A072405 -id:A072405 - cp's OEIS frontend

A006355 Number of binary vectors of length n containing no singletons.

1, 0, 2, 2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050, 242786, 392836, 635622, 1028458, 1664080, 2692538, 4356618, 7049156, 11405774, 18454930, 29860704, 48315634

Offset: 0

David M. Bloom

Number of cvtemplates at n-2 letters given <= 2 consecutive consonants or vowels (n >= 4).
Number of (n,2) Freiman-Wyner sequences.
Diagonal sums of the Riordan array ((1-x+x^2)/(1-x), x/(1-x)), A072405 (where this begins 1,0,1,1,1,1,...). - Paul Barry, May 04 2005
Central terms of the triangle in A094570. - Reinhard Zumkeller, Mar 22 2011
Pisano period lengths: 1, 1, 8, 3, 20, 8, 16, 6, 24, 20, 10, 24, 28, 16, 40, 12, 36, 24, 18, 60, ... . - R. J. Mathar, Aug 10 2012
Also the number of matchings in the (n-2)-pan graph for n >= 5. - Eric W. Weisstein, Oct 03 2017
a(n) is the number of bimultus bitstrings of length n. A bitstring is bimultus if each of its 1's possess at least one neighboring 1 and each of its 0's possess at least one neighboring 0. - Steven Finch, May 26 2020

			a(6)=10 because we have: 000000, 000011, 000111, 001100, 001111, 110000, 110011, 111000, 111100, 111111. - _Geoffrey Critzer_, Jan 26 2014

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 16, 51.

Charles R Greathouse IV, Table of n, a(n) for n = 0..4786 (next term has 1001 digits)
Kassie Archer and Aaron Geary, Powers of permutations that avoid chains of patterns, arXiv:2312.14351 [math.CO], 2023. See p. 15.
J. Berman and P. Koehler, Cardinalities of finite distributive lattices, Mitteilungen aus dem Mathematischen Seminar Giessen, 121 (1976), 103-124. [Annotated scanned copy]
Ian F. Blake, The enumeration of certain run length sequences, Information and Control, 55 (1982), 222-237.
Alexander Burstein, Sergey Kitaev, and Toufik Mansour, Partially ordered patterns and their combinatorial interpretations, PU. M. A. Vol. 19 (2008), No. 2-3, pp. 27-38.
Steven Finch, Variance of longest run duration in a random bitstring, arXiv:2005.12185 [math.CO], 2020.
Enoch Haga, Room for Expansion, Word Ways, 33 (No. 2, 2000), pp. 106-113 (see p. 110).
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 898.
Sergey Kitaev and Jeffrey Remmel, (a,b)-rectangle patterns in permutations and words, arXiv:1304.4286 [math.CO], 2013.
Thor Martinsen, Non-Fisherian generalized Fibonacci numbers, Notes Num. Theor. Disc. Math. (2025) Vol. 31, No. 2, 370-389. See p. 389.
Noriaki Sannomiya, Hosho Katsura, and Yu Nakayama, Supersymmetry breaking and Nambu-Goldstone fermions with cubic dispersion, arXiv preprint arXiv:1612.02285 [cond-mat.str-el], 2016-2017. See Table II, line 2.
Eric Weisstein's World of Mathematics, Independent Edge Set, Matching and Pan Graph.
Index entries for linear recurrences with constant coefficients, signature (1,1).

