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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A377148 a(n) = Sum_{k=0..n} binomial(k+3,3) * binomial(k,n-k)^2.

Original entry on oeis.org

1, 4, 14, 60, 225, 796, 2764, 9304, 30580, 98700, 313422, 981548, 3037473, 9301620, 28222000, 84927760, 253699285, 752863840, 2220831160, 6515581600, 19021079866, 55276625304, 159967084164, 461150383400, 1324652146775, 3792447499916, 10824189204014
Offset: 0

Views

Author

Seiichi Manyama, Oct 18 2024

Keywords

Crossrefs

Programs

  • Magma
    [&+[Binomial(k+3,3)*Binomial(k, n-k)^2: k in [0..n]]: n in [0..30]]; // Vincenzo Librandi, May 12 2025
  • Mathematica
    Table[Sum[Binomial[k+3,3]*Binomial[k, n-k]^2,{k,0,n}],{n,0,30}] (* Vincenzo Librandi, May 12 2025 *)
  • PARI
    a(n) = sum(k=0, n, binomial(k+3, 3)*binomial(k, n-k)^2);
    
  • PARI
    a089627(n, k) = n!/((n-2*k)!*k!^2);
    my(N=3, M=30, x='x+O('x^M), X=1-x-x^2, Y=3); Vec(sum(k=0, N\2, a089627(N, k)*X^(N-2*k)*x^(Y*k))/(X^2-4*x^Y)^(N+1/2))
    

Formula

G.f.: (1-x-x^2) * ((1-x-x^2)^2 + 6*x^3) / ((1-x-x^2)^2 - 4*x^3)^(7/2).

A377145 a(n) = Sum_{k=0..n} binomial(k+2,2) * binomial(k,n-k)^2.

Original entry on oeis.org

1, 3, 9, 34, 111, 351, 1103, 3384, 10224, 30536, 90222, 264186, 767663, 2215623, 6356907, 18143300, 51540885, 145801395, 410888595, 1153964520, 3230723826, 9019081038, 25112021154, 69750583164, 193303849531, 534602071341, 1475644537323, 4065845732794
Offset: 0

Views

Author

Seiichi Manyama, Oct 17 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(k+2, 2)*binomial(k, n-k)^2);
    
  • PARI
    a089627(n, k) = n!/((n-2*k)!*k!^2);
    my(N=2, M=30, x='x+O('x^M), X=1-x-x^2, Y=3); Vec(sum(k=0, N\2, a089627(N, k)*X^(N-2*k)*x^(Y*k))/(X^2-4*x^Y)^(N+1/2))

Formula

G.f.: ((1-x-x^2)^2 + 2*x^3) / ((1-x-x^2)^2 - 4*x^3)^(5/2).

A078698 Number of ways to lace a shoe that has n pairs of eyelets such that each eyelet has at least one direct connection to the opposite side.

Original entry on oeis.org

1, 2, 20, 396, 14976, 907200, 79315200, 9551001600, 1513528934400, 305106949324800, 76296489615360000, 23175289163980800000, 8404709419090575360000, 3587225703492542791680000, 1779970753996760560435200000, 1016036270188884847558656000000, 661106386935312429191528448000000
Offset: 1

Views

Author

Hugo Pfoertner, Dec 18 2002

Keywords

Comments

The lace is "directed": reversing the order of eyelets along the path counts as a different solution. It must begin and end at the extreme pair of eyelets,

Examples

			a(3) = 20: label the eyelets 1,2,3 from front to back on the left side then 4,5,6 from back to front on the right side. The lacings are: 124356 154326 153426 142536 145236 135246 and the following and their mirror images: 125346 124536 125436 152346 153246 152436 154236.
Examples for n=2,3,4 can be found following the FORTRAN program at given link.
		

References

  • C. A. Pickover, The Math Book, Sterling, NY, 2009; see p. 494.

Crossrefs

Programs

  • Fortran
    c Program provided at Pfoertner link
  • Mathematica
    a[n_] := (n-1)!^2 Sum[Binomial[n-k, k]^2, {k, 0, n/2}];
    Array[a, 17] (* Jean-François Alcover, Jul 20 2018 *)

Formula

Conjecture: a(n) = (n-1)!^2*A051286(n). - Vladeta Jovovic, Sep 14 2005 (correct, see the Khrabrov/Kokhas reference, Joerg Arndt, May 26 2015)

Extensions

Terms a(9) and beyond (using A051286) from Joerg Arndt, May 26 2015

A181546 a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)^4.

