A000071 -id:A000071 - cp's OEIS frontend

A384893 Triangle read by rows where T(n,k) is the number of subsets of {1..n} with k maximal anti-runs (increasing by more than 1).

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 7, 5, 2, 1, 1, 12, 10, 6, 2, 1, 1, 20, 20, 13, 7, 2, 1, 1, 33, 38, 29, 16, 8, 2, 1, 1, 54, 71, 60, 39, 19, 9, 2, 1, 1, 88, 130, 122, 86, 50, 22, 10, 2, 1, 1, 143, 235, 241, 187, 116, 62, 25, 11, 2, 1, 1, 232, 420, 468, 392, 267, 150, 75, 28, 12, 2, 1

Offset: 0

Gus Wiseman, Jun 21 2025

			The subset {3,6,7,9,11,12} has maximal anti-runs ((3,6),(7,9,11),(12)), so is counted under T(12,3).
The subset {3,6,7,9,10,12} has maximal anti-runs ((3,6),(7,9),(10,12)), so is counted under T(12,3).
Row n = 5 counts the following subsets:
  {}  {1}      {1,2}    {1,2,3}    {1,2,3,4}  {1,2,3,4,5}
      {2}      {2,3}    {2,3,4}    {2,3,4,5}
      {3}      {3,4}    {3,4,5}
      {4}      {4,5}    {1,2,3,5}
      {5}      {1,2,4}  {1,2,4,5}
      {1,3}    {1,2,5}  {1,3,4,5}
      {1,4}    {1,3,4}
      {1,5}    {1,4,5}
      {2,4}    {2,3,5}
      {2,5}    {2,4,5}
      {3,5}
      {1,3,5}
Triangle begins:
   1
   1   1
   1   2   1
   1   4   2   1
   1   7   5   2   1
   1  12  10   6   2   1
   1  20  20  13   7   2   1
   1  33  38  29  16   8   2   1
   1  54  71  60  39  19   9   2   1
   1  88 130 122  86  50  22  10   2   1
   1 143 235 241 187 116  62  25  11   2   1
   1 232 420 468 392 267 150  75  28  12   2   1
   1 376 744 894 806 588 363 188  89  31  13   2   1

Column k = 1 is A000071.
Row sums are A000079.
Column k = 2 is A001629.
For runs instead of anti-runs we have A034839, for strict partitions A116674.
The case containing n is A053538.
For integer partitions instead of subsets we have A268193, strict A384905.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.
Cf. A010027, A384177, A384879, A384889, A384890.

Table[Length[Select[Subsets[Range[n]],Length[Split[#,#2!=#1+1&]]==k&]],{n,0,10},{k,0,n}]

A204922 Ordered differences of Fibonacci numbers.

Original entry on oeis.org

1, 2, 1, 4, 3, 2, 7, 6, 5, 3, 12, 11, 10, 8, 5, 20, 19, 18, 16, 13, 8, 33, 32, 31, 29, 26, 21, 13, 54, 53, 52, 50, 47, 42, 34, 21, 88, 87, 86, 84, 81, 76, 68, 55, 34, 143, 142, 141, 139, 136, 131, 123, 110, 89, 55, 232, 231, 230, 228, 225, 220, 212, 199, 178

Offset: 1

Clark Kimberling, Jan 21 2012

For a guide to related sequences, see A204892. For numbers not in A204922, see A050939.
From Emanuele Munarini, Mar 29 2012: (Start)
Diagonal elements = Fibonacci numbers F(n+1) (A000045)
First column = Fibonacci numbers - 1 (A000071);
Second column = Fibonacci numbers - 2 (A001911);
Row sums = n*F(n+3) - F(n+2) + 2 (A014286);
Central coefficients = F(2*n+1) - F(n+1) (A096140).
(End)

			a(1) = s(2) - s(1) = F(3) - F(2) = 2-1 = 1, where F=A000045;
a(2) = s(3) - s(1) = F(4) - F(2) = 3-1 = 2;
a(3) = s(3) - s(2) = F(4) - F(3) = 3-2 = 1;
a(4) = s(4) - s(1) = F(5) - F(2) = 5-1 = 4.
From _Emanuele Munarini_, Mar 29 2012: (Start)
Triangle begins:
   1;
   2,  1;
   4,  3,  2;
   7,  6,  5,  3;
  12, 11, 10,  8,  5;
  20, 19, 18, 16, 13,  8;
  33, 32, 31, 29, 26, 21, 13;
  54, 53, 52, 50, 47, 42, 34, 21;
  88, 87, 86, 84, 81, 76, 68, 55, 34;
  ... (End)

G. C. Greubel, Rows n=1..100 of triangle, flattened

Cf. A204924, A204892.

/* As triangle */ [[Fibonacci(n+2)-Fibonacci(k+1) : k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Aug 04 2015

Mathematica
```
(See the program at A204924.)
```

create_list(fib(n+3)-fib(k+2),n,0,20,k,0,n); /* Emanuele Munarini, Mar 29 2012 */

{T(n,k) = fibonacci(n+2) - fibonacci(k+1)};
for(n=1,15, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 03 2019

[[fibonacci(n+2) - fibonacci(k+1) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Feb 03 2019

From Emanuele Munarini, Mar 29 2012: (Start)
T(n,k) = Fibonacci(n+2) - Fibonacci(k+1).
T(n,k) = Sum_{i=k..n} Fibonacci(i+1). (End)

A004070 Table of Whitney numbers W(n,k) read by antidiagonals, where W(n,k) is maximal number of pieces into which n-space is sliced by k hyperplanes, n >= 0, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 4, 4, 1, 1, 2, 4, 7, 5, 1, 1, 2, 4, 8, 11, 6, 1, 1, 2, 4, 8, 15, 16, 7, 1, 1, 2, 4, 8, 16, 26, 22, 8, 1, 1, 2, 4, 8, 16, 31, 42, 29, 9, 1, 1, 2, 4, 8, 16, 32, 57, 64, 37, 10, 1, 1, 2, 4, 8, 16, 32, 63, 99, 93, 46, 11, 1, 1, 2, 4, 8, 16, 32, 64, 120, 163

