cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A047999 Sierpiński's [Sierpinski's] triangle (or gasket): triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 2.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Restored the alternative spelling of Sierpinski to facilitate searching for this triangle using regular-expression matching commands in ASCII. - N. J. A. Sloane, Jan 18 2016
Also triangle giving successive states of cellular automaton generated by "Rule 60" and "Rule 102". - Hans Havermann, May 26 2002
Also triangle formed by reading triangle of Eulerian numbers (A008292) mod 2. - Philippe Deléham, Oct 02 2003
Self-inverse when regarded as an infinite lower triangular matrix over GF(2).
Start with [1], repeatedly apply the map 0 -> [00/00], 1 -> [10/11] [Allouche and Berthe]
Also triangle formed by reading triangles A011117, A028338, A039757, A059438, A085881, A086646, A086872, A087903, A104219 mod 2. - Philippe Deléham, Jun 18 2005
J. H. Conway writes (in Math Forum): at least the first 31 rows give odd-sided constructible polygons (sides 1, 3, 5, 15, 17, ... see A001317). The 1's form a Sierpiński sieve. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 19 2005
When regarded as an infinite lower triangular matrix, its inverse is a (-1,0,1)-matrix with zeros undisturbed and the nonzero entries in every column form the Prouhet-Thue-Morse sequence (1,-1,-1,1,-1,1,1,-1,...) A010060 (up to relabeling). - David Callan, Oct 27 2006
Triangle read by rows: antidiagonals of an array formed by successive iterates of running sums mod 2, beginning with (1, 1, 1, ...). - Gary W. Adamson, Jul 10 2008
T(n,k) = A057427(A143333(n,k)). - Reinhard Zumkeller, Oct 24 2010
The triangle sums, see A180662 for their definitions, link Sierpiński’s triangle A047999 with seven sequences, see the crossrefs. The Kn1y(n) and Kn2y(n), y >= 1, triangle sums lead to the Sierpiński-Stern triangle A191372. - Johannes W. Meijer, Jun 05 2011
Used to compute the total Steifel-Whitney cohomology class of the Real Projective space. This was an essential component of the proof that there are no product operations without zero divisors on R^n for n not equal to 1, 2, 4 or 8 (real numbers, complex numbers, quaternions, Cayley numbers), proved by Bott and Milnor. - Marcus Jaiclin, Feb 07 2012
T(n,k) = A134636(n,k) mod 2. - Reinhard Zumkeller, Nov 23 2012
T(n,k) = 1 - A219463(n,k), 0 <= k <= n. - Reinhard Zumkeller, Nov 30 2012
From Vladimir Shevelev, Dec 31 2013: (Start)
Also table of coefficients of polynomials s_n(x) of degree n which are defined by formula s_n(x) = Sum_{i=0..n} (binomial(n,i) mod 2)*x^k. These polynomials we naturally call Sierpiński's polynomials. They also are defined by the recursion: s_0(x)=1, s_(2*n+1)(x) = (x+1)*s_n(x^2), n>=0, and s_(2*n)(x) = s_n(x^2), n>=1.
