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.

A198256 Row sums of A197653.

Original entry on oeis.org

1, 5, 46, 485, 5626, 69062, 882540, 11614437, 156343330, 2142556130, 29791689148, 419260001030, 5960334608788, 85469709312860, 1234797737654296, 17955907741675749, 262607675818816050, 3860239468267647914, 57002176852356800700, 845159480056345448610
Offset: 0

Views

Author

Susanne Wienand, Oct 22 2011

Keywords

Comments

Number of meanders of length (n+1)*4 which are composed by arcs of equal length and a central angle of 90 degrees.
Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated a value of dir,
(c) consecutive Ls increment the index of dir,
(d) consecutive Rs decrement the index of dir,
(e) the integer m > 0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 4.
The terms are proved by brute force for 0 <= n <= 8, but not yet in general. - Susanne Wienand, Oct 29 2011

Examples

			Some examples of list S and allocated values of dir if n = 4:
Length(S) = (4+1)*4 = 20.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L
dir: 1,2,3,0,1,2,3,0,1,2,3,0,1,2,3,0,1,2,3,0
  S: L,L,L,L,R,L,R,R,L,R,R,R,L,R,L,L,R,L,L,L
dir: 1,2,3,0,0,0,0,3,3,3,2,1,1,1,1,2,2,2,3,0
  S: L,L,R,L,L,L,L,R,R,L,R,R,L,R,L,L,L,L,R,R
dir: 1,2,2,2,3,0,1,1,0,0,0,3,3,3,3,0,1,2,2,1
Each value of dir occurs 20/4 = 5 times.

Links

Crossrefs

Cf. A197653, A198060, A198257, A198258.
Cf. A005260, A181069.
Cf. A181067, A374614.

Programs

  • Mathematica
    A198256[n_] := Sum[Sum[Sum[(-1)^(j + i)* Binomial[i, j]*Binomial[n, k]^4*(n + 1)^j*(k + 1)^(3 - j)/(k + 1)^3, {i, 0, 3}], {j, 0, 3}], {k, 0, n}]; Table[A198256[n], {n, 0, 16}] (* Peter Luschny, Nov 02 2011 *)
  • PARI
    A198256(n) = {sum(k=0,n,if(n == 1+2*k,4,(1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n))*binomial(n,k)^4)} \\ Peter Luschny, Nov 24 2011
  • Sage
    from mpmath import mp, hyper
def A198256(n) : return hyper([1-n, 1-n, 1-n, 1-n], [3, 3, 3], 1)*(n^4-n^6)/4 + hyper([-n, -n, -n, -n], [2, 2, 2], 1)*(1+n+n^2+n^3) + hyper([2, 1-n, 1-n, 1-n, 1-n], [1, 3, 3, 3], 1)*(n^4+n^5)/4
mp.dps = 32
for n in (0..19) : print(int(A198256(n)))  # Peter Luschny, Oct 24 2011

Formula

a(n) = Sum_{k=0..n} Sum_{j=0..3} Sum_{i=0..3} (-1)^(j+i)*C(i,j)*C(n,k)^4*(n+1)^j*(k+1)^(3-j)/(k+1)^3. - Peter Luschny, Nov 02 2011
a(n) = Sum_{k=0..n} h(n,k)*binomial(n,k)^4, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 4. - Peter Luschny, Nov 24 2011
From Peter Bala, Mar 21 2023: (Start)
Conjecture 1: a(n) = Sum_{k = 0..n} binomial(n+1,k)^2*binomial(n,k)^2.
If true, then we have the third-order recurrence equation
n^2*(n + 1)^3*P(n-1)*a(n) = 2*n^2*(400*n^8 - 1260*n^7 + 20*n^6 + 3020*n^5 - 1646*n^4 - 1951*n^3 + 1142*n^2 + 465*n - 290)*a(n-1) + 4*(n - 1)*(2800*n^9 - 15420*n^8 + 30620*n^7 - 23710*n^6 + 808*n^5 + 6863*n^4 - 1309*n^3 - 1218*n^2 + 496*n - 60)*a(n-2) + 8*(n - 2)^2*(2*n - 3)*(4*n - 5)*(4*n - 7)*P(n)*a(n-3) with a(0) = 1, a(1) = 5 and a(2) = 46 and where P(n) = 100*n^5 - 65*n^4 - 35*n^3 + 25*n^2 + 6*n - 5.
Conjecture 2: working with offset 1, that is, a(1) = 1, a(2) = 5, ..., then the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5. (End)
a(n) ~ 2^(4*n + 5/2) / (Pi*n)^(3/2). - Vaclav Kotesovec, Apr 17 2023
Peter Bala's conjecture 1 can equivalently written: a(n) = hypergeom([-n - 1, -n - 1, -n, -n], [1, 1, 1], 1). - Detlef Meya, May 28 2024
a(n) = Sum_{k=0..n+1} (k/(n+1))^2 * binomial(n+1,k)^4. - Seiichi Manyama, Jul 14 2024

A181067 a(n) = Sum_{k=0..n-1} binomial(n-1,k)^2 * binomial(n,k).

