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 1-3 of 3 results.

A121860 a(n) = Sum_{d|n} n!/(d!*(n/d)!).

Original entry on oeis.org

1, 2, 2, 8, 2, 122, 2, 1682, 10082, 30242, 2, 7318082, 2, 17297282, 3632428802, 36843206402, 2, 2981705126402, 2, 1690185726028802, 3379030566912002, 28158588057602, 2, 76941821303636889602, 1077167364120207360002
Offset: 1

Views

Author

Vladeta Jovovic, Sep 09 2006

Keywords

Comments

a(n) = 2 iff n is prime.
a(468) has 1007 decimal digits. - Michael De Vlieger, Sep 12 2018
From Gus Wiseman, Jan 10 2019: (Start)
Number of matrices whose entries are 1,...,n, up to row and column permutations. For example, inequivalent representatives of the a(4) = 8 matrices are:
[1 2 3 4]
.
[1 2] [1 2] [1 3] [1 3] [1 4] [1 4]
[3 4] [4 3] [2 4] [4 2] [2 3] [3 2]
.
[1]
[2]
[3]
[4]
(End)
Conjecture: the sequence a(n) taken modulo a positive integer k >= 3 eventually becomes constant equal to 2. For example, the sequence taken modulo 11 is [1, 2, 2, 8, 2, 1, 2, 10, 6, 3, 2, 2, 2, 2, 2, 2, ...]. - Peter Bala, Aug 08 2025

Crossrefs

Programs

  • Maple
    with(numtheory): seq(n!*add(1/(d!*(n/d)!), d in divisors(n)), n = 1..25); # Peter Bala, Aug 04 2025
  • Mathematica
    f[n_] := Block[{d = Divisors@n}, Plus @@ (n!/(d! (n/d)!))]; Array[f, 25] (* Robert G. Wilson v, Sep 11 2006 *)
    Table[DivisorSum[n, n!/(#!*(n/#)!) &], {n, 25}] (* Michael De Vlieger, Sep 12 2018 *)
  • PARI
    a(n) = sumdiv(n, d, n!/(d!*(n/d)!)); \\ Michel Marcus, Sep 13 2018

Formula

E.g.f.: Sum_{k>0} (exp(x^k)-1)/k!.

Extensions

More terms from Robert G. Wilson v, Sep 11 2006

A258900 E.g.f.: S(x) = Series_Reversion( Integral 1/(1-x^4)^(1/4) dx ), where the constant of integration is zero.

Original entry on oeis.org

1, -6, -1764, -7700616, -147910405104, -8310698364852576, -1085420895640591777344, -284168646775526186095019136, -134459287943928269154814258953984, -106506405136317713669903020280294647296
Offset: 0

Views

Author

Paul D. Hanna, Jun 14 2015

Keywords

Examples

			E.g.f. with offset 0 is C(x) and e.g.f. with offset 1 is S(x) where:
C(x) = 1 - 6*x^4/4! - 1764*x^8/8! - 7700616*x^12/12! - 147910405104*x^16/16! -...
S(x) = x - 6*x^5/5! - 1764*x^9/9! - 7700616*x^13/13! - 147910405104*x^17/17! -...
such that C(x)^4 + S(x)^4 = 1:
C(x)^4 = 1 - 24*x^4/4! + 8064*x^8/8! + 2128896*x^12/12! + 52932870144*x^16/16! +...
S(x)^4 = 24*x^4/4! - 8064*x^8/8! - 2128896*x^12/12! - 52932870144*x^16/16! -...
Related Expansions.
(1) The series reversion of S(x) is Integral 1/(1-x^4)^(1/4) dx:
Series_Reversion(S(x)) = x + 6*x^5/5! + 6300*x^9/9! + 56133000*x^13/13! +...
1/(1-x^4)^(1/4) = 1 + 6*x^4/4! + 6300*x^8/8! + 56133000*x^12/12! + 1992160170000*x^16/16! +...+ A258899(n)*x^(4*n)/(4*n)! +...
(2) d/dx S(x)/C(x) = 1/C(x)^4:
1/C(x)^4 = 1 + 24*x^4/4! + 32256*x^8/8! + 285272064*x^12/12! +...
S(x)/C(x) = x + 24*x^5/5! + 32256*x^9/9! + 285272064*x^13/13! + 8967114326016*x^17/17! +...+ A258901(n)*x^(4*n+1)/(4*n+1)! +...
where
Series_Reversion(S(x)/C(x)) = x - x^5/5 + x^9/9 - x^13/13 + x^17/17 - x^21/21 +...
		