Except for initial term, = 2*Fibonacci numbers (A000045).
Essentially the same as A047992, A054886, A055389, A068922, and A090991.
Column 2 in A265584.
Cf. A094570, A097925, A097926, A118658, A119457.

a006355 n = a006355_list !! n
a006355_list = 1 : fib2s where
   fib2s = 0 : map (+ 1) (scanl (+) 1 fib2s)
-- Reinhard Zumkeller, Mar 20 2013

[1] cat [Lucas(n) - Fibonacci(n): n in [1..50]]; // Vincenzo Librandi, Aug 02 2014

a:= n-> if n=0 then 1 else (Matrix([[2,-2]]). Matrix([[1,1], [1,0]])^n) [1,1] fi: seq(a(n), n=0..38); # Alois P. Heinz, Aug 18 2008
a := n -> ifelse(n=0, 1, -2*I^n*ChebyshevU(n-2, -I/2)):
seq(simplify(a(n)), n = 0..38);  # Peter Luschny, Dec 03 2023

Join[{1}, Last[#] - First[#] & /@ Partition[Fibonacci[Range[-1, 40]], 4, 1]] (* Harvey P. Dale, Sep 30 2011 *)
Join[{1}, LinearRecurrence[{1, 1}, {0, 2}, 38]] (* Jean-François Alcover, Sep 23 2017 *)
(* Programs from Eric W. Weisstein, Oct 03 2017 *)
Join[{1}, Table[2 Fibonacci[n], {n, 0, 40}]]
Join[{1}, 2 Fibonacci[Range[0, 40]]]
CoefficientList[Series[(1-x+x^2)/(1-x-x^2), {x, 0, 40}], x] (* End *)

a(n)=if(n,2*fibonacci(n-1),1) \\ Charles R Greathouse IV, Mar 14 2012

my(x='x+O('x^50)); Vec((1-x+x^2)/(1-x-x^2)) \\ Altug Alkan, Nov 01 2015

def A006355(n): return 2*fibonacci(n-1) - int(n==0)
print([A006355(n) for n in range(51)]) # G. C. Greubel, Apr 18 2025

a(n+2) = F(n-1) + F(n+2), for n > 0.
G.f.: (1-x+x^2)/(1-x-x^2). - Paul Barry, May 04 2005
a(n) = A119457(n-1,n-2) for n > 2. - Reinhard Zumkeller, May 20 2006
a(n) = 2*F(n-1) for n > 0, F(n)=A000045(n) and a(0)=1. - Mircea Merca, Jun 28 2012
G.f.: 1 - x + x*Q(0), where Q(k) = 1 + x^2 + (2*k+3)*x - x*(2*k+1 + x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 05 2013
a(n) = A118658(n) - 0^n. - M. F. Hasler, Nov 05 2014
a(n) = 2^(-n)*((1+r)*(1-r)^n - (1-r)*(1+r)^n)/r for n > 0, where r=sqrt(5). - Colin Barker, Jan 28 2017
a(n) = a(n-1) + a(n-2) for n >= 3. - Armend Shabani, Nov 25 2020
E.g.f.: 2*exp(x/2)*(5*cosh(sqrt(5)*x/2) - sqrt(5)*sinh(sqrt(5)*x/2))/5 - 1. - Stefano Spezia, Apr 18 2022
a(n) = F(n-3) + F(n-2) + F(n-1) for n >= 3, where F(n)=A000045(n). - Gergely Földvári, Aug 03 2024

Corrected by T. D. Noe, Oct 31 2006

A028262 Elements in 3-Pascal triangle (by row).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 4, 4, 1, 1, 5, 8, 5, 1, 1, 6, 13, 13, 6, 1, 1, 7, 19, 26, 19, 7, 1, 1, 8, 26, 45, 45, 26, 8, 1, 1, 9, 34, 71, 90, 71, 34, 9, 1, 1, 10, 43, 105, 161, 161, 105, 43, 10, 1, 1, 11, 53, 148, 266, 322, 266, 148, 53, 11, 1, 1, 12, 64, 201, 414, 588, 588, 414, 201, 64, 12, 1

Offset: 0

Mohammad K. Azarian

			Triangle begins:
  1;
  1 1;
  1 3 1;
  1 4 4 1;
  1 5 8 5 1;
  ...

