A107066 -id:A107066 - cp's OEIS frontend

A008937 a(n) = Sum_{k=0..n} T(k) where T(n) are the tribonacci numbers A000073.

0, 1, 2, 4, 8, 15, 28, 52, 96, 177, 326, 600, 1104, 2031, 3736, 6872, 12640, 23249, 42762, 78652, 144664, 266079, 489396, 900140, 1655616, 3045153, 5600910, 10301680, 18947744, 34850335, 64099760, 117897840, 216847936, 398845537, 733591314, 1349284788

Offset: 0

N. J. A. Sloane, Alejandro Teruel (teruel(AT)usb.ve)

a(n+1) is the number of n-bit sequences that avoid 1100. - David Callan, Jul 19 2004 [corrected by Kent E. Morrison, Jan 08 2019]. Also the number of n-bit sequences avoiding one of the patterns 1000, 0011, 1110, ... or any binary string of length 4 without overlap at beginning and end. Strings where it is not true are: 1111, 1010, 1001, ... and their bitwise complements. - Alois P. Heinz, Jan 09 2019
Row sums of Riordan array (1/(1-x), x(1+x+x^2)). - Paul Barry, Feb 16 2005
Diagonal sums of Riordan array (1/(1-x)^2, x(1+x)/(1-x)), A104698.
A shifted version of this sequence can be found in Eqs. (4) and (3) on p. 356 of Dunkel (1925) with r = 3. (Equation (3) follows equation (4) in the paper!) The whole paper is a study of the properties of this and other similar sequences indexed by the parameter r. For r = 2, we get a shifted version of A000071. For r = 4, we get a shifted version of A107066. For r = 5, we get a shifted version of A001949. For r = 6, we get a shifted version of A172316. See also the table in A172119. - Petros Hadjicostas, Jun 14 2019
Officially, to match A000073, this should start with a(0)=a(1)=0, a(2)=1. - N. J. A. Sloane, Sep 12 2020
Numbers with tribonacci representation that is a prefix of 100100100100... . - Jeffrey Shallit, Jul 10 2024

			G.f. = x + 2*x^2 + 4*x^3 + 8*x^4 + 15*x^5 + 28*x^6 + 52*x^7 + 96*x^8 + 177*x^9 + ... [edited by _Petros Hadjicostas_, Jun 12 2019]

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 41.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Isha Agarwal, Matvey Borodin, Aidan Duncan, Kaylee Ji, Tanya Khovanova, Shane Lee, Boyan Litchev, Anshul Rastogi, Garima Rastogi, and Andrew Zhao, From Unequal Chance to a Coin Game Dance: Variants of Penney's Game, arXiv:2006.13002 [math.HO], 2020.
Kassie Archer and Noel Bourne, Pattern avoidance in compositions and powers of permutations, arXiv:2505.05218 [math.CO], 2025. See pp. 6, 9.
Erik Bates, Blan Morrison, Mason Rogers, Arianna Serafini, and Anav Sood, A new combinatorial interpretation of partial sums of m-step Fibonacci numbers, arXiv:2503.11055 [math.CO], 2025. See p. 3.
Otto Dunkel, Solutions of a probability difference equation, Amer. Math. Monthly, 32 (1925), 354-370; see pp. 356 and 369.
Jia Huang, A coin flip game and generalizations of Fibonacci numbers, arXiv:2501.07463 [math.CO], 2025. See p. 7.
Thomas Langley, Jeffrey Liese, and Jeffrey Remmel, Generating Functions for Wilf Equivalence Under Generalized Factor Order, J. Int. Seq. 14 (2011), Article # 11.4.2.
William Verreault, MacMahon Partition Analysis: a discrete approach to broken stick problems, arXiv:2107.10318 [math.CO], 2021.
Index entries for linear recurrences with constant coefficients, signature (2,0,0,-1).