Note that: s_n(1) = A001316(n),
s_n(2) = A001317(n),
s_n(3) = A100307(n),
s_n(4) = A001317(2*n),
s_n(5) = A100308(n),
s_n(6) = A100309(n),
s_n(7) = A100310(n),
s_n(8) = A100311(n),
s_n(9) = A100307(2*n),
s_n(10) = A006943(n),
s_n(16) = A001317(4*n),
s_n(25) = A100308(2*n), etc.
The equality s_n(10) = A006943(n) means that sequence A047999 is obtained from A006943 by the separation by commas of the digits of its terms. (End)
Comment from N. J. A. Sloane, Jan 18 2016: (Start)
Take a diamond-shaped region with edge length n from the top of the triangle, and rotate it by 45 degrees to get a square S_n. Here is S_6:
[1, 1, 1, 1, 1, 1]
[1, 0, 1, 0, 1, 0]
[1, 1, 0, 0, 1, 1]
[1, 0, 0, 0, 1, 0]
[1, 1, 1, 1, 0, 0]
[1, 0, 1, 0, 0, 0].
Then (i) S_n contains no square (parallel to the axes) with all four corners equal to 1 (cf. A227133); (ii) S_n can be constructed by using the greedy algorithm with the constraint that there is no square with that property; and (iii) S_n contains A064194(n) 1's. Thus A064194(n) is a lower bound on A227133(n). (End)
See A123098 for a multiplicative encoding of the rows, i.e., product of the primes selected by nonzero terms; e.g., 1 0 1 => 2^1 * 3^0 * 5^1. - M. F. Hasler, Sep 18 2016
From Valentin Bakoev, Jul 11 2020: (Start)
The Sierpinski's triangle with 2^n rows is a part of a lower triangular matrix M_n of dimension 2^n X 2^n. M_n is a block matrix defined recursively: M_1= [1, 0], [1, 1], and for n>1, M_n = [M_(n-1), O_(n-1)], [M_(n-1), M_(n-1)], where M_(n-1) is a block matrix of the same type, but of dimension 2^(n-1) X 2^(n-1), and O_(n-1) is the zero matrix of dimension 2^(n-1) X 2^(n-1). Here is how M_1, M_2 and M_3 look like:
1 0 1 0 0 0 1 0 0 0 0 0 0 0
1 1 1 1 0 0 1 1 0 0 0 0 0 0 - It is seen the self-similarity of the
1 0 1 0 1 0 1 0 0 0 0 0 matrices M_1, M_2, ..., M_n, ...,
1 1 1 1 1 1 1 1 0 0 0 0 analogously to the Sierpinski's fractal.
1 0 0 0 1 0 0 0
1 1 0 0 1 1 0 0
1 0 1 0 1 0 1 0
1 1 1 1 1 1 1 1
M_n can also be defined as M_n = M_1 X M_(n-1) where X denotes the Kronecker product. M_n is an important matrix in coding theory, cryptography, Boolean algebra, monotone Boolean functions, etc. It is a transformation matrix used in computing the algebraic normal form of Boolean functions. Some properties and links concerning M_n can be seen in LINKS. (End)
Sierpinski's gasket has fractal (Hausdorff) dimension log(A000217(2))/log(2) = log(3)/log(2) = 1.58496... (and cf. A020857). This gasket is the first of a family of gaskets formed by taking the Pascal triangle (A007318) mod j, j >= 2 (see CROSSREFS). For prime j, the dimension of the gasket is log(A000217(j))/log(j) = log(j(j + 1)/2)/log(j) (see Reiter and Bondarenko references). - Richard L. Ollerton, Dec 14 2021

