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 17 results. Next

A001879 a(n) = (2n+2)!/(n!*2^(n+1)).

Original entry on oeis.org

1, 6, 45, 420, 4725, 62370, 945945, 16216200, 310134825, 6547290750, 151242416325, 3794809718700, 102776096548125, 2988412653476250, 92854250304440625, 3070380543400170000, 107655217802968460625, 3989575718580595893750, 155815096120119939628125
Offset: 0

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Keywords

Comments

From Wolfdieter Lang, Oct 06 2008: (Start)
a(n) is the denominator of the n-th approximant to the continued fraction 1^2/(6+3^2/(6+5^2/(6+... for Pi-3. W. Lang, Oct 06 2008, after an e-mail from R. Rosenthal. Cf. A142970 for the corresponding numerators.
The e.g.f. g(x)=(1+x)/(1-2*x)^(5/2) satisfies (1-4*x^2)*g''(x) - 2*(8*x+3)*g'(x) -9*g(x) = 0 (from the three term recurrence given below). Also g(x)=hypergeom([2,3/2],[1],2*x). (End)
Number of descents in all fixed-point-free involutions of {1,2,...,2(n+1)}. A descent of a permutation p is a position i such that p(i) > p(i+1). Example: a(1)=6 because the fixed-point-free involutions 2143, 3412, and 4321 have 2, 1, and 3 descents, respectively. - Emeric Deutsch, Jun 05 2009
First differences of A193651. - Vladimir Reshetnikov, Apr 25 2016
a(n-2) is the number of maximal elements in the absolute order of the Coxeter group of type D_n. - Jose Bastidas, Nov 01 2021

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 77 (Problem 10, values of Bessel polynomials).
  • 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

Second column of triangle A001497. Equals (A001147(n+1)-A001147(n))/2.
Equals row sums of A163938.

Programs

  • Magma
    [Factorial(2*n+2)/(Factorial(n)*2^(n+1)): n in [0..20]]; // Vincenzo Librandi, Nov 22 2011
  • Maple
    restart: G(x):=(1-x)/(1-2*x)^(1/2): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1],x) od:x:=0:seq(f[n],n=2..20); # Zerinvary Lajos, Apr 04 2009
  • Mathematica
    Table[(2n+2)!/(n!2^(n+1)),{n,0,20}] (* Vincenzo Librandi, Nov 22 2011 *)
  • PARI
    a(n)=if(n<0,0,(2*n+2)!/n!/2^(n+1))
    

Formula

E.g.f.: (1+x)/(1-2*x)^(5/2).
a(n)*n = a(n-1)*(2n+1)*(n+1); a(n) = a(n-1)*(2n+4)-a(n-2)*(2n-1), if n>0. - Michael Somos, Feb 25 2004
From Wolfdieter Lang, Oct 06 2008: (Start)
a(n) = (n+1)*(2*n+1)!! with the double factorials (2*n+1)!!=A001147(n+1).
D-finite with recurrence a(n) = 6*a(n-1) + ((2*n-1)^2)*a(n-2), a(-1)=0, a(0)=1. (End)
With interpolated 0's, e.g.f.: B(A(x)) where B(x)= x exp(x) and A(x)=x^2/2.
E.g.f.: -G(0)/2 where G(k) = 1 - (2*k+3)/(1 - x/(x - (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 06 2012
G.f.: (1-x)/(2*x^2*Q(0)) - 1/(2*x^2), where Q(k) = 1 - x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 20 2013
From Karol A. Penson, Jul 12 2013: (Start)
Integral representation as n-th moment of a signed function w(x) of bounded variation on (0,infinity),
w(x) = -(1/4)*sqrt(2)*sqrt(x)*(1-x)*exp(-x/2)/sqrt(Pi):
a(n) = Integral_{x>=0} x^n*w(x), n>=0.
For x>1, w(x)>0. w(0)=w(1)=limit(w(x),x=infinity)=0. For x<1, w(x)<0.
Asymptotics: a(n)->(1/576)*2^(1/2+n)*(1152*n^2+1680*n+505)*exp(-n)*(n)^(n), for n->infinity. (End)
G.f.: 2F0(3/2,2;;2x). - R. J. Mathar, Aug 08 2015