Crossrefs

Programs

  • PARI
    /* E.g.f. Series_Reversion(Integral 1/(1-x^4)^(1/4) dx): */
    {a(n)=local(S=x); S = serreverse( intformal(  1/(1-x^4 +x*O(x^(4*n)))^(1/4) )); (4*n+1)!*polcoeff(S,4*n+1)}
    for(n=0,15,print1(a(n),", "))
    
  • PARI
    /* E.g.f. C(x) with offset 0: */
    {a(n)=local(S=x, C=1+x); for(i=1, n, S=intformal(C +x*O(x^(4*n))); C=1-intformal(S^3/C^2 +x*O(x^(4*n))); ); (4*n)!*polcoeff(C, 4*n)}
    for(n=0, 15, print1(a(n), ", "))
    
  • PARI
    /* E.g.f. S(x) with offset 1: */
    {a(n)=local(S=x, C=1+x); for(i=1, n+1, S=intformal(C +x*O(x^(4*n))); C=1-intformal(S^3/C^2 +x*O(x^(4*n+1))); ); (4*n+1)!*polcoeff(S, 4*n+1)}
    for(n=0, 15, print1(a(n), ", "))

Formula

Let e.g.f. C(x) = Sum_{n>=0} a(n)*x^(4*n)/(4*n)! and e.g.f. S(x) = Sum_{n>=0} a(n)*x^(4*n+1)/(4*n+1)!, then C(x) and S(x) satisfy:
(1) C(x)^4 + S(x)^4 = 1,
(2) S'(x) = C(x),
(3) C'(x) = -S(x)^3/C(x)^2,
(4) C(x)^3 * C'(x) + S(x)^3 * S'(x) = 0,
(5) S(x)/C(x) = Integral 1/C(x)^4 dx,
(6) S(x)/C(x) = Series_Reversion( Integral 1/(1+x^4) dx ) = Series_Reversion( Sum_{n>=0} (-1)^n * x^(4*n+1)/(4*n+1) ).
(7) S(x)^2/C(x)^2 = tan( 2 * Integral S(x)/C(x) dx ).
(8) C(x)^2 + I*S(x)^2 = exp( 2*I * Integral S(x)/C(x) dx ).

A258903 E.g.f.: 2 - exp(2) + Sum_{n>=1} 2^n * exp(3*x^n) / n!.

Original entry on oeis.org

1, 6, 30, 78, 426, 582, 12450, 4758, 407010, 2218182, 19172370, 360438, 4755166050, 3213222, 85631151090, 5099958831318, 54483404779650, 258673542, 11939347971403410, 2326095798, 5556296851712151330, 35398724239897109862, 10235928407592878130, 188311523478, 758680053859872239555010
Offset: 0

Views

Author

Paul D. Hanna, Jun 20 2015

Keywords

Comments

Conjecture: the sequence a(n) taken modulo a positive integer k is eventually periodic with the period dividing phi(k). For example, the sequence taken modulo 7 is [6, 2, 1, 6, 1, 4, 5, 2, 1, 0, 1, 5, 5, 2, 1, 0, 1, 5, 5, 2, 1, 0, 1, 5, 5, 2, 1, 0, 1, 5, ...] with an apparent period of 6 (= phi(7)) starting at n = 7. - Peter Bala, Aug 08 2025

Examples

			E.g.f.: A(x) = 1 + 6*x + 30*x^2/2! + 78*x^3/3! + 426*x^4/4! + 582*x^5/5! +...
where
A(x) = 2 - exp(2) + 2*exp(3*x) + 2^2*exp(3*x^2)/2! + 2^3*exp(3*x^3)/3! + 2^4*exp(3*x^4)/4! + 2^5*exp(3*x^5)/5! +...
A(x) = 2 - exp(3) + 3*exp(2*x) + 3^2*exp(2*x^2)/2! + 3^3*exp(2*x^3)/3! + 3^4*exp(2*x^4)/4! + 3^5*exp(2*x^5)/5! +...
		

Crossrefs

Cf. A258899.

Programs

  • Maple
    with(numtheory): seq(n!*add(2^d*3^(n/d)/(d!*(n/d)!), d in divisors(n)), n = 1..25); # Peter Bala, Aug 08 2025
  • PARI
    {a(n) = local(A=1); A = 2-exp(2) + sum(m=1,n,2^m/m!*exp(3*x^m +x*O(x^n))); if(n==0,1, n!*polcoeff(A,n))}
    for(n=0,30, print1(a(n),", "))
    
  • PARI
    {a(n) = local(A=1); A = 2-exp(3) + sum(m=1,n,3^m/m!*exp(2*x^m +x*O(x^n))); if(n==0,1, n!*polcoeff(A,n))}
    for(n=0,30, print1(a(n),", "))

Formula

E.g.f.: 2 - exp(3) + Sum_{n>=1} 3^n * exp(2*x^n) / n!.
For n >= 1, a(n) = Sum_{d divides n} 2^d * 3^(n/d) * n!/(d!*(n/d)!). - Peter Bala, Aug 08 2025
Showing 1-3 of 3 results.