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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A113861 a(n) = (1/9)*((6*n - 7)*2^(n-1) - (-1)^n).

Original entry on oeis.org

0, 1, 5, 15, 41, 103, 249, 583, 1337, 3015, 6713, 14791, 32313, 70087, 151097, 324039, 691769, 1470919, 3116601, 6582727, 13864505, 29127111, 61050425, 127693255, 266571321, 555512263, 1155763769, 2401006023, 4980969017, 10319851975, 21355531833, 44142719431
Offset: 1

Views

Author

N. J. A. Sloane, Jan 25 2006

Keywords

Comments

This sequence is connected with the Collatz problem (see the sequences A045883 and A001045). - Michel Lagneau, Jan 13 2012

Crossrefs

Programs

Formula

a(n+1) - 2*a(n) = A001045(n+2), Jacobsthal numbers. - Paul Curtz, Jul 05 2008
3*a(n) - a(n+1) = -1, -2, 4*a(n). - Paul Curtz, Jul 05 2008
From R. J. Mathar, Nov 11 2008: (Start)
G.f.: x^2*(1+2*x)/((1+x)*(1-2*x)^2).
a(n) + a(n+1) = A014480(n-1). (End)
a(n) = 4*a(n-1) - 4*a(n-2) + (-1)^(n+1), n>2. - Gary Detlefs, Dec 19 2010
a(n) = 3*a(n-1) - 4*a(n-3), n>3. - Gary Detlefs, Dec 19 2010
a(n) = n*2^n - A045883(n). - Michel Lagneau, Jan 13 2012
Starting with "1" = triangle A059260 * A016813 as a vector, where A016813 = (4n + 1): [ 1, 5, 9, 13, ...]. - Gary W. Adamson, Mar 06 2012

A052543 Expansion of (1-x)/(1 - 3*x - 2*x^2 + 2*x^3).

Original entry on oeis.org

1, 2, 8, 26, 90, 306, 1046, 3570, 12190, 41618, 142094, 485138, 1656366, 5655186, 19308014, 65921682, 225070702, 768439442, 2623616366, 8957586578, 30583113582, 104417281170, 356502897518, 1217177027730, 4155702315886
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

From Andrew Woods, Jun 03 2013: (Start)
a(n) is the number of ways to tile a 2 X n square grid with 1 X 1, 1 X 2, 2 X 1, and 2 X 2 tiles. Solutions for a(2)=8:
| | | | || | | |__| |_| || | ||_| |||
||| ||| |_| |_| ||| ||| |_| |||
(End)

Crossrefs

Programs

  • GAP
    a:=[1,2,8];; for n in [4..30] do a[n]:=3*a[n-1]+2*a[n-2]-2*a[n-3]; od; a; # G. C. Greubel, May 09 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1-x)/(1 -3*x-2*x^2+2*x^3) )); // G. C. Greubel, May 09 2019
    
  • Maple
    spec := [S,{S=Sequence(Prod(Union(Z,Z),Union(Z,Sequence(Z))))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    CoefficientList[Series[(1-x)/(1-3x-2x^2+2x^3),{x,0,30}],x] (* or *) LinearRecurrence[{3,2,-2},{1,2,8},30] (* Harvey P. Dale, Jan 23 2013 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-x)/(1-3*x-2*x^2+2*x^3)) \\ G. C. Greubel, May 09 2019
    
  • Sage
    ((1-x)/(1-3*x-2*x^2+2*x^3)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 09 2019
    

Formula

G.f.: (1-x)/(1 - 3*x - 2*x^2 + 2*x^3).
a(n) = 3*a(n-1) + 2*a(n-2) - 2*a(n-3), with a(0)=1, a(1)=2, a(2)=8.
a(n) = Sum_{alpha = RootOf(1 -3*x -2*x^2 +2*x^3)} (1/98)*(13 + 25*alpha - 16*alpha^2)*alpha^(-n-1).
Equals triangle A059260 * the Pell sequence [1, 2, 5, 12, ...] as a vector. - Gary W. Adamson, Mar 06 2012
a(n) = A214997(n) - A214996(n). - Clark Kimberling, Nov 28 2012

Extensions

More terms from James Sellers, Jun 06 2000

A128494 Coefficient table for sums of Chebyshev's S-Polynomials.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, -1, 1, 1, 1, -1, -2, 1, 1, 1, 2, -2, -3, 1, 1, 0, 2, 4, -3, -4, 1, 1, 0, -2, 4, 7, -4, -5, 1, 1, 1, -2, -6, 7, 11, -5, -6, 1, 1, 1, 3, -6, -13, 11, 16, -6, -7, 1, 1, 0, 3, 9, -13, -24, 16, 22, -7, -8, 1, 1, 0, -3, 9, 22, -24, -40, 22, 29, -8, -9, 1, 1, 1, -3, -12, 22, 46, -40, -62, 29, 37, -9, -10, 1, 1, 1, 4, -12
Offset: 0

