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 1-4 of 4 results.

A186284 Self-convolution square equals A127776.

Original entry on oeis.org

1, 2, 48, 1704, 71490, 3291780, 160844160, 8189867280, 429832053840, 23088359467040, 1263134996327680, 70138971602098560, 3942799810867610280, 223942062435751452240, 12831882367225056387840, 740872398293620831990080
Offset: 0

Views

Author

Paul D. Hanna, Feb 16 2011

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 48*x^2 + 1704*x^3 + 71490*x^4 + 3291780*x^5 +...
Related expansions.
The g.f. of A127776 equals A(x)^2:
A(x)^2 = 1 + 4*x + 100*x^2 + 3600*x^3 + 152100*x^4 + 7033104*x^5 +...+ A004981(n)^2*x^n +...
The g.f. of A002897 equals A(x)^4:
A(x)^4 = 1 + 8*x + 216*x^2 + 8000*x^3 + 343000*x^4 + 16003008*x^5 +...+ A000984(n)^3*x^n +...
The g.f. of A004981 begins:
1/(1-8*x)^(1/4) = 1 + 2*x + 10*x^2 + 60*x^3 + 390*x^4 + 2652*x^5 +...
where A004981(n) = (2^n/n!)*Product_{k=0..n-1} (4k + 1).
The g.f. of A000984 begins:
1/(1-4*x)^(1/2) = 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 +...
where A000984(n) = (2n)!/(n!)^2 forms the central binomial coefficients.
		

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Sqrt[Hypergeometric2F1[ 1/4, 1/4, 1, 64*x]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Apr 10 2018 *)
  • PARI
    {a(n)=local(A004981=1/(1-8*x+x*O(x^n))^(1/4),A=sum(m=0,n,polcoeff(A004981,m)^2*x^m+x*O(x^n))^(1/2));polcoeff(A,n)}
    
  • PARI
    {a(n)=local(A000984=1/(1-4*x+x*O(x^n))^(1/2),A=sum(m=0,n,polcoeff(A000984,m)^3*x^m+x*O(x^n))^(1/4));polcoeff(A,n)}

Formula

Self-convolution 4th power equals A002897.
G.f.: sqrt( K(k)/(Pi/2) ) in powers of (kk'/4)^2, where K(k) is complete elliptic integral of first kind evaluated at modulus k. [From a formula by Michael Somos in A002897]
G.f.: sqrt( 1/AGM(1, (1-16x)^(1/2)) ) in powers of x(1-16x) where AGM() is the arithmetic-geometric mean. [From a formula by Michael Somos in A004981]
a(n) ~ Pi^(3/4) * 2^(6*n - 1/2) / (Gamma(1/4)^3 * n^(3/2)). - Vaclav Kotesovec, Apr 10 2018

A004981 a(n) = (2^n/n!) * Product_{k=0..n-1} (4*k + 1).

Original entry on oeis.org

1, 2, 10, 60, 390, 2652, 18564, 132600, 961350, 7049900, 52169260, 388898120, 2916735900, 21987701400, 166478310600, 1265235160560, 9647418099270, 73774373700300, 565603531702300, 4346216612028200, 33465867912617140, 258165266754475080, 1994913424920943800
Offset: 0

Views

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

The convolution of this sequence with itself yields A059304. - T. D. Noe, Jun 11 2002
Conjecture: a(p*n) = a(n) (mod p^2) for prime p = 1 (mod 4) and all positive integers n. Cf. A004982 and A298799. - Peter Bala, Dec 22 2019

Crossrefs

Programs

  • GAP
    List([0..25], n-> 2^n*Product([0..n-1], k-> 4*k+1)/Factorial(n) ); # G. C. Greubel, Aug 22 2019
  • Magma
    [1] cat [2^n*&*[4*k+1: k in [0..n-1]]/Factorial(n): n in [1..25]]; // G. C. Greubel, Aug 22 2019
    
