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.

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A016116 a(n) = 2^floor(n/2).

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, 64, 128, 128, 256, 256, 512, 512, 1024, 1024, 2048, 2048, 4096, 4096, 8192, 8192, 16384, 16384, 32768, 32768, 65536, 65536, 131072, 131072, 262144, 262144, 524288, 524288, 1048576, 1048576, 2097152
Offset: 0

Views

Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

Powers of 2 doubled up. The usual OEIS policy is to omit the duplicates in such cases (when this would become A000079). This is an exception.
Number of symmetric compositions of n: e.g., 5 = 2+1+2 = 1+3+1 = 1+1+1+1+1 so a(5) = 4; 6 = 3+3 = 2+2+2 = 1+4+1 = 2+1+1+2 = 1+2+2+1 = 1+1+2+1+1 = 1+1+1+1+1+1 so a(6) = 8. - Henry Bottomley, Dec 10 2001
This sequence is the number of digits of each term of A061519. - Dmitry Kamenetsky, Jan 17 2009
Starting with offset 1 = binomial transform of [1, 1, -1, 3, -7, 17, -41, ...]; where A001333 = (1, 1, 3, 7, 17, 41, ...). - Gary W. Adamson, Mar 25 2009
a(n+1) is the number of symmetric subsets of [n]={1,2,...,n}. A subset S of [n] is symmetric if k is an element of S implies (n-k+1) is an element of S. - Dennis P. Walsh, Oct 27 2009
INVERT and inverse INVERT transforms give A006138, A039834(n-1).
The Kn21 sums, see A180662, of triangle A065941 equal the terms of this sequence. - Johannes W. Meijer, Aug 15 2011
First differences of A027383. - Jason Kimberley, Nov 01 2011
Run lengths in A079944. - Jeremy Gardiner, Nov 21 2011
Number of binary palindromes (A006995) between 2^(n-1) and 2^n (for n>1). - Hieronymus Fischer, Feb 17 2012
Pisano period lengths: 1, 1, 4, 1, 8, 4, 6, 1, 12, 8, 20, 4, 24, 6, 8, 1, 16, 12, 36, 8, ... . - R. J. Mathar, Aug 10 2012
Range of row n of the Circular Pascal array of order 4. - Shaun V. Ault, May 30 2014
a(n) is the number of permutations of length n avoiding both 213 and 312 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
Also, the decimal representation of the diagonal from the origin to the corner (and from the corner to the origin except for the initial term) of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 190", based on the 5-celled von Neumann neighborhood when initialized with a single black (ON) cell at stage zero. - Robert Price, May 10 2017
a(n + 1) + n - 1, n > 0, is the number of maximal subsemigroups of the monoid of partial order-preserving or -reversing mappings on a set with n elements. See the East et al. link. - James Mitchell and Wilf A. Wilson, Jul 21 2017
Number of symmetric stairs with n cells. A stair is a snake polyomino allowing only two directions for adjacent cells: east and north. See A005418. - Christian Barrientos, May 11 2018
For n >= 4, a(n) is the exponent of the group of the Gaussian integers in a reduced system modulo (1+i)^(n+2). See A302254. - Jianing Song, Jun 27 2018
a(n) is the number of length-(n+1) binary sequences, denoted , with s(1)=1 and with s(i+1)=s(i) for odd i. - Dennis P. Walsh, Sep 06 2018
a(n+1) is the number of subsets of {1,2,..,n} in which all differences between successive elements of subsets are even. For example, for n = 7, a(6) = 8 and the 8 subsets are {7}, {1,7}, {3,7}, {5,7}, {1,3,7}, {1,5,7}, {3,5,7}, {1,3,5,7}. For odd differences between elements see Comment in A000045 (Fibonacci numbers). - Enrique Navarrete, Jul 01 2020
Also, the number of walks of length n on the graph x--y--z, starting at x. - Sean A. Irvine, May 30 2025

Examples

			For n=5 the a(5)=4 symmetric subsets of [4] are {1,4}, {2,3}, {1,2,3,4} and the empty set. - _Dennis P. Walsh_, Oct 27 2009
For n=5 the a(5)=4 length-6 binary sequences are <1,1,0,0,0,0>, <1,1,0,0,1,1>, <1,1,1,1,0,0> and <1,1,1,1,1,1>. - _Dennis P. Walsh_, Sep 06 2018

Links

Crossrefs

Cf. A006995, A057148, A079944, A112030, A112033.
a(n) = A094718(3, n).
Cf. A001333.
See A052955 for partial sums (without the initial term).
A000079 gives the odd-indexed terms of a(n).
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Programs

