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.

A229451 G.f.: exp( Sum_{n>=1} (3*n)!/n!^3 * x^n/n ).

Original entry on oeis.org

1, 6, 63, 866, 13899, 246366, 4676768, 93322596, 1934035965, 41286407510, 902562584556, 20119266633060, 455832458083577, 10470568749165246, 243361203186769659, 5714294570067499930, 135377464019074334826, 3232534121305720233264, 77726654423445817800164
Offset: 0

Views

Author

Paul D. Hanna, Sep 23 2013

Keywords

Comments

The sixth root of the o.g.f. A(x)^(1/6) = 1 + x + 8*x^2 + 101*x^3 + 1569*x^4 + 27445*x^5 + ... appears to have integer coefficients. See A229452. More generally, if A(m,x) := exp( Sum_{n >= 1} (m*n)!/n!^m * x^n/n ), m = 1,2,3,..., then it can be shown that the expansion of A(m,x) has integer coefficients. We conjecture that the expansion of A(m,x)^(1/m!) also has integer coefficients. - Peter Bala, Feb 16 2020

Examples

			G.f.: A(x) = 1 + 6*x + 63*x^2 + 866*x^3 + 13899*x^4 + 246366*x^5 +...,
where
log(A(x)) = 6*x + 90*x^2/2 + 1680*x^3/3 + 34650*x^4/4 + 756756*x^5/5 +...+ A006480(n)*x^n/n + ....
		

Crossrefs

Cf. A229452, A006480 (de Bruijn's S(3,n)), A061401, A333042, A333043, A370288, A362732, A370289, A370293.

Programs

  • Mathematica
    CoefficientList[Series[Exp[6*x*HypergeometricPFQ[{1,1,4/3,5/3},{2,2,2},27*x]],{x,0,20}],x] (* Vaclav Kotesovec, Dec 25 2013 *)
  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n,(3*k)!/k!^3*x^k/k) +x*O(x^n)),n)}
    for(n=0,25,print1(a(n),", "))

Formula

a(n) ~ c * 3^(3*n)/n^2, where c = 2^11 * 3^(7/2) * Pi^5 * A370293^6 = 0.406436497... - Vaclav Kotesovec, Dec 25 2013, updated Feb 14 2024
a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} A006480(k) * a(n-k). - Seiichi Manyama, Feb 09 2024
From Peter Bala, Oct 24 2024: (Start)
Series reversion of x*A(-x) = x + 6*x^2 + 9*x^3 + 56*x^4 - 300*x^5 + 3942*x^6 - ... is the g.f. of A061401.
The g.f. A(x) satisfies [x^n] A(x)^n = A362732(n). (End)

A370295 G.f.: exp(Sum_{k>=1} (5*k)!/(5!*k!^5) * x^k/k).

Original entry on oeis.org

1, 1, 473, 467606, 637121154, 1039792179805, 1905441263652576, 3785382599457953517, 7981116324798212651066, 17613760342120835610374245, 40303877398793645855018120732, 94970269248783993542201925505548, 229287077006842005926064077532676555, 565001770629439341048001870559581136157
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 14 2024

Keywords

Comments

In general, for m>=2, if g.f. = exp(Sum_{k>=1} (m*k)!/(m!*k!^m) * x^k/k), then a(n,m) ~ c(m) * m^(m*n) / n^((m+1)/2), where c(m) = exp(HypergeometricPFQ[{1, 1, (m+1)/m, (m+2)/m, ... , (2*m-1)/m}, {2, 2, ...m-times... 2, 2}, 1] / m^m) / (m! * (2*Pi)^((m-1)/2) / sqrt(m)).
Limit_{m->oo} c(m) / (exp(m)/(m^m*(2*Pi)^(m/2))) = 1.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[Exp[Sum[(5*k)!/(5!*k!^5)*x^k/k, {k, 1, 20}]], {x, 0, 20}], x]
    CoefficientList[Series[Exp[x*HypergeometricPFQ[{1, 1, 6/5, 7/5, 8/5, 9/5}, {2, 2, 2, 2, 2}, 3125*x]], {x, 0, 20}], x]

Formula

G.f. A(x) = G(x)^(1/120), where G(x) is the g.f. for A333043.
a(n) ~ c * 5^(5*n)/n^3, where c = exp(HypergeometricPFQ[{1, 1, 6/5, 7/5, 8/5, 9/5}, {2, 2, 2, 2, 2}, 1] / 3125) / (96*sqrt(5)*Pi^2) = 0.00047219161473962545263459216995582653262467228952818554361164671183728...

A333042 G.f.: exp(Sum_{k>=1} (4*k)!/k!^4 * x^k/k).

Original entry on oeis.org

1, 24, 1548, 155744, 19893054, 2937661200, 477691374152, 83161733788992, 15230338934722749, 2900395347525785464, 569718535329796732476, 114759815105897160007392, 23602808330272138320592494, 4940203531008336735249385488, 1049571237547858314991495867848
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 06 2020

Keywords

Comments

From Peter Bala, Feb 08 2023: (Start)
Let A(x) denote the o.g.f. of the sequence. The sequence defined by b(n) := [x^n] A(x)^n for n >= 1 begins [24, 3672, 703968, 149835864, 33911355024, 7993981771488, 1940145241321920, ...]. We conjecture that b(n) satisfies the supercongruences b(n*p^r) == b(n*p^(r-1)) ( mod p^(3*r) ) for prime p >= 5 and all positive integers n and r.
More generally, for a positive integer m, set A_m(x) = exp( Sum_{n >= 1} (m*n)!/(n!^m) * x^n/n ) and define a sequence {b_m(n): n >= 1} by b_m(n) := [x^n] A_m(x)^n. Then we conjecture that b_m(n) is an integer sequence satisfying the same supercongruences. (End)

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[Exp[Sum[(4*k)!/k!^4*x^k/k, {k, 1, 20}]], {x, 0, 20}], x]
    CoefficientList[Series[Exp[24*x*HypergeometricPFQ[{1, 1, 5/4, 3/2, 7/4}, {2, 2, 2, 2}, 256*x]], {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 09 2024 *)

Formula

a(n) ~ c * 4^(4*n)/n^(5/2), where c = exp(3*HypergeometricPFQ[{1, 1, 5/4, 3/2, 7/4}, {2, 2, 2, 2}, 1] / 32) / (sqrt(2)*Pi^(3/2)) = 0.14496966... - Vaclav Kotesovec, Mar 06 2020, updated Feb 16 2024
a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} A008977(k) * a(n-k). - Seiichi Manyama, Feb 09 2024

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

Views

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.
Showing 1-4 of 4 results.