A027926 -id:A027926 - cp's OEIS frontend

A027935 Triangular array T read by rows: T(n,k)=t(n,2k), t given by A027926; 0 <= k <= n, n >= 0.

1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 2, 5, 7, 1, 1, 2, 5, 12, 11, 1, 1, 2, 5, 13, 26, 16, 1, 1, 2, 5, 13, 33, 51, 22, 1, 1, 2, 5, 13, 34, 79, 92, 29, 1, 1, 2, 5, 13, 34, 88, 176, 155, 37, 1, 1, 2, 5, 13, 34, 89, 221, 365, 247, 46, 1, 1, 2, 5, 13, 34, 89, 232, 530, 709, 376, 56, 1

Offset: 0

Clark Kimberling

			Triangle begins:
  1;
  1, 1;
  1, 2, 1;
  1, 2, 4,  1:
  1, 2, 5,  7,  1;
  1, 2, 5, 12, 11,  1;
  1, 2, 5, 13, 26, 16, 1;
  ...

G. C. Greubel, Rows n = 0..100 of triangle, flattened

The row sums of this bisection of the "Fibonacci array" A027926 are powers of 2, see A027948 for the other bisection.

Flat(List([0..12], n-> List([0..n], k-> Sum([0..Int((2*n-2*k+1)/2) ], j-> Binomial(n-j, 2*(n-k-j))) ))); # G. C. Greubel, Sep 27 2019

T:= func< n,k | &+[Binomial(n-j, 2*(n-k-j)) : j in [0..Floor((2*n -2*k+1)/2)]] >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 27 2019

T:= proc(n, k) option remember;
     add( binomial(n-j, 2*(n-k-j)), j=0..floor((2*n - 2*k+1)/2))
   end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Sep 27 2019

T[n_, k_]:= Sum[Binomial[n-j, 2*(n-k-j)], {j,0,Floor[(2*n-2*k+1)/2]}]; Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 27 2019 *)

T(n,k) = sum(j=0,(2*n-2*k+1)\2, binomial(n-j, 2*(n-k-j)));
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Sep 27 2019

[[ sum(binomial(n-j, 2*(n-k-j)) for j in (0..floor((2*n-2*k+1)/2)) ) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Sep 27 2019

T(n,k) = Sum_{j=0..floor((2*n-2*k-1)/2)} binomial(n-j, 2*(n-k-j)). - G. C. Greubel, Sep 27 2019

A027948 Triangular array T read by rows: T(n,k) = t(n,2k+1) for 0 <= k <= n, T(n,n)=1, t given by A027926, n >= 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 3, 7, 4, 1, 1, 3, 8, 14, 5, 1, 1, 3, 8, 20, 25, 6, 1, 1, 3, 8, 21, 46, 41, 7, 1, 1, 3, 8, 21, 54, 97, 63, 8, 1, 1, 3, 8, 21, 55, 133, 189, 92, 9, 1, 1, 3, 8, 21, 55, 143, 309, 344, 129, 10, 1, 1, 3, 8, 21, 55, 144, 364, 674, 591, 175, 11, 1

Offset: 0

Clark Kimberling

			Triangle begins with:
  1;
  1, 1;
  1, 2, 1;
  1, 3, 3,  1;
  1, 3, 7,  4,  1;
  1, 3, 8, 14,  5,  1;
  1, 3, 8, 20, 25,  6, 1;
  1, 3, 8, 21, 46, 41, 7, 1; ...

G. C. Greubel, Rows n = 0..100 of triangle, flattened

The row sums of this (slightly extended) bisection of the "Fibonacci array" A027926 are powers of 2, see A027935 for the other bisection.

T:= function(n,k)
    if k=n then return 1;
    else return Sum([0..n-k], j-> Binomial(n-j, 2*(n-k-j)-1) );
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Sep 29 2019

T:= func< n,k | k eq n select 1 else &+[Binomial(n-j, 2*(n-k-j) -1): j in [0..n-k]] >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 29 2019

T:= proc(n, k)
      if k=n then 1
      else add(binomial(n-j, 2*(n-k-j)-1), j=0..n-k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Sep 29 2019

T[n_, k_]:= If[k==n, 1, Sum[Binomial[n-j, 2*(n-k-j)-1], {j, 0, n-k}]];
Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 29 2019 *)

T(n,k) = if(k==n, 1, sum(j=0,n-k, binomial(n-j, 2*(n-k-j)-1)) );
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Sep 29 2019

def T(n, k):
    if (k==n): return 1
    else: return sum(binomial(n-j, 2*(n-k-j)-1) for j in (0..n-k))
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Sep 29 2019

T(n,k) = Sum_{j=0..n-k} binomial(n-j, 2*(n-k-j) -1) with T(n,n)=1 in the region n >= 0, 0 <= k <= n. - G. C. Greubel, Sep 29 2019

Name edited by G. C. Greubel, Sep 29 2019

A027992 a(n) = 1T(n,0) + 2T(n,1) + ... + (2n+1)*T(n,2n), T given by A027926.