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
László Németh, Tetrahedron trinomial coefficient transform, Integers (2019) Vol. 19, Article A41.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A028275, A072405.

a028262 n k = a028262_tabl !! n !! k
a028262_row n = a028262_tabl !! n
a028262_tabl = [1] : [1,1] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,3,1]
-- Reinhard Zumkeller, Aug 02 2012

T:= func< n,k | n lt 2 select 1 else Binomial(n, k) + Binomial(n-2, k-1) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 28 2021

T[n_, k_]:= If[n==1, 1, Binomial[n, k] + Binomial[n-2, k-1]]; Table[T[n, k], {n, 0, 11}, {k, 0, n}]//Flatten (* Jean-François Alcover, Jan 28 2015 *)

def T(n,k): return 1 if n<2 else binomial(n,k) + binomial(n-2,k-1)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 28 2021

After the 3rd row, use Pascal's rule.
From Ralf Stephan, Jan 31 2005: (Start)
T(n, k) = C(n, k) + C(n-2, k-1).
G.f.: (1 + x^2*y)/(1 - x*(1+y)). (End)
T(n+2,k+1) = A007318(n,k) - A007318(n+2,k+1); 0 < k < n. - Reinhard Zumkeller, Aug 02 2012
Sum_{k=0..n} T(n,k) = (n+1)*[n<2] + 5*2^(n-2)*[n>=2]. - G. C. Greubel, Apr 28 2021

More terms from James Sellers

A051597 Rows of triangle formed using Pascal's rule except begin and end n-th row with n+1.

Original entry on oeis.org

1, 2, 2, 3, 4, 3, 4, 7, 7, 4, 5, 11, 14, 11, 5, 6, 16, 25, 25, 16, 6, 7, 22, 41, 50, 41, 22, 7, 8, 29, 63, 91, 91, 63, 29, 8, 9, 37, 92, 154, 182, 154, 92, 37, 9, 10, 46, 129, 246, 336, 336, 246, 129, 46, 10, 11, 56, 175, 375, 582, 672, 582, 375, 175, 56, 11

Offset: 0

Asher Auel

Row sums give A033484(n).
The number of spotlight tilings of an (m+1) X (n+1) rectangle, read by antidiagonals. - Bridget Tenner, Nov 09 2007
T(n,k) = A134636(n,k) - A051601(n,k). - Reinhard Zumkeller, Nov 23 2012
T(n,k) = A209561(n+2,k+1), 0 <= k <= n. - Reinhard Zumkeller, Dec 26 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

			Triangle begins as:
  1;
  2,  2;
  3,  4,  3;
  4,  7,  7,  4;
  5, 11, 14, 11, 5;

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
B. E. Tenner, Spotlight tiling, Ann. Combin. 14 (4) (2010) 553.
Index entries for triangles and arrays related to Pascal's triangle

Stripped variant of A072405, A122218.

Cf. A007318, A228196, A228576.

T:= function(n,k)
    if k<0 or k>n then return 0;
    elif k=0 or k=n then return n+1;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 18 2019

a051597 n k = a051597_tabl !! n !! k
a051597_row n = a051597_tabl !! n
a051597_tabl = iterate (\row -> zipWith (+) ([1] ++ row) (row ++ [1])) [1]
-- Reinhard Zumkeller, Nov 23 2012

function T(n,k)
    if k lt 0 or k gt n then return 0;
  elif k eq 0 or k eq n then return n+1;
  else return T(n-1,k-1) + T(n-1,k);
  end if;
  return T;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 18 2019

T:= proc(n, k) option remember;
      `if`(k<0 or k>n, 0,
      `if`(k=0 or k=n, n+1,
         T(n-1, k-1) + T(n-1, k) ))
    end:
seq(seq(T(n, k), k=0..n), n=0..14);  # Alois P. Heinz, May 27 2013