Extensions

Entry revised Aug 31 2004 (thanks to Ralf Stephan and Michael Somos)
E.g.f. in comment line corrected by Wolfdieter Lang, Nov 21 2011

A001877 Number of divisors of n of the form 5k+2; a(0) = 0.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 2, 0, 2, 0, 1, 1, 1, 0, 1, 1, 2, 0, 2, 0, 1, 1, 2, 0, 1, 0, 2, 0, 2, 1, 2, 1, 1, 0, 1, 0, 3, 0, 2, 0, 1, 1, 2, 1, 1, 1, 2, 0, 2, 0, 2, 1, 1, 0, 2, 0, 2, 1, 2, 0, 2, 1, 2, 0, 2, 0, 3, 0, 2, 0, 1, 2, 1, 0, 1, 1, 2, 0, 4, 1, 1, 1
Offset: 0

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Programs

Formula

G.f.: Sum_{n>=0} x^(5n+2)/(1-x^(5n+2)).
G.f.: Sum_{n>=1} x^(2*n)/(1-x^(5*n)). - Joerg Arndt, Jan 30 2011
Sum_{k=1..n} a(k) = n*log(n)/5 + c*n + O(n^(1/3)*log(n)), where c = gamma(2,5) - (1 - gamma)/5 = A256780 - (1 - A001620)/5 = 0.105832... (Smith and Subbarao, 1981). - Amiram Eldar, Nov 25 2023

A001878 Number of divisors of n of the form 5k+3; a(0) = 0.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 2, 0, 0, 1, 0, 1, 2, 0, 1, 1, 1, 0, 1, 0, 1, 2, 0, 0, 2, 0, 1, 2, 1, 0, 1, 1, 0, 1, 1, 0, 3, 0, 0, 1, 1, 1, 2, 0, 2, 1, 1, 0, 1, 0, 0, 2, 1, 1, 2, 0, 1, 2, 0, 0, 3, 1, 0, 1, 1, 0, 3, 0, 1, 1, 0, 1, 2, 0, 1, 1
Offset: 0

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Programs

Formula

G.f.: Sum_{n>=0} x^(5*n+3)/(1 - x^(5*n+3)).
G.f.: Sum_{k>=1} x^(3*k)/(1 - x^(5*k)). - Ilya Gutkovskiy, Sep 11 2019
Sum_{k=1..n} a(k) = n*log(n)/5 + c*n + O(n^(1/3)*log(n)), where c = gamma(3,5) - (1 - gamma)/5 = A256848 - (1 - A001620)/5 = -0.0983206... (Smith and Subbarao, 1981). - Amiram Eldar, Nov 25 2023

A001899 Number of divisors of n of the form 5k+4; a(0) = 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 2, 0, 0, 1, 2, 1, 0, 0, 1, 0, 1, 0, 2, 0, 1, 1, 1, 0, 1, 0, 2, 1, 0, 0, 2, 1, 0, 0, 1, 0, 2, 0, 2, 1, 1, 1, 1, 0, 0, 1, 2, 0, 0, 0, 2, 1, 1, 0, 3, 0, 1, 0, 2, 0, 1, 1, 1, 1, 0, 0, 3, 0, 0
Offset: 0

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Author

Keywords

Crossrefs

Programs

  • Mathematica
    Join[{0}, Table[d = Divisors[n]; Length[Select[d, Mod[#, 5] == 4 &]], {n, 100}]] (* T. D. Noe, Aug 10 2012 *)
  • PARI
    a(n) = if (n==0, 0, sumdiv(n, d, (d % 5)==4)); \\ Michel Marcus, Feb 28 2021

Formula

G.f.: Sum_{n>=0} x^(5*n+4)/(1 - x^(5*n+4)).
G.f.: Sum_{k>=1} x^(4*k)/(1 - x^(5*k)). - Ilya Gutkovskiy, Sep 11 2019
Sum_{k=1..n} a(k) = n*log(n)/5 + c*n + O(n^(1/3)*log(n)), where c = gamma(4,5) - (1 - gamma)/5 = A256849 - (1 - A001620)/5 = -0.213442... (Smith and Subbarao, 1981). - Amiram Eldar, Nov 25 2023

Extensions

Better definition from Michael Somos, Aug 31 2004

A279061 Number of divisors of n of the form 7*k + 1.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 2, 3, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2
Offset: 0

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Author

Ilya Gutkovskiy, Dec 05 2016

Keywords

Comments

Möebius transform is a period-7 sequence {1, 0, 0, 0, 0, 0, 0, ...}.