Examples

			Triangle begins:
              1,
             1,1,
            1,0,1,
           1,1,1,1,
          1,0,0,0,1,
         1,1,0,0,1,1,
        1,0,1,0,1,0,1,
       1,1,1,1,1,1,1,1,
      1,0,0,0,0,0,0,0,1,
     1,1,0,0,0,0,0,0,1,1,
    1,0,1,0,0,0,0,0,1,0,1,
   1,1,1,1,0,0,0,0,1,1,1,1,
  1,0,0,0,1,0,0,0,1,0,0,0,1,
  ...

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Brand, Neal; Das, Sajal; Jacob, Tom. The number of nonzero entries in recursively defined tables modulo primes. Proceedings of the Twenty-first Southeastern Conference on Combinatorics, Graph Theory, and Computing (Boca Raton, FL, 1990). Congr. Numer. 78 (1990), 47--59. MR1140469 (92h:05004).
  • John W. Milnor and James D. Stasheff, Characteristic Classes, Princeton University Press, 1974, pp. 43-49 (sequence appears on p. 46).
  • H.-O. Peitgen, H. Juergens and D. Saupe: Chaos and Fractals (Springer-Verlag 1992), p. 408.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; Chapter 3.

Links

Crossrefs

Sequences based on the triangles formed by reading Pascal's triangle mod m: (this sequence) (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).
Other versions: A090971, A038183.
Cf. A007318, A054431, A001317, A008292, A083093, A034931, A034930, A008975, A034932, A166360, A249133, A064194, A227133.
From Johannes W. Meijer, Jun 05 2011: (Start)
A106344 is a skew version of this triangle.
Triangle sums (see the comments): A001316 (Row1; Related to Row2), A002487 (Related to Kn11, Kn12, Kn13, Kn21, Kn22, Kn23), A007306 (Kn3, Kn4), A060632 (Fi1, Fi2), A120562 (Ca1, Ca2), A112970 (Gi1, Gi2), A127830 (Ze3, Ze4). (End)

Programs

  • Haskell
    import Data.Bits (xor)
a047999 :: Int -> Int -> Int
a047999 n k = a047999_tabl !! n !! k
a047999_row n = a047999_tabl !! n
a047999_tabl = iterate (\row -> zipWith xor ([0] ++ row) (row ++ [0])) [1]
-- Reinhard Zumkeller, Dec 11 2011, Oct 24 2010
  • Magma
    A047999:= func< n,k | BitwiseAnd(n-k, k) eq 0 select 1 else 0 >;
[A047999(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Dec 03 2024
  • Maple
    # Maple code for first M rows (here M=10) - N. J. A. Sloane, Feb 03 2016
ST:=[1,1,1]; a:=1; b:=2; M:=10;
for n from 2 to M do ST:=[op(ST),1];
for i from a to b-1 do ST:=[op(ST), (ST[i+1]+ST[i+2]) mod 2 ]; od:
ST:=[op(ST),1];
a:=a+n; b:=a+n; od:
ST; # N. J. A. Sloane
# alternative
A047999 := proc(n,k)
    modp(binomial(n,k),2) ;
end proc:
seq(seq(A047999(n,k),k=0..n),n=0..12) ; # R. J. Mathar, May 06 2016
  • Mathematica
    Mod[ Flatten[ NestList[ Prepend[ #, 0] + Append[ #, 0] &, {1}, 13]], 2] (* Robert G. Wilson v, May 26 2004 *)
rows = 14; ca = CellularAutomaton[60, {{1}, 0}, rows-1]; Flatten[ Table[ca[[k, 1 ;; k]], {k, 1, rows}]] (* Jean-François Alcover, May 24 2012 *)
Mod[#,2]&/@Flatten[Table[Binomial[n,k],{n,0,20},{k,0,n}]] (* Harvey P. Dale, Jun 26 2019 *)
A047999[n_,k_]:= Boole[BitAnd[n-k,k]==0];
Table[A047999[n,k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 03 2025 *)
  • PARI
    \\ Recurrence for Pascal's triangle mod p, here p = 2.
p = 2; s=13; T=matrix(s,s); T[1,1]=1;
for(n=2,s, T[n,1]=1; for(k=2,n, T[n,k] = (T[n-1,k-1] + T[n-1,k])%p ));
for(n=1,s,for(k=1,n,print1(T[n,k],", "))) \\ Gerald McGarvey, Oct 10 2009
  • PARI
    A011371(n)=my(s);while(n>>=1,s+=n);s
T(n,k)=A011371(n)==A011371(k)+A011371(n-k) \\ Charles R Greathouse IV, Aug 09 2013
  • PARI
    T(n,k)=bitand(n-k,k)==0 \\ Charles R Greathouse IV, Aug 11 2016
  • Python
    def A047999_T(n,k):
    return int(not ~n & k) # Chai Wah Wu, Feb 09 2016