Partial sums of A000073. Cf. A000213, A018921, A027084, A077908, A209972.
Row sums of A055216.
Column k = 1 of A140997 and second main diagonal of A140994.

a:=[0,1,1];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # G. C. Greubel, Sep 13 2019

a008937 n = a008937_list !! n
a008937_list = tail $ scanl1 (+) a000073_list
-- Reinhard Zumkeller, Apr 07 2012

[ n eq 1 select 0 else n eq 2 select 1 else n eq 3 select 2 else n eq 4 select 4 else 2*Self(n-1)-Self(n-4): n in [1..40] ]; // Vincenzo Librandi, Aug 21 2011

A008937 := proc(n) option remember; if n <= 3 then 2^n else 2*procname(n-1)-procname(n-4) fi; end;
a:= n-> (Matrix([[1,1,0,0], [1,0,1,0], [1,0,0,0], [1,0,0,1]])^n)[4,1]: seq(a(n), n=0..50); # Alois P. Heinz, Jul 24 2008

CoefficientList[Series[x/(1-2x+x^4), {x, 0, 40}], x]
Accumulate[LinearRecurrence[{1,1,1},{0,1,1},40]] (* Harvey P. Dale, Dec 04 2017 *)
LinearRecurrence[{2, 0, 0, -1},{0, 1, 2, 4},40] (* Ray Chandler, Mar 01 2024 *)

{a(n) = if( n<0, polcoeff( - x^3 / (1 - 2*x^3 + x^4) + x * O(x^-n), -n), polcoeff( x / (1 - 2*x + x^4) + x * O(x^n), n))}; /* Michael Somos, Aug 23 2014 */

a(n) = sum(j=0, n\2, sum(k=0, j, binomial(n-2*j,k+1)*binomial(j,k)*2^k)); \\ Michel Marcus, Sep 08 2017

def A008937_list(prec):
    P = PowerSeriesRing(ZZ, 'x', prec)
    x = P.gen().O(prec)
    return (x/(1-2*x+x^4)).list()
A008937_list(40) # G. C. Greubel, Sep 13 2019

a(n) = A018921(n-2) = A027084(n+1) + 1.
a(n) = (A000073(n+2) + A000073(n+4) - 1)/2.
From Mario Catalani (mario.catalani(AT)unito.it), Aug 09 2002: (Start)
G.f.: x/((1-x)*(1-x-x^2-x^3)).
a(n) = 2*a(n-1) - a(n-4), a(0) = 0, a(1) = 1, a(2) = 2, a(3) = 4. (End)
a(n) = 1 + a(n-1) + a(n-2) + a(n-3). E.g., a(11) = 1 + 600 + 326 + 177 = 1104. - Philippe LALLOUET (philip.lallouet(AT)orange.fr), Oct 29 2007
a(n) = term (4,1) in the 4 X 4 matrix [1,1,0,0; 1,0,1,0; 1,0,0,0; 1,0,0,1]^n. - Alois P. Heinz, Jul 24 2008
a(n) = -A077908(-n-3). - Alois P. Heinz, Jul 24 2008
a(n) = (A000213(n+2) - 1) / 2. - Reinhard Zumkeller, Apr 07 2012
a(n) = Sum_{j=0..floor(n/2)} Sum_{k=0..j} binomial(n-2j,k+1) *binomial(j,k)*2^k. - Tony Foster III, Sep 08 2017
a(n) = Sum_{k=0..floor(n/2)} (n-2*k)*hypergeom([-k,-n+2*k+1], [2], 2). - Peter Luschny, Nov 09 2017
a(n) = 2^(n-1)*hypergeom([1-n/4, 1/4-n/4, 3/4-n/4, 1/2-n/4], [1-n/3, 1/3-n/3, 2/3-n/3], 16/27) for n > 0. - Peter Luschny, Aug 20 2020
a(n-1) = T(n) + T(n-3) + T(n-6) + ... + T(2+((n-2) mod 3)), for n >= 4, where T is A000073(n+1). - Jeffrey Shallit, Dec 24 2020

