)*(3+j))));
    for(n=0, 18, print1(b(n)*c(n),",")) \\ Petros Hadjicostas, Jun 11 2020

Formula

a(n) = Sum_{k=0..n} (-1)^(n+k)*binomial(k+2, 2)*3^k*Stirling1(n+2, k+2). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
E.g.f.: (1 - 7*log(1 - x) + 6*log(1 - x)^2)/(1 - x)^5. - Vladeta Jovovic, Mar 01 2004
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k, i)*Product_{j=0..k-1} (-a-j), then a(n-2) = |f(n,2,3)|, for n >= 2. [Milan Janjic, Dec 21 2008]
Conjecture: a(n) + 3*(-n-3)*a(n-1) + (3*n^2 + 15*n + 19)*a(n-2) - (n+2)^3*a(n-3)=0. - R. J. Mathar, Jun 09 2018
From Petros Hadjicostas, Jun 11 2020: (Start)
a(n) = [x^2] Product_{r=0}^{n+1} (x + 3 + r) = (Product_{r=0}^{n+1} (r+3)) * Sum_{0 <= i < j <= n+1} 1/((3+i)*(3+j)).
Since a(n) = R_{n+2}^2(a=-3, b=-1) and A001711(n) = R_{n+1}^1(a=-3,b=-1), the equation R_{n+2}^2(a=-3,b=-1) = R_{n+1}^1(a=-3,b=-1) + (n+4)*R_{n+1}^2(a=-3,b=-1) implies the following:
(i) a(n) = A001711(n) + (n+4)*a(n-1) for n >= 1.
(ii) a(n) = (n+2)!/2 + (2*n+7)*a(n-1) - (n+3)^2*a(n-2) for n >= 2.
(iii) R. J. Mathar's recurrence above. (End)

Extensions

More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004

A163927 Numerators of the higher order exponential integral constants alpha(k,4).

Original entry on oeis.org

1, 49, 1897, 69553, 2515513, 90663937, 3264855049, 117543378001, 4231639039705, 152339702576545, 5484235568128681, 197432536935184369, 7107571838026381177, 255872590744254526273, 9211413307971174616393
Offset: 0

Views

Author

Johannes W. Meijer and Nico Baken, Aug 13 2009, Aug 17 2009

Keywords

Comments

The higher order exponential integrals, see A163931, are defined by E(x,m,n) = x^(n-1)*Integral_{t>=x} E(t,m-1,n)/t^n for m >= 1 and n >= 1, with E(x,m=0,n) = exp(-x).
The series expansions of the higher order exponential integrals are dominated by the alpha(k,n) and the gamma(k,n) constants, see A090998.
The first Maple program uses the alpha(k,n) formula and the second the GF(z,n) to generate the alpha(k,n) coefficients in each column.
Appears to equal the numerator of the multiple harmonic (star) sum Sum_{1 <= k_1 <= ... <= k_n <= 3} 1/(k_1^2*...*k_n^2). If true, then a(n) = numerator( 3/2 - 3/(5*4^n) + 1/(10*9^n) ). - Peter Bala, Jan 31 2019

Examples

			a(k=0,n=4) = 1, a(k=1,4) = 49/36, a(k=2,4) = 1897/1296, a(k=3,4) = 69553/46656.
		

Crossrefs

Cf. A163931 (E(x,m,n)), A090998 (gamma(k,n)).
a(k,1) = A000007(k)
a(k,2) = A000012(k) = 1^k.
a(k,3) = A002450(k+1)/A000302(k) with A000302(k) = 4^k.
a(k,4) = A163927(k)/A009980(k) with A009980(k) = 36^k.
The GF(z,n) lead to A008955.
The denominators of a(1,n), n >= 2, lead to A007407.

Programs

  • Maple
    coln := 4; nmax := 15; kmax := nmax: k:=0: for n from 1 to nmax do alpha(k, n) := 1 od: for k from 1 to kmax do for n from 1 to nmax do alpha(k, n) := (1/k)*sum(sum(p^(-2*(k-i)), p=0..n-1)*alpha(i, n), i=0..k-1) od; od: seq(alpha(k, coln), k=0..nmax-1);
    # End program 1
    coln:=4; nmax1 := 16; for n from 0 to nmax1 do A008955(n, 0):=1 end do: for n from 0 to nmax1 do A008955(n, n) := (n!)^2 end do: for n from 1 to nmax1 do for m from 1 to n-1 do A008955(n, m) := A008955(n-1, m-1)*n^2 + A008955(n-1, m) end do: end do: m:=coln-1: f(m):=0: for n from 0 to m do f(m) := f(m) + (-1)^(n + m)*A008955(m, n)*z^(2*m-2*n) od: GF(z,coln) := m!^2/f(m): GF(z,coln):=series(GF(z,coln), z, nmax1);
    # End program 2

Formula

alpha(k,n) = (1/k) * Sum_{i=0..k-1} (Sum_{p=0..n-1}(p^(2*i-2*k))*alpha(i, n)) with alpha(0,n) = 1, k >= 0 and n >= 1.
alpha(k,n) = alpha(k,n+1) -alpha(k-1,n+1)/n^2.
GF(z,n) = product((1-(z/k)^2)^(-1), k = 1..n-1) = (Pi*z/sin(Pi*z))/(Beta(n+z,n-z)/Beta(n,n)).