Original entry on oeis.org

1, 3, 16, 95, 606, 4032, 27616, 193167, 1372930, 9881498, 71846160, 526764680, 3889340560, 28888634400, 215680108416, 1617467908751, 12177754012458, 92004463332486, 697263463622080, 5298985086555090, 40371796982444356
Offset: 1

Views

Author

Paul D. Hanna, Oct 03 2010

Keywords

Examples

			L.g.f.: L(x) = x + 3*x^2/2 + 16*x^3/3 + 95*x^4/4 + 606*x^5/5 + ...
which equals the series:
  L(x) = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ...)*x
  + (1 + 2^3*x +  3^3*x^2 +  4^3*x^3 +   5^3*x^4 +   6^3*x^5 + ...)*x^2/2
  + (1 + 3^3*x +  6^3*x^2 + 10^3*x^3 +  15^3*x^4 +  21^3*x^5 + ...)*x^3/3
  + (1 + 4^3*x + 10^3*x^2 + 20^3*x^3 +  35^3*x^4 +  56^3*x^5 + ...)*x^4/4
  + (1 + 5^3*x + 15^3*x^2 + 35^3*x^3 +  70^3*x^4 + 126^3*x^5 + ...)*x^5/5
  + (1 + 6^3*x + 21^3*x^2 + 56^3*x^3 + 126^3*x^4 + 252^3*x^5 + ...)*x^6/6 + ...
Exponentiation yields the g.f. of A181066:
  exp(L(x)) = 1 + x + 2*x^2 + 7*x^3 + 31*x^4 + 157*x^5 + 865*x^6 + ... + A181066(n)*x^n + ...

Links

Crossrefs

Cf. A181066 (exp), A181069 (variant).

Programs

  • Magma
    [(&+[ ((n-k)/n)^2*Binomial(n,k)^3 : k in [0..n-1]]): n in [1..25]]; // G. C. Greubel, Apr 05 2021
  • Maple
    A181067:= n-> add(((n-k)/n)^2*binomial(n,k)^3, k=0..n-1); seq(A181067(n), n=1..25); # G. C. Greubel, Apr 05 2021
  • Mathematica
    Table[Sum[Binomial[n-1,k]^2*Binomial[n,k],{k,0,n-1}], {n,1,20}] (* Vaclav Kotesovec, Oct 24 2012 *)
Table[HypergeometricPFQ[{1-n, 1-n, -n}, {1, 1}, -1], {n,1,20}] (* Pierre-Louis Giscard, Jul 20 2013 *)
  • Maxima
    makelist(hypergeometric([-n+1,-n+1,-n],[1,1],-1),n,0,12); /* Emanuele Munarini, Sep 27 2016 */
  • PARI
    {a(n)=sum(k=0, n-1, binomial(n-1, k)^3*n/(n-k))}
  • PARI
    {a(n)=n*polcoeff(sum(m=1, n, sum(k=0, n, binomial(m+k-1,k)^3*x^k)*x^m/m)+x*O(x^n), n)}
  • Sage
    [sum( ((n-k)/n)^2*binomial(n,k)^3 for k in (0..n-1) ) for n in (1..25)] # G. C. Greubel, Apr 05 2021

Formula

L.g.f.: Sum_{n>=1} [ Sum_{k>=0} binomial(n+k-1,k)^3 *x^k ] *x^n/n.
Logarithmic derivative of A181066.
Recurrence: n^2*a(n) = - (n^2-17*n+10)*a(n-1) + 48*(n^2-3*n+1)*a(n-2) + 16*(n-3)*(11*n-36)*a(n-3) + 128*(n-4)^2*a(n-4). - Vaclav Kotesovec, Oct 24 2012
a(n) ~ sqrt(3)*8^n/(6*Pi*n). - Vaclav Kotesovec, Oct 24 2012
a(n) = 3F2([1-n, 1-n, -n], [1, 1], -1). - Pierre-Louis Giscard, Jul 20 2013
a(n) = n * hypergeometric([-n+1,-n+1,-n+1], [1,2], -1) for n > 0. - Emanuele Munarini, Sep 27 2016
a(n) = Sum_{k=0..n-1} ((n-k)/n)^2 * binomial(n,k)^3. - G. C. Greubel, Apr 05 2021

A181068 Expansion of g.f.: exp( Sum_{n>=1} [ Sum_{k>=0} C(n+k-1,k)^4 *x^k ] *x^n/n ).