A140996 Triangle G(n, k) read by rows for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) + G(n+3, m) + G(n+4, m) for n >= 0 for m = 1..(n+1).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 4, 2, 1, 1, 16, 8, 4, 2, 1, 1, 31, 17, 8, 4, 2, 1, 1, 60, 35, 17, 8, 4, 2, 1, 1, 116, 72, 35, 17, 8, 4, 2, 1, 1, 224, 148, 72, 35, 17, 8, 4, 2, 1, 1, 432, 303, 149, 72, 35, 17, 8, 4, 2, 1, 1, 833, 618, 308, 149, 72, 35, 17, 8, 4, 2, 1, 1, 1606, 1257, 636, 308, 149, 72, 35, 17, 8, 4, 2, 1

Offset: 0

Juri-Stepan Gerasimov, Jul 08 2008

From Petros Hadjicostas, Jun 12 2019: (Start)
This is a mirror image of the triangular array A140995. The current array has index of asymmetry s = 3 and index of obliqueness (obliquity) e = 0. Array A140995 has the same index of asymmetry, but has index of obliqueness e = 1. (In other related sequences, the author uses the letter y for the index of asymmetry and the letter z for the index of obliqueness, but on the stone slab that appears over a tomb in a picture that he posted in those sequences, the letters s and e are used instead. See, for example, the documentation for sequences A140998, A141065, A141066, and A141067.)
In general, if the index of asymmetry (from the Pascal triangle A007318) is s, then the order of the recurrence is s + 2 (because the recurrence of the Pascal triangle has order 2). There are also s + 2 infinite sets of initial conditions (as opposed to the Pascal triangle, which has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0).
Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), arrays A140997 and A140994 have s = 2 (with e = 0 and e = 1, respectively), and arrays A141020 and A141021 have s = 4 (with e = 0 and e = 1, respectively).
(End)

			Triangle (with rows n >= 0 and columns k >= 0) begins as follows:
  1
  1   1
  1   2   1
  1   4   2   1
  1   8   4   2   1
  1  16   8   4   2  1
  1  31  17   8   4  2  1
  1  60  35  17   8  4  2  1
  1 116  72  35  17  8  4  2 1
  1 224 148  72  35 17  8  4 2 1
  1 432 303 149  72 35 17  8 4 2 1
  1 833 618 308 149 72 35 17 8 4 2 1
  ...

Robert Price, Table of n, a(n) for n = 0..5150
Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A007318, A107066, A140993, A140994, A140995, A140997, A140998, A141020, A141021, A141031, A141065, A141066, A141067, A141068, A141069, A141070, A141072, A141073, A309462.

nlim = 100;
For[n = 0, n <= nlim, n++, G[n, 0] = 1];
For[n = 1, n <= nlim, n++, G[n, n] = 1];
For[n = 2, n <= nlim, n++, G[n, n-1] = 2];
For[n = 3, n <= nlim, n++, G[n, n-2] = 4];
For[n = 4, n <= nlim, n++, G[n, n-3] = 8];
For[n = 5, n <= nlim, n++, For[k = 1, k < n - 3, k++,
   G[n, k] = G[n-4, k-1] + G[n-4, k] + G[n-3, k] + G[n-2, k] + G[n-1, k]]];
A140996 = {}; For[n = 0, n <= nlim, n++,
For[k = 0, k <= n, k++, AppendTo[A140996, G[n, k]]]];
A140996 (* Robert Price, Jul 03 2019 *)
G[n_, k_] := G[n, k] = Which[k < 0 || k > n, 0, k == 0 || k == n, 1, k == n - 1, 2, k == n - 2, 4, k == n - 3, 8, True, G[n - 1, k] + G[n - 2, k] + G[n - 3, k] + G[n - 4, k] + G[n - 4, k - 1]];
Table[G[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 28 2024 *)