Offset: 0

N. J. A. Sloane

As a number triangle, this is given by T(n,k)=sum{j=0..n, C(n,j)(-1)^(n-j)sum{i=0..j, C(j+k,i-k)}}. - Paul Barry, Aug 23 2004
As a number triangle, this is the Riordan array (1/(1-x), x(1+x)) with T(n,k)=sum{i=0..n, binomial(k,i-k)}. Diagonal sums are then A023434(n+1). - Paul Barry, Feb 16 2005
Form partial sums across rows of square array of binomial coefficients A026729; see also A008949. - Philippe Deléham, Aug 28 2005
Square array A026729 -> Partial sums across rows
1 0 0 0 0 0 0 . . . . 1 1 1 1 1 1 1 . . . . . .
1 1 0 0 0 0 0 . . . . 1 2 2 2 2 2 2 . . . . . .
1 2 1 0 0 0 0 . . . . 1 3 4 4 4 4 4 . . . . . .
1 3 3 1 0 0 0 . . . . 1 4 7 8 8 8 8 . . . . . .
For other Whitney numbers see A007799.
W(n,k) is the number of length k binary sequences containing no more than n 1's. - Geoffrey Critzer, Mar 15 2010
From Emeric Deutsch, Jun 15 2010: (Start)
Viewed as a number triangle, T(n,k) is the number of internal nodes of the Fibonacci tree of order n+2 at level k. A Fibonacci tree of order n (n>=2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node.
(End)
Named after the American mathematician Hassler Whitney (1907-1989). - Amiram Eldar, Jun 13 2021

			Table W(n,k) begins:
  1 1 1 1  1  1  1 ...
  1 2 3 4  5  6  7 ...
  1 2 4 7 11 16 22 ...
  1 2 4 8 15 26 42 ...
W(2,4) = 11 because there are 11 length 4 binary sequences containing no more than 2 1's: {0, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}, {0, 0, 1, 1}, {0, 1, 0, 0}, {0, 1, 0, 1}, {0, 1, 1, 0}, {1, 0, 0, 0}, {1, 0, 0, 1}, {1, 0, 1, 0}, {1, 1, 0, 0}. - _Geoffrey Critzer_, Mar 15 2010
Table T(n, k) begins:
  1
  1  1
  1  2  1
  1  2  3  1
  1  2  4  4  1
  1  2  4  7  5  1
  1  2  4  8 11  6  1
...

Donald E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.

Gustav Burosch, Hans-Dietrich O.F. Gronau, Jean-Marie Laborde and Ingo Warnke, On posets of m-ary words, Discrete Math., Vol. 152, No. 1-3 (1996), pp. 69-91. MR1388633 (97e:06002)
Matteo Cervetti and Luca Ferrari, Pattern avoidance in the matching pattern poset, arXiv:2009.01024 [math.CO], 2020.
Matteo Cervetti and Luca Ferrari, Enumeration of Some Classes of Pattern Avoiding Matchings, with a Glimpse into the Matching Pattern Poset, Annals of Combinatorics (2022).
Richard K. Guy, Letter to N. J. A. Sloane, Apr 1975.
Yasuichi Horibe, An entropy view of Fibonacci trees, Fibonacci Quarterly, Vol. 20, No. 2 (1982), pp. 168-178.
Robin Pemantle and Mark C. Wilson, Twenty Combinatorial Examples of Asymptotics Derived from Multivariate Generating Functions, SIAM Rev., Vol. 50, No. 2 (2008), pp. 199-272. See p. 233.

Cf. A007799. As a triangle, mirror A052509.
Rows converge to powers of two (A000079). Subdiagonals include A000225, A000295, A002662, A002663, A002664, A035038, A035039, A035040, A035041, A035042. Antidiagonal sums are A000071.
Rows are: A000012, A000027, A000124, A000125, A000127, A006261, A008859, A008860, A008861, A008862, A008863.
Cf. A178522, A178524.

Transpose[ Table[Table[Sum[Binomial[n, k], {k, 0, m}], {m, 0, 15}], {n, 0, 15}]] // Grid (* Geoffrey Critzer, Mar 15 2010 *)
T[ n_, k_] := Sum[ Binomial[n, j] (-1)^(n - j) Sum[ Binomial[j + k, i - k], {i, 0, j}], {j, 0, n}]; (* Michael Somos, May 31 2016 *)

/* array read by antidiagonals up coordinate index functions */
t1(n) = binomial(floor(3/2 + sqrt(2+2*n)), 2) - (n+1); /* A025581 */
t2(n) = n - binomial(floor(1/2 + sqrt(2+2*n)), 2); /* A002262 */
/* define the sequence array function for A004070 */
W(n, k) = sum(i=0, n, binomial(k, i));
/* visual check ( origin 0,0 ) */
printp(matrix(7, 7, n, k, W(n-1, k-1)));
/* print the sequence entries by antidiagonals going up ( origin 0,0 ) */
print1("S A004070 "); for(n=0, 32, print1(W(t1(n), t2(n))","));
print1("T A004070 "); for(n=33, 61, print1(W(t1(n), t2(n))","));
print1("U A004070 "); for(n=62, 86, print1(W(t1(n), t2(n))",")); /* Michael Somos, Apr 28 2000 */

T(n, k)=sum(m=0, n-k, binomial(k, m)) \\ Jianing Song, May 30 2022

W(n, k) = Sum_{i=0..n} binomial(k, i). - Bill Gosper
W(n, k) = if k=0 or n=0 then 1 else W(n, k-1)+W(n-1, k-1). - David Broadhurst, Jan 05 2000
The table W(n,k) = A000012 * A007318(transform), where A000012 = (1; 1,1; 1,1,1; ...). - Gary W. Adamson, Nov 15 2007
E.g.f. for row n: (1 + x + x^2/2! + ... + x^n/n!)* exp(x). - Geoffrey Critzer, Mar 15 2010
G.f.: 1 / (1 - x - x*y*(1 - x^2)) = Sum_{0 <= k <= n} x^n * y^k * T(n, k). - Michael Somos, May 31 2016
W(n, n) = 2^n. - Michael Somos, May 31 2016
From Jianing Song, May 30 2022: (Start)
T(n, 0) = T(n, n) = 1 for n >= 0; T(n, k) = T(n-1, k-1) + T(n-2, k-1) for k=1, 2, ..., n-1, n >= 2.
T(n, k) = Sum_{m=0..n-k} binomial(k, m).
T(n,k) = 2^k for 0 <= k <= floor(n/2). (End)