Original entry on oeis.org

1, 6, 22, 66, 178, 450, 1090, 2562, 5890, 13314, 29698, 65538, 143362, 311298, 671746, 1441794, 3080194, 6553602, 13893634, 29360130, 61865986, 130023426, 272629762, 570425346, 1191182338, 2483027970, 5167382530, 10737418242

Offset: 0

Clark Kimberling

nonn

Also total sum of squares of parts in all compositions of n (offset 1). Total sum of cubes of parts in all compositions of n is (13*n-36)*2^(n-1)+6*n+18 with g.f. x*(1+4x+x^2)/((2x-1)(1-x))^2, A271638; total sum of fourth powers of parts in all compositions of n is (75*n-316)*2^(n-1)+12*n^2+72*n+158 with g.f. x*(1+x)*(x^2+10*x+1)/((2*x-1)^2*(1-x)^3); total sum of fifth powers of parts in all compositions of n is (541*n-3060)*2^(n-1)+20*n^3+180*n^2+790*n+1530. - Vladeta Jovovic, Mar 18 2005

Let M = the 3 X 3 matrix [(1,0,0),(1,2,0),(1,3,2)] and column vector V = [1,1,1]. a(n) is the lower term in the product M^n * V.

Alejandro Erickson and Mark Schurch, Monomer-dimer tatami tilings of square regions, arXiv preprint arXiv:1110.5103 [math.CO], 2011.
Alejandro Erickson and Mark Schurch, Enumerating tatami mat arrangements of square grids, in 22nd International Workshop on Combinatorial Algorithms, University of Victoria, June 20-22, volume 7056 of Lecture Notes in Computer Science (LNCS), Springer Berlin / Heidelberg, 2011, pp. 223-235
K. Kimura, S. Higuchi, Monte Carlo estimation of the number of tatami tilings, arXiv:1509.05983 [cond-mat.stat-mech], 2015-2016, eq. (2).

Cf. A027926, A066183.

M = {{1, 0, 0}, {1, 2, 0}, {1, 3, 2}};
a[n_] := MatrixPower[M, n].{1, 1, 1} // Last;
Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Aug 12 2018, from PARI *)

vector(40, n, n--; ([1,0,0;1,2,0;1,3,2]^n*[1,1,1]~)[3]) \\ Michel Marcus, Aug 06 2015

a(n) = 2^n*(3n-1)+2 = A048496(n+1)-1 = A053565(n+1)+2. - Ralf Stephan, Jan 15 2004

a(n) = 5*a(n-1)-8*a(n-2)+4*a(n-3). G.f.: (1+x)/((1-x)*(1-2*x)^2). - Colin Barker, Apr 04 2012

A027933 a(n) = T(n, 2*n-10), T given by A027926.

Original entry on oeis.org

1, 2, 5, 13, 34, 89, 232, 596, 1490, 3588, 8273, 18228, 38403, 77533, 150438, 281403, 509015, 892926, 1523117, 2532359, 4112704, 6536993, 10186540, 15586342, 23449376, 34731776, 50700937, 73018870, 103843433, 145950389, 202879594, 279108997, 380260541

Offset: 5

Clark Kimberling

Colin Barker, Table of n, a(n) for n = 5..1000
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Cf. A027926, A228074.

List([5..40], n-> Sum([0..5], k-> Binomial(n-k, 10-2*k)) ); # G. C. Greubel, Sep 27 2019

[&+[Binomial(n-k, 10-2*k): k in [0..5]] : n in [5..40]]; // G. C. Greubel, Sep 27 2019

seq(add(binomial(n-k, 10-2*k), k=0..5), n=5..40); # G. C. Greubel, Sep 27 2019

Table[Sum[Binomial[n-k, 10-2k], {k,0,5}], {n,5,40}] (* or *)
Drop[#, 5] &@ CoefficientList[Series[x^5(1-x+x^2)(1-5x+9x^2-5x^3+x^4)(1- 3x+5x^2-3x^3+x^4)/(1-x)^11, {x, 0, 37}], x] (* Michael De Vlieger, Feb 17 2016 *)