From Petros Hadjicostas, Jun 12 2019: (Start)
G(n, k) = A140995(n, n - k) for 0 <= k <= n.
Bivariate g.f.: Sum_{n,k >= 0} G(n, k)*x^n*y^k = (1 - x - x^2 - x^3 - x^4 + x^2*y + x^3*y + x^5*y)/((1 - x) * (1 - x*y) * (1 - x - x^2 - x^3 - x^4 - x^4*y)).
If we take the first derivative of the bivariate g.f. w.r.t. y and set y = 0, we get the g.f. of column k = 1: x/((1 - x) * (1 - x - x^2 - x^3 - x^4)). This is the g.f. of a shifted version of sequence A107066.
Substituting y = 1 in the above bivariate function and simplifying, we get the g.f. of row sums: 1/(1 - 2*x). Hence, the row sums are powers of 2; i.e., A000079.
(End)

Name edited by Petros Hadjicostas, Jun 12 2019

A172119 Sum the k preceding elements in the same column and add 1 every time.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 4, 2, 1, 1, 5, 7, 4, 2, 1, 1, 6, 12, 8, 4, 2, 1, 1, 7, 20, 15, 8, 4, 2, 1, 1, 8, 33, 28, 16, 8, 4, 2, 1, 1, 9, 54, 52, 31, 16, 8, 4, 2, 1, 1, 10, 88, 96, 60, 32, 16, 8, 4, 2, 1, 1, 11, 143, 177, 116, 63, 32, 16, 8, 4, 2, 1, 1, 12, 232, 326, 224, 124, 64, 32, 16

Offset: 0

Mats Granvik, Jan 26 2010

Columns are related to Fibonacci n-step numbers. Are there closed forms for the sequences in the columns?
We denote by a(n,k) the number which is in the (n+1)-th row and (k+1)-th-column. With help of the definition, we also have the recurrence relation: a(n+k+1, k) = 2*a(n+k, k) - a(n, k). We see on the main diagonal the numbers 1,2,4, 8, ..., which is clear from the formula for the general term d(n)=2^n. - Richard Choulet, Jan 31 2010
Most of the paper by Dunkel (1925) is a study of the columns of this table. - Petros Hadjicostas, Jun 14 2019

			Triangle begins:
n\k|....0....1....2....3....4....5....6....7....8....9...10
---|-------------------------------------------------------
0..|....1
1..|....1....1
2..|....1....2....1
3..|....1....3....2....1
4..|....1....4....4....2....1
5..|....1....5....7....4....2....1
6..|....1....6...12....8....4....2....1
7..|....1....7...20...15....8....4....2....1
8..|....1....8...33...28...16....8....4....2....1
9..|....1....9...54...52...31...16....8....4....2....1
10.|....1...10...88...96...60...32...16....8....4....2....1

Erik Bates, Blan Morrison, Mason Rogers, Arianna Serafini, and Anav Sood, A new combinatorial interpretation of partial sums of m-step Fibonacci numbers, arXiv:2503.11055 [math.CO], 2025. See p. 3.
Otto Dunkel, Solutions of a probability difference equation, Amer. Math. Monthly, 32 (1925), 354-370; see p. 356.
Thomas Langley, Jeffrey Liese, and Jeffrey Remmel, Generating Functions for Wilf Equivalence Under Generalized Factor Order, J. Int. Seq. 14 (2011), # 11.4.2.
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.
Wikipedia, Fibonacci number.

Cf. A000071 (col. 3), A008937 (col. 4), A107066 (col. 5), A001949 (col. 6), A172316 (col. 7), A172317 (col. 8), A172318 (col. 9), A172319 (col. 10), A172320 (col. 11), A144428.

Cf. (1-((-1)^T(n, k)))/2 = A051731, see formula by Hieronymus Fischer in A022003.

T:= function(n,k)
    if k=0 and k=n then return 1;
    elif k<0 or k>n then return 0;
    else return 1 + Sum([1..k], j-> T(n-j,k));
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Jul 27 2019

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

for k from 0 to 20 do for n from 0 to 20 do b(n):=sum((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j),j=0..floor(n/(k+1))):od: seq(b(n),n=0..20):od; # Richard Choulet, Jan 31 2010
A172119 := proc(n,k)
    option remember;
    if k = 0 then
        1;
    elif k > n then
        0;
    else
        1+add(procname(n-k+i,k),i=0..k-1) ;
    end if;
end proc:
seq(seq(A172119(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Sep 16 2017