NestList[Append[ Prepend[Map[Apply[Plus, #] &, Partition[#, 2, 1]], #[[1]] + 1], #[[1]] + 1] &, {1}, 10] // Grid  (* Geoffrey Critzer, May 26 2013 *)
T[n_, k_] := T[n, k] = If[k<0 || k>n, 0, If[k==0 || k==n, n+1, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n, 0, 14}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 09 2016, after Alois P. Heinz *)

T(n,k) = if(k<0 || k>n, 0, if(k==0 || k==n, n+1, T(n-1, k-1) + T(n-1, k) ));
for(n=0, 12, for(k=0, n, print1(T(n,k), ", "))) \\ G. C. Greubel, Nov 18 2019

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==0 or k==n): return n+1
    else: return T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 18 2019

T(2n,n) = A051924(n+1). - Philippe Deléham, Nov 26 2006
T(m,n) = binomial(m+n,m) - binomial(m+n-2,m-1) (correct up to offset and transformation of square indices to triangular indices). - Bridget Tenner, Nov 09 2007
T(0,n) = T(n,0) = n+1, T(n,k) = T(n-1,k) + T(n-1,k-1), 0 < k < n.
From Peter Bala, Feb 28 2013: (Start)
T(n,k) = binomial(n,k-1) + binomial(n,k) + binomial(n,k+1) for 0 <= k <= n.
O.g.f.: (1 - xt^2)/((1 - t)(1 - xt)(1 - (1+x)t)) = 1 + (2 + 2x)t + (3 + 4x + 3x^2)t^2 + ....
Row polynomials: ((1+x+x^2)*(1+x)^n - 1 - x^(n+2))/x. (End)

A173117 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 1, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 4, 4, 1, 1, 5, 8, 5, 1, 1, 6, 14, 14, 6, 1, 1, 7, 20, 28, 20, 7, 1, 1, 8, 27, 49, 49, 27, 8, 1, 1, 9, 35, 76, 98, 76, 35, 9, 1, 1, 10, 44, 111, 175, 175, 111, 44, 10, 1, 1, 11, 54, 155, 286, 350, 286, 155, 54, 11, 1

Offset: 0

Roger L. Bagula, Feb 10 2010

			Triangle begins as:
  1;
  1,  1;
  1,  3,  1;
  1,  4,  4,   1;
  1,  5,  8,   5,   1;
  1,  6, 14,  14,   6,   1;
  1,  7, 20,  28,  20,   7,   1;
  1,  8, 27,  49,  49,  27,   8,   1;
  1,  9, 35,  76,  98,  76,  35,   9,  1;
  1, 10, 44, 111, 175, 175, 111,  44, 10,  1;
  1, 11, 54, 155, 286, 350, 286, 155, 54, 11, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A007318 (q=0), A072405 (q= -1), this sequence (q=1), A173118 (q=2), A173119 (q=3), A173120 (q= -4), A173122.

T[n_, k_, q_]:= If[k==0 || k==n, 1, q*Boole[n==2] + Sum[q^j*Binomial[n-2*j, k-j]*Boole[n>2*j], {j,0,5}]];
Table[T[n,k,1], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Apr 27 2021 *)

@CachedFunction
def T(n,k,q): return 1 if (k==0 or k==n) else q*bool(n==2) + sum( q^j*binomial(n-2*j, k-j)*bool(n>2*j) for j in (0..5) )
flatten([[T(n,k,1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 27 2021

T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 1.

Sum_{k=0..n} T(n, k, q) = [n=0] + q*[n=2] + Sum_{j=0..5} q^j*2^(n-2*j)*[n > 2*j] for q = 1. - G. C. Greubel, Apr 27 2021

Edited by G. C. Greubel, Apr 27 2021

A173118 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 5, 5, 1, 1, 6, 10, 6, 1, 1, 7, 20, 20, 7, 1, 1, 8, 27, 40, 27, 8, 1, 1, 9, 35, 75, 75, 35, 9, 1, 1, 10, 44, 110, 150, 110, 44, 10, 1, 1, 11, 54, 154, 276, 276, 154, 54, 11, 1, 1, 12, 65, 208, 430, 552, 430, 208, 65, 12, 1

