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

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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.
		

Crossrefs

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

A181069 Expansion of l.g.f. Sum_{n>=1} [ Sum_{k>=0} C(n+k-1,k)^4 *x^k ] *x^n/n.

Original entry on oeis.org

1, 3, 28, 275, 3126, 37632, 475056, 6192531, 82754650, 1127504378, 15603575208, 218727171104, 3099183987004, 44315462038200, 638663235342528, 9267264584278419, 135279095477748642, 1985221072388231742
Offset: 1

Views

Author

Paul D. Hanna, Oct 08 2010

Keywords

Examples

			L.g.f.: L(x) = x + 3*x^2/2 + 28*x^3/3 + 275*x^4/4 + 3126*x^5/5 +...
which equals the series:
  L(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 + ...
Exponentiation yields the g.f. of A181068:
  exp(L(x)) = 1 + x + 2*x^2 + 11*x^3 + 80*x^4 + 714*x^5 + 7095*x^6 +...
		

Crossrefs

Cf. A181067 (variant), A181068.

Programs

  • Magma
    [(&+[Binomial(n,k)*Binomial(n-1, k)^3: k in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Apr 05 2021
    
  • Maple
    A181069:= n-> add( binomial(n,k)*binomial(n-1,k)^3, k=0..n-1); seq(A181069(n), n=1..20); # G. C. Greubel, Apr 05 2021
  • Mathematica
    Table[Sum[Binomial[n-1,k]^3 * Binomial[n,k],{k,0,n-1}],{n,1,20}] (* Vaclav Kotesovec, Mar 06 2014 *)
    a[n_] := HypergeometricPFQ[{-n + 1, -n + 1, -n + 1, -n}, {1, 1, 1}, 1];Table[a[n], {n, 1, 18}] (* Detlef Meya, May 28 2024 *)
  • PARI
    {a(n)=sum(k=0, n-1, binomial(n-1, k)^4*n/(n-k))}
    
  • PARI
    {a(n)=n*polcoeff(sum(m=1, n, sum(k=0, n, binomial(m+k-1,k)^4*x^k)*x^m/m)+x*O(x^n), n)}
    for(n=1,20,print1(a(n),", "))
    
  • Sage
    [sum( binomial(n,k)*binomial(n-1,k)^3 for k in (0..n-1) ) for n in (1..20)] # G. C. Greubel, Apr 05 2021

Formula

a(n) = Sum_{k=0..n-1} binomial(n-1,k)^3 * binomial(n,k).
From Vaclav Kotesovec, Mar 06 2014: (Start)
Recurrence: (n-1)^2*n^3*(10*n^2 - 25*n + 16)*a(n) = 2*(n-1)^2*(60*n^5 - 240*n^4 + 341*n^3 - 225*n^2 + 90*n - 16)*a(n-1) + 4*(n-2)^2*n*(4*n - 7)*(4*n - 5)*(10*n^2 - 5*n + 1)*a(n-2).
a(n) ~ 2^(4*n-5/2) / (Pi*n)^(3/2). (End)
a(n) = hypergeom([-n + 1, -n + 1, -n + 1, -n], [1, 1, 1], 1). - Detlef Meya, May 28 2024
a(n) = Sum_{k=0..n} (k/n)^3 * binomial(n,k)^4. - Seiichi Manyama, Jul 14 2024

A181066 Expansion of g.f.: exp( Sum_{n>=1} [ Sum_{k>=0} C(n+k-1,k)^3 *x^k ] *x^n/n ).

Original entry on oeis.org

1, 1, 2, 7, 31, 157, 865, 5051, 30774, 193669, 1250319, 8240232, 55239187, 375624781, 2585449450, 17982937876, 126222946496, 893073250063, 6363674671524, 45631735776036, 329065051395940, 2385126419825231
Offset: 0

Views

Author

Paul D. Hanna, Oct 03 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 + 7*x^3 + 31*x^4 + 157*x^5 + 865*x^6 +...
The logarithm begins:
log(A(x)) = x + 3*x^2/2 + 16*x^3/3 + 95*x^4/4 + 606*x^5/5 + 4032*x^6/6 +...+ A181067(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 + 2^3*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
+ (1 + 7^3*x + 28^3*x^2 + 84^3*x^3 + 210^3*x^4 + 462^3*x^5 + ...)*x^7/7 + ...
		

Crossrefs

Cf. A000108, A181067 (log), A181068 (variant).

Programs

  • Magma
    m:=30;
    R:=PowerSeriesRing(Integers(), m);
    Coefficients(R!( Exp( (&+[ (&+[ Binomial(n+k-1,k)^3*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]^3*x^k*x^n/n, {k, 0, m+2}], {n, 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)^3*x^k)*x^m/m)+x*O(x^n)), n)}
    
  • Sage
    m=30;
    def A181066_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( exp( sum( sum( binomial(n+k-1,k)^3*x^(n+k)/n for k in (0..m+2) ) for n in (1..m+1)) ) ).list()
    A181066_list(m) # G. C. Greubel, Apr 05 2021

A374614 a(n) = Sum_{k=0..n} (k/n)^2 * binomial(n,k)^5.

Original entry on oeis.org

1, 9, 136, 2585, 54126, 1227492, 29226688, 723533337, 18438032890, 480994824134, 12787403151744, 345355150592036, 9451729196625184, 261628075707534720, 7313361005558843136, 206190939973811373593, 5857313490484652859282
Offset: 1

Views

Author

Seiichi Manyama, Jul 14 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[(k/n)^2 Binomial[n,k]^5,{k,0,n}],{n,20}] (* Harvey P. Dale, Jun 16 2025 *)
  • PARI
    a(n) = sum(k=0, n-1, binomial(n-1, k)^2*binomial(n, k)^3);

Formula

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