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

T(n,k) = if(k<0 || k>n, 0, k==1 && k==n, 1, 1 + sum(j=1,k, T(n-j,k)));
for(n=1,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 27 2019

@CachedFunction
def T(n, k):
    if (k==0 and k==n): return 1
    elif (k<0 or k>n): return 0
    else: return 1 + sum(T(n-j, k) for j in (1..k))
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jul 27 2019

T(n,0) = 1.
T(n,1) = n.
T(n,2) = A000071(n+1).
T(n,3) = A008937(n-2).
The general term in the n-th row and k-th column is given by: a(n, k) = Sum_{j=0..floor(n/(k+1))} ((-1)^j binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j)). For example: a(5,3) = binomial(5,5)*2^5 - binomial(2,1)*2^1 = 28. The generating function of the (k+1)-th column satisfies: psi(k)(z)=1/(1-2*z+z^(k+1)) (for k=0 we have the known result psi(0)(z)=1/(1-z)). - Richard Choulet, Jan 31 2010 [By saying "(k+1)-th column" the author actually means "k-th column" for k = 0, 1, 2, ... - Petros Hadjicostas, Jul 26 2019]

A001949 Solutions of a fifth-order probability difference equation.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 4, 8, 16, 32, 63, 124, 244, 480, 944, 1856, 3649, 7174, 14104, 27728, 54512, 107168, 210687, 414200, 814296, 1600864, 3147216, 6187264, 12163841, 23913482, 47012668, 92424472, 181701728, 357216192, 702268543, 1380623604, 2714234540

Offset: 0

N. J. A. Sloane

This sequence is the case r = 5 in the solution to an r-th order probability difference equation that can be found in Eqs. (4) and (3) on p. 356 of Dunkel (1925). (Equation (3) follows equation (4) in the paper!) For r = 2, we get a shifted version of A000071. For r = 3, we get a shifted version of A008937. For r = 4, we get a shifted version of A107066. For r = 6, we get a shifted version of A172316. See also the table in A172119. - Petros Hadjicostas, Jun 15 2019

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
O. Dunkel, Solutions of a probability difference equation, Amer. Math. Monthly, 32 (1925), 354-370; see pp. 356 and 369.
T. Langley, J. Liese, and J. Remmel, Generating Functions for Wilf Equivalence Under Generalized Factor Order, J. Int. Seq. 14 (2011), Article #11.4.2.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,0,-1)

Column k = 1 of A141020 (with a different offset) and second main diagonal of A141021 (with no zeros).
Column k = 5 of A172119.
Partial sums of A001591.

A001949:=1/(z-1)/(z**5+z**4+z**3+z**2+z-1); # Simon Plouffe in his 1992 dissertation

t={0,0,0,0,0};Do[AppendTo[t,t[[-5]]+t[[-4]]+t[[-3]]+t[[-2]]+t[[-1]]+1],{n,40}];t (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
LinearRecurrence[{2,0,0,0,0,-1},{0,0,0,0,0,1},40] (* Harvey P. Dale, Jan 17 2015 *)

a(n):=sum(sum((-1)^j*binomial(n-5*j-5,k-1)*binomial(n-k-5*j-4,j),j,0,(n-k-4)/5),k,1,n-4); /* Vladimir Kruchinin, Oct 19 2011 */

x='x+O('x^99); concat(vector(5), Vec(x^5/((x-1)*(x^5+x^4+x^3+x^2+x-1)))) \\ Altug Alkan, Oct 04 2017