Examples

			a(8) = 2 because 8 has 4 divisors {1,2,4,8} among which 2 divisors {1,8} are of the form 7*k + 1.
		

Crossrefs

Programs

  • Maple
    N:= 200: # to get a(0)..a(N)
    V:= Vector(N):
    for k from 1 to N do
      R:= [seq(i,i=k..N,7*k)];
      V[R]:= map(`+`,V[R],1);
    od:
    0,seq(V[i],i=1..N); # Robert Israel, Dec 05 2016
  • Mathematica
    nmax = 120; CoefficientList[Series[Sum[x^k/(1 - x^(7 k)), {k, 1, nmax}], {x, 0, nmax}], x]
    nmax = 120; CoefficientList[Series[Sum[x^(7 k + 1)/(1 - x^(7 k + 1)), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[Count[Divisors[n],?(IntegerQ[(#-1)/7]&)],{n,0,100}] (* _Harvey P. Dale, Nov 08 2022 *)
  • PARI
    concat([0], Vec(sum(k=1, 100, x^k / (1 - x^(7*k))) + O(x^101))) \\ Indranil Ghosh, Mar 29 2017

Formula

G.f.: Sum_{k>=1} x^k/(1 - x^(7*k)).
G.f.: Sum_{k>=0} x^(7*k+1)/(1 - x^(7*k+1)).
Sum_{k=1..n} a(k) = n*log(n)/7 + c*n + O(n^(1/3)*log(n)), where c = gamma(1,7) - (1 - gamma)/7 = 0.713612..., gamma(1,7) = -(psi(1/7) + log(7))/7 is a generalized Euler constant, and gamma is Euler's constant (A001620) (Smith and Subbarao, 1981). - Amiram Eldar, Nov 25 2023

A279060 Number of divisors of n of the form 6*k + 1.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 3, 2, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 4, 1, 2, 1, 2, 1, 2, 3, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 1, 2, 2, 1, 2, 1, 2, 1
Offset: 0

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Author

Ilya Gutkovskiy, Dec 05 2016

Keywords

Comments

Möbius transform is the period-6 sequence {1, 0, 0, 0, 0, 0, ...}.

Examples

			a(14) = 2 because 14 has 4 divisors {1,2,7,14} among which 2 divisors {1,7} are of the form 6*k + 1.
		

Crossrefs

Programs

  • Mathematica
    nmax = 120; CoefficientList[Series[Sum[x^k/(1 - x^(6 k)), {k, 1, nmax}], {x, 0, nmax}], x]
    nmax = 120; CoefficientList[Series[Sum[x^(6 k + 1)/(1 - x^(6 k + 1)), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[Count[Divisors[n],?(Mod[#,6]==1&)],{n,0,120}] (* _Harvey P. Dale, Apr 27 2018 *)
  • PARI
    A279060(n) = if(!n,n,sumdiv(n, d, (1==(d%6)))); \\ Antti Karttunen, Jul 09 2017
    
  • Python
    from sympy import divisors
    def A279060(n): return sum(d%6 == 1 for d in divisors(n)) # David Radcliffe, Jun 19 2025

Formula

G.f.: Sum_{k>=1} x^k/(1 - x^(6*k)).
G.f.: Sum_{k>=0} x^(6*k+1)/(1 - x^(6*k+1)).
From Antti Karttunen, Oct 03 2018: (Start)
a(n) = A320001(n) + [1 == n (mod 6)], where [ ] is the Iverson bracket, giving 1 only when n = 1 mod 6, and 0 otherwise.
a(n) = A035218(n) - A319995(n). (End)
a(n) = (A035218(n) + A035178(n)) / 2. - David Radcliffe, Jun 19 2025
Sum_{k=1..n} a(k) = n*log(n)/6 + c*n + O(n^(1/3)*log(n)), where c = gamma(1,6) - (1 - gamma)/6 = 0.686263..., gamma(1,6) = -(psi(1/6) + log(6))/6 is a generalized Euler constant, and gamma is Euler's constant (A001620) (Smith and Subbarao, 1981). - Amiram Eldar, Nov 25 2023

A359233 Number of divisors of 5*n-1 of form 5*k+1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 3, 2, 1, 1, 3, 1, 3, 1, 2, 1, 1, 1, 4, 1, 1, 2, 2, 1, 3, 1, 3, 1, 1, 2, 4, 1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 1, 1, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1
Offset: 1

Views

Author

Seiichi Manyama, Dec 22 2022

Keywords

Comments

Also number of divisors of 5*n-1 of form 5*k+4.