Formula

Lucas's Theorem is that T(n,k) = 1 if and only if the 1's in the binary expansion of k are a subset of the 1's in the binary expansion of n; or equivalently, k AND NOT n is zero, where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 09 2016 and N. J. A. Sloane, Feb 10 2016
Sum_{k>=0} T(n, k) = A001316(n) = 2^A000120(n).
T(n,k) = T(n-1,k-1) XOR T(n-1,k), 0 < k < n; T(n,0) = T(n,n) = 1. - Reinhard Zumkeller, Dec 13 2009
T(n,k) = (T(n-1,k-1) + T(n-1,k)) mod 2 = |T(n-1,k-1) - T(n-1,k)|, 0 < k < n; T(n,0) = T(n,n) = 1. - Rick L. Shepherd, Feb 23 2018
From Vladimir Shevelev, Dec 31 2013: (Start)
For polynomial {s_n(x)} we have
s_0(x)=1; for n>=1, s_n(x) = Product_{i=1..A000120(n)} (x^(2^k_i) + 1),
if the binary expansion of n is n = Sum_{i=1..A000120(n)} 2^k_i;
G.f. Sum_{n>=0} s_n(x)*z^n = Product_{k>=0} (1 + (x^(2^k)+1)*z^(2^k)) (0
Let x>1, t>0 be real numbers. Then
Sum_{n>=0} 1/s_n(x)^t = Product_{k>=0} (1 + 1/(x^(2^k)+1)^t);
Sum_{n>=0} (-1)^A000120(n)/s_n(x)^t = Product_{k>=0} (1 - 1/(x^(2^k)+1)^t).
In particular, for t=1, x>1, we have
Sum_{n>=0} (-1)^A000120(n)/s_n(x) = 1 - 1/x. (End)
From Valentin Bakoev, Jul 11 2020: (Start)
(See my comment about the matrix M_n.) Denote by T(i,j) the number in the i-th row and j-th column of M_n (0 <= i, j < 2^n). When i>=j, T(i,j) is the j-th number in the i-th row of the Sierpinski's triangle. For given i and j, we denote by k the largest integer of the type k=2^m and k
T(i,0) = T(i,i) = 1, or
T(i,j) = 0 if i < j, or
T(i,j) = T(i-k,j), if j < k, or
T(i,j) = T(i-k,j-k), if j >= k.
Thus, for given i and j, T(i,j) can be computed in O(log_2(i)) steps. (End)

Extensions

Additional links from Lekraj Beedassy, Jan 22 2004

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

Original entry on oeis.org

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024
Offset: 1

Views

Author

Boris Putievskiy, Aug 15 2013

Keywords

Comments

The third row is (n^4 - n^2 + 24*n + 24)/12.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

Examples

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;

Links

Crossrefs

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).
Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

Programs

  • GAP
    T:= function(n,k)
    if k=0 then return n^2;
    elif k=n then return 2^n;
    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 12 2019
  • Maple
    T:= proc(n, k) option remember;
      if k=0 then n^2
    elif k=n then 2^k
    else T(n-1, k-1) + T(n-1, k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019
  • Mathematica
    T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)
  • PARI
    T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019
  • Python
    def funcL(n):
   q = n**2
   return q
def funcR(n):
   q = 2**n
   return q
for n in range (1,9871):
   t=int((math.sqrt(8*n-7) - 1)/ 2)
   i=n-t*(t+1)/2-1
   j=(t*t+3*t+4)/2-n-1
   sum1=0
   sum2=0
   for m1 in range (1,i+1):
      sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
   for m2 in range (1,j+1):
      sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
   sum=sum1+sum2
  • Sage
    @CachedFunction
def T(n, k):
    if (k==0): return n^2
    elif (k==n): return 2^n
    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 12 2019

Formula

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Extensions

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A051601 Rows of triangle formed using Pascal's rule except we begin and end the n-th row with n.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 4, 4, 3, 4, 7, 8, 7, 4, 5, 11, 15, 15, 11, 5, 6, 16, 26, 30, 26, 16, 6, 7, 22, 42, 56, 56, 42, 22, 7, 8, 29, 64, 98, 112, 98, 64, 29, 8, 9, 37, 93, 162, 210, 210, 162, 93, 37, 9, 10, 46, 130, 255, 372, 420, 372, 255, 130, 46, 10
Offset: 0