Formula

a(n) = a(n-1)*a(n-2)/a(n-3) = 2*a(n-2) = 2^A004526(n).
G.f.: (1+x)/(1-2*x^2).
a(n) = (1/2 + sqrt(1/8))*sqrt(2)^n + (1/2 - sqrt(1/8))*(-sqrt(2))^n. - Ralf Stephan, Mar 11 2003
E.g.f.: cosh(sqrt(2)*x) + sinh(sqrt(2)*x)/sqrt(2). - Paul Barry, Jul 16 2003
The signed sequence (-1)^n*2^floor(n/2) has a(n) = (sqrt(2))^n(1/2 - sqrt(2)/4) + (-sqrt(2))^n(1/2 + sqrt(2)/4). It is the inverse binomial transform of A000129(n-1). - Paul Barry, Apr 21 2004
Diagonal sums of A046854. a(n) = Sum_{k=0..n} binomial(floor(n/2), k). - Paul Barry, Jul 07 2004
a(n) = a(n-2) + 2^floor((n-2)/2). - Paul Barry, Jul 14 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(floor(n/2), floor(k/2)). - Paul Barry, Jul 15 2004
E.g.f.: cosh(asinh(1) + sqrt(2)*x)/sqrt(2). - Michael Somos, Feb 28 2005
a(n) = Sum_{k=0..n} A103633(n,k). - Philippe Deléham, Dec 03 2006
a(n) = 2^(n/2)*((1 + (-1)^n)/2 + (1-(-1)^n)/(2*sqrt(2))). - Paul Barry, Nov 12 2009
a(n) = 2^((2*n - 1 + (-1)^n)/4). - Luce ETIENNE, Sep 20 2014

A005043 Riordan numbers: a(n) = (n-1)*(2*a(n-1) + 3*a(n-2))/(n+1).