Views

Author

Wolfdieter Lang, Apr 04 2007

Keywords

Comments

See A049310 for the coefficient table of Chebyshev's S(n,x)=U(n,x/2) polynomials.
This is a 'repetition triangle' based on a signed version of triangle A059260: a(2*p,2*k) = a(2*p+1,2*k) = A059260(p+k,2*k)*(-1)^(p+k) and a(2*p+1,2*k+1) = a(2*p+2,2*k+1) = A059260(p+k+1,2*k+1)*(-1)^(p+k), k >= 0.

Examples

			The triangle a(n,m) begins:
  n\m  0   1   2   3   4   5   6   7   8   9  10
   0:  1
   1:  1   1
   2:  0   1   1
   3:  0  -1   1   1
   4:  1  -1  -2   1   1
   5:  1   2  -2  -3   1   1
   6:  0   2   4  -3  -4   1   1
   7:  0  -2   4   7  -4  -5   1   1
   8:  1  -2  -6   7  11  -5  -6   1   1
   9:  1   3  -6 -13  11  16  -6  -7   1   1
  10:  0   3   9 -13 -24  16  22  -7  -8   1   1
... reformatted by _Wolfdieter Lang_, Oct 16 2012
Row polynomial S(1;4,x) = 1 - x - 2*x^2 + x^3 + x^4 = Sum_{k=0..4} S(k,x).
S(4,y)*S(5,y)/y = 3 - 13*y^2 + 16*y^4 - 7*y^6 + y^8, with y=sqrt(2+x) this becomes S(1;4,x).
From _Wolfdieter Lang_, Oct 16 2012: (Start)
S(1;4,x) = (1 - (S(5,x) - S(4,x)))/(2-x) = (1-x)*(2-x)*(1+x)*(1-x-x^2)/(2-x) = (1-x)*(1+x)*(1-x-x^2).
S(5,x) - S(4,x) = R(11,sqrt(2+x))/sqrt(2+x) = -1 + 3*x + 3*x^2 - 4*x^3 - x^4 + x^5. (End)
		

Crossrefs

Row sums (signed): A021823(n+2). Row sums (unsigned): A070550(n).
Cf. A128495 for S(2; n, x) coefficient table.
The column sequences (unsigned) are, for m=0..4: A021923, A002265, A008642, A128498, A128499.
For m >= 1 the column sequences (without leading zeros) are of the form a(m, 2*k) = a(m, 2*k+1) = ((-1)^k)*b(m, k) with the sequences b(m, k), given for m=1..11 by A008619, A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808.

Formula

S(1;n,x) = Sum_{k=0..n} S(k,x) = Sum_{m=0..n} a(n,m)*x^m, n >= 0.
a(n,m) = [x^m](S(n,y)*S(n+1,y)/y) with y:=sqrt(2+x).
G.f. for column m: (x^m)/((1-x)*(1+x^2)^(m+1)), which shows that this is a lower diagonal matrix of the Riordan type, named (1/((1+x^2)*(1-x)), x/(1+x^2)).
From Wolfdieter Lang, Oct 16 2012: (Start)
a(n,m) = [x^m](1- (S(n+1,x) - S(n,x)))/(2-x). From the Binet - de Moivre formula for S and use of the geometric sum.
a(n,m) = [x^m](1- R(2*n+3,sqrt(2+x))/sqrt(2+x))/(2-x) with the monic integer T-polynomials R with coefficient triangle given in A127672. From the odd part of the bisection of the T-polynomials. (End)

A235501 Riordan array (1/(1-2*x^2), x/(1-x)).

Original entry on oeis.org

1, 0, 1, 2, 1, 1, 0, 3, 2, 1, 4, 3, 5, 3, 1, 0, 7, 8, 8, 4, 1, 8, 7, 15, 16, 12, 5, 1, 0, 15, 22, 31, 28, 17, 6, 1, 16, 15, 37, 53, 59, 45, 23, 7, 1, 0, 31, 52, 90, 112, 104, 68, 30, 8, 1, 32, 31, 83, 142, 202, 216, 172, 98, 38, 9, 1, 0, 63, 114, 225
Offset: 0

Views

Author

Philippe Deléham, Jan 11 2014

Keywords

Comments

Row sums are A007179(n+1).