More terms from Larry Reeves (larryr(AT)acm.org), Mar 20 2000

A048004 Triangular array read by rows: T(n,k) = number of binary vectors of length n whose longest run of consecutive 1's has length k, for n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 7, 5, 2, 1, 1, 12, 11, 5, 2, 1, 1, 20, 23, 12, 5, 2, 1, 1, 33, 47, 27, 12, 5, 2, 1, 1, 54, 94, 59, 28, 12, 5, 2, 1, 1, 88, 185, 127, 63, 28, 12, 5, 2, 1, 1, 143, 360, 269, 139, 64, 28, 12, 5, 2, 1, 1, 232, 694, 563, 303, 143, 64, 28, 12, 5, 2, 1, 1, 376, 1328, 1167, 653, 315, 144, 64, 28, 12, 5, 2, 1

Offset: 0

Clark Kimberling

Equivalently, number of compositions of n+1 having largest part (exactly) k+1. Example: T(4,2)=5 because we have 3+2, 2+3, 3+1+1, 1+3+1 and 1+1+3. - Emeric Deutsch, Apr 01 2005
Here is a bijection between the binary words and the compositions: prefix the vector with a 0, place a comma before each 0, then read the lengths of the runs. Example: 1100 -> 01100 -> ,011,0,0 -> 311 -> 3+1+1. - N. J. A. Sloane, Apr 03 2011
A formula based on the conjugates of the partitions of n with largest part k is given as a Sage program below. Note that it gives the compositions in the natural enumeration 'n with largest part k'. The 'conjugate' formula leads to A097805. - Peter Luschny, Jul 13 2015

			Triangle begins:
1;
1,  1;
1,  2,   1;
1,  4,   2,   1;
1,  7,   5,   2,  1;
1, 12,  11,   5,  2,  1;
1, 20,  23,  12,  5,  2,  1;
1, 33,  47,  27, 12,  5,  2, 1;
1, 54,  94,  59, 28, 12,  5, 2, 1;
1, 88, 185, 127, 63, 28, 12, 5, 2, 1;
...
Example: T(4,2) = 5 because we have 1100, 1101, 0110, 0011, 1011.

J. Kappraff, Beyond Measure, World Scientific, 2002; see pp. 471-472.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 155.

Alois P. Heinz, Rows n = 0..140, flattened
Richard Southern, Python program
J. L. Yucas, Counting special sets of binary Lyndon words, Ars Combin., 31 (1991), 21-29. (Annotated scanned copy)

See A126198 and A048887 for closely related arrays.
T(n,2) = Fibonacci(n+2) - 1, A000071, T(n,3) = b(n) for n=3, 4, ..., where b=A000100, T(n,4) = c(n) for n = 4, 5, ..., where c=A000102.
Nonnegative elements of columns approach A045623.
Cf. A048003, A097805.

tri n k | (k < 0) || (k > n) = 0
        | (k == 0) || (k == n) = 1
        | otherwise = 2*tri (n-1) k + tri (n-1) (k-1) - 2*tri (n-2) (k-1)
                            + tri (n-k-1) (k-1) - tri (n-k-2) k
-- Valentin Hübner, Jul 20 2017, after David W. Wilson

G:=k->(1-x)^2*x^k/(1-2*x+x^(k+1))/(1-2*x+x^(k+2)): for k from 0 to 14 do g[k]:=series(G(k),x=0,15) od: 1,seq(seq(coeff(g[k],x^n),k=0..n),n=1..12); # Emeric Deutsch, Apr 01 2005
# second Maple program:
B:= proc(n, k) option remember; `if`(n=0 or k=1, 1,
      add (B(n-j, k), j=1..min(n, k)))
    end:
T:= (n, k)-> B(n+1, k+1)-B(n+1, k):
seq(seq(T(n, k), k=0..n), n=0..14); # Alois P. Heinz, May 21 2013

nn=10;f[list_]:=Select[list,#>0&];Map[f,Transpose[Table[ CoefficientList[ Series[(1-x^k)/(1-2x+x^(k+1))-(1-x^(k-1))/ (1-2x+x^k),{x,0,nn}],x],{k,1,nn}]]]//Grid  (* Geoffrey Critzer, Jan 13 2013 *)
B[n_, k_] := B[n, k] = If[n==0 || k==1, 1, Sum[B[n-j, k], {j, 1, Min[n, k]} ]]; T[n_, k_] := B[n+1, k+1] - B[n+1, k]; Table[T[n, k], {n, 0, 14}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 01 2015, after Alois P. Heinz *)

Python
```
# See Richard Southern link.
```

# Computes the triangle obtained by augmenting T(n,k) by appending the column
# 1,0,0,0,... on the left. Illustrates a basic partition formula, is not
# efficient as a program for large n.
def A048004_row(n):
    r = []
    for k in (0..n):
        s = 0
        for p in Partitions(n, max_part=k, inner=[k]):
            q = p.conjugate()
            s += mul(binomial(q[j],q[j+1]) for j in range(len(q)-1))
        r.append(s)
    return r
[A048004_row(n) for n in (0..9)] # Peter Luschny, Jul 13 2015