Original entry on oeis.org

1, 0, 1, 1, 3, 6, 15, 36, 91, 232, 603, 1585, 4213, 11298, 30537, 83097, 227475, 625992, 1730787, 4805595, 13393689, 37458330, 105089229, 295673994, 834086421, 2358641376, 6684761125, 18985057351, 54022715451, 154000562758, 439742222071, 1257643249140
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Also called Motzkin summands or ring numbers.
The old name was "Motzkin sums", used in certain publications. The sequence has the property that Motzkin(n) = A001006(n) = a(n) + a(n+1), e.g., A001006(4) = 9 = 3 + 6 = a(4) + a(5).
Number of 'Catalan partitions', that is partitions of a set 1,2,3,...,n into parts that are not singletons and whose convex hulls are disjoint when the points are arranged on a circle (so when the parts are all pairs we get Catalan numbers). - Aart Blokhuis (aartb(AT)win.tue.nl), Jul 04 2000
Number of ordered trees with n edges and no vertices of outdegree 1. For n > 1, number of dissections of a convex polygon by nonintersecting diagonals with a total number of n+1 edges. - Emeric Deutsch, Mar 06 2002
Number of Motzkin paths of length n with no horizontal steps at level 0. - Emeric Deutsch, Nov 09 2003
Number of Dyck paths of semilength n with no peaks at odd level. Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUDUDD, where U=(1,1), D=(1,-1). Number of Dyck paths of semilength n with no ascents of length 1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUUDDD. - Emeric Deutsch, Dec 05 2003
Arises in Schubert calculus as follows. Let P = complex projective space of dimension n+1. Take n projective subspaces of codimension 3 in P in general position. Then a(n) is the number of lines of P intersecting all these subspaces. - F. Hirzebruch, Feb 09 2004
Difference between central trinomial coefficient and its predecessor. Example: a(6) = 15 = 141 - 126 and (1 + x + x^2)^6 = ... + 126*x^5 + 141*x^6 + ... (Catalan number A000108(n) is the difference between central binomial coefficient and its predecessor.) - David Callan, Feb 07 2004
a(n) = number of 321-avoiding permutations on [n] in which each left-to-right maximum is a descent (i.e., is followed by a smaller number). For example, a(4) counts 4123, 3142, 2143. - David Callan, Jul 20 2005
The Hankel transform of this sequence give A000012 = [1, 1, 1, 1, 1, 1, 1, ...]; example: Det([1, 0, 1, 1; 0, 1, 1, 3; 1, 1, 3, 6; 1, 3, 6, 15]) = 1. - Philippe Deléham, May 28 2005
The number of projective invariants of degree 2 for n labeled points on the projective line. - Benjamin J. Howard (bhoward(AT)ima.umn.edu), Nov 24 2006
Define a random variable X=trA^2, where A is a 2 X 2 unitary symplectic matrix chosen from USp(2) with Haar measure. The n-th central moment of X is E[(X+1)^n] = a(n). - Andrew V. Sutherland, Dec 02 2007
Let V be the adjoint representation of the complex Lie algebra sl(2). The dimension of the invariant subspace of the n-th tensor power of V is a(n). - Samson Black (sblack1(AT)uoregon.edu), Aug 27 2008
Starting with offset 3 = iterates of M * [1,1,1,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 08 2009
a(n) has the following standard-Young-tableaux (SYT) interpretation: binomial(n+1,k)*binomial(n-k-1,k-1)/(n+1)=f^(k,k,1^{n-2k}) where f^lambda equals the number of SYT of shape lambda. - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 02 2010
a(n) is also the sum of the numbers of standard Young tableaux of shapes (k,k,1^{n-2k}) for all 1 <= k <= floor(n/2). - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 10 2010
a(n) is the number of derangements of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(3)=1 because p=231=(123) is the only derangement of {1,2,3} with genus 0. Indeed, cp'=231*312=123=(1)(2)(3) and so g(p) = (1/2)(3+1-1-3)=0. - Emeric Deutsch, May 29 2010
Apparently: Number of Dyck 2n-paths with all ascents length 2 and no descent length 2. - David Scambler, Apr 17 2012
This is true. Proof: The mapping "insert a peak (UD) after each upstep (U)" is a bijection from all Dyck n-paths to those Dyck (2n)-paths in which each ascent is of length 2. It sends descents of length 1 in the n-path to descents of length 2 in the (2n)-path. But Dyck n-paths with no descents of length 1 are equinumerous with Riordan n-paths (Motzkin n-paths with no flatsteps at ground level) as follows. Given a Dyck n-path with no descents of length 1, split it into consecutive step pairs, then replace UU with U, DD with D, UD with a blue flatstep (F), DU with a red flatstep, and concatenate the new steps to get a colored Motzkin path. Each red F will be (immediately) preceded by a blue F or a D. In the latter case, transfer the red F so that it precedes the matching U of the D. Finally, erase colors to get the required Riordan path. For example, with lowercase f denoting a red flatstep, U^5 D^2 U D^4 U^4 D^3 U D^2 -> (U^2, U^2, UD, DU, D^2, D^2, U^2, U^2 D^2, DU, D^2) -> UUFfDDUUDfD -> UUFFDDUFUDD. - David Callan, Apr 25 2012
From Nolan Wallach, Aug 20 2014: (Start)
Let ch[part1, part2] be the value of the character of the symmetric group on n letters corresponding to the partition part1 of n on the conjucgacy class given by part2. Let A[n] be the set of (n+1) partitions of 2n with parts 1 or 2. Then deleting the first term of the sequence one has a(n) = Sum_{k=1..n+1} binomial(n,k-1)*ch[[n,n], A[n][[k]]])/2^n. This via the Frobenius Character Formula can be interpreted as the dimension of the SL(n,C) invariants in tensor^n (wedge^2 C^n).
Explanation: Let p_j denote sum (x_i)^j the sum in k variables. Then the Frobenius formula says then (p_1)^j_1 (p_2)^j_2 ... (p_r)^j_r is equal to sum(lambda, ch[lambda, 1^j_12^j_2 ... r^j_r] S_lambda) with S_lambda the Schur function corresponding to lambda. This formula implies that the coefficient of S([n,n]) in (((p_1)^1+p_2)/2)^n in its expansion in terms of Schur functions is the right hand side of our formula. If we specialize the number of variables to 2 then S[n,n](x,y)=(xy)^n. Which when restricted to y=x^(-1) is 1. That is it is 1 on SL(2).
On the other hand ((p_1)^2+p_2)/2 is the complete homogeneous symmetric function of degree 2 that is tr(S^2(X)). Thus our formula for a(n) is the same as that of Samson Black above since his V is the same as S^2(C^2) as a representation of SL(2). On the other hand, if we multiply ch(lambda) by sgn you get ch(Transpose(lambda)). So ch([n,n]) becomes ch([2,...,2]) (here there are n 2's). The formula for a(n) is now (1/2^n)*Sum_{j=0..n} ch([2,..,2], 1^(2n-2j) 2^j])*(-1)^j)*binomial(n,j), which calculates the coefficient of S_(2,...,2) in (((p_1)^2-p_2)/2)^n. But ((p_1)^2-p_2)/2 in n variables is the second elementary symmetric function which is the character of wedge^2 C^n and S_(2,...,2) is 1 on SL(n).
(End)
a(n) = number of noncrossing partitions (A000108) of [n] that contain no singletons, also number of nonnesting partitions (A000108) of [n] that contain no singletons. - David Callan, Aug 27 2014
From Tom Copeland, Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Various relations among the o.g.f.s may be easily constructed, such as Fib[-Mot(-x)] = -P[P(-x)] = x/(1-2*x) a generating fct for 2^n.
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1]. (End)
Consistent with David Callan's comment above, A249548, provides a refinement of the Motzkin sums into the individual numbers for the non-crossing partitions he describes. - Tom Copeland, Nov 09 2014
The number of lattice paths from (0,0) to (n,0) that do not cross below the x-axis and use up-step=(1,1) and down-steps=(1,-k) where k is a positive integer. For example, a(4) = 3: [(1,1)(1,1)(1,-1)(1,-1)], [(1,1)(1,-1)(1,1)(1,-1)] and [(1,1)(1,1)(1,1)(1,-3)]. - Nicholas Ham, Aug 19 2015
A series created using 2*(a(n) + a(n+1)) + (a(n+1) + a(n+2)) has Hankel transform of F(2n), offset 3, F being a Fibonacci number, A001906 (Empirical observation). - Tony Foster III, Jul 30 2016
The series a(n) + A001006(n) has Hankel transform F(2n+1), offset n=1, F being the Fibonacci bisection A001519 (empirical observation). - Tony Foster III, Sep 05 2016
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n. Moreover, the number of such algebras having the double centraliser property with respect to a minimal faithful projective-injective module equals the number of Riordan paths, that is, Motzkin paths without level-steps at height zero, of length n." - Eric M. Schmidt, Dec 16 2017
A connection to the Thue-Morse sequence: (-1)^a(n) = (-1)^A010060(n) * (-1)^A010060(n+1) = A106400(n) * A106400(n+1). - Vladimir Reshetnikov, Jul 21 2019
Named by Bernhart (1999) after the American mathematician John Riordan (1903-1988). - Amiram Eldar, Apr 15 2021