Original entry on oeis.org

1, 1, 2, 17, 83, 338, 1923, 11553, 63028, 359203, 2172469, 13026034, 78106885, 478415635, 2957675956, 18321372721, 114301292581, 718253640196, 4531427831111, 28699590926291, 182566373639352, 1165539703613397
Offset: 0

Views

Author

Paul D. Hanna, Oct 29 2010

Keywords

Comments

Conjecture: Given F(n,L) = Sum_{k=0..[n/2]} C(n-k,k)^L, then lim_{n->oo} F(n+1,L)/F(n,L) = (Fibonacci(L)*sqrt(5) + Lucas(L))/2 for L>=0 where Fibonacci(n) = A000045(n) and Lucas(n) = A000032(n).
For this sequence (L=4): lim_{n->oo} a(n+1)/a(n) = (3*sqrt(5)+7)/2 = 6.8541...
Diagonal of the rational function 1 / ((1 - x)*(1 - y)*(1 - z)*(1 - w) - (x*y*z*w)^2). - Ilya Gutkovskiy, Apr 23 2025

Examples

			G.f. A(x) = 1 + x + 2*x^2 + 17*x^3 + 83*x^4 + 338*x^5 + 1923*x^6 +...
The terms begin:
a(0) = a(1) = 1^4;
a(2) = 1^4 + 1^4 = 2;
a(3) = 1^4 + 2^4 = 17;
a(4) = 1^4 + 3^4 + 1^4 = 83;
a(5) = 1^4 + 4^4 + 3^4 = 338;
a(6) = 1^4 + 5^4 + 6^4 + 1^4 = 1923;
a(7) = 1^4 + 6^4 + 10^4 + 4^4 = 11553; ...
		

Crossrefs

Cf. variants: A181545, A181547, A051286.

Programs

  • Mathematica
    Table[Sum[Binomial[n-k,k]^4,{k,0,Floor[n/2]}],{n,0,30}] (* Harvey P. Dale, May 22 2021 *)
  • PARI
    {a(n)=sum(k=0,n\2,binomial(n-k,k)^4)}

A185828 Half the number of n X 2 binary arrays with every element equal to exactly one or two of its horizontal and vertical neighbors.

Original entry on oeis.org

1, 3, 10, 23, 61, 162, 421, 1103, 2890, 7563, 19801, 51842, 135721, 355323, 930250, 2435423, 6376021, 16692642, 43701901, 114413063, 299537290, 784198803, 2053059121, 5374978562, 14071876561, 36840651123, 96450076810, 252509579303
Offset: 1

Views

Author

R. H. Hardin, Feb 05 2011

Keywords

Comments

Column 2 of A185835.

Examples

			Some solutions for 4 X 2 with a(1,1)=0:
  0 0   0 1   0 0   0 0   0 1   0 0   0 0   0 0   0 0   0 0
  1 1   0 1   0 1   1 1   0 1   1 0   0 1   1 1   1 0   0 1
  0 1   0 0   0 1   0 1   1 0   1 0   1 1   1 1   1 1   0 1
  0 0   1 1   0 0   0 1   1 0   0 0   0 0   0 0   0 0   0 1
The logarithmic g.f. begins:
L(x) = x + 3*x^2/2 + 10*x^3/3 + 23*x^4/4 + 61*x^5/5 + 162*x^6/6 + ..., where
exp(L(x)) = 1 + x + 2*x^2 + 5*x^3 + 11*x^4 + 26*x^5 + 63*x^6 + ... + A051286(n)*x^n/n + ... - _Paul D. Hanna_, Mar 19 2011
		

Crossrefs

Cf. A051286 (exp), A180662 (Fi1).