Vec(x^5*(1-x+x^2)*(1-5*x+9*x^2-5*x^3+x^4)*(1-3*x+5*x^2-3*x^3+x^4) / (1-x)^11 + O(x^40)) \\ Colin Barker, Feb 17 2016

vector(40, n, sum(k=0,5, binomial(n+4-k, 10-2*k)) ) \\ G. C. Greubel, Sep 27 2019

[sum(binomial(n-k, 10-2*k) for k in (0..5)) for n in (5..40)] # G. C. Greubel, Sep 27 2019

a(n) = Sum_{k=0..5} binomial(n-k, 10-2*k). - Len Smiley, Oct 20 2001
a(n) = 34 -9161*n/280 -101897*n^3/20160 +794293*n^2/50400 -287*n^5/1280 +438209*n^4/362880 +5593*n^6/172800 -47*n^7/13440 -n^9/80640 +n^8/3780 +n^10/3628800. - R. J. Mathar, Oct 05 2009
G.f.: x^5*(1-x+x^2)*(1-5*x+9*x^2-5*x^3+x^4)*(1-3*x+5*x^2-3*x^3+x^4) / (1-x)^11. - Colin Barker, Feb 17 2016

A027988 Greatest number in row n of array T given by A027926.

Original entry on oeis.org

1, 1, 2, 4, 7, 14, 26, 51, 97, 189, 365, 709, 1383, 2683, 5270, 10220, 20175, 39130, 77533, 150438, 298925, 580328, 1155661, 2245004, 4478413, 8705686, 17390359, 33828704, 67650909, 131901368, 263589730, 515037942, 1028483089

Offset: 0

Clark Kimberling

nonn

Also greatest number in row n of the Fibonacci-Pascal matrix A105809. - Jason Riggle (jriggle(AT)uchicago.edu), Aug 22 2006

For n > 0 also largest term in row n of the triangle in A228074. - Reinhard Zumkeller, Aug 15 2013

a027988 = maximum . a027926_row  -- Reinhard Zumkeller, Aug 15 2013

A027989 a(n) = self-convolution of row n of array T given by A027926.

Original entry on oeis.org

1, 3, 10, 33, 105, 324, 977, 2895, 8462, 24465, 70101, 199368, 563425, 1583643, 4430290, 12342849, 34262337, 94800780, 261545777, 719697255, 1975722326, 5412138033, 14796520365, 40380240528, 110016825025, 299285288499, 813011578522, 2205652007265, 5976479585817

Offset: 0

Clark Kimberling

a(n) is the number of all columns in stack polyominoes of perimeter 2n+4. - Emanuele Munarini, Apr 07 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1).

Cf. A027926, A054142, A172991, A188648.

Table[((5+4n)Fibonacci[1+2n]-(1+2n)Fibonacci[2n])/5,{n,0,28}] (* Emanuele Munarini, Apr 07 2011 *)

makelist(((5+4*n)*fib(1+2*n)-(1+2*n)*fib(2*n))/5,n,0,20); /* Emanuele Munarini, Apr 07 2011 */

Vec((1-3*x+3*x^2)/(1-3*x+x^2)^2+O(x^66)) /* Joerg Arndt, Apr 08 2011 */

a(n) = (2/5)*(n + 1)*F(2*n+3) + (1/5)*F(2*n+2) - (4/5)*(n + 1)*F(2*n), where F(n) = A000045(n). - Ralf Stephan, May 13 2004
From Emanuele Munarini, Apr 07 2011: (Start)
a(n) = ((4*n + 5)*F(2*n+1) - (2*n + 1)*F(2*n))/5, where F(n) = A000045(n).
a(n) = Sum_{k=0..n} binomial(2*n-k, k)*(k + 1).
G.f.: (1 - 3*x + 3*x^2)/(1 - 3*x + x^2)^2.
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). (End)

A027991 a(n) = Sum{T(n,k)*T(n,2n-k)}, 0<=k<=n-1, T given by A027926.

Original entry on oeis.org

1, 3, 12, 40, 130, 404, 1227, 3653, 10720, 31090, 89316, 254568, 720757, 2029095, 5684340, 15855964, 44061862, 122032508, 336966015, 927953705, 2549229256, 6987648358, 19115124552, 52194037200, 142274514025, 387215773899