Views

Author

Asher Auel

Keywords

Comments

The number of spotlight tilings of an m X n rectangle missing the southeast corner. E.g., there are 2 spotlight tilings of a 2 X 2 square missing its southeast corner. - Bridget Tenner, Nov 10 2007
T(n,k) = A134636(n,k) - A051597(n,k). - Reinhard Zumkeller, Nov 23 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

Examples

			From _Roger L. Bagula_, Feb 17 2009: (Start)
Triangle begins:
   0;
   1,  1;
   2,  2,   2;
   3,  4,   4,   3;
   4,  7,   8,   7,    4;
   5, 11,  15,  15,   11,    5;
   6, 16,  26,  30,   26,   16,   6;
   7, 22,  42,  56,   56,   42,   22,    7;
   8, 29,  64,  98,  112,   98,   64,   29,   8;
   9, 37,  93, 162,  210,  210,  162,   93,   37,   9;
  10, 46, 130, 255,  372,  420,  372,  255,  130,  46,  10;
  11, 56, 176, 385,  627,  792,  792,  627,  385, 176,  56, 11;
  12, 67, 232, 561, 1012, 1419, 1584, 1419, 1012, 561, 232, 67, 12. ... (End)

Links

Crossrefs

Row sums give A000918(n+1).
Cf. A007318, A224791, A228196, A228576.
Columns from 2 to 9, respectively: A000124; A000125, A055795, A027660, A055796, A055797, A055798, A055799 (except 1 for the last seven). [Bruno Berselli, Aug 02 2013]
Cf. A001477, A162551 (central terms).

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k->  Binomial(n, k+1) + Binomial(n, n-k+1) ))); # G. C. Greubel, Nov 12 2019
  • Haskell
    a051601 n k = a051601_tabl !! n !! k
a051601_row n = a051601_tabl !! n
a051601_tabl = iterate
               (\row -> zipWith (+) ([1] ++ row) (row ++ [1])) [0]
-- Reinhard Zumkeller, Nov 23 2012
  • Magma
    /* As triangle: */ [[Binomial(n,m+1)+Binomial(n,n-m+1): m in [0..n]]: n in [0..12]]; // Bruno Berselli, Aug 02 2013
  • Maple
    seq(seq(binomial(n,k+1) + binomial(n, n-k+1), k=0..n), n=0..12); # G. C. Greubel, Nov 12 2019
  • Mathematica
    T[n_, k_]:= T[n, k] = Binomial[n, k+1] + Binomial[n, n-k+1];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* Roger L. Bagula, Feb 17 2009; modified by G. C. Greubel, Nov 12 2019 *)
  • PARI
    T(n,k) = binomial(n, k+1) + binomial(n, n-k+1);
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Nov 12 2019
  • Sage
    [[binomial(n, k+1) + binomial(n, n-k+1) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

Formula

T(m,n) = binomial(m+n,m) - 2*binomial(m+n-2,m-1), up to offset and transformation of array to triangular indices. - Bridget Tenner, Nov 10 2007
T(n,k) = binomial(n, k+1) + binomial(n, n-k+1). - Roger L. Bagula, Feb 17 2009
T(0,n) = T(n,0) = n, T(n,k) = T(n-1,k) + T(n-1,k-1), 0 < k < n.

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

Views

Author

Asher Auel

Keywords

Comments

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

Examples

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

Links

Crossrefs

Stripped variant of A072405, A122218.
Cf. A007318, A228196, A228576.

Programs

  • GAP
    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
  • Haskell
    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
  • Magma
    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
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    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
  • Sage
    @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

Formula

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)

A048487 a(n) = T(4,n), array T given by A048483.