Programs

  • Maple
    a := proc(n): n*add(binomial(2*n-2*k, 2*k)/(n-k), k=0..n-1) end: seq(a(n), n=1..28); # Johannes W. Meijer, Jun 18 2018
  • PARI
    {a(n)=n*sum(k=0, n-1, binomial(2*n-2*k, 2*k)/(n-k))} /* Paul D. Hanna, Mar 19 2011 */
    
  • PARI
    {a(n)=n*polcoeff(-log( (1+x+x^2)*(1-3*x+x^2) +x*O(x^n))/2, n)} /* Paul D. Hanna, Mar 19 2011 */

Formula

Empirical: a(n) = 2*a(n-1) + a(n-2) + 2*a(n-3) - a(n-4).
a(n) = n*Sum_{k=0..n-1} C(2n-2k, 2k)/(n-k). - Paul D. Hanna, Mar 19 2011
L.g.f.: Sum_{n>=1} a(n)*x^n/n = -log((1+x+x^2)*(1-3*x+x^2))/2. - Paul D. Hanna, Mar 19 2011
Logarithmic derivative of A051286, which is the Whitney number of level n of the lattice of the ideals of the fence of order 2n. - Paul D. Hanna, Mar 19 2011
Empirical g.f.: x*(1+x+3*x^2-2*x^3)/(1+x+x^2)/(1-3*x+x^2). - Colin Barker, Feb 22 2012
Empirical: a(n) = Sum_{k=0..floor(n/2)} A084534(n, 2*k). - Johannes W. Meijer, Jun 17 2018
Empirical: a(n) = A100886(2n). - Wojciech Florek, Jan 26 2020

A377152 a(n) = Sum_{k=0..n} binomial(k+4,4) * binomial(k,n-k)^2.

Original entry on oeis.org

1, 5, 20, 95, 400, 1561, 5915, 21610, 76585, 265075, 898622, 2992235, 9810290, 31727815, 101379175, 320464280, 1003259080, 3113576320, 9586763720, 29305985800, 88997753446, 268642069750, 806394498200, 2408144329250, 7157177344225, 21177323087891
Offset: 0

Views

Author

Seiichi Manyama, Oct 18 2024

Keywords

Crossrefs

Programs

  • Maple
    f:= proc(n) local k; add(binomial(k+4,4)*binomial(k,n-k)^2,k=0..n) end proc:
    map(f, [$0..50]); # Robert Israel, Dec 05 2024
  • PARI
    a(n) = sum(k=0, n, binomial(k+4, 4)*binomial(k, n-k)^2);
    
  • PARI
    a089627(n, k) = n!/((n-2*k)!*k!^2);
    my(N=4, M=30, x='x+O('x^M), X=1-x-x^2, Y=3); Vec(sum(k=0, N\2, a089627(N, k)*X^(N-2*k)*x^(Y*k))/(X^2-4*x^Y)^(N+1/2))

Formula

G.f.: (Sum_{k=0..2} A089627(4,k) * (1-x-x^2)^(4-2*k) * x^(3*k)) / ((1-x-x^2)^2 - 4*x^3)^(9/2).

A384747 Triangle read by rows: T(n,k) is the number of rooted ordered trees with node weights summing to n, where the root has weight 0, non-root node weights are in {1,..,k}, and no nodes have the same weight as their parent node.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 5, 6, 0, 1, 11, 15, 16, 0, 1, 26, 39, 43, 44, 0, 1, 63, 110, 123, 127, 128, 0, 1, 153, 308, 358, 371, 375, 376, 0, 1, 376, 869, 1046, 1096, 1109, 1113, 1114, 0, 1, 931, 2499, 3098, 3278, 3328, 3341, 3345, 3346, 0, 1, 2317, 7238, 9283, 9904, 10084, 10134, 10147, 10151, 10152
Offset: 0

Views

Author

John Tyler Rascoe, Jun 09 2025

Keywords

Examples

			Triangle begins:
    k=0  1    2     3     4     5     6     7     8     9
 n=0 [1]
 n=1 [0, 1]
 n=2 [0, 1,   2]
 n=3 [0, 1,   5,    6]
 n=4 [0, 1,  11,   15,   16]
 n=5 [0, 1,  26,   39,   43,   44]
 n=6 [0, 1,  63,  110,  123,  127,  128]
 n=7 [0, 1, 153,  308,  358,  371,  375,  376]
 n=8 [0, 1, 376,  869, 1046, 1096, 1109, 1113, 1114]
 n=9 [0, 1, 931, 2499, 3098, 3278, 3328, 3341, 3345, 3346]
...
T(3,3) = 6 counts:
  o    o    o      o        o        __o__
  |    |    |     / \      / \      /  |  \
 (3)  (2)  (1)  (1) (2)  (2) (1)  (1) (1) (1)
       |    |
      (1)  (2)
		

Crossrefs

Cf. A051286 (column k=2), A382096 (column k=3), A384748 (main diagonal).