Crossrefs

Programs

  • Mathematica
    a[n_] := DivisorSum[5*n-1, 1 &, Mod[#, 5] == 1 &]; Array[a, 100] (* Amiram Eldar, Aug 23 2023 *)
  • PARI
    a(n) = sumdiv(5*n-1, d, d%5==1);
    
  • PARI
    a(n) = sumdiv(5*n-1, d, d%5==4);
    
  • PARI
    my(N=100, x='x+O('x^N)); Vec(sum(k=1, N, x^k/(1-x^(5*k-1))))
    
  • PARI
    my(N=100, x='x+O('x^N)); Vec(sum(k=1, N, x^(4*k-3)/(1-x^(5*k-4))))

Formula

a(n) = A001876(5*n-1) = A001899(5*n-1).
G.f.: Sum_{k>0} x^k/(1 - x^(5*k-1)).
G.f.: Sum_{k>0} x^(4*k-3)/(1 - x^(5*k-4)).

A359236 Number of divisors of 5*n-2 of form 5*k+1.

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Dec 22 2022

Keywords

Comments

Also number of divisors of 5*n-2 of form 5*k+3.

Crossrefs

Programs

  • Mathematica
    a[n_] := DivisorSum[5*n-2, 1 &, Mod[#, 5] == 1 &]; Array[a, 100] (* Amiram Eldar, Aug 23 2023 *)
  • PARI
    a(n) = sumdiv(5*n-2, d, d%5==1);
    
  • PARI
    a(n) = sumdiv(5*n-2, d, d%5==3);
    
  • PARI
    my(N=100, x='x+O('x^N)); Vec(sum(k=1, N, x^k/(1-x^(5*k-2))))
    
  • PARI
    my(N=100, x='x+O('x^N)); Vec(sum(k=1, N, x^(3*k-2)/(1-x^(5*k-4))))

Formula

a(n) = A001876(5*n-2) = A001878(5*n-2).
G.f.: Sum_{k>0} x^k/(1 - x^(5*k-2)).
G.f.: Sum_{k>0} x^(3*k-2)/(1 - x^(5*k-4)).

A359237 Number of divisors of 5*n-3 of form 5*k+1.

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Dec 22 2022

Keywords

Comments

Also number of divisors of 5*n-3 of form 5*k+2.

Crossrefs

Programs

  • Mathematica
    a[n_] := DivisorSum[5*n-3, 1 &, Mod[#, 5] == 1 &]; Array[a, 100] (* Amiram Eldar, Aug 23 2023 *)
  • PARI
    a(n) = sumdiv(5*n-3, d, d%5==1);
    
  • PARI
    a(n) = sumdiv(5*n-3, d, d%5==2);
    
  • PARI
    my(N=100, x='x+O('x^N)); Vec(sum(k=1, N, x^k/(1-x^(5*k-3))))
    
  • PARI
    my(N=100, x='x+O('x^N)); Vec(sum(k=1, N, x^(2*k-1)/(1-x^(5*k-4))))

Formula

a(n) = A001876(5*n-3) = A001877(5*n-3).
G.f.: Sum_{k>0} x^k/(1 - x^(5*k-3)).
G.f.: Sum_{k>0} x^(2*k-1)/(1 - x^(5*k-4)).

A359238 Number of divisors of 5*n-4 of form 5*k+1.

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Dec 22 2022

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := DivisorSum[5*n-4, 1 &, Mod[#, 5] == 1 &]; Array[a, 100] (* Amiram Eldar, Aug 23 2023 *)
  • PARI
    a(n) = sumdiv(5*n-4, d, d%5==1);
    
  • PARI
    my(N=100, x='x+O('x^N)); Vec(sum(k=1, N, x^k/(1-x^(5*k-4))))

Formula

a(n) = A001876(5*n-4).
G.f.: Sum_{k>0} x^k/(1 - x^(5*k-4)).
Showing 1-10 of 17 results. Next