Examples

			a(5)=6 because the only dissections of a polygon with a total number of 6 edges are: five pentagons with one of the five diagonals and the hexagon with no diagonals.
G.f. = 1 + x^2 + x^3 + 3*x^4 + 6*x^5 + 15*x^6 + 36*x^7 + 91*x^8 + 232*x^9 + ...
From _Gus Wiseman_, Nov 15 2022: (Start)
The a(0) = 1 through a(6) = 15 lone-child-avoiding (no vertices of outdegree 1) ordered rooted trees with n + 1 vertices (ranked by A358376):
  o  .  (oo)  (ooo)  (oooo)   (ooooo)   (oooooo)
                     ((oo)o)  ((oo)oo)  ((oo)ooo)
                     (o(oo))  ((ooo)o)  ((ooo)oo)
                              (o(oo)o)  ((oooo)o)
                              (o(ooo))  (o(oo)oo)
                              (oo(oo))  (o(ooo)o)
                                        (o(oooo))
                                        (oo(oo)o)
                                        (oo(ooo))
                                        (ooo(oo))
                                        (((oo)o)o)
                                        ((o(oo))o)
                                        ((oo)(oo))
                                        (o((oo)o))
                                        (o(o(oo)))
(End)

References

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

Links

Crossrefs

Row sums of triangle A020474, first differences of A082395.
First diagonal of triangular array in A059346.
Binomial transform of A126930. - Philippe Deléham, Nov 26 2009
The Hankel transform of a(n+1) is A128834. The Hankel transform of a(n+2) is floor((2*n+4)/3) = A004523(n+2). - Paul Barry, Mar 08 2011
The Kn11 triangle sums of triangle A175136 lead to A005043(n+2), while the Kn12(n) = A005043(n+4)-2^(n+1), Kn13(n) = A005043(n+6)-(n^2+9*n+56)*2^(n-2) and the Kn4(n) = A005043(2*n+2) = A099251(n+1) triangle sums are related to the sequence given above. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, May 06 2011
Cf. A187306 (self-convolution), A348210 (column 1).
Cf. A000045, A000108, A000957, A001006, A005717, A005773, A007317, A057078, A091867, A104597, A126120, A178514, A249548, A309303.
Bisections: A099251, A099252.

Programs

  • Haskell
    a005043 n = a005043_list !! n
a005043_list = 1 : 0 : zipWith div
   (zipWith (*) [1..] (zipWith (+)
       (map (* 2) $ tail a005043_list) (map (* 3) a005043_list))) [3..]
-- Reinhard Zumkeller, Jan 31 2012
  • Maple
    A005043 := proc(n) option remember; if n <= 1 then 1-n else (n-1)*(2*A005043(n-1)+3*A005043(n-2))/(n+1); fi; end;
Order := 20: solve(series((x-x^2)/(1-x+x^2),x)=y,x); # outputs g.f.
  • Mathematica
    a[0]=1; a[1]=0; a[n_]:= a[n] = (n-1)*(2*a[n-1] + 3*a[n-2])/(n+1); Table[ a[n], {n, 0, 30}] (* Robert G. Wilson v, Jun 14 2005 *)
Table[(-3)^(1/2)/6 * (-1)^n*(3*Hypergeometric2F1[1/2,n+1,1,4/3]+ Hypergeometric2F1[1/2,n+2,1,4/3]), {n,0,32}] (* cf. Mark van Hoeij in A001006 *) (* Wouter Meeussen, Jan 23 2010 *)
RecurrenceTable[{a[0]==1,a[1]==0,a[n]==(n-1) (2a[n-1]+3a[n-2])/(n+1)},a,{n,30}] (* Harvey P. Dale, Sep 27 2013 *)
a[ n_]:= SeriesCoefficient[2/(1+x +Sqrt[1-2x-3x^2]), {x, 0, n}]; (* Michael Somos, Aug 21 2014 *)
a[ n_]:= If[n<0, 0, 3^(n+3/2) Hypergeometric2F1[3/2, n+2, 2, 4]/I]; (* Michael Somos, Aug 21 2014 *)
Table[3^(n+3/2) CatalanNumber[n] (4(5+2n)Hypergeometric2F1[3/2, 3/2, 1/2-n, 1/4] -9 Hypergeometric2F1[3/2, 5/2, 1/2 -n, 1/4])/(4^(n+3) (n+1)), {n, 0, 31}] (* Vladimir Reshetnikov, Jul 21 2019 *)
Table[Sqrt[27]/8 (3/4)^n CatalanNumber[n] Hypergeometric2F1[1/2, 3/2, 1/2 - n, 1/4], {n, 0, 31}] (* Jan Mangaldan, Sep 12 2021 *)
  • Maxima
    a[0]:1$
a[1]:0$
a[n]:=(n-1)*(2*a[n-1]+3*a[n-2])/(n+1)$
makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( (x - x^3) / (1 + x^3) + x * O(x^n)), n))}; /* Michael Somos, May 31 2005 */
  • PARI
    my(N=66); Vec(serreverse(x/(1+x*sum(k=1,N,x^k))+O(x^N))) \\ Joerg Arndt, Aug 19 2012
  • Python
    from functools import cache
@cache
def A005043(n: int) -> int:
    if n <= 1: return 1 - n
    return (n - 1) * (2 * A005043(n - 1) + 3 * A005043(n - 2)) // (n + 1)
print([A005043(n) for n in range(32)]) # Peter Luschny, Nov 20 2022
  • Sage
    A005043 = lambda n: (-1)^n*jacobi_P(n,1,-n-3/2,-7)/(n+1)
[simplify(A005043(n)) for n in (0..29)]
# Peter Luschny, Sep 23 2014
  • Sage
    def ms():
    a, b, c, d, n = 0, 1, 1, -1, 1
    yield 1
    while True:
        yield -b + (-1)^n*d
        n += 1
        a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)/((n+1)*(n-1))
        c, d = d, (3*(n-1)*c-(2*n-1)*d)/n
A005043 = ms()
print([next(A005043) for  in range(32)]) # _Peter Luschny, May 16 2016