Programs

  • PARI
    b(i,j,k,N) = {if(k>N,1, 1/( 1  - sum(u=1,j, if(u==i,0,x^u * b(u,j,k+1,N-u+1)))))}
    Gx(k,N) = {my(x='x+O('x^(N+1))); Vec(1/(1 - sum(i=1,k, b(i,k,1,N)*x^i)))}
    T(max_row) = { my( N = max_row+1, v = vector(N, i, if(i==1, 1, 0))~); for(k=1, N, v=matconcat([v, Gx(k,N)~])); vector(N, n, vector(n, k, v[n, k]))}
    T(9)

Formula

T(n,k) = T(n,n) for k > n.

A077419 Largest Whitney number of Fibonacci lattices J(Z_n).

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 5, 7, 11, 17, 26, 40, 63, 97, 153, 238, 376, 587, 931, 1458, 2317, 3640, 5794, 9124, 14545, 22951, 36631, 57904, 92512, 146461, 234205, 371281, 594169, 943045, 1510192, 2399460, 3844787, 6114555, 9802895, 15603339, 25027296
Offset: 0

Views

Author

N. J. A. Sloane, Jan 19 2003

Keywords

Comments

A051286 and A051291, interleaved. a(n) is the maximal element in the n-th row of A079487 or A123245 and in the (n+2)-th row of A078807 or A078808. - Andrey Zabolotskiy, Sep 21 2017

Programs

  • Maple
    with(FormalPowerSeries): with(LREtools): # requires Maple 2022
    gf:= (1 + 2*x + 2*x^4 - x^6 - (1-x^2)*sqrt(1 - 2*x^2 - x^4 - 2*x^6 + x^8))/(2*x*sqrt(1 - 2*x^2 - x^4 - 2*x^6 + x^8));
    re:= FindRE(gf,x,a(n));
    inits:= {seq(a(i-1)=[1,1,1,2,2,3,5,7,11,17,26,40,63,97, 153][i],i=1..14)};
    rm:=  (n+1)*a(n) +(n-2)*a(n-1) +2*(-n+1)*a(n-2) +2*(-n+1)*a(n-3) +(-n-3)*a(n-4) +(-n+8)*a(n-5) +2*(-n+6)*a(n-6) +2*(-n+7)*a(n-7) +(n-9)*a(n-8) +(n-10)*a(n-9)=0;
    minre:= MinimalRecurrence(re, a(n), inits); minrm:= MinimalRecurrence(rm, a(n), inits); # shows that Mathar's recurrence is equivalent
    f:= REtoproc(re,a(n),inits); seq(f(n),n=0..40); # Georg Fischer, Oct 22 2022
  • Mathematica
    gf[x_] = (1 + 2 x + 2 x^4 - x^6 - (1 - x^2) Sqrt[1 - 2 x^2 - x^4 - 2 x^6 + x^8])/(2 x Sqrt[1 - 2 x^2 - x^4 - 2 x^6 + x^8]);
    Table[SeriesCoefficient[gf[x], {x, 0, n}], {n, 0, 40}] (* Hugo Pfoertner, Oct 22 2022 *)

Formula

G.f.: (1 + 2 x + 2 x^4 - x^6 - (1-x^2) sqrt(1 - 2 x^2 - x^4 - 2 x^6 + x^8) )/(2x sqrt(1 - 2 x^2 - x^4 - 2 x^6 + x^8)). - Emanuele Munarini, Mar 05 2007
a(n) ~ phi^(n+2) / (5^(1/4) * sqrt(2*Pi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Sep 22 2017
D-finite with recurrence: (n+1)*a(n) +(n-2)*a(n-1) +2*(-n+1)*a(n-2) +2*(-n+1)*a(n-3) +(-n-3)*a(n-4) +(-n+8)*a(n-5) +2*(-n+6)*a(n-6) +2*(-n+7)*a(n-7) +(n-9)*a(n-8) +(n-10)*a(n-9)=0. - R. J. Mathar, Nov 19 2019

Extensions

More terms from Emanuele Munarini, Mar 05 2007

A108488 Expansion of 1/sqrt(1 -2*x -3*x^2 -4*x^3 +4*x^4).