For n >= 6, a(n+1) = 2*a(n) - a(n-5).
G.f.: x^5 / ( (x-1)*(x^5 + x^4 + x^3 + x^2 + x - 1) ).
a(n) = Sum_{k=1..n-4} Sum_{j=0..floor((n-k-4)/5)} (-1)^j*binomial(n-5*j-5, k-1)*binomial(n-k-5*j-4, j). - Vladimir Kruchinin, Oct 19 2011
4*a(n) = A000322(n+1) - 1. - R. J. Mathar, Aug 16 2017
From Petros Hadjicostas, Jun 15 2019: (Start)
a(n) = 1 + a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) for n >= 5. (See Eq. (4) and the Theorem with r = 5 on p. 356 of Dunkel (1925).)
a(n) = T(n - 5, 5) for n >= 5, where T(n, k) = Sum_{j = 0..floor(n/(k+1))} (-1)^j * binomial(n - k*j, n - (k+1)*j) * 2^(n - (k+1)*j) for 0 <= k <= n. This is Richard Choulet's formula in A172119.
(End)

Name edited by Petros Hadjicostas, Jun 15 2019

A209972 Number of binary words of length n avoiding the subword given by the binary expansion of k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 2, 3, 4, 1, 1, 1, 2, 4, 5, 5, 1, 1, 1, 2, 4, 7, 8, 6, 1, 1, 1, 2, 4, 7, 12, 13, 7, 1, 1, 1, 2, 4, 7, 12, 20, 21, 8, 1, 1, 1, 2, 4, 7, 12, 21, 33, 34, 9, 1, 1, 1, 2, 4, 8, 13, 20, 37, 54, 55, 10, 1, 1, 1, 2, 4, 8, 15, 24, 33, 65, 88, 89, 11, 1, 1

Offset: 0

Alois P. Heinz, Mar 16 2012

			Square array begins:
  1,  1,  1,   1,   1,   1,   1,   1,   1, ...
  1,  1,  2,   2,   2,   2,   2,   2,   2, ...
  1,  1,  3,   3,   4,   4,   4,   4,   4, ...
  1,  1,  4,   5,   7,   7,   7,   7,   8, ...
  1,  1,  5,   8,  12,  12,  12,  13,  15, ...
  1,  1,  6,  13,  20,  21,  20,  24,  28, ...
  1,  1,  7,  21,  33,  37,  33,  44,  52, ...
  1,  1,  8,  34,  54,  65,  54,  81,  96, ...
  1,  1,  9,  55,  88, 114,  88, 149, 177, ...

Alois P. Heinz, Antidiagonals n = 0..150, flattened

Columns give: 0, 1: A000012, 2: A001477(n+1), 3: A000045(n+2), 4, 6: A000071(n+3), 5: A005251(n+3), 7: A000073(n+3), 8, 12, 14: A008937(n+1), 9, 11, 13: A049864(n+2), 10: A118870, 15: A000078(n+4), 16, 20, 24, 26, 28, 30: A107066, 17, 19, 23, 25, 29: A210003, 18, 22: A209888, 21: A152718(n+3), 27: A210021, 31: A001591(n+5), 32: A001949(n+5), 33, 35, 37, 39, 41, 43, 47, 49, 53, 57, 61: A210031.

Main diagonal equals A234005 or column k=0 of A233940.

A[n_, k_] := Module[{bb, cnt = 0}, Do[bb = PadLeft[IntegerDigits[j, 2], n]; If[SequencePosition[bb, IntegerDigits[k, 2], 1]=={}, cnt++], {j, 0, 2^n-1 }]; cnt];
Table[A[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Nov 01 2021 *)

A126198 Triangle read by rows: T(n,k) (1 <= k <= n) = number of compositions of n into parts of size <= k.

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 1, 5, 7, 8, 1, 8, 13, 15, 16, 1, 13, 24, 29, 31, 32, 1, 21, 44, 56, 61, 63, 64, 1, 34, 81, 108, 120, 125, 127, 128, 1, 55, 149, 208, 236, 248, 253, 255, 256, 1, 89, 274, 401, 464, 492, 504, 509, 511, 512, 1, 144, 504, 773, 912, 976, 1004, 1016, 1021, 1023, 1024

Offset: 1

N. J. A. Sloane, Mar 09 2007

Also has an interpretation as number of binary vectors of length n-1 in which the length of the longest run of 1's is <= k (see A048004). - N. J. A. Sloane, Apr 03 2011

