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.

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A269123 Number of sequences with n copies each of 1,2,...,5 avoiding the pattern 12...5.

Original entry on oeis.org

1, 119, 102011, 117392909, 142951955371, 173996758190594, 208728647384065181, 246211478304046636024, 285867869484243410805931, 327305762373962750213344469, 370243779118368267931211789686, 414473407439255725429093382135614, 459837370068489293091980501717140861
Offset: 0

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Author

Alois P. Heinz, Feb 19 2016

Keywords

Crossrefs

Row n=5 of A269129.

Formula

a(n) = A008978(n) - A268841(n).

A377219 Expansion of the o.g.f. A(x) defined by [x^n] A(x)^(120*n) = (5*n)!/n!^5 for n >= 0.

Original entry on oeis.org

1, 1, 353, 318986, 408941594, 633438203535, 1105336091531052, 2093867978990821853, 4212168629863126220194, 8871676970891643267231886, 19375253437183554713216237582, 43574669954100844749472466829032, 100404408695672206422230611142618195, 236114213302057579962294974098604849352, 564982003808755415617353442524468859709030
Offset: 0

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Author

Peter Bala, Oct 20 2024

Keywords

Comments

Compare with A000984(n) = [x^n] (1 + x)^(2*n) = (2*n)!/n!^2.
The central binomial coefficients A000984(n) satisfy the supercongruences u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) for all primes p >= 5 and positive integers n and k.
More generally, for positive integers r and s, the sequence {u(r,s; n) : n >= 0} defined by u(r,s; n) = [x^(s*n)] (1 + x)^(r*n) = binomial(r*n, s*n) satisfies the same supercongruences (Meštrović, Section 6, equation 39).
Conjecture: for positive integers r and s, the sequence {v(r,s; n) : n >= 0} defined by v(r,s; n) = [x^(s*n)] A(x)^(r*n) also satisfies the same supercongruences.

Crossrefs

Programs

  • Maple
    Order := 25:
    E(x) := exp(add((5*n)!/n!^5 * x^n/n, n = 1..25)):
    solve(series(x*E(x),x) = y, x):
    convert(%, polynom):
    g := taylor(y/%, y = 0, 25):
    seq(coeftayl(g^(1/120), y = 0,  n), n = 0..20);

Formula

O.g.f.: A(x) = ( x/(x * series_reversion(E(x)))^(1/120), where E(x) = exp(Sum_{n >= 1} (5*n)!/n!^5 *x^n/n) is the o.g.f. of A333043.

A307618 A Calabi-Yau period integral: a(n) = C(4*n,2*n)*C(2*n,n)^3.

Original entry on oeis.org

1, 48, 15120, 7392000, 4414410000, 2956651746048, 2133278987583744, 1621682968820428800, 1281351259836532170000, 1043032815185819858400000, 869343653096068540955685120, 738637974389826550020188712960, 637665137404661719206664998969600
Offset: 0

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Author

Bradley Klee, Jun 04 2019

Keywords

Comments

Entry number six in the "Big Table" of Almkvist et al. (see links). The period T(x) = Sum_{n>=0} a(n)*x^(2*n) is also the first x-derivative of the 6-volume associated to the algebraic variety V6 = P1 & P2 & P3, with P1 : X1^2 + Y1^2 = X2^2 + Y2^2, P2 : X2^2 + Y2^2 = X3^2 + Y3^2, P3 : x=(X1^2 + X2^2 + X3^2 + Y1^2 + Y2^2 + Y3^2)^3*(1 - X1*X2*X3*Y1*Y2*Y3). The small x limit reduces V6 to a 6-ball with 6-volume proportional to x. Similar constructions are known to exist for a few other geometries on Almkvist's list, most notably #3: A186420, and #16: A039699.

Crossrefs

Hadamard Factors: A000984, A002894, A002897, A001448, A000897, A008977.
Calabi-Yau Periods: A008978, A186420, A268553, A039699.

Programs

  • Mathematica
    Binomial[4*#,2*#]*Binomial[2*#,#]^3&/@Range[0,10]

Formula

G.f.: 4F3({1/4, 3/4, 1/2, 1/2}, {1, 1, 1}, 1024*x).
Define the period integral:
dt(x) = dz1*dz2*dz3/sqrt(1-32*x*cos(z1)*cos(z2)*cos(z3)).
T(x)=1/(2*Pi)^3*Integral_{0..2*Pi,0..2*Pi,0..2*Pi} dt(x),
the Picard-Fuchs coefficients:(c0,c1,c2,c3,c4)=
(768*x, 14592*x^2-1, x*(25344*x^2-7), 2*x^2*(5120*x^2-3), x^3*(32*x-1)*(32*x+1)),
and the certificate function:
G(z1,z2,z3)=(16*sin(z1)*(
48*x*cos(z1)
+ cos(z2)*cos(z3)
+ 48*x*cos(z1)*(cos(z3)^2 + cos(z2)^2)
+ 2304*x^2*cos(z1)^2*cos(z2)*cos(z3)
+ 80*x*cos(z1)*cos(z2)^2*cos(z3)^2
+ 384*x^2*cos(z1)^2*(cos(z2)*cos(z3)^3 + cos(z2)^3*cos(z3))
+ 256*x^2*cos(z1)^2*cos(z2)^3*cos(z3)^3)
)/(3*(1 - 32*x*cos(z1)*cos(z2)*cos(z3))^(7/2)),
Then: 0 = Sum_{n=0..4}cn*d^n/dx^n dt(x) + d/dz1 G(z1,z2,z3) + d/dz2 G(z2,z3,z1) + d/dz3 G(z3,z1,z2), thus: 0 = Sum_{n=0..4} cn*d^n/dx^n T(x).
Furthermore, let (a1,a2,a3)=(c1,c2,c3)/c0, then also: 0 = (1/2)*a2*a3 - (1/8)*a3^3 + d/dx(a2) - (3/4)*a3*d/dx(a3) - (1/2)*d^2/dx^2(a3) - a1.
D-finite with recurrence: n^4*a(n) -16*(4*n-1)*(4*n-3)*(-1+2*n)^2*a(n-1)=0. - R. J. Mathar, Jan 27 2020
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