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k). a(n) = (1/(n+1)) * Sum_{k=0..ceiling(n/2)} binomial(n+1, k)*binomial(n-k-1, k-1), for n > 1. - Len Smiley. [Comment from Amitai Regev (amitai.regev(AT)weizmann.ac.il), Mar 02 2010: the latter sum should be over the range k=1..floor(n/2).]
G.f.: (1 + x - sqrt(1-2*x-3*x^2))/(2*x*(1+x)).
G.f.: 2/(1+x+sqrt(1-2*x-3*x^2)). - Paul Peart (ppeart(AT)fac.howard.edu), May 27 2000
a(n+1) + (-1)^n = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). - Bernhart
a(n) = (1/(n+1)) * Sum_{i} (-1)^i*binomial(n+1, i)*binomial(2*n-2*i, n-i). - Bernhart
G.f. A(x) satisfies A = 1/(1+x) + x*A^2.
E.g.f.: exp(x)*(BesselI(0, 2*x) - BesselI(1, 2*x)). - Vladeta Jovovic, Apr 28 2003
a(n) = A001006(n-1) - a(n-1).
a(n+1) = Sum_{k=0..n} (-1)^k*A026300(n, k), where A026300 is the Motzkin triangle.
a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005
a(n) = Sum_{k>=0} A086810(n-k, k). - Philippe Deléham, May 30 2005
a(n+2) = Sum_{k>=0} A064189(n-k, k). - Philippe Deléham, May 31 2005
Moment representation: a(n) = (1/(2*Pi))*Int(x^n*sqrt((1+x)(3-x))/(1+x),x,-1,3). - Paul Barry, Jul 09 2006
Inverse binomial transform of A000108 (Catalan numbers). - Philippe Deléham, Oct 20 2006
a(n) = (2/Pi)* Integral_{x=0..Pi} (4*cos(x)^2-1)^n*sin(x)^2 dx. - Andrew V. Sutherland, Dec 02 2007
G.f.: 1/(1-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). - Paul Barry, Jan 22 2009
G.f.: 1/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, May 16 2009
G.f.: 1/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-... (continued fraction). - Paul Barry, Mar 02 2010
a(n) = -(-1)^n * hypergeom([1/2, n+2],[2],4/3) / sqrt(-3). - Mark van Hoeij, Jul 02 2010
a(n) = (-1)^n*hypergeometric([-n,1/2],[2],4). - Peter Luschny, Aug 15 2012
Let A(x) be the g.f., then x*A(x) is the reversion of x/(1 + x^2*Sum_{k>=0} x^k); see A215340 for the correspondence to Dyck paths without length-1 ascents. - Joerg Arndt, Aug 19 2012 and Apr 16 2013
a(n) ~ 3^(n+3/2)/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 02 2012
G.f.: 2/(1+x+1/G(0)), where G(k) = 1 + x*(2+3*x)*(4*k+1)/( 4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 05 2013
D-finite (an alternative): (n+1)*a(n) = 3*(n-2)*a(n-3) + (5*n-7)*a(n-2) + (n-2)*a(n-1), n >= 3. - Fung Lam, Mar 22 2014
Asymptotics: a(n) = (3^(n+2)/(sqrt(3*n*Pi)*(8*n)))*(1-21/(16*n) + O(1/n^2)) (with contribution by Vaclav Kotesovec). - Fung Lam, Mar 22 2014
a(n) = T(2*n-1,n)/n, where T(n,k) = triangle of A180177. - Vladimir Kruchinin, Sep 23 2014
a(n) = (-1)^n*JacobiP(n,1,-n-3/2,-7)/(n+1). - Peter Luschny, Sep 23 2014
a(n) = Sum_{k=0..n} C(n,k)*(C(k,n-k)-C(k,n-k-1)). - Peter Luschny, Oct 01 2014
Conjecture: a(n) = A002426(n) - A005717(n), n > 0. - Mikhail Kurkov, Feb 24 2019 [The conjecture is true. - Amiram Eldar, May 17 2024]
a(n) = A309303(n) + A309303(n+1). - Vladimir Reshetnikov, Jul 22 2019
From Peter Bala, Feb 11 2022: (Start)
a(n) = A005773(n+1) - 2*A005717(n) for n >= 1.
Conjectures: for n >= 1, n divides a(2*n+1) and 2*n-1 divides a(2*n). (End)