Higher Order Fibonacci numbers: A126198(n,k) = Sum_{h=0..k} A048004(n,h); for example, A126198(7,3) = Sum_{h=0..3} A048004(7,h) or A126198(7,3) = 1 + 33 + 47 + 27 = 108, the 7th tetranacci number. A048004 row(7) produces A126198 row(7) list of 1,34,81,108,120,125,127,128 which are 1, the 7th Fibonacci, the 7th tribonacci, ... 7th octanacci numbers. - Richard Southern, Aug 04 2017

			Triangle begins:
  1;
  1,  2;
  1,  3,  4;
  1,  5,  7,  8;
  1,  8, 13, 15, 16;
  1, 13, 24, 29, 31, 32;
  1, 21, 44, 56, 61, 63, 64;
Could also be extended to a square array:
  1  1  1  1  1  1  1 ...
  1  2  2  2  2  2  2 ...
  1  3  4  4  4  4  4 ...
  1  5  7  8  8  8  8 ...
  1  8 13 15 16 16 16 ...
  1 13 24 29 31 32 32 ...
  1 21 44 56 61 63 64 ...
which when read by antidiagonals (downwards) gives A048887.

J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 154-155.

Alois P. Heinz, Rows n = 1..141

Rows are partial sums of rows of A048004. Cf. A048887, A092921 for other versions.
2nd column = Fibonacci numbers, next two columns are A000073, A000078; last three diagonals are 2^n, 2^n-1, 2^n-3.
Cf. A082267.

A126198 := proc(n,k) coeftayl( x*(1-x^k)/(1-2*x+x^(k+1)),x=0,n); end: for n from 1 to 11 do for k from 1 to n do printf("%d, ",A126198(n,k)); od; od; # R. J. Mathar, Mar 09 2007
# second Maple program:
T:= proc(n, k) option remember;
      if n=0 or k=1 then 1
    else add(T(n-j, k), j=1..min(n, k))
      fi
    end:
seq(seq(T(n, k), k=1..n), n=1..15);  # Alois P. Heinz, Oct 23 2011

rows = 11; t[n_, k_] := Sum[ (-1)^i*2^(n-i*(k+1))*Binomial[ n-i*k, i], {i, 0, Floor[n/(k+1)]}] - Sum[ (-1)^i*2^((-i)*(k+1)+n-1)*Binomial[ n-i*k-1, i], {i, 0, Floor[(n-1)/(k+1)]}]; Flatten[ Table[ t[n, k], {n, 1, rows}, {k, 1, n}]](* Jean-François Alcover, Nov 17 2011, after Max Alekseyev *)

G.f. for column k: (x-x^(k+1))/(1-2*x+x^(k+1)). [Riordan]
T(n,3) = A008937(n) - A008937(n-3) for n>=3. T(n,4) = A107066(n-1) - A107066(n-5) for n>=5. T(n,5) = A001949(n+4) - A001949(n-1) for n>=5. - R. J. Mathar, Mar 09 2007
T(n,k) = A181695(n,k) - A181695(n-1,k). - Max Alekseyev, Nov 18 2010
Conjecture: Sum_{k=1..n} T(n,k) = A039671(n), n>0. - L. Edson Jeffery, Nov 29 2013

More terms from R. J. Mathar, Mar 09 2007

A172316 7th column of the array A172119.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 127, 252, 500, 992, 1968, 3904, 7744, 15361, 30470, 60440, 119888, 237808, 471712, 935680, 1855999, 3681528, 7302616, 14485344, 28732880, 56994048, 113052416, 224248833, 444816138, 882329660

Offset: 0

Richard Choulet, Jan 31 2010

			a(3) = binomial(3,3)*2^3 = 8.
a(7) = binomial(7,7)*2^7 - binomial(1,0)*2^0 = 127.

O. Dunkel, Solutions of a probability difference equation, Amer. Math. Monthly, 32 (1925), 354-370; see p. 356 with r = 6.
Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,0,0,-1)

Cf. A000071, A001949, A008937, A107066, A172119.