Offset: 0

Roger L. Bagula, Feb 10 2010

			Triangle begins as:
  1;
  1,  1;
  1,  4,  1;
  1,  5,  5,   1;
  1,  6, 10,   6,   1;
  1,  7, 20,  20,   7,   1;
  1,  8, 27,  40,  27,   8,   1;
  1,  9, 35,  75,  75,  35,   9,   1;
  1, 10, 44, 110, 150, 110,  44,  10,  1;
  1, 11, 54, 154, 276, 276, 154,  54, 11,  1;
  1, 12, 65, 208, 430, 552, 430, 208, 65, 12, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A007318 (q=0), A072405 (q= -1), A173117 (q=1), this sequence (q=2), A173119 (q=3), A173120 (q= -4), A173122.

T[n_, k_, q_]:= If[k==0 || k==n, 1, q*Boole[n==2] + Sum[q^j*Binomial[n-2*j, k-j]*Boole[n>2*j], {j,0,5}]];
Table[T[n,k,2], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Apr 27 2021 *)

@CachedFunction
def T(n,k,q): return 1 if (k==0 or k==n) else q*bool(n==2) + sum( q^j*binomial(n-2*j, k-j)*bool(n>2*j) for j in (0..5) )
flatten([[T(n,k,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 27 2021

T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 2.

Sum_{k=0..n} T(n, k, q) = [n=0] + q*[n=2] + Sum_{j=0..5} q^j*2^(n-2*j)*[n > 2*j] for q = 2. - G. C. Greubel, Apr 27 2021

Edited by G. C. Greubel, Apr 27 2021

A173119 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 3, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 6, 6, 1, 1, 7, 12, 7, 1, 1, 8, 28, 28, 8, 1, 1, 9, 36, 56, 36, 9, 1, 1, 10, 45, 119, 119, 45, 10, 1, 1, 11, 55, 164, 238, 164, 55, 11, 1, 1, 12, 66, 219, 483, 483, 219, 66, 12, 1, 1, 13, 78, 285, 702, 966, 702, 285, 78, 13, 1

Offset: 0

Roger L. Bagula, Feb 10 2010

			Triangle begins as:
  1;
  1,  1;
  1,  5,  1;
  1,  6,  6,   1;
  1,  7, 12,   7,   1;
  1,  8, 28,  28,   8,   1;
  1,  9, 36,  56,  36,   9,   1;
  1, 10, 45, 119, 119,  45,  10,   1;
  1, 11, 55, 164, 238, 164,  55,  11,  1;
  1, 12, 66, 219, 483, 483, 219,  66, 12,  1;
  1, 13, 78, 285, 702, 966, 702, 285, 78, 13, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A007318 (q=0), A072405 (q= -1), A173117 (q=1), A173118 (q=2), this sequence (q=3), A173120 (q= -4), A173122.

T[n_, k_, q_]:= If[k==0 || k==n, 1, q*Boole[n==2] + Sum[q^j*Binomial[n-2*j, k-j]*Boole[n>2*j], {j,0,5}]];
Table[T[n,k,3], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Apr 27 2021 *)

@CachedFunction
def T(n,k,q): return 1 if (k==0 or k==n) else q*bool(n==2) + sum( q^j*binomial(n-2*j, k-j)*bool(n>2*j) for j in (0..5) )
flatten([[T(n,k,3) for k in (0..n)] for n in (0..10)]) # G. C. Greubel, Apr 27 2021

T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 3.

Sum_{k=0..n} T(n, k, q) = [n=0] + q*[n=2] + Sum_{j=0..5} q^j*2^(n-2*j)*[n > 2*j] for q = 3. - G. C. Greubel, Apr 27 2021

Edited by G. C. Greubel, Apr 27 2021

A173120 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = -4, read by rows.