Extensions

Thanks to Laura L. M. Yang (yanglm(AT)hotmail.com) for a correction, Aug 29 2004
Name changed to Riordan numbers following a suggestion from Ira M. Gessel. - N. J. A. Sloane, Jul 24 2020

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

A008288 Square array of Delannoy numbers D(i,j) (i >= 0, j >= 0) read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 13, 7, 1, 1, 9, 25, 25, 9, 1, 1, 11, 41, 63, 41, 11, 1, 1, 13, 61, 129, 129, 61, 13, 1, 1, 15, 85, 231, 321, 231, 85, 15, 1, 1, 17, 113, 377, 681, 681, 377, 113, 17, 1, 1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1, 1, 21, 181, 833, 2241, 3653, 3653
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

In the Formula section, some contributors use T(n,k) = D(n-k, k) (for 0 <= k <= n), which is the triangular version of the square array (D(n,k): n,k >= 0). Conversely, D(n,k) = T(n+k,k) for n,k >= 0. - Petros Hadjicostas, Aug 05 2020
Also called the tribonacci triangle [Alladi and Hoggatt (1977)]. - N. J. A. Sloane, Mar 23 2014
D(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0), (0,1), (1,1). - Joerg Arndt, Jul 01 2011 [Corrected by N. J. A. Sloane, May 30 2020]
Or, triangle read by rows of coefficients of polynomials P[n](x) defined by P[0] = 1, P[1] = x+1; for n >= 2, P[n] = (x+1)*P[n-1] + x*P[n-2].
D(n, k) is the number of k-matchings of a comb-like graph with n+k teeth. Example: D(1, 3) = 7 because the graph consisting of a horizontal path ABCD and the teeth Aa, Bb, Cc, Dd has seven 3-matchings: four triples of three teeth and the three triples {Aa, Bb, CD}, {Aa, Dd, BC}, {Cc, Dd, AB}. Also D(3, 1)=7, the 1-matchings of the same graph being the seven edges: {AB}, {BC}, {CD}, {Aa}, {Bb}, {Cc}, {Dd}. - Emeric Deutsch, Jul 01 2002
Sum of n-th antidiagonal of the array D is A000129(n+1). - Reinhard Zumkeller, Dec 03 2004 [Edited by Petros Hadjicostas, Aug 05 2020 so that the counting of antidiagonals of D starts at n = 0. That is, the sum of the terms in the n-th row of the triangles T is A000129(n+1).]
The A-sequence for this Riordan type triangle (see one of Paul Barry's comments under Formula) is A112478 and the Z-sequence the trivial: {1, 0, 0, 0, ...}. See the W. Lang link under A006232 for Sheffer a- and z-sequences where also Riordan A- and Z-sequences are explained. This leads to the recurrence for the triangle given below. - Wolfdieter Lang, Jan 21 2008
The triangle or chess sums, see A180662 for their definitions, link the Delannoy numbers with twelve different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 and Kn15 have been added. It is remarkable that all knight sums are related to the tribonacci numbers, that is, A000073 and A001590, but none of the others. - Johannes W. Meijer, Sep 22 2010
This sequence, A008288, is jointly generated with A035607 as an array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) and v(n,x) = 2*x*u(n-1,x) + v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 09 2012
Row n, for n > 0, of Roger L. Bagula's triangle in the Example section shows the coefficients of the polynomial u(n) = c(0) + c(1)*x + ... + c(n)*x^n which is the numerator of the n-th convergent of the continued fraction [k, k, k, ...], where k = sqrt(x) + 1/sqrt(x); see A230000. - Clark Kimberling, Nov 13 2013
In an n-dimensional hypercube lattice, D(n,k) gives the number of nodes situated at a Minkowski (Manhattan) distance of k from a given node. In cellular automata theory, the cells at Manhattan distance k are called the von Neumann neighborhood of radius k. For k=1, see A005843. - Dmitry Zaitsev, Dec 10 2015
These numbers appear as the coefficients of series relating spherical and bispherical harmonics, in the solutions of Laplace's equation in 3D. [Majic 2019, Eq. 22] - Matt Majic, Nov 24 2019
From Peter Bala, Feb 19 2020: (Start)
The following remarks assume an offset of 1 in the row and column indices of the triangle.
The sequence of row polynomials T(n,x), beginning with T(1,x) = x, T(2,x) = x + x^2, T(3,x) = x + 3*x^2 + x^3, ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd(T(n,x), T(m,x)) = T(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the sequence (T(n,x): n >= 1) is a divisibility sequence in the polynomial ring Z[x]; that is, if n divides m then T(n,x) divides T(m,x) in Z[x].
Let S(x) = 1 + 2*x + 6*x^2 + 22*x^3 + ... denote the o.g.f. for the large Schröder numbers A006318. The power series (x*S(x))^n, n = 2, 3, 4, ..., can be expressed as a linear combination with polynomial coefficients of S(x) and 1: (x*S(x))^n = T(n-1,-x) - T(n,-x)*S(x). The result can be extended to negative integer n if we define T(0,x) = 0 and T(-n,x) = (-1)^(n+1) * T(n,x)/x^n. Cf. A115139.
[In the previous two paragraphs, D(n,x) was replaced with T(n,x) because the contributor is referring to the rows of the triangle T(n,k), not the rows of the array D(n,k). - Petros Hadjicostas, Aug 05 2020] (End)
Named after the French amateur mathematician Henri-Auguste Delannoy (1833-1915). - Amiram Eldar, Apr 15 2021
D(i,j) = D(j,i). With this and Dmitry Zaitsev's Dec 10 2015 comment, D(i,j) can be considered the number of points at L1 distance <= i in Z^j or the number of points at L1 distance <= j in Z^i from any given point. The rows and columns of D(i,j) are the crystal ball sequences on cubic lattices. See the first example below. The n-th term in the k-th crystal ball sequence can be considered the number of points at distance <= n from any point in a k-dimensional cubic lattice, or the number of points at distance <= k from any point in an n-dimensional cubic lattice. - Shel Kaphan, Jan 01 2023 and Jan 07 2023
Dimensions of hom spaces Hom(R^{(i)}, R^{(j)}) in the Delannoy category attached to the oligomorphic group of order preserving self-bijections of the real line. - Noah Snyder, Mar 22 2023