A163938 Triangle related to the o.g.f.s. of the right hand columns of A163932 (E(x, m=3, n)).

Original entry on oeis.org

1, 3, 3, 11, 28, 6, 50, 225, 135, 10, 274, 1858, 2092, 486, 15, 1764, 16464, 29148, 13482, 1491, 21, 13068, 158352, 398640, 301220, 70485, 4152, 28, 109584, 1655172, 5552724, 6132780, 2432070, 322971, 10863, 36
Offset: 1

Views

Author

Johannes W. Meijer, Aug 13 2009

Keywords

Comments

The asymptotic expansions of the higher order exponential integral E(x, m=3, n) lead to triangle A163932, see A163931 for information on the E(x,m,n). The o.g.f.s. of the right hand columns of triangle A163932 have a nice structure Gf(p) = W3(z,p)/(1-z)^(2*p+1) with p = 1 for the first right hand column, p = 2 for the second right hand column, etc. The coefficients of the W3(z,p) polynomials lead to the triangle given above, n >= 1 and 1 <= m <= n. The row sums of this triangle lead to A001879, see A163936 for more information.

Examples

			The first few W3(z,p) polynomials are:
W3(z,p=1) = 1/(1-z)^3
W3(z,p=2) = (3 + 3*z)/(1-z)^5
W3(z,p=3) = (11 + 28*z + 6*z^2)/(1-z)^7
W3(z,p=4) = (50 + 225*z + 135*z^2 + 10*z^3)/(1-z)^9
		

Crossrefs

Row sums equal A001879.
A000254 equals the first left hand column.
A000217 equals the first right hand column.
Cf. A163931 (E(x,m,n)) and A163932.
Cf. A163936 (E(x,m=1,n)), A163937 (E(x,m=2,n)) and A163939 (E(x,m=4,n)).

Programs

  • Maple
    with(combinat): a := proc(n, m): add((-1)^(n+k+1)*((m-k+1)*(m-k)/2!)*binomial(2*n+1, k)*stirling1(m+n-k, m-k+1), k=0..m-1) end: seq(seq(a(n, m), m=1..n), n=1..8); # Johannes W. Meijer, revised Nov 27 2012
  • Mathematica
    Table[Sum[(-1)^(n + k + 1)*Binomial[m - k + 1, 2]*Binomial[2*n + 1, k]*StirlingS1[m + n - k, m - k + 1], {k, 0, m - 1}], {n, 1, 50}, {m, 1, n}] // Flatten (* G. C. Greubel, Aug 13 2017 *)
  • PARI
    for(n=1,10, for(m=1,n, print1(sum(k=0,m-1, (-1)^(n+k+1)* binomial(m-k+1,2)*binomial(2*n+1,k) *stirling(m+n-k,m-k+1, 1)) ,", "))) \\ G. C. Greubel, Aug 13 2017

Formula

a(n,m) = Sum_{k=0..(m-1)} (-1)^(n+k+1)*binomial(m-k+1,2) *binomial(2*n+1,k) *stirling1(m+n-k,m-k+1), for 1 <= m <= n.

A163939 Triangle related to the o.g.f.s. of the right hand columns of A163934 (E(x,m=4,n)).

Original entry on oeis.org

1, 6, 4, 35, 60, 10, 225, 690, 325, 20, 1624, 7588, 6762, 1316, 35, 13132, 85288, 120358, 46928, 4508, 56, 118124, 1004736, 2028660, 1298860, 265365, 13896, 84, 1172700, 12529400, 33896400, 31862400, 11077255, 1313610, 39915, 120
Offset: 1

Views

Author

Johannes W. Meijer, Aug 13 2009

Keywords

Comments

The asymptotic expansions of the higher order exponential integral E(x,m=4,n) lead to triangle A163934, see A163931 for information on the E(x,m,n). The o.g.f.s. of the right hand columns of triangle A163934 have a nice structure Gf(p) = W4(z,p)/(1-z)^(2*p+2) with p = 1 for the first right hand column, p = 2 for the second right hand column, etc.. The coefficients of the W4(z,p) polynomials lead to the triangle given above, n >= 1 and 1 <= m <= n. The row sums of this triangle lead to A000457, see A163936 for more information.