Offset: 1

Clark Kimberling

nonn

From Wolfdieter Lang, Jan 02 2012: (Start)
a(n) = A024458(2*n-1), n>=1 (bisection, odd arguments).
chate(n):=a(n+1), n>=0, is the even part of the bisection of the half-convolution of the sequence A000045(n+1), n>=0, with itself. See a comment on A201204 for the definition of half-convolution. There one finds also the rule for the o.g.f.s of the bisection. Here the o.g.f. of the sequence chate(n), n>=0, is Chate(x):= (Ce(x)+U2(x))/2 with Ce(x)=(1-x+x^2)/(1-3*x+x^2)^2, the o.g.f. of A054444(n), and
  U2(x)=(1-x)/((1+x)*(1-3*x+x^2)), the o.g.f. of A007598(n+1), n>=0. This results (after multiplying with x) in the o.g.f. given below in the formula section. It is equivalent to the explicit formula given there, as can be seen after a partial fraction decomposition of the o.g.f.
(End)

Cf. A027926, A024458, A201204, A007598.

a(n) = (1/5)[n*F(2n+2) - n*F(2n-2) + F(2n-1) - (-1)^n], F(n)=A000045(n).

O.g.f.: x*(1-2*x+2*x^2)/((1-3*x+x^2)^2*(1+x)). See the comment above. - Wolfdieter Lang, Jan 02 2012

A027928 a(n) = T(n, 2*n-5), T given by A027926.

Original entry on oeis.org

1, 3, 8, 20, 46, 97, 189, 344, 591, 967, 1518, 2300, 3380, 4837, 6763, 9264, 12461, 16491, 21508, 27684, 35210, 44297, 55177, 68104, 83355, 101231, 122058, 146188, 174000, 205901, 242327, 283744, 330649, 383571, 443072

Offset: 3

Clark Kimberling

Vincenzo Librandi, Table of n, a(n) for n = 3..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A228074, A000045.

List([3..40], n-> (n-2)*(n^4 -8*n^3 +39*n^2 -92*n +180)/120); # G. C. Greubel, Sep 06 2019

[(n-2)*(n^4-8*n^3+39*n^2-92*n+180)/120: n in [3..40]]; // Vincenzo Librandi, Apr 22 2012

seq(binomial(n,n-1)+binomial(n+1,n-2)+binomial(n+2,n-3), n=1..35); # Zerinvary Lajos, May 29 2007

CoefficientList[Series[(1-3*x+5*x^2-3*x^3+x^4)/(1-x)^6,{x,0,40}],x] (* Vincenzo Librandi, Apr 22 2012 *)

vector(40, n, m=n+2; n*(m^4 -8*m^3 +39*m^2 -92*m +180)/120) \\ G. C. Greubel, Sep 06 2019

[(n-2)*(n^4 -8*n^3 +39*n^2 -92*n +180)/120 for n in (3..40)] # G. C. Greubel, Sep 06 2019

a(n) = (n-2)*(n^4 - 8*n^3 + 39*n^2 - 92*n + 180)/120.
a(n) = C(n,n-1) + C(n+1,n-2) + C(n+2,n-3) with offset 1. - Zerinvary Lajos, May 29 2007
G.f.: x^3*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)/(1-x)^6. - Colin Barker, Mar 18 2012
E.g.f.: 3 + x -(360 - 240*x + 60*x^2 - 20*x^3 - x^5)*exp(x)/120. - G. C. Greubel, Sep 06 2019

A027929 a(n) = T(n, 2*n-6), T given by A027926.

Original entry on oeis.org

1, 2, 5, 13, 33, 79, 176, 365, 709, 1300, 2267, 3785, 6085, 9465, 14302, 21065, 30329, 42790, 59281, 80789, 108473, 143683, 187980, 243157, 311261, 394616, 495847, 617905, 764093, 938093, 1143994, 1386321, 1670065, 2000714

Offset: 3

Clark Kimberling

Vincenzo Librandi, Table of n, a(n) for n = 3..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A228074.

List([3..40], n-> (3600 -3420*n +1684*n^2 -525*n^3 +115*n^4 -15*n^5 +n^6)/720); G. C. Greubel, Sep 06 2019

[(3600 -3420*n +1684*n^2 -525*n^3 +115*n^4 -15*n^5 +n^6)/720: n in [3..40]]; // G. C. Greubel, Sep 06 2019

seq((3600 -3420*n +1684*n^2 -525*n^3 +115*n^4 -15*n^5 +n^6)/720, n=3..40); # G. C. Greubel, Sep 06 2019

CoefficientList[Series[(1-x+x^2)(1-4x+7x^2-4x^3+x^4)/(1-x)^7, {x, 0, 40}], x] (* Vincenzo Librandi, Oct 18 2013 *)

vector(40, n, m=n+2; (3600 -3420*m +1684*m^2 -525*m^3 +115*m^4 -15*m^5 +m^6)/720) \\ G. C. Greubel, Sep 06 2019