Original entry on oeis.org

1, 1, 1, 1, -2, 1, 1, -1, -1, 1, 1, 0, -2, 0, 1, 1, 1, 14, 14, 1, 1, 1, 2, 15, 28, 15, 2, 1, 1, 3, 17, -21, -21, 17, 3, 1, 1, 4, 20, -4, -42, -4, 20, 4, 1, 1, 5, 24, 16, 210, 210, 16, 24, 5, 1, 1, 6, 29, 40, 226, 420, 226, 40, 29, 6, 1

Offset: 0

Roger L. Bagula, Feb 10 2010

			Triangle begins as:
  1;
  1,  1;
  1, -2,  1;
  1, -1, -1,   1;
  1,  0, -2,   0,   1;
  1,  1, 14,  14,   1,   1;
  1,  2, 15,  28,  15,   2,   1;
  1,  3, 17, -21, -21,  17,   3,  1;
  1,  4, 20,  -4, -42,  -4,  20,  4,  1;
  1,  5, 24,  16, 210, 210,  16, 24,  5, 1;
  1,  6, 29,  40, 226, 420, 226, 40, 29, 6, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A007318 (q=0), A072405 (q= -1), A173117 (q=1), A173118 (q=2), A173119 (q=3), this sequence (q= -4), A173122.

T[n_, k_, q_]:= If[k==0 || k==n, 1, q*Boole[n==2] + Sum[q^j*Binomial[n-2*j, k-j]*Boole[n>2*j], {j,0,5}]];
Table[T[n,k,-4], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Apr 27 2021 *)

@CachedFunction
def T(n,k,q): return 1 if (k==0 or k==n) else q*bool(n==2) + sum( q^j*binomial(n-2*j, k-j)*bool(n>2*j) for j in (0..5) )
flatten([[T(n,k,-4) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 27 2021

T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = -4.

Sum_{k=0..n} T(n, k, q) = [n=0] + q*[n=2] + Sum_{j=0..5} q^j*2^(n-2*j)*[n > 2*j] for q = -4. - G. C. Greubel, Apr 27 2021

Edited by G. C. Greubel, Apr 27 2021

A163733 Number of n X 2 binary arrays with all 1's connected, all corners 1, and no 1 having more than two 1's adjacent.

Original entry on oeis.org

1, 1, 2, 2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050, 242786, 392836, 635622, 1028458, 1664080, 2692538, 4356618, 7049156, 11405774, 18454930, 29860704, 48315634, 78176338

Offset: 1

R. H. Hardin, Aug 03 2009

nonn

Same recurrence for A163695.
Same recurrence for A163714.
Appears to coincide with diagonal sums of A072405. - Paul Barry, Aug 10 2009
From Gary W. Adamson, Sep 15 2016: (Start)
Let the sequence prefaced with a 1: (1, 1, 1, 2, 2, 4, 6, ...) equate to r(x). Then (r(x) * r(x^2) * r(x^4) * r(x^8) * ...) = the Fibonacci sequence, (1, 1, 2, 3, 5, ...). Let M = the following production matrix:
  1, 0, 0, 0, 0, ...
  1, 0, 0, 0, 0, ...
  1, 1, 0, 0, 0, ...
  2, 1, 0, 0, 0, ...
  2, 1, 1, 0, 0, ...
  4, 2, 1, 0, 0, ...
  6, 2, 1, 1, 0, ...
  ...
Limit of the matrix power M^k as k->infinity results in a single column vector equal to the Fibonacci sequence. (End)
Apparently a(n) = A128588(n-2) for n > 3. - Georg Fischer, Oct 14 2018