Partial sums of A001592.

for k from 0 to 20 do for n from 0 to 30 do b(n):=sum((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j),j=0..floor(n/(k+1))):od:k: seq(b(n),n=0..30):od; k:=6:taylor(1/(1-2*z+z^(k+1)),z=0,30);

G.f.: 1/(1 - 2*z + z^7).
Recurrence formula: a(n+7) = 2*a(n+6) - a(n).
a(n) = Sum_{j=0..floor(n/(k+1))} ((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j)) with k=6.

A172317 8th column of A172119.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 255, 508, 1012, 2016, 4016, 8000, 15936, 31744, 63233, 125958, 250904, 499792, 995568, 1983136, 3950336, 7868928, 15674623, 31223288, 62195672, 123891552, 246787536, 491591936, 979233536

Offset: 0

Richard Choulet, Jan 31 2010

			a(4) = binomial(4,4)*2^4 = 16.
a(9) = binomial(9,9)*2^9 - binomial(2,1)*2^1 = 512 - 4 = 508.

Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,0,0,0,-1).

Cf. A172316, A001949, A107066, A008937, A172119.

Partial sums of A066178.

k:=7:taylor(1/(1-2*z+z^(k+1)),z=0,30); for k from 0 to 20 do for n from 0 to 30 do b(n):=sum((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j),j=0..floor(n/(k+1))):od:k: seq(b(n),n=0..30):od;

The generating function is f such that: f(z)=1/(1-2*z+z^8). Recurrence relation: a(n+8)=2*a(n+7)-a(n). General term: a(n) = Sum_{j=0..floor(n/(k+1))} ((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j)) with k=7.

A209888 Number of binary words of length n containing no subword 01101.

Original entry on oeis.org

1, 2, 4, 8, 16, 31, 60, 116, 225, 436, 845, 1637, 3172, 6146, 11909, 23075, 44711, 86633, 167863, 325256, 630226, 1221144, 2366125, 4584673, 8883398, 17212733, 33351899, 64623621, 125216632, 242623433, 470114310, 910907331, 1765000872, 3419917668, 6626533192

Offset: 0

Alois P. Heinz, Mar 14 2012

nonn

Notice that the proper suffix 01 of 01101 is also a prefix of 01101. If instead of 01101 subword 01011 is not allowed, we get A107066 with A107066(n) < a(n) for all n >= 8. Word 01101101 of length 8 is the smallest binary word having two or more copies of 01101.

			a(6) = 60 because among the 2^6 = 64 binary words of length 6 only 4, namely 001101, 011010, 011011 and 101101 contain the subword 01101.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-1,2,-1).

Column 22 of A209972.
Column k=0 of A277751.
Cf. A107066, A317669.

a:= n-> (Matrix(5, (i, j)-> `if`(i=j-1, 1, `if`(i=5, [-1, 2, -1, 0, 2][j], 0)))^n. <<1, 2, 4, 8, 16>>)[1, 1]: seq(a(n), n=0..40);

CoefficientList[Series[(x + 1)*(x^2 - x + 1)/(x^5 - 2*x^4 + x^3 - 2*x + 1), {x, 0, 40}], x] (* Wesley Ivan Hurt, Apr 28 2017 *)
LinearRecurrence[{2,0,-1,2,-1},{1,2,4,8,16},40] (* Harvey P. Dale, Sep 17 2017 *)

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

a(n) = 2^n if n<5, and a(n) = 2*(a(n-1)+a(n-4)) -a(n-3) -a(n-5) otherwise.

cp's OEIS Frontend

A008937 a(n) = Sum_{k=0..n} T(k) where T(n) are the tribonacci numbers A000073.

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A140996 Triangle G(n, k) read by rows for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) + G(n+3, m) + G(n+4, m) for n >= 0 for m = 1..(n+1).

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A172119 Sum the k preceding elements in the same column and add 1 every time.

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A001949 Solutions of a fifth-order probability difference equation.

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A209972 Number of binary words of length n avoiding the subword given by the binary expansion of k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A126198 Triangle read by rows: T(n,k) (1 <= k <= n) = number of compositions of n into parts of size <= k.

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