Examples

			The first few W4(z,p) polynomials are:
W4(z,p=1) = 1/(1-z)^4
W4(z,p=2) = (6+4*z)/(1-z)^6
W4(z,p=3) = (35+60*z+10*z^2)/(1-z)^8
W4(z,p=4) = (225+690*z+325*z^2+20*z^3)/(1-z)^10
		

Crossrefs

Row sums equal A000457.
A000399 equals the first left hand column.
A000292 equals the first right hand column.
Cf. A163931 (E(x,m,n)) and A163934.
Cf. A163936 (E(x,m=1,n)), A163937 (E(x,m=2,n)) and A163938 (E(x,m=3,n)).

Programs

  • Maple
    with(combinat): a := proc(n, m): add((-1)^(n+k+1)*((m-k+2)*(m-k+1)*(m-k)/3!)*binomial(2*n+2, k)*stirling1(m+n-k+1, m-k+2), k=0..m-1) end: seq(seq(a(n, m), m=1..n), n=1..8); # Johannes W. Meijer, revised Nov 27 2012
  • Mathematica
    Table[Sum[(-1)^(n + k + 1)*Binomial[m - k + 2, 3]*Binomial[2*n + 2, k]*StirlingS1[m + n - k + 1, m - k + 2], {k, 0, m - 1}], {n, 1, 50}, {m, 1, n}] // Flatten (* G. C. Greubel, Aug 13 2017 *)
  • PARI
    for(n=1,10, for(m=1,n, print1(sum(k=0, m-1, (-1)^(n+k+1)* binomial(m-k+2,3)* binomial(2*n+2,k)*stirling(m+n-k+1,m-k+2,1)), ", "))) \\ G. C. Greubel, Aug 13 2017

Formula

a(n,m) = Sum_{k=0..(m-1)} (-1)^(n+k+1)*binomial(m-k+2,3)* binomial(2*n+2,k)*stirling1(m+n-k+1,m-k+2), for 1<= m <=n.

A274181 Decimal expansion of Phi(1/2, 2, 2), where Phi is the Lerch transcendent.

Original entry on oeis.org

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

Views

Author

Johannes W. Meijer and N. H. G. Baken, Jun 17 2016, Jul 08 2016

Keywords

Comments

The exponential integral distribution is defined by p(x, m, n, mu) = ((n+mu-1)^m * x^(mu-1) / (mu-1)!) * E(x, m, n), see A163931 and the Meijer link. The moment generating function of this probability distribution function is M(a, m, n, mu) = Sum_{k>=0}(((mu+k-1)!/((mu-1)!*k!)) * ((n+mu-1) / (n+mu+k-1))^m * a^k).
In the special case that mu=1 we get p(x, m, n, mu=1) = n^m * E(x, m, n) and M(a, m, n, mu=1) = n^m * Phi(a, m, n), with Phi the Lerch transcendent. If n=1 and mu=1 we get M(a, m, n=1, mu=1) = polylog(m, a)/a = Li_m(a)/a.

Examples

			0.32896210586005002361062528063872043497679389922...
		

References

  • William Feller, An introduction to probability theory and its applications, Vol. 1. p. 285, 1968.

Crossrefs

Cf. A163931, A002162 (Phi(1/2, 1, 1)/2), A076788 (Phi(1/2, 2, 1)/2), A112302, A008276.

Programs

  • Maple
    Digits := 101; c := evalf(LerchPhi(1/2, 2, 2));
  • Mathematica
    N[HurwitzLerchPhi[1/2, 2, 2], 25] (* G. C. Greubel, Jun 19 2016 *)
  • PARI
    Pi^2/3 - 2*log(2)^2 - 2 \\ Altug Alkan, Jul 08 2016
    
  • PARI
    lerchphi(.5,2,2) \\ Charles R Greathouse IV, Jan 30 2025
    
  • Python
    from mpmath import mp, lerchphi
    mp.dps=102
    print([int(d) for d in list(str(lerchphi(1/2, 2, 2))[2:-1])]) # Indranil Ghosh, Jul 04 2017

Formula

Equals Phi(1/2, 2, 2) with Phi the Lerch transcendent.
Equals Sum_{k>=0}(1/((2+k)^2*2^k)).
Equals 4 * polylog(2, 1/2) - 2.
Equals Pi^2/3 - 2*log(2)^2 - 2.
Equals Integral_{x=0..oo} x*exp(-x)/(exp(x)-1/2) dx. - Amiram Eldar, Aug 24 2020

A001713 Generalized Stirling numbers.