Examples

			Triangle begins (0<=k<=n):
1
0, 1
2, 1, 1
0, 3, 2, 1
4, 3, 5, 3, 1
0, 7, 8, 8, 4, 1
8, 7, 15, 16, 12, 5, 1
0, 15, 22, 31, 28, 17, 6, 1
		

Crossrefs

Cf. Columns: A077957, A052551, A077866.
Diagonals: A000012, A001477, A022856.
Cf. Similar sequences: A059260, A191582.

Formula

T(n,n)=1, T(2n,0)=2^n, T(2n+1,0)=0, T(n,k)=T(n-1,k-1)+T(n-1,k) for 0
T(n,k)=T(n-1,k)+T(n-1,k-1)+2*T(n-2,k)-T(n-3,k)-2*T(n-3,k-1), T(0,0)=1, T(1,0)=0, T(1,1)=1, T(n,k)=0 if k<0 or if k>n.
T(n,n)=1, T(n+1,n)=n, T(n+2,n)=n*(n+1)/2 + 2.
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(3*x + 2*x^2/2! + x^3/3!) = 3*x + 8*x^2/2! + 16*x^3/3! + 28*x^4/4! + 45*x^5/5! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A239230 Expansion of -x*log'(-sqrt(12*x+2*sqrt(1-4*x)+2)/4+sqrt(1-4*x)/4+5/4).

Original entry on oeis.org

0, 1, 1, 4, 9, 36, 112, 428, 1505, 5692, 21026, 79806, 301488, 1151866, 4403778, 16929474, 65204353, 251947668, 975366094, 3784197606, 14705937794, 57242631464, 223121176224, 870805992278, 3402485053664, 13308485156086, 52104519751272, 204176144516818
Offset: 0

Author

Vladimir Kruchinin, May 22 2014

Keywords

Crossrefs

Cf. A055113.

Programs

  • Maple
    a:= n-> add((-1)^(k+n)*binomial(2*n-k-1, n-1)*hypergeom([k-n, (n+1)/2, n/2], [n, n+1], 4), k=1..n);
    seq(round(evalf(a(n), 32)), n=0..24); # Peter Luschny, May 22 2014
  • Maxima
    a(n):=n*sum(sum(binomial(n+2*j-1,j+n-1)*(-1)^(k+j+n)*binomial(2*n-k,j+n), j,0,n-k)/(2*n-k), k,1,n);

Formula

G.f.: A(x) = x*F'(x)/(1-F(x)), where F(x) is g.f. of A055113.
a(n) = n * Sum_{k=1..n} (Sum_{j=0..n-k} C(n+2*j-1,j+n-1) * (-1)^(k+j+n) * C(2*n-k,j+n)) / (2*n-k).
a(n) = Sum_{k=1..n} (-1)^(k+n) * C(2*n-k-1,n-1) * hypergeom([k-n, n/2+1/2, n/2], [n, n+1], 4). - Peter Luschny, May 22 2014
Conjecture D-finite with recurrence +24*(365025561*n-1672569283)*(n-1)*(n-2)*(2*n-1)*a(n) -4*(n-2)*(27129169947*n^3-209577621466*n^2+463278020461*n-314084557758)*a(n-1) +2*(-28823853823*n^4+487259692534*n^3-3105214937957*n^2+8814274338098*n-9143920331436)*a(n-2) +4*(276083065830*n^4-4172118623320*n^3+24824880820695*n^2-69263721795041*n+75832154222148)*a(n-3) +(-587491214125*n^4+9941738070620*n^3-75070680472775*n^2+281912285021344*n-413197788157152)*a(n-4) +2*(-1186924847911*n^4+26108844767699*n^3-211936472383904*n^2+757584729548632*n-1009721693733312)*a(n-5) +4*(2*n-11)*(328530544924*n^3-5280431217363*n^2+28334632524947*n-50473913356356)*a(n-6) -72*(4143100547*n-18456753180)*(n-6)*(2*n-11)*(2*n-13)*a(n-7)=0. - R. J. Mathar, Jul 27 2022
Conjecture: a(n) = Sum_{i=0..n-1} A059260(2*(n-1),n+i-1)*A009766(n+i-1,i)*(-1)^(n+i-1) = n*Sum_{i=0..n-1} binomial(n+2*i, i)/(n+2*i)*(-1)^(n+i-1)*[x^(n+i-1)] (1+(x+1)^(2*n-1))/(x+2) for n >= 0. - Mikhail Kurkov, Feb 18 2025
Previous Showing 21-25 of 25 results.