Examples

			The square array D(i,j) (i >= 0, j >= 0) begins:
  1, 1,  1,   1,   1,   1,    1,    1,    1,    1, ... = A000012
  1, 3,  5,   7,   9,  11,   13,   15,   17,   19, ... = A005408
  1, 5, 13,  25,  41,  61,   85,  113,  145,  181, ... = A001844
  1, 7, 25,  63, 129, 231,  377,  575,  833, 1159, ... = A001845
  1, 9, 41, 129, 321, 681, 1289, 2241, 3649, 5641, ... = A001846
  ...
For D(2,5) = 61, which is seen above in the row labeled A001844, we calculate the sum (9 + 11 + 41) of the 3 nearest terms above and/or to the left. - _Peter Munn_, Jan 01 2023
D(2,5) = 61 can also be obtained from the row labeled A005408 using a recurrence mentioned in the formula section:  D(2,5) = D(1,5) + 2*Sum_{k=0..4} D(1,k), so D(2,5) = 11 + 2*(1+3+5+7+9) = 11 + 2*25. - _Shel Kaphan_, Jan 01 2023
As a triangular array (on its side) this begins:
   0,   0,   0,   0,   1,   0,  11,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   1,   0,   3,   0,  13,   0,  63,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   0,   0,   1,   0,  11,   0, ...
   [Edited by _Shel Kaphan_, Jan 01 2023]
From _Roger L. Bagula_, Dec 09 2008: (Start)
As a triangle T(n,k) (with rows n >= 0 and columns k = 0..n), this begins:
   1;
   1,  1;
   1,  3,   1;
   1,  5,   5,   1;
   1,  7,  13,   7,    1;
   1,  9,  25,  25,    9,    1;
   1, 11,  41,  63,   41,   11,    1;
   1, 13,  61, 129,  129,   61,   13,   1;
   1, 15,  85, 231,  321,  231,   85,  15,   1;
   1, 17, 113, 377,  681,  681,  377, 113,  17,  1;
   1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1;
   ... (End)
Triangle T(n,k) recurrence: 63 = T(6,3) = 25 + 13 + 25 = T(5,2) + T(4,2) + T(5,3).
Triangle T(n,k) recurrence with A-sequence A112478: 63 = T(6,3) = 1*25 + 2*25 - 2*9 + 6*1 (T entries from row n = 5 only). [Here the formula T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j) is used with n = 6 and k = 3; i.e., T(6,3) = Sum_{j=0..3} A111478(j) * T(5, 2+j). - _Petros Hadjicostas_, Aug 05 2020]
From _Philippe Deléham_, Mar 29 2012: (Start)
Subtriangle of the triangle given by (1, 0, 1, -1, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   1,  0;
   1,  1,  0;
   1,  3,  1,  0;
   1,  5,  5,  1,  0;
   1,  7, 13,  7,  1,  0;
   1,  9, 25, 25,  9,  1, 0;
   1, 11, 41, 63, 41, 11, 1, 0;
   ...
Subtriangle of the triangle given by (0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   0, 1;
   0, 1,  1;
   0, 1,  3,  1;
   0, 1,  5,  5,  1;
   0, 1,  7, 13,  7,  1;
   0, 1,  9, 25, 25,  9,  1;
   0, 1, 11, 41, 63, 41, 11, 1;
   ... (End)

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 593.
  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
  • L. Moser and W. Zayachkowski, Lattice paths with diagonal steps, Scripta Mathematica, 26 (1963), 223-229.
  • G. Picou, Note #2235, L'Intermédiaire des Mathématiciens, 8 (1901), page 281. - N. J. A. Sloane, Mar 02 2022
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 28.