Original entry on oeis.org

1, 18, 245, 3135, 40369, 537628, 7494416, 109911300, 1698920916, 27679825272, 474957547272, 8572072384512, 162478082312064, 3229079010579072, 67177961946534528, 1460629706845766400, 33139181950164806400, 783398920650352012800, 19268391564147377318400
Offset: 0

Views

Author

Keywords

Comments

The asymptotic expansion of the higher order exponential integral E(x,m=4,n=3) ~ exp(-x)/x^4*(1 - 18/x + 245/x^2 - 3135/x^3 + 40369/x^4 - 537628/x^5 + ...) leads to the sequence given above. See A163931 and A163934 for more information. - Johannes W. Meijer, Oct 20 2009
From Petros Hadjicostas, Jun 12 2020: (Start)
For nonnegative integers n, m and complex numbers a, b (with b <> 0), the numbers R_n^m(a,b) were introduced by Mitrinovic (1961) and Mitrinovic and Mitrinovic (1962) using slightly different notation.
These numbers are defined via the g.f. Product_{r=0..n-1} (x - (a + b*r)) = Sum_{m=0..n} R_n^m(a,b)*x^m for n >= 0.
As a result, R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b) for n >= m >= 1 with R_0^0(a,b) = 1, R_1^0(a,b) = a, R_1^1(a,b) = 1, and R_n^m(a,b) = 0 for n < m.
With a = 0 and b = 1, we get the Stirling numbers of the first kind S1(n,m) = R_n^m(a=0, b=1) = A048994(n,m) for n, m >= 0.
We have R_n^m(a,b) = Sum_{k=0}^{n-m} (-1)^k * a^k * b^(n-m-k) * binomial(m+k, k) * S1(n, m+k) for n >= m >= 0.
For the current sequence, a(n) = R_{n+3}^3(a=-3, b=-1) for n >= 0. (End)

References

  • 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

  • Mathematica
    nn = 23; t = Range[0, nn]! CoefficientList[Series[-Log[1 - x]^3/(6*(1 - x)^3), {x, 0, nn}], x]; Drop[t, 3] (* T. D. Noe, Aug 09 2012 *)
  • PARI
    a(n) = sum(k=0, n, (-1)^(n+k)*binomial(k+3, 3)*3^k*stirling(n+3, k+3, 1)); \\ Michel Marcus, Jan 20 2016
    
  • PARI
    b(n) = prod(r=0, n+2, r+3);
    c(n) = sum(i=0, n+2, sum(j=i+1, n+2, sum(k=j+1, n+2, 1/((3+i)*(3+j)*(3+k)))));
    for(n=0, 18, print1(b(n)*c(n), ", ")) \\ Petros Hadjicostas, Jun 12 2020

Formula

E.g.f.: Sum_{n>=0} a(n)*x^(n+3)/(n+3)! = (log(1 - x)/(x - 1))^3/6. - Vladeta Jovovic, May 05 2003 [Edited by Petros Hadjicostas, Jun 13 2020]
a(n) = Sum_{k=0..n} (-1)^(n+k) * binomial(k+3, 3) * 3^k * Stirling1(n+3, k+3). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k) * Stirling1(n-k,i) * Product_{j=0..k-1} (-a-j), then a(n-3) = |f(n,3,3)| for n >= 3. - Milan Janjic, Dec 21 2008
From Petros Hadjicostas, Jun 12 2020: (Start)
a(n) = [x^3] Product_{r=0}^{n+2} (x + 3 + r) = (Product_{r=0}^{n+2} (r+3)) * Sum_{0 <= i < j < k <= n+2} 1/((3+i)*(3+j)*(3+k)).
Since a(n) = R_{n+3}^3(a=-3, b=-1), A001712(n) = R_{n+2}^2(a=-3,b=-1), and A001711(n) = R_{n+1}^1(a=-3, b=-1), the equation R_{n+3}^3(a=-3,b=-1) = R_{n+2}^2(a=-3,b=-1) + (n+5)*R_{n+2}^3(a=-3,b=-1) implies the following:
(i) a(n) = A001712(n) + (n+5)*a(n-1) for n >= 1.
(ii) a(n) = A001711(n) + (2*n+9)*a(n-1) - (n+4)^2*a(n-2) for n >= 2.
(iii) a(n) = (n+2)!/2 + 3*(n+4)*a(n-1) - (3*n^2+21*n+37)*a(n-2) + (n+3)^3*a(n-3) for n >= 3.
(iv) a(n) = 2*(2*n+7)*a(n-1) - (6*n^2+36*n+55)*a(n-2) + (2*n^2+10*n+13)*(2*n+5)*a(n-3) - (n+2)^4*a(n-4) for n >= 4. (End)

Extensions

More terms from Vladeta Jovovic, May 05 2003
Previous Showing 41-50 of 74 results. Next