Original entry on oeis.org

1, 1, 2, 11, 80, 714, 7095, 76206, 864590, 10227727, 125001862, 1568419058, 20108619244, 262510020319, 3479914302802, 46742907726147, 635092339459857, 8716058291255777, 120686879727465365, 1684357785848110976
Offset: 0

Views

Author

Paul D. Hanna, Oct 08 2010

Keywords

Comments

Compare g.f. to a g.f. of the Catalan numbers (A000108):
. exp( Sum_{n>=1} [ Sum_{k>=0} C(n+k-1,k)^2 *x^k ] *x^n/n ).

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 11*x^3 + 80*x^4 + 714*x^5 + 7095*x^6 +...
The logarithm begins:
log(A(x)) = x + 3*x^2/2 + 28*x^3/3 + 275*x^4/4 + 3126*x^5/5 + 37632*x^6/6 +...+ A181069(n)*x^n/n +...
which equals the series:
log(A(x)) = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 +...)*x
+ (1 + 2^4*x +  3^4*x^2 +  4^4*x^3 +   5^4*x^4 +   6^4*x^5 +...)*x^2/2
+ (1 + 3^4*x +  6^4*x^2 + 10^4*x^3 +  15^4*x^4 +  21^4*x^5 +...)*x^3/3
+ (1 + 4^4*x + 10^4*x^2 + 20^4*x^3 +  35^4*x^4 +  56^4*x^5 +...)*x^4/4
+ (1 + 5^4*x + 15^4*x^2 + 35^4*x^3 +  70^4*x^4 + 126^4*x^5 +...)*x^5/5
+ (1 + 6^4*x + 21^4*x^2 + 56^4*x^3 + 126^4*x^4 + 252^4*x^5 +...)*x^6/6
+ (1 + 7^4*x + 28^4*x^2 + 84^4*x^3 + 210^4*x^4 + 462^4*x^5 +...)*x^7/7 + ...

Links

Crossrefs

Cf. A000108, A181069 (log), A181066 (variant).

Programs

  • Magma
    m:=30;
R:=PowerSeriesRing(Integers(), m);
Coefficients(R!( Exp( (&+[ (&+[ Binomial(n+k-1,k)^4*x^(n+k)/n : k in [0..m+2]]): n in [1..m+1]]) ) )); // G. C. Greubel, Apr 05 2021
  • Mathematica
    With[{m=30}, CoefficientList[Series[Exp[Sum[Sum[Binomial[n+k-1, k]^4*x^(n+k)/n, {k,0,m+2}], {n,1,m+1}]], {x,0,m}], x]] (* G. C. Greubel, Apr 05 2021 *)
  • PARI
    {a(n)=polcoeff(exp(sum(m=1, n, sum(k=0, n, binomial(m+k-1,k)^4*x^k)*x^m/m)+x*O(x^n)), n)}
  • Sage
    m=30;
def A181068_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( exp( sum( sum( binomial(n+k-1,k)^4*x^(n+k)/n for k in (0..m+2) ) for n in (1..m+1)) ) ).list()
A181068_list(m) # G. C. Greubel, Apr 05 2021

Formula

a(n) ~ c * 16^n / n^(5/2), where c = 0.034183651246881715583041336040447549489320454248320978... - Vaclav Kotesovec, Apr 05 2021

A374615 a(n) = Sum_{k=0..n} (k/n)^3 * binomial(n,k)^5.

Original entry on oeis.org

1, 5, 82, 1421, 29626, 657662, 15528640, 381137549, 9656742322, 250689517130, 6638957500924, 178721359853390, 4878005765458528, 134712060315562784, 3758101325718600832, 105769714118196065933, 3000003700599260555650, 85677293959381174518986
Offset: 1

Views

Author

Seiichi Manyama, Jul 14 2024

Keywords

Crossrefs

Cf. A005261, A374614, A374616.
Cf. A181069.

Programs

  • PARI
    a(n) = sum(k=0, n-1, binomial(n-1, k)^3*binomial(n, k)^2);

Formula

a(n) = Sum_{k=0..n-1} binomial(n-1,k)^3 * binomial(n,k)^2.
Showing 1-4 of 4 results.

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