T(n, k) = 0 if k < 0 or k > n, 1 if k = 0 or k = n, 2T(n-1, k) + T(n-1, k-1) - 2T(n-2, k-1) + T(n-k-1, k-1) - T(n-k-2, k) otherwise. - David W. Wilson
T(n, k) = A048887(n+1, k+1) - A048887(n+1, k). - Henry Bottomley, Oct 29 2002
G.f. for column k: (1-x)^2*x^k/((1-2*x+x^(k+1))*(1-2*x+x^(k+2))). - Emeric Deutsch, Apr 01 2005
From Gary W. Adamson, Jun 23 2012: (Start)
Create an array of rows such that row 0 is the INVERT transform of (1,0,0,0,...); row 1 is the INVERT transform of (1,1,0,0,0,...); row 2 is the INVERT transform of (1,1,1,0,0,0,...) and so on:
  1, 1, 1, 1,  1,  1, ...
  1, 2, 3, 5,  8, 13, ...
  1, 2, 4, 7, 13, 24, ...
  1, 2, 4, 8, 15, 29, ...
  ... Then, take finite differences of column terms from the top -> down. Row terms of the triangle are finite differences of the array columns. (End)
T(n,k) = A126198(n+1,k+1) - A126198(n+1,k). - L. Edson Jeffery, May 21 2013
Recurrence: T(n+1,k) = Sum_{h=0..k} T(n-k, h) + Sum_{i=n-k+1..n} T(i, k); for example, T(7,3) = Sum_{h=0..3} T(3,h) + Sum_{i=4..6} T(i,3) or T(7,3) = (1+4+2+1) + (2+5+12) = 27. Example: T(4,2) = (1+1) + (1+2) = 5. - Richard Southern, Jul 09 2017
Difference between higher order Fibonacci numbers is equal to recurrence. T(n+1,k) = A126198 (n+1,k) - A126198 (n+1,k-1) = Sum_{i=n-k..n} A126198 (i,k) - Sum_{i=n-k+1..n} A126198 (i,k-1) = A126198 (n-k,k) + Sum_{i=n-k+1..n} (A126198 (i,k) - A126198 (i,k-1)) = Sum_{h=0..k} T(n-k, h) + Sum_{i=n-k+1..n} T(i, k). For example T(7,3) = A126198 (7,3) - A126198 (7,2) = 108 - 81 = (8+15+29+56) - (13+24+44) = 8 + (15-13) + (29-24) + (56-44) = 8 + (2+5+12) = (1+4+2+1) + (2+5+12). - Richard Southern, Aug 04 2017

More terms from Emeric Deutsch, Apr 01 2005

Edited by N. J. A. Sloane, Apr 03 2011

A052509 Knights-move Pascal triangle: T(n,k), n >= 0, 0 <= k <= n; T(n,0) = T(n,n) = 1, T(n,k) = T(n-1,k) + T(n-2,k-1) for k = 1,2,...,n-1, n >= 2.

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, 11, 8, 4, 2, 1, 1, 7, 16, 15, 8, 4, 2, 1, 1, 8, 22, 26, 16, 8, 4, 2, 1, 1, 9, 29, 42, 31, 16, 8, 4, 2, 1, 1, 10, 37, 64, 57, 32, 16, 8, 4, 2, 1, 1, 11, 46, 93, 99, 63, 32, 16, 8, 4, 2, 1

Offset: 0

N. J. A. Sloane, Mar 17 2000

Also square array T(n,k) (n >= 0, k >= 0) read by antidiagonals: T(n,k) = Sum_{i=0..k} binomial(n,i).
As a number triangle read by rows, this is T(n,k) = Sum_{i=n-2*k..n-k} binomial(n-k,i), with T(n,k) = T(n-1,k) + T(n-2,k-1). Row sums are A000071(n+2). Diagonal sums are A023435(n+1). It is the reverse of the Whitney triangle A004070. - Paul Barry, Sep 04 2005
Also, twice number of orthants intersected by a generic k-dimensional subspace of R^n [Naiman and Scheinerman, 2017]. - N. J. A. Sloane, Mar 03 2018

			Triangle begins:
[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, 11,  8,  4,  2, 1;
[7] 1, 7, 16, 15,  8,  4, 2, 1;
[8] 1, 8, 22, 26, 16,  8, 4, 2, 1;
[9] 1, 9, 29, 42, 31, 16, 8, 4, 2, 1;
As a square array, this begins:
  1  1  1  1  1  1 ...
  1  2  2  2  2  2 ...
  1  3  4  4  4  4 ...
  1  4  7  8  8  8 ...
  1  5 11 15 16 ...
  1  6 16 26 31 32 ...

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Clark Kimberling, Path-counting and Fibonacci numbers, Fib. Quart. 40 (4) (2002) 328-338, Example 1C.
Daniel Q. Naiman and Edward R. Scheinerman, Arbitrage and Geometry, arXiv:1709.07446 [q-fin.MF], 2017 [Contains the square array multiplied by 2].
Richard L. Ollerton and Anthony G. Shannon, Some properties of generalized Pascal squares and triangles, Fib. Q., 36 (1998), 98-109. See Tables 5 and 14.
D. J. Price, Some unusual series occurring in n-dimensional geometry, Math. Gaz., 30 (1946), 149-150.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A054123, A054124, A007318, A008949.
Row sums A000071; Diagonal sums A023435; Mirror A004070.
Columns give A000027, A000124, A000125, A000127, A006261, ...
Cf. A052509, A054123, A054124, A007318, A008949, A052553.
Partial sums across rows of (extended) Pascal's triangle A052553.

A052509:=Flat(List([0..100],n->List([0..n],k->Sum([0..n],m->Binomial(n-k,k-m))))); # Muniru A Asiru, Sat Feb 17 2018

a052509 n k = a052509_tabl !! n !! k
a052509_row n = a052509_tabl !! n
a052509_tabl = [1] : [1,1] : f [1] [1,1] where
   f row' row = rs : f row rs where
     rs = zipWith (+) ([0] ++ row' ++ [1]) (row ++ [0])
-- Reinhard Zumkeller, Nov 22 2012

[[(&+[Binomial(n-k, k-j): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, May 13 2019

a := proc(n::nonnegint, k::nonnegint) option remember: if k=0 then RETURN(1) fi: if k=n then RETURN(1) fi: a(n-1,k)+a(n-2,k-1) end: for n from 0 to 11 do for k from 0 to n do printf(`%d,`,a(n,k)) od: od: # James Sellers, Mar 17 2000
with(combinat): for s from 0 to 11 do for n from s to 0 by -1 do if n=0 or s-n=0 then printf(`%d,`,1) else printf(`%d,`,sum(binomial(n, i), i=0..s-n)) fi; od: od: # James Sellers, Mar 17 2000

Table[Sum[Binomial[n-k, k-m], {m, 0, n}], {n, 0, 10}, {k, 0, n}]
T[n_, k_] := Hypergeometric2F1[-k, -n + k, -k, -1];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Peter Luschny, Nov 28 2021 *)

T(n,k)=sum(m=0,n,binomial(n-k,k-m));
for(n=0,10,for(k=0,n,print1(T(n,k),", "););print();); /* show triangle */