Original entry on oeis.org

1, 1, 3, 9, 23, 69, 203, 601, 1815, 5493, 16731, 51225, 157367, 485093, 1499499, 4646233, 14427095, 44880981, 139849979, 436419737, 1363713015, 4266417221, 13362194571, 41891406681, 131452430999, 412835452213, 1297543367835
Offset: 0

Views

Author

Paul Barry, Jun 04 2005

Keywords

Comments

In general, Sum_{k=0..n} C(n-k,k)^2*a^k*b^(n-k) has the expansion 1/sqrt(1 -2*b*x -(2*a*b -b^2)*x^2 -2*a*b^2*x^3 +(a*b)^2*x^4).
Diagonal of the rational function 1 / ((1 - x)*(1 - y) - 2*(x*y)^2). - Ilya Gutkovskiy, Apr 23 2025

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n-k,k]^2*2^k,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jul 24 2013 *)
    CoefficientList[Series[1/Sqrt[1-2x-3x^2-4x^3+4x^4],{x,0,30}],x] (* Harvey P. Dale, Apr 06 2023 *)
  • PARI
    {a(n)=polcoeff( exp(sum(m=1, n, sum(k=0, m, binomial(2*m, 2*k) * 2^k * x^k) *x^m/m) +x*O(x^n)), n)}
    for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 31 2014

Formula

a(n) = Sum_{k=0..n} C(n-k, k)^2*2^k.
a(n) ~ ((4*sqrt(2)-1)/62)^(1/4) * (1+2*sqrt(2)+sqrt(1+4*sqrt(2)))^(n+1) /(sqrt(Pi*n)*2^(n+2)). - Vaclav Kotesovec, Jul 24 2013
D-finite with recurrence: n*a(n) +(-2*n+1)*a(n-1) +3*(-n+1)*a(n-2) +2*(-2*n+3)*a(n-3) +4*(n-2)*a(n-4)=0. - R. J. Mathar, Aug 06 2013
G.f.: exp( Sum_{n>=1} (x^n/n) * Sum_{k=0..n} C(2*n,2*k) * 2^k * x^k ). - Paul D. Hanna, Aug 31 2014

A174882 A (3/2,-1) Somos-4 sequence.

Original entry on oeis.org

1, 1, -2, -8, -16, -16, 32, 128, 256, 256, -512, -2048, -4096, -4096, 8192, 32768, 65536, 65536, -131072, -524288, -1048576, -1048576, 2097152, 8388608, 16777216, 16777216, -33554432, -134217728, -268435456, -268435456
Offset: 0

Views

Author

Paul Barry, Mar 31 2010

Keywords

Comments

Hankel transform of A051286. a(n+2) = -(-1)^floor(n/4) * 2^A098181(n).

Examples

			G.f. = 1 + x - 2*x^2 - 8*x^3 - 16*x^4 - 16*x^5 + 32*x^6 + 128*x^7 + ...
		

Crossrefs

Programs

  • Magma
    Q:=Rationals(); R:=PowerSeriesRing(Q, 40); Coefficients(R!((1-2*x)*(4*x^2+3*x+1)/(1+16*x^4))) // G. C. Greubel, Feb 21 2018
  • Mathematica
    a[ n_] := (-1)^Quotient[n + 2, 4] 2^(n - Mod[ Quotient[n + 1, 2], 2]); (* Michael Somos, Sep 18 2014 *)
    CoefficientList[Series[(1-2*x)*(4*x^2+3*x+1)/(1+16*x^4), {x,0,50}], x] (* G. C. Greubel, Feb 21 2018 *)
  • PARI
    {a(n) = (-1)^((n+2) \ 4) * 2^(n - ((n+1) \ 2 % 2))}; /* Michael Somos, Jan 06 2011 */
    
  • PARI
    x='x+O('x^30); Vec((1-2*x)*(4*x^2+3*x+1)/(1+16*x^4)) \\ G. C. Greubel, Feb 21 2018
    

Formula

a(n) = ((3/2)*a(n-1)*a(n-3) - a(n-2)^2)/a(n-4), n>3.
a(-n) = a(n-1) / 2^(2*n - 1) for all n in Z. - Michael Somos, Jan 06 2011
0 = a(n)*(+2*a(n+4)) + a(n+1)*(-3*a(n+3)) + a(n+2)*(+2*a(n+2)) for all n in Z. - Michael Somos, Sep 18 2014
a(n+4) = -16 * a(n) for all n in Z. - Michael Somos, Sep 02 2015
G.f.: -(2*x-1)*(4*x^2+3*x+1)/(1+16*x^4) . - R. J. Mathar, Aug 18 2017
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