A089023 Duplicate of A035610.
1, 4, 28, 232, 2092, 19864, 195352, 1970896, 20275660, 211823800, 2240795848
Offset: 0
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.
Triangle T(n, k) begins: n\k 0 1 2 3 4 5 6 7 8 9 0: 1 1: 1 1 2: 2 3 1 3: 5 9 5 1 4: 14 28 20 7 1 5: 42 90 75 35 9 1 6: 132 297 275 154 54 11 1 7: 429 1001 1001 637 273 77 13 1 8: 1430 3432 3640 2548 1260 440 104 15 1 9: 4862 11934 13260 9996 5508 2244 663 135 17 1 ... Reformatted by _Wolfdieter Lang_, Dec 21 2015 From _Paul Barry_, Feb 17 2011: (Start) Production matrix begins 1, 1, 1, 2, 1, 0, 1, 2, 1, 0, 0, 1, 2, 1, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 0, 1, 2, 1 (End) From _Wolfdieter Lang_, Sep 20 2013: (Start) Example for rho(N) = 2*cos(Pi/N) powers: n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) = 2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)
/* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 16 2015
T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, May 06 2006 T := proc(n, k) option remember; if k = n then 1 elif k > n then 0 elif k = 0 then T(n-1, 0) + T(n-1,1) else T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1) fi end: seq(seq(T(n, k), k = 0..n), n = 0..9) od; # Peter Luschny, Feb 14 2023
Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* Geoffrey Critzer, Dec 18 2011 *) Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* Harvey P. Dale, Dec 18 2011 *) Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* Jayanta Basu, Apr 30 2013 *)
a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k) trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print("")) trianglerows(10) \\ Felix Fröhlich, Jun 24 2016
# Algorithm of L. Seidel (1877) # Prints the first n rows of the triangle def A039599_triangle(n) : D = [0]*(n+2); D[1] = 1 b = True ; h = 1 for i in range(2*n-1) : if b : for k in range(h,0,-1) : D[k] += D[k-1] h += 1 else : for k in range(1,h, 1) : D[k] += D[k+1] if b : print([D[z] for z in (1..h-1)]) b = not b A039599_triangle(10) # Peter Luschny, May 01 2012
Triangle begins: 1; 0, 1; 0, 1, 1; 0, 2, 2, 1; 0, 5, 5, 3, 1; 0, 14, 14, 9, 4, 1; 0, 42, 42, 28, 14, 5, 1; 0, 132, 132, 90, 48, 20, 6, 1; From _Paul Barry_, Sep 28 2009: (Start) Production array is 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1 (End)
A106566:= func< n,k | n eq 0 select 1 else (k/n)*Binomial(2*n-k-1, n-k) >; [A106566(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 06 2021
A106566 := proc(n,k) if n = 0 then 1; elif k < 0 or k > n then 0; else binomial(2*n-k-1,n-k)*k/n ; end if; end proc: # R. J. Mathar, Mar 01 2015
T[n_, k_] := Binomial[2n-k-1, n-k]*k/n; T[0, 0] = 1; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 18 2017 *) (* The function RiordanArray is defined in A256893. *) RiordanArray[1&, #(1-Sqrt[1-4#])/(2#)&, 11] // Flatten (* Jean-François Alcover, Jul 16 2019 *)
{T(n, k) = if( k<=0 || k>n, n==0 && k==0, binomial(2*n - k, n) * k/(2*n - k))}; /* Michael Somos, Oct 01 2022 */
def A106566(n, k): return 1 if (n==0) else (k/n)*binomial(2*n-k-1, n-k) flatten([[A106566(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 06 2021
A(2,2) = 6, because 6 words of length 4 can be built over 2-letter alphabet {a, b} by repeatedly inserting doublets (words with two equal letters) into the initially empty word: aaaa, aabb, abba, baab, bbaa, bbbb. Square array A(n,k) begins: 1, 1, 1, 1, 1, 1, ... 0, 1, 2, 3, 4, 5, ... 0, 1, 6, 15, 28, 45, ... 0, 1, 20, 87, 232, 485, ... 0, 1, 70, 543, 2092, 5725, ... 0, 1, 252, 3543, 19864, 71445, ...
A:= proc(n, k) local j; if n=0 then 1 else k/n *add(binomial(2*n,j) *(n-j) *(k-1)^j, j=0..n-1) fi end: seq(seq(A(n, d-n), n=0..d), d=0..10);
A[, 1] = 1; A[n, k_] := If[n == 0, 1, k/n*Sum[Binomial[2*n, j]*(n - j)*(k - 1)^j, {j, 0, n - 1}]]; Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Dec 27 2013, translated from Maple *)
Triangle begins: 1; 0, 1; 0, -1, 2; 0, 2, -6, 5; 0, -5, 20, -28, 14; ...
h := n -> GAMMA(n+1/2)/GAMMA(n+2)*hypergeom([2, 1-n], [n+2], -3): a := n -> 3-(-3)^n-5^n+(1/sqrt(Pi))*add(12^(k+1)*binomial(n, 2*k)*h(k), k=1..n/2): seq(simplify(a(n)), n=0..26); # Peter Luschny, Oct 25 2019
a(n)={my(p=3/(1+2*sqrt(1-12*x+O(x*x^(n\2))))); sum(k=0, n\2, binomial(n, 2*k)*polcoef(p,k))} \\ Andrew Howroyd, Oct 25 2019
library("freealg") g <- function(p,string){constant(as.freealg(string)^p)} sapply(0:7,g,"1+x+y+X+Y")
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