			All solutions for n=8:
   1 1   1 1   1 1   1 1   1 1   1 1   1 1   1 1   1 1   1 1
   0 1   1 0   1 0   1 0   1 0   1 0   0 1   0 1   0 1   0 1
   0 1   1 0   1 0   1 0   1 1   1 0   0 1   0 1   1 1   0 1
   0 1   1 0   1 0   1 1   0 1   1 0   0 1   0 1   1 0   1 1
   0 1   1 0   1 1   0 1   0 1   1 0   0 1   1 1   1 0   1 0
   0 1   1 0   0 1   0 1   0 1   1 1   1 1   1 0   1 0   1 0
   0 1   1 0   0 1   0 1   0 1   0 1   1 0   1 0   1 0   1 0
   1 1   1 1   1 1   1 1   1 1   1 1   1 1   1 1   1 1   1 1
------
   1 1   1 1   1 1   1 1   1 1   1 1
   0 1   0 1   0 1   1 0   1 0   1 0
   1 1   1 1   0 1   1 0   1 1   1 1
   1 0   1 0   1 1   1 1   0 1   0 1
   1 1   1 0   1 0   0 1   0 1   1 1
   0 1   1 1   1 1   1 1   1 1   1 0
   0 1   0 1   0 1   1 0   1 0   1 0
   1 1   1 1   1 1   1 1   1 1   1 1

R. H. Hardin, Table of n, a(n) for n=1..100

Cf. A118658, A055389, A006355, A006355, A169985, A000045.

Cf. A128588. - Georg Fischer, Oct 14 2018

Join[{1, 1}, Table[2*Fibonacci[n], {n, 70}]] (* Vladimir Joseph Stephan Orlovsky, Feb 10 2012 *)
Table[Round[GoldenRatio^(k-1)] - Round[GoldenRatio^(k-1)/Sqrt[5]], {k, 1, 70}] (* Federico Provvedi, Mar 26 2013 *)

Empirical: a(n) = a(n-1) + a(n-2) for n >= 5.
G.f.: (1-x^3)/(1-x-x^2) (conjecture). - Paul Barry, Aug 10 2009
a(n) = round(phi^(k-1)) - round(phi^(k-1)/sqrt(5)), phi = (1 + sqrt(5))/2 (conjecture). - Federico Provvedi, Mar 26 2013
G.f.: 1 + 2*x - x*Q(0), where Q(k) = 1 + x^2 - (2*k+1)*x + x*(2*k-1 - x)/Q(k+1); (conjecture), (continued fraction). - Sergei N. Gladkovskii, Oct 05 2013
G.f.: If prefaced with a 1, (1, 1, 1, 2, 2, 4, ...): (1 - x^2 - x^4)/(1 - x - x^2); where the modified sequence satisfies A(x)/A(x^2), A(x) is the Fibonacci sequence. - Gary W. Adamson, Sep 15 2016

A173122 Irregular triangle T(n) = coefficients of Sum_{k=0..n} t(n,k,q) for powers of q, where t(n,k,q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with t(n,0,q) = t(n,n,q) = 1, read by rows.

Original entry on oeis.org

1, 2, 4, 1, 8, 2, 16, 4, 32, 8, 2, 64, 16, 4, 128, 32, 8, 2, 256, 64, 16, 4, 512, 128, 32, 8, 2, 1024, 256, 64, 16, 4, 2048, 512, 128, 32, 8, 2, 4096, 1024, 256, 64, 16, 4, 8192, 2048, 512, 128, 32, 8, 16384, 4096, 1024, 256, 64, 16, 32768, 8192, 2048, 512, 128, 32, 65536, 16384, 4096, 1024, 256, 64

Offset: 0

Roger L. Bagula, Feb 10 2010

			Irregular triangle begins as:
     1;
     2;
     4,   1;
     8,   2;
    16,   4;
    32,   8,  2;
    64,  16,  4;
   128,  32,  8,  2;
   256,  64, 16,  4;
   512, 128, 32,  8, 2;
  1024, 256, 64, 16, 4;

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

Cf. A007318, A072405, A173117, A173118, A173119, A173120.