[[sum(binomial(n-k, k-j) for j in (0..n)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, May 13 2019

T(n, k) = Sum_{m=0..n} binomial(n-k, k-m). - Wouter Meeussen, Oct 03 2002
From Werner Schulte, Feb 15 2018: (Start)
Referring to the square array T(i,j):
G.f. of row n: Sum_{k>=0} T(n,k) * x^k = (1+x)^n / (1-x).
G.f. of T(i,j): Sum_{k>=0, n>=0} T(n,k) * x^k * y^n = 1 / ((1-x)*(1-y-x*y)).
Let a_i(n) be multiplicative with a_i(p^e) = T(i, e), p prime and e >= 0, then Sum_{n>0} a_i(n)/n^s = (zeta(s))^(i+1) / (zeta(2*s))^i for i >= 0.
(End)
T(n, k) = hypergeom([-k, -n + k], [-k], -1). - Peter Luschny, Nov 28 2021
From Jianing Song, May 30 2022: (Start)
Referring to the triangle, G.f.: Sum_{n>=0, 0<=k<=n} T(n,k) * x^n * y^k = 1 / ((1-x*y)*(1-x-x^2*y)).
T(n,k) = 2^(n-k) for ceiling(n/2) <= k <= n. (End)

More terms from James Sellers, Mar 17 2000
Entry formed by merging two earlier entries. - N. J. A. Sloane, Jun 17 2007
Edited by Johannes W. Meijer, Jul 24 2011

A140993 Triangle G(n, k) read by rows, for 1 <= k <= n, where G(n, n) = G(n+1, 1) = 1, G(n+2, 2) = 2, G(n+3, m) = G(n+1, m-1) + G(n+1, m-2) + G(n+2, m-1) for n >= 1 and m = 3..(n+2).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 2, 5, 7, 1, 1, 2, 5, 11, 12, 1, 1, 2, 5, 12, 23, 20, 1, 1, 2, 5, 12, 28, 46, 33, 1, 1, 2, 5, 12, 29, 63, 89, 54, 1, 1, 2, 5, 12, 29, 69, 137, 168, 88, 1, 1, 2, 5, 12, 29, 70, 161, 289, 311, 143, 1, 1, 2, 5, 12, 29, 70, 168, 367, 594, 567, 232, 1, 1, 2, 5, 12, 29, 70, 169, 399, 817, 1194, 1021, 376, 1

Offset: 1

Juri-Stepan Gerasimov, Jul 08 2008

From Petros Hadjicostas, Jun 10 2019: (Start)
Let b(m) = lim_{n -> infinity} G(n, m) for each m >= 1. Then b(1) = 1, b(2) = 2, and b(m) = 2*b(m-1) + b(m-2) for m >= 3. (The existence of the limit can be proved by induction on m.) This means b(m) = A000129(m) for m >= 1 (known as the Pell numbers).
If we want to get the second main diagonal, we let c(n) = G(n+1, n) for n >= 1. Then c(n+2) = G(n+3, n+2) = G(n+1, n+1) + G(n+1, n) + G(n+2, n+1) = 1 + c(n) + c(n+1) with c(1) = G(2, 1) = 1 and c(2) = G(3, 2) = 2, which implies that c(n) = A000071(n+2) = Fibonacci(n+2) - 1 for n >= 1.
This array is the mirror image of A140998 (except for a shifting of the indices by 1). Thus, G(n, k) = A140998(n - 1, n - k) for 1 <= k <= n. This array has index of obliqueness e = 1, while array A140998 has index of obliqueness e = 0. Both arrays have the same index of asymmetry (s = 1). (End)
From Petros Hadjicostas, Feb 09 2021: (Start)
One of the two rectangular versions, say (RA(n,k): n,k >= 0), of this triangular array (G(n,k): 1 <= k <= n) is given by RA(n,k) = G(n+k-1,k) for n,k >= 1. Conversely, G(n,k) = RA(n-k+1, k) for 1 <= k <= n. (This assumes that the triangle G(n,k) is read from the array RA(n,k) by ascending antidiagonals.)
Note that [o.g.f of RA](x,y) = x*[o.g.f. of G](x, y/x) and [o.g.f of G](x,y) = x^(-1)*[o.g.f of RA](x,x*y).
The other rectangular version, say (RD(n,k): n,k >= 0), of this triangular array (G(n,k): 1 <= k <= n) is given by RD(n,k) = RA(k,n) = G(n+k-1,n) for n,k >= 1. Conversely, G(n,k) = RD(k,n-k+1) for 1 <= k <= n. (This assumes that the triangle G(n,k) is read from the array RD(n,k) by descending antidiagonals.)
Note that [o.g.f of RD](x,y) = y*[o.g.f. of G](y,x/y) and [o.g.f of G](x,y) = x^(-1)*[o.g.f of RD](x*y, x). (End)

			Triangle G(n,k) (with rows for n >= 1 and columns for 1 <= k <= n) begins:
  1
  1 1
  1 2 1
  1 2 4  1
  1 2 5  7  1
  1 2 5 11 12  1
  1 2 5 12 23 20   1
  1 2 5 12 28 46  33   1
  1 2 5 12 29 63  89  54   1
  1 2 5 12 29 69 137 168  88    1
  1 2 5 12 29 70 161 289 311  143    1
  1 2 5 12 29 70 168 367 594  567  232    1
  1 2 5 12 29 70 169 399 817 1194 1021  376   1
  1 2 5 12 29 70 169 407 934 1778 2355 1820 609 1
  ...
From _Petros Hadjicostas_, Feb 09 2021: (Start)
Rectangular array RA(n,k) (with rows for n >= 1 and columns for k >= 1) begins:
  1, 1, 1,  1,  1,  1,   1,   1,   1,    1, ...
  1, 2, 4,  7, 12, 20,  33,  54,  88,  143, ...
  1, 2, 5, 11, 23, 46,  89, 168, 311,  567, ...
  1, 2, 5, 12, 28, 63, 137, 289, 594, 1194, ...
  1, 2, 5, 12, 29, 69, 161, 367, 817, 1778, ...
  1, 2, 5, 12, 29, 70, 168, 399, 934, 2150, ...
  1, 2, 5, 12, 29, 70, 169, 407, 975, 2316, ...
  1, 2, 5, 12, 29, 70, 169, 408, 984, 2367, ...
  1, 2, 5, 12, 29, 70, 169, 408, 985, 2377, ...
  1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, ...
  ...
Reading the array RA(n,k) by ascending antidiagonals, we get triangle G(n,k) above. (End)

Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A000071, A000079, A000129, A007318, A140994, A140995, A140996, A140997, A140998, A141020, A141021.