  • Maple
    A004981 := n -> (-8)^n*binomial(-1/4, n):
    seq(A004981(n), n=0..25); # Peter Luschny, Oct 23 2018
  • Mathematica
    CoefficientList[Series[(1-8x)^(-1/4), {x, 0, 25}], x] (* Vincenzo Librandi, Mar 16 2014 *)
    Table[8^n*Pochhammer[1/4, n]/n!, {n,0,25}] (* G. C. Greubel, Aug 22 2019 *)
  • Maxima
    a(n):=if n=0 then 1 else (sum(m*sum(binomial(-m+2*k-1,k-1) *2^(n+m-k)*binomial(2*n-k-1,n-1),k,m,n),m,1,n))/(n); /* Vladimir Kruchinin, Dec 26 2011 */
    
  • PARI
    a(n)=if(n<0,0,prod(k=1,n,(8*k-6)/k))
    
  • PARI
    {a(n)=if(n<0, 0, polcoeff( (1-8*x+x*O(x^n))^(-1/4), n))} /* Michael Somos, Jan 31 2007 */
    
  • Sage
    [8^n*rising_factorial(1/4, n)/factorial(n) for n in (0..25)] # G. C. Greubel, Aug 22 2019
    

Formula

a(n) ~ Gamma(1/4)^-1*n^(-3/4)*2^(3*n)*{1 - 3/32*n^-1 - ...}
G.f.: (1-8*x)^(-1/4).
A002897(n) = Sum_{k=0..n} a(k)^2*a(n-k)^2. - Michael Somos, Jan 31 2007
a(n) = (Sum_{m=1..n} m*Sum_{k=m..n} binomial(-m+2*k-1,k-1)*2^(n+m-k)*binomial(2*n-k-1,n-1))/n, n>0, a(0)=1. - Vladimir Kruchinin, Dec 26 2011
D-finite with recurrence: n*a(n) = 2*(4*n-3)*a(n-1). - R. J. Mathar, Mar 14 2014
From Karol A. Penson, Dec 19 2015: (Start)
a(n) = (-8)^n*binomial(-1/4,n).
E.g.f.: is the hypergeometric function of type 1F1, in Maple notation hypergeom([1/4], [1], 8*x).
Representation as n-th moment of a positive function on (0, 8): a(n)=int(x^n*(sqrt(2)/(16*Pi*(x/8)^(3/4)*(1-x/8)^(1/4))), x=0..8), n=0, 1, ... . This function is the solution of the Hausdorff moment problem on (0, 8) with moments equal to a(n). As a consequence this representation is unique. (End)

Extensions

More terms from James Sellers, May 01 2000

A091401 Numbers n such that genus of group Gamma_0(n) is zero.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 16, 18, 25
Offset: 1

Views

Author

N. J. A. Sloane, Mar 02 2004

Keywords

Comments

Equivalently, numbers n such that genus of modular curve X_0(n) is zero.

References

  • G. Shimura, Introduction to the Arithmetic Theory of Automorphic Functions, Princeton, 1971, see Prop. 1.40 and 1.43.

Crossrefs

The table below is a consequence of Theorem 7.3 in Maier's paper.
N EntryID K alpha
1
2 A127776 4096 1
3 A276018 729 1
4 A002894 256 1
5 A276019 125 4
6 A093388 72 1
7 A276021 49 9
8 A081085 32 1
9 A006077 27 1
10 A276020 20 2
12 A276022 12 1
13 A276177 13 36
16 A276178 8 1
18 A276179 6 1
25 A276180 5 4

Programs

Formula

Numbers n such that A001617(n) = 0.

A302577 Convolution square root of A186284 multiplied by 2^n.

Original entry on oeis.org

1, 2, 94, 6628, 554246, 50936956, 4971074892, 505747739784, 53048521913478, 5695802803696236, 622942370315360004, 69155891028297395448, 7772714892571857579036, 882718626126348791323992, 101137353917153181195426264, 11676481964194514316750017040
Offset: 0

Views

Author

Vaclav Kotesovec, Apr 10 2018

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Hypergeometric2F1[1/4, 1/4, 1, 64*x]^(1/4), {x, 0, nmax}], x] * 2^Range[0, nmax]

Formula

a(n) ~ Pi^(9/8) * 2^(7*n - 5/4) / (Gamma(1/4)^(7/2) * n^(3/2)).
Showing 1-4 of 4 results.