Original entry on oeis.org

1, 6, 16, 36, 76, 156, 316, 636, 1276, 2556, 5116, 10236, 20476, 40956, 81916, 163836, 327676, 655356, 1310716, 2621436, 5242876, 10485756, 20971516, 41943036, 83886076, 167772156, 335544316, 671088636, 1342177276, 2684354556, 5368709116, 10737418236, 21474836476
Offset: 0

Views

Author

Clark Kimberling

Keywords

Comments

Row sums of triangle A131113. - Gary W. Adamson, Jun 15 2007
a(n) = sum of (n+1)-th row terms of triangle A134636. This sequence is the binomial transform of 1, 5, 5, (5 continued). - Gary W. Adamson, Nov 04 2007
Row sums of triangle A135856. - Gary W. Adamson, Dec 01 2007

Links

Crossrefs

Cf. A010716 (n-th difference of a(n), a(n-1), ..., a(0)).
Diagonal of A062001.
A column of A119726.
Cf. A048483, A123208, A131113, A134636, A135856.

Programs

Formula

a(n) = 5*2^n - 4. - Henry Bottomley, May 29 2001
a(n) = 2*a(n-1) + 4 for n > 0 with a(0) = 1. - Paul Barry, Aug 25 2004
From Colin Barker, Sep 13 2012: (Start)
a(n) = 3*a(n-1) - 2*a(n-2) for n >= 2.
G.f.: (1 + 3*x)/((1 - x)*(1 - 2*x)). (End)
a(n) = A123208(2*n). - Philippe Deléham, Apr 15 2013
E.g.f.: exp(x)*(5*exp(x) - 4). - Stefano Spezia, Oct 03 2023

A227550 A triangle formed like Pascal's triangle, but with factorial(n) on the borders instead of 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 6, 4, 4, 6, 24, 10, 8, 10, 24, 120, 34, 18, 18, 34, 120, 720, 154, 52, 36, 52, 154, 720, 5040, 874, 206, 88, 88, 206, 874, 5040, 40320, 5914, 1080, 294, 176, 294, 1080, 5914, 40320, 362880, 46234, 6994, 1374, 470, 470, 1374, 6994, 46234, 362880, 3628800
Offset: 0

Views

Author

Vincenzo Librandi, Aug 04 2013

Keywords

Comments

A003422 gives the second column (after 0).

Examples

			Triangle begins:
       1;
       1,     1;
       2,     2,    2;
       6,     4,    4,    6;
      24,    10,    8,   10,  24;
     120,    34,   18,   18,  34, 120;
     720,   154,   52,   36,  52, 154,  720;
    5040,   874,  206,   88,  88, 206,  874, 5040;
   40320,  5914, 1080,  294, 176, 294, 1080, 5914, 40320;
  362880, 46234, 6994, 1374, 470, 470, 1374, 6994, 46234, 362880;

Links

Crossrefs

Cf. similar triangles with t on the borders: A007318 (t = 1), A028326 (t = 2), A051599 (t = prime(n)), A051601 (t = n), A051666 (t = n^2), A108617 (t = fibonacci(n)), A134636 (t = 2n+1), A137688 (t = 2^n), A227075 (t = 3^n).
Cf. A003422.
Cf. A227791 (central terms), A001563, A074911.

Programs

  • Haskell
    a227550 n k = a227550_tabl !! n !! k
a227550_row n = a227550_tabl !! n
a227550_tabl = map fst $ iterate
   (\(vs, w:ws) -> (zipWith (+) ([w] ++ vs) (vs ++ [w]), ws))
   ([1], a001563_list)
-- Reinhard Zumkeller, Aug 05 2013
  • Magma
    function T(n,k)
  if k eq 0 or k eq n then return Factorial(n);
  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, May 02 2021
  • Mathematica
    t = {}; Do[r = {}; Do[If[k == 0||k == n, m = n!, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]
  • Sage
    def T(n,k): return factorial(n) if (k==0 or k==n) else T(n-1, k-1) + T(n-1, k)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 02 2021

Formula

From G. C. Greubel, May 02 2021: (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = T(n, n) = n!.
Sum_{k=0..n} T(n, k) = 2^n * (1 +Sum_{j=1..n-1} j*j!/2^j) = A140710(n). (End)
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