[(3600 -3420*n +1684*n^2 -525*n^3 +115*n^4 -15*n^5 +n^6)/720 for n in (3..40)] # G. C. Greubel, Sep 06 2019

a(n) = Sum_{k=0..3} binomial(n-k, 6-2*k). - Len Smiley, Oct 20 2001
From Colin Barker, May 01 2012: (Start)
a(n) = (3600 -3420*n +1684*n^2 -525*n^3 +115*n^4 -15*n^5 +n^6)/720.
G.f.: x^3*(1-x+x^2)*(1-4*x+7*x^2-4*x^3+x^4)/(1-x)^7. (End)
E.g.f.: (3600 - 2160*x + 720*x^2 - 120*x^3 + 30*x^4 + x^6)*exp(x)/720 - 5 + 2*x - x^2/2. - G. C. Greubel, Sep 06 2019

A027930 a(n) = T(n, 2*n-7), T given by A027926.

Original entry on oeis.org

1, 3, 8, 21, 54, 133, 309, 674, 1383, 2683, 4950, 8735, 14820, 24285, 38587, 59652, 89981, 132771, 192052, 272841, 381314, 524997, 712977, 956134, 1267395, 1662011, 2157858, 2775763, 3539856, 4477949, 5621943, 7008264, 8678329, 10679043, 13063328, 15890685

Offset: 4

Clark Kimberling

G. C. Greubel, Table of n, a(n) for n = 4..1003
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A228074.

List([4..40], n-> Binomial(n-1, n-7) + (n-3)*((n-3)^4 +15*(n-3)^2 +104)/120); # G. C. Greubel, Sep 06 2019

[Binomial(n-1, n-7) + (n-3)*((n-3)^4 +15*(n-3)^2 +104)/120: n in [4..40]]; // G. C. Greubel, Sep 06 2019

seq(binomial(n-3,n-4)+binomial(n-2,n-5)+binomial(n-1,n-6)+binomial(n,n-7) , n=4..50); # Zerinvary Lajos, May 29 2007

Table[Total[Binomial[First[#],Last[#]]&/@Table[{n+i,n-1-i},{i,0,3}]],{n,35}] (* or *) LinearRecurrence[{8,-28,56,-70,56,-28,8,-1}, {1,3,8,21,54,133,309,674}, 35] (* Harvey P. Dale, Jun 23 2011 *)

vector(40, n, binomial(n+3, n-4) + n*(n^4 +15*n^2 +104)/120) \\ G. C. Greubel, Sep 06 2019

[binomial(n-1, n-7) + (n-3)*((n-3)^4 +15*(n-3)^2 +104)/120 for n in (4..40)] # G. C. Greubel, Sep 06 2019

a(n) = Sum_{k=0..3} binomial(n-k, 7-2k). - Len Smiley, Oct 20 2001
a(n) = C(n-3,n-4)+C(n-2,n-5)+C(n-1,n-6)+C(n,n-7). - Zerinvary Lajos, May 29 2007
From R. J. Mathar, Oct 05 2009: (Start)
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).
G.f.: x^4*(1 - x + x^2)*(1 - 4*x + 7*x^2 - 4*x^3 + x^4)/(1-x)^8. (End)
From G. C. Greubel, Sep 06 2019: (Start)
a(n) = binomial(n-1, n-7) + (n-3)*((n-3)^4 + 15*(n-3)^2 + 104)/120.
E.g.f.: x*(5040 + 2520*x + 1680*x^2 + 630*x^3 + 168*x^4 + 21*x^5 + x^6)*exp(x)/5040. (End)

cp's OEIS Frontend

A027935 Triangular array T read by rows: T(n,k)=t(n,2k), t given by A027926; 0 <= k <= n, n >= 0.

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A027948 Triangular array T read by rows: T(n,k) = t(n,2k+1) for 0 <= k <= n, T(n,n)=1, t given by A027926, n >= 0.

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A027992 a(n) = 1*T(n,0) + 2*T(n,1) + ... + (2n+1)*T(n,2n), T given by A027926.

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A027933 a(n) = T(n, 2*n-10), T given by A027926.

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Sage

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A027988 Greatest number in row n of array T given by A027926.

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Programs

Haskell

A027989 a(n) = self-convolution of row n of array T given by A027926.

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Mathematica

Maxima

PARI

Formula

A027991 a(n) = Sum{T(n,k)*T(n,2n-k)}, 0<=k<=n-1, T given by A027926.

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A027992 a(n) = 1T(n,0) + 2T(n,1) + ... + (2n+1)*T(n,2n), T given by A027926.