A140993 := proc(n,k) if k = n then 1; elif k = 1 then 1; elif k = 2 then 2; else procname(n-2,k-1)+procname(n-2,k-2)+procname(n-1,k-1) ; end if; end proc: seq(seq(A140993(n,k),k=1..n),n=1..15) ; # R. J. Mathar, Apr 28 2010

t[n_, k_] := If[k == n, 1, If[k == 1, 1, If[k == 2, 2, t[n - 2, k - 1] + t[n - 2, k - 2] + t[n - 1, k - 1]]]]; Flatten[Table[ t[n, k], {n, 13}, {k, n}]] (* Robert G. Wilson v, Dec 22 2011 *)

From Petros Hadjicostas, Jun 10 2019: (Start)
G(n, k) = A140998(n - 1, n - k) for 1 <= k <= n.
Bivariate o.g.f.: Sum_{n >= 1, k >= 1} G(n, k)*x^n*y^k = x*y*(1 - x*y -x^2*y^2 + x^3*y^2)/((1 - x) * (1 - x*y) * (1 - x*y - x^2*y - x^2*y^2)). (Here, we let G(n, k) = 0 when 1 <= n < k.)
To get the row sums, we let y = 1 in the above bivariate g.f. and simplify. We get x/(1 - 2*x), which is the g.f. of sequence (A000079(n-1): n >= 1) = (2^(n-1): n >= 1). (End)
From Petros Hadjicostas, Feb 10 2021: (Start)
We give formulas about the rectangular array RA(n,k).
Initial conditions: RA(1,n) = RA(n+1,1) = 1 and RA(n+1,2) = 2 for n >= 1.
Recurrence: RA(n,k) = RA(n-1,k-1) + RA(n,k-2) + RA(n,k-1) for n >= 2 and k >= 3.
The main diagonal of the array is RA(n,n) = A000129(n) (Pell numbers).
Bivariate o.g.f: Sum_{n,k >= 1} RA(n,m)*x^n*y^k = x*y*(x*y^2 - y^2 - y + 1)/((1 - x)*(1 - y)*(-x*y - y^2 - y + 1)).
To obtain formulas about the other rectangular array, RD(n,k), we use the equations RD(n,k) = RA(k,n) for n,k >= 1 and [o.g.f. of RD](x,y) = [o.g.f. of RA](y,x). (End)

Entries checked by R. J. Mathar, Apr 28 2010

Name and offset edited by Petros Hadjicostas, Jun 10 2019

A140998 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, and G(n+3, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) for n >= 0 and m = 1..n+1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 7, 5, 2, 1, 1, 12, 11, 5, 2, 1, 1, 20, 23, 12, 5, 2, 1, 1, 33, 46, 28, 12, 5, 2, 1, 1, 54, 89, 63, 29, 12, 5, 2, 1, 1, 88, 168, 137, 69, 29, 12, 5, 2, 1, 1, 143, 311, 289, 161, 70, 29, 12, 5, 2, 1, 1, 232, 567, 594, 367, 168, 70, 29, 12, 5, 2, 1

Offset: 0

Juri-Stepan Gerasimov, Jul 08 2008

From Petros Hadjicostas, Jun 10 2019: (Start)
According to the attached picture, the index of asymmetry here is s = 1 and the index of obliqueness (or obliquity) is e = 0.
In the picture, the equation G(n, e*n) = 1 becomes G(n, 0) = 1, while the equations G(n+x+1, n-e*n+e*x-e+1) = 2^x for 0 <= x < s = 1 become G(n+1, n+1) = 1 and G(n+2, n+1) = 2.
Also, in the picture, the recurrence G(n+s+2, k) = G(n+1, k-e*s+e-1) + Sum_{m=1..s+1} G(n+m, k-e*s+m*e-2*e) for k = 1..n+1 becomes G(n+3, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) for k = 1..n+1.
Except for a shifting of the indices by 1, this array is a mirror image of array A140993. We have G(n, k) = A140993(n+1, n-k+1) for 0 <= k <= n. Triangular array A140993 has the same index of asymmetry (i.e., s = 1) but index of obliqueness e = 1.
(End)

			Triangle begins (with rows for n >= 0 and columns for k >= 0):
  1;
  1,   1;
  1,   2,   1;
  1,   4,   2,   1;
  1,   7,   5,   2,   1;
  1,  12,  11,   5,   2,   1;
  1,  20,  23,  12,   5,   2,   1;
  1,  33,  46,  28,  12,   5,   2,   1;
  1,  54,  89,  63,  29,  12,   5,   2,   1;
  1,  88, 168, 137,  69,  29,  12,   5,   2,   1;
  1, 143, 311, 289, 161,  70,  29,  12,   5,   2,   1;

G. C. Greubel, Rows n = 0..100 of triangle, flatten
Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A000071, A007318, A140992, A140993, A140994, A140995, A140996, A140997, A141020, A141021.

G[n_,k_] := G[n,k] = Which[k==0 || k==n, 1, k==n-1, 2, True, G[n-2,k-1] + G[n-2,k] + G[n-1,k]]; Table[G[n,k], {n,0,12}, {k,0,n}] (* Jean-François Alcover, Jun 09 2019 *)

G(n,k) = if(k==0 || k==n, 1, if(k==n-1, 2, G(n-1, k) + G(n-2, k) + G(n-2, k-1)));
for(n=0,12, for(k=0,n, print1(G(n,k), ", "))) \\ G. C. Greubel, Jun 09 2019

def G(n,k):
    if (k==0 or k==n): return 1
    elif (k==n-1): return 2
    else: return G(n-1, k) + G(n-2, k) + G(n-2, k-1)
[[G(n,k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jun 09 2019