t[n_, k_, q_]:= If[k==0 || k==n, 1, q*Boole[n==2] + Sum[q^j*Binomial[n-2*j, k-j] *Boole[n>2*j], {j, 0, 5}]];
T[n_]:= CoefficientList[Series[Sum[t[n,k,q], {k,0,n}], {q,0,n}], q];
Table[T[n], {n, 0, 12}]//Flatten (* modified by G. C. Greubel, Apr 29 2021 *)

@CachedFunction
def t(n, k, x): return 1 if (k==0 or k==n) else x*bool(n==2) + sum( x^j*binomial(n-2*j, k-j)*bool(n>2*j) for j in (0..5) )
def s(n,x): return sum( t(n,k,x) for k in (0..n) )
flatten([taylor(s(n,x), x, 0, n).list() for n in (0..12)]) # G. C. Greubel, Apr 29 2021

T(n) = coefficients of Sum_{k=0..n} t(n,k,q) for powers of q, where t(n,k,q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with t(n,0,q) = t(n,n,q) = 1.

More terms and edited by G. C. Greubel, Apr 29 2021

A106509 Riordan array ((1+x)/(1+x+x^2), x/(1+x)), read by rows.

Original entry on oeis.org

1, 0, 1, -1, -1, 1, 1, 0, -2, 1, 0, 1, 2, -3, 1, -1, -1, -1, 5, -4, 1, 1, 0, 0, -6, 9, -5, 1, 0, 1, 0, 6, -15, 14, -6, 1, -1, -1, 1, -6, 21, -29, 20, -7, 1, 1, 0, -2, 7, -27, 50, -49, 27, -8, 1, 0, 1, 2, -9, 34, -77, 99, -76, 35, -9, 1, -1, -1, -1, 11, -43, 111, -176, 175, -111, 44, -10, 1

Offset: 0

Paul Barry, May 04 2005

Row sums are A106510.
Diagonal sums are A106511.
Inverse of A072405 (when this starts 1, 0, 1, ...).

			Triangle begins:
   1;
   0,  1;
  -1, -1,  1;
   1,  0, -2,  1;
   0,  1,  2, -3,  1;
  -1, -1, -1,  5, -4,  1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.

Cf. A072405, A106510, A106511.

T:= func< n,k | (&+[ (-1)^j*Binomial(2*n-k-j, j): j in [0..n-k]]) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 28 2021

(* The function RiordanArray is defined in A256893. *)
RiordanArray[(1 + #)/(1 + # + #^2)&, #/(1 + #)&, 12] // Flatten (* Jean-François Alcover, Jul 19 2019 *)

def T(n,k): return sum( (-1)^j*binomial(2*n-k-j, j) for j in (0..n-k))
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 28 2021

T(n, k) = Sum_{j=0..n-k} (-1)^j*binomial(2n-k-j, j).
T(n,k) = T(n-1,k-1) - 2*T(n-1,k) + T(n-2,k-1) - 2*T(n-2,k) + T(n-3,k-1) - T(n-3,k), T(0,0) = T(1,1) = T(2,2) = 1, T(1,0) = 0, T(2,1) = T(2,0) = -1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 12 2014
Sum_{k=0..n} T(n,k) = A106510(n). - G. C. Greubel, Apr 28 2021

cp's OEIS Frontend

A006355 Number of binary vectors of length n containing no singletons.

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A028262 Elements in 3-Pascal triangle (by row).

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A051597 Rows of triangle formed using Pascal's rule except begin and end n-th row with n+1.

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A173117 Triangle T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 1, read by rows.

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A173118 Triangle T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 2, read by rows.

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A173119 Triangle T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 3, read by rows.

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A173117 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 1, read by rows.

A173118 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 2, read by rows.

A173119 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 3, read by rows.

A173120 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = -4, read by rows.

A173122 Irregular triangle T(n) = coefficients of Sum_{k=0..n} t(n,k,q) for powers of q, where t(n,k,q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with t(n,0,q) = t(n,n,q) = 1, read by rows.