From Petros Hadjicostas, Jun 10 2019: (Start)
G(n, k) = A140993(n+1, n-k+1) for 0 <= k <= n.
Let A(x,y) = Sum_{n,k >= 0} G(n, k)*x^n*y^k and B(x,y) = Sum_{n,k >= 1} A140993(n, k). Then A(x, y) = x^(-1) * B(x*y, y^(-1)). Thus, the g.f. of the current array is A(x, y) = (1 - x - x^2 + x^3*y)/((1 - x) * (1 - x*y) * (1 - x - x^2 - x^2*y)).
To find the g.f. of the k-th column (where k >= 0), we differentiate A(x, y) k times with respect to y, divide by k!, and substitute y = 0. For example, differentiating A(x, y) once w.r.t. y and setting y = 0, we get the g.f. of the k = 1 column: x/((1 - x)*(1 - x - x^2)). This is the g.f. of sequence (A000071(n+2): n >= 0) = (Fibonacci(n+2) - 1: n >= 0).
G.f. of column k = 2 is x^2*(1 - x + x^3)/((1 - x)*(1 - x - x^2)^2). Thus, column k = 2 is a shifted version of (A140992(n): n >= 0).
(End)

Indices in the definition corrected by R. J. Mathar, Aug 02 2009

Name edited by Petros Hadjicostas, Jun 10 2019

A006327 a(n) = Fibonacci(n) - 3. Number of total preorders.

Original entry on oeis.org

0, 2, 5, 10, 18, 31, 52, 86, 141, 230, 374, 607, 984, 1594, 2581, 4178, 6762, 10943, 17708, 28654, 46365, 75022, 121390, 196415, 317808, 514226, 832037, 1346266, 2178306, 3524575, 5702884, 9227462, 14930349, 24157814, 39088166, 63245983, 102334152, 165580138

Offset: 4

N. J. A. Sloane, Simon Plouffe

Minimal cost of maximum height Huffman tree of size n. - Alex Vinokur (alexvn(AT)barak-online.net), Oct 25 2004

			G.f. = 2*x^5 + 5*x^6 + 10*x^7 + 18*x^8 + 31*x^9 + 52*x^10 + 86*x^11 + ...

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

T. D. Noe, Table of n, a(n) for n = 4..1000
G. Kreweras, Les préordres totaux compatibles avec un ordre partiel, Math. Sci. Humaines No. 53 (1976), 5-30.
G. Kreweras, Les préordres totaux compatibles avec un ordre partiel, Math. Sci. Humaines No. 53 (1976), 5-30. (Annotated scanned copy)
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
A. Sapounakis, I. Tasoulas and P. Tsikouras, On the Dominance Partial Ordering of Dyck Paths, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.5.
A. B. Vinokur, Huffman trees and Fibonacci numbers, Kibernetika Issue 6 (1986) 9-12 (in Russian); English translation in Cybernetics 21, Issue 6 (1986), 692-696.
Alex Vinokur, Fibonacci connection between Huffman codes and Wythoff array, arXiv:cs/0410013 [cs.DM], 2004-2005.
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

A diagonal of A079502.

Cf. A000045, A001611, A000071, A157725, A001911, A157726, A006327, A157727, A157728, A157729, A167616. [Added by N. J. A. Sloane, Jun 25 2010 in response to a comment from Aviezri S. Fraenkel]

List([4..45], n-> Fibonacci(n)-3) # G. C. Greubel, Jul 13 2019

[Fibonacci(n)-3: n in [4..45]]; // G. C. Greubel, Jul 13 2019

with(combinat):a:=n->sum(fibonacci(j),j=3..n): seq(a(n),n=2..40); # Zerinvary Lajos, Oct 03 2007
A006327:=(2+z)/(z-1)/(z**2+z-1); # conjectured by Simon Plouffe in his 1992 dissertation

Fibonacci[Range[4, 45]] - 3 (* Vladimir Joseph Stephan Orlovsky, Mar 19 2010 *)

a(n)=fibonacci(n)-3 \\ Charles R Greathouse IV, Feb 03 2014

[fibonacci(n)-3 for n in (4..45)] # G. C. Greubel, Jul 13 2019

G.f.: x^5*(2 + x)/((1-x)*(1-x-x^2)).
a(n) = a(n-1) + a(n-2) + 3.
a(n+3) = Sum_{k=-n+1..n} F(abs(n)+1). - Paul Barry, Oct 24 2007
a(n) = F(4*n) mod F(n+1) = F(n) - (F(n+4)^2 - F(n)^2)/F(2*n+4). - Gary Detlefs, Apr 02 2012

Offset corrected by Gary Detlefs, Apr 02 2012

A056830 Alternate digits 1 and 0.

Original entry on oeis.org

0, 1, 10, 101, 1010, 10101, 101010, 1010101, 10101010, 101010101, 1010101010, 10101010101, 101010101010, 1010101010101, 10101010101010, 101010101010101, 1010101010101010, 10101010101010101, 101010101010101010

Offset: 0

Henry Bottomley, Aug 30 2000

Fibonacci bit-representations of numbers for which there is only one possible representation and for which the maximal and minimal bit-representations (A104326 and A014417) are equal. The numbers represented are equal to the numbers in A000071 (subtract the first term of that sequence). For example, 10101 = 12 because 8+5+1. - Casey Mongoven, Mar 19 2006
Sequence A000975 written in base 2. - Jaroslav Krizek, Aug 05 2009
The absolute value of alternating sum of the first n repunits: a(n) = abs(Sum_{k=0..n} (-1)^k*A002275(n)). - Ilya Gutkovskiy, Dec 02 2015
Binary representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - Robert Price, Dec 05 2016

			n  a(n)             A000975(n)   (If a(n) is interpreted in base 2.)
------------------------------
0  0 ....................... 0
1  1 ....................... 1
2  10 ...................... 2 = 2^1
3  101 ..................... 5
4  1010 ................... 10 = 2^1 + 2^3
5  10101 .................. 21
6  101010 ................. 42 = 2^1 + 2^3 + 2^5
7  1010101 ................ 85
8  10101010 .............. 170 = 2^1 + 2^3 + 2^5 + 2^7
9  101010101 ............. 341
10 1010101010 ............ 682 = 2^1 + 2^3 + 2^5 + 2^7 + 2^9
11 10101010101 .......... 1365
12 101010101010 ......... 2730 = 2^1 + 2^3 + 2^5 + 2^7 + 2^9 + 2^11, etc.
- _Bruno Berselli_, Dec 02 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (10,1,-10).
Index entries for 2-automatic sequences.

Partial sums of A015585.

Cf. A000975, A033113, A033114, A033115, A033116, A033117, A033118, A033119, A059848, A062864.

List([0..30], n-> Int(10^(n+1)/99) ); # G. C. Greubel, Jul 14 2019

[Round((20*10^n-11)/198) : n in [0..30]]; // Vincenzo Librandi, Jun 25 2011

A056830 := proc(n) floor(10^(n+1)/99) ; end proc:

CoefficientList[Series[x/((1-x^2)*(1-10*x)), {x,0,30}], x] (* G. C. Greubel, Sep 26 2017 *)

Vec(x/((1-x)*(1+x)*(1-10*x))+O(x^30)) \\ Charles R Greathouse IV, Feb 13 2017

[floor(10^(n+1)/99) for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = +10*a(n-1) + a(n-2) - 10*a(n-3).
a(n) = floor(10^(n+1)/(10^2-1)) = a(n-2)+10^(n-1) = 10*a(n-1) + (1 - (-1)^n)/2.
From Paul Barry, Nov 12 2003: (Start)
a(n+1) = Sum_{k=0..floor(n/2)} 10^(n-2*k).
a(n+1) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(j+k)*10^j.
G.f.: x/((1-x)*(1+x)*(1-10*x)).
a(n) = 9*a(n-1) + 10*a(n-2) + 1.
a(n) = 10^(n+1)/99 - (-1)^n/22 - 1/18. (End)
a(n) = A007088(A107909(A104161(n))) = A007088(A000975(n)). - Reinhard Zumkeller, May 28 2005
a(n) = round((20*10^n-11)/198) = floor((10*10^n-1)/99) = ceiling((10*10^n-10)/99) = round((10*10^n-10)/99). - Mircea Merca, Dec 27 2010
From Daniel Forgues, Sep 20 2018: (Start)
If a(n) is interpreted in base 2:
a(2n) = Sum_{k=1..n} 2^(2n-1), n >= 0; a(2n-1) = a(2n)/2, n >= 1.
a(2n) = A020988(n), n >= 0.
a(0) = 0; a(2n) = 4*a(2n-2) + 2, n >= 1. (End)

More terms from Casey Mongoven, Mar 19 2006

A083047 Square table read by antidiagonals forms a permutation of the natural numbers: T(n,0) = floor(nx/(x-1))+1, T(n,k+1) = ceiling(xT(n,k)), where x = (sqrt(5)+1)/2, n>=0, k>=0.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 10, 8, 12, 15, 17, 13, 11, 20, 25, 28, 22, 18, 14, 33, 41, 46, 36, 30, 23, 16, 54, 67, 75, 59, 49, 38, 26, 19, 88, 109, 122, 96, 80, 62, 43, 31, 21, 143, 177, 198, 156, 130, 101, 70, 51, 34, 24, 232, 287, 321, 253, 211, 164, 114, 83, 56, 39, 27, 376

Offset: 0

Paul D. Hanna, Apr 18 2003

First row is A000071 offset by 2, first column is A026352, main diagonal is A083048, antidiagonal sums give A083049.

A083047 is an interspersion (hence a dispersion), with fractal sequence A167198. See A167198 for a construction of A083047 that does not refer to (1+sqrt(5))/2. - Clark Kimberling, Oct 30 2009

			Table begins:
   1  2  4   7  12  20  33  54   88  143  232  376 ...
   3  5  9  15  25  41  67 109  177  287  465  753 ...
   6 10 17  28  46  75 122 198  321  520  842 1363 ...
   8 13 22  36  59  96 156 253  410  664 1075 1740 ...
  11 18 30  49  80 130 211 342  554  897 1452 2350 ...
  14 23 38  62 101 164 266 431  698 1130 1829 2960 ...
  16 26 43  70 114 185 300 486  787 1274 2062 3337 ...
  19 31 51  83 135 219 355 575  931 1507 2439 3947 ...
  21 34 56  91 148 240 389 630 1020 1651 2672 4324 ...
  24 39 64 104 169 274 444 719 1164 1884 3049 4934 ...
  27 44 72 117 190 308 499 808 1308 2117 3426 5544 ...

Cf. A026352, A083048, A083049, A083044, A083050.

Cf. A167198. - Clark Kimberling, Oct 30 2009

t[n_, 0] = Floor[n*GoldenRatio/(GoldenRatio - 1) + 1];
t[n_, k_] := t[n, k] = Ceiling[GoldenRatio*t[n, k-1]];
Flatten[Table[t[k-1, n-k ], {n, 12}, {k, n}] ][[;; 67]]
(* Jean-François Alcover, Jul 13 2011 *)

cp's OEIS Frontend

A384893 Triangle read by rows where T(n,k) is the number of subsets of {1..n} with k maximal anti-runs (increasing by more than 1).

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

A204922 Ordered differences of Fibonacci numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

PARI

Sage

Formula

A004070 Table of Whitney numbers W(n,k) read by antidiagonals, where W(n,k) is maximal number of pieces into which n-space is sliced by k hyperplanes, n >= 0, k >= 0.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

Extensions

A048004 Triangular array read by rows: T(n,k) = number of binary vectors of length n whose longest run of consecutive 1's has length k, for n >= 0, 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Python

Sage

Formula

Extensions

A052509 Knights-move Pascal triangle: T(n,k), n >= 0, 0 <= k <= n; T(n,0) = T(n,n) = 1, T(n,k) = T(n-1,k) + T(n-2,k-1) for k = 1,2,...,n-1, n >= 2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A140993 Triangle G(n, k) read by rows, for 1 <= k <= n, where G(n, n) = G(n+1, 1) = 1, G(n+2, 2) = 2, G(n+3, m) = G(n+1, m-1) + G(n+1, m-2) + G(n+2, m-1) for n >= 1 and m = 3..(n+2).

Views

Author

Keywords

Comments

Examples

A083047 Square table read by antidiagonals forms a permutation of the natural numbers: T(n,0) = floor(nx/(x-1))+1, T(n,k+1) = ceiling(xT(n,k)), where x = (sqrt(5)+1)/2, n>=0, k>=0.