Links

Crossrefs

Sums of antidiagonals: A000129 (Pell numbers).
Main diagonal: A001850 (central Delannoy numbers), which has further information and references.
A002002, A026002, and A190666 are +-k-diagonals for k=1, 2, 3 resp. - Shel Kaphan, Jan 01 2023
Rows 0..10: A000012, A005408, A001844, A001845, A001846, A001847, A001848, A001849, A008417, A008419, A008421.
See also A027618.
Cf. A059446.
Has same main diagonal as A064861. Different from A100936.
Cf. A101164, A101167, A128966, A131887, A131935.
Cf. A035607, A108625, A142979, A142992, A143007.
Read mod small primes: A211312, A211313, A211314, A211315.
Triangle sums (see the comments): A000129 (Row1); A056594 (Row2); A000073 (Kn11 & Kn21); A089068 (Kn12 & Kn22); A180668 (Kn13 & Kn23); A180669 (Kn14 & Kn24); A180670 (Kn15 & Kn25); A099463 (Kn3 & Kn4); A116404 (Fi1 & Fi2); A006498 (Ca1 & Ca2); A006498(3*n) (Ca3 & Ca4); A079972 (Gi1 & Gi2); A079972(4*n) (Gi3 & Gi4); A079973(3*n) (Ze1 & Ze2); A079973(2*n) (Ze3 & Ze4).
Cf. A102413, A128966. (D(n,1)) = A005843. Cf. A115139.

Programs

  • Haskell
    a008288 n k = a008288_tabl !! n !! k
a008288_row n = a008288_tabl !! n
a008288_tabl = map fst $ iterate
    (\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
                       zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
-- Reinhard Zumkeller, Jul 21 2013
  • Maple
    A008288 := proc(n, k) option remember; if k = 0 then 1 elif n=k then 1 else procname(n-1, k-1) + procname(n-2, k-1) + procname(n-1, k) end if; end proc: seq(seq(A008288(n,k),k=0..n), n=0..10); # triangular indices n and k
P[0]:=1; P[1]:=x+1; for n from 2 to 12 do P[n]:=expand((x+1)*P[n-1]+x*P[n-2]); lprint(P[n]); lprint(seriestolist(series(P[n],x,200))); end do:
  • Mathematica
    (* Next, A008288 jointly generated with A035607 *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A008288 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A035607 *)
(* Clark Kimberling, Mar 09 2012 *)
d[n_, k_] := Binomial[n+k, k]*Hypergeometric2F1[-k, -n, -n-k, -1]; A008288 = Flatten[Table[d[n-k, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Apr 05 2012, after 3rd formula *)
  • Python
    from functools import cache
@cache
def delannoy_row(n: int) -> list[int]:
    if n == 0: return [1]
    if n == 1: return [1, 1]
    rov = delannoy_row(n - 2)
    row = delannoy_row(n - 1) + [1]
    for k in range(n - 1, 0, -1):
        row[k] += row[k - 1] + rov[k - 1]
    return row
for n in range(10): print(delannoy_row(n))  # Peter Luschny, Jul 30 2023
  • Sage
    for k in range(8):  # seen as an array, read row by row
    a = lambda n: hypergeometric([-n, -k], [1], 2)
    print([simplify(a(n)) for n in range(11)]) # Peter Luschny, Nov 19 2014

Formula

D(n, 0) = 1 = D(0, n) for n >= 0; D(n, k) = D(n, k-1) + D(n-1, k-1) + D(n-1, k).
Bivariate o.g.f.: Sum_{n >= 0, k >= 0} D(n, k)*x^n*y^k = 1/(1 - x - y - x*y).
D(n, k) = Sum_{d = 0..min(n,k)} binomial(k, d)*binomial(n+k-d, k) = Sum_{d=0..min(n,k)} 2^d*binomial(n, d)*binomial(k, d). [Edited by Petros Hadjicostas, Aug 05 2020]
Seen as a triangle read by rows: T(n, 0) = T(n, n) = 1 for n >= 0 and T(n, k) = T(n-1, k-1) + T(n-2, k-1) + T(n-1, k), 0 < k < n and n > 1. - Reinhard Zumkeller, Dec 03 2004
Read as a number triangle, this is the Riordan array (1/(1-x), x(1+x)/(1-x)) with T(n, k) = Sum_{j=0..n-k} C(n-k, j) * C(k, j) * 2^j. - Paul Barry, Jul 18 2005
T(n,k) = Sum_{j=0..n-k} C(k,j)*C(n-j,k). - Paul Barry, May 21 2006
Let y^k(n) be the number of Khalimsky-continuous functions f from [0,n-1] to Z such that f(0) = 0 and f(n-1) = k. Then y^k(n) = D(i,j) for i = (1/2)*(n-1-k) and j = (1/2)*(n-1+k) where n-1+k belongs to 2Z. - Shiva Samieinia (shiva(AT)math.su.se), Oct 08 2007
Recurrence for triangle from A-sequence (see the Wolfdieter Lang comment above): T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j), n >= 1, k >= 1. [For k > n, the sum is empty, in which case T(n,k) = 0.]
From Peter Bala, Jul 17 2008: (Start)
The n-th row of the square array is the crystal ball sequence for the product lattice A_1 x ... x A_1 (n copies). A035607 is the table of the associated coordination sequences for these lattices.
The polynomial p_n(x) := Sum {k = 0..n} 2^k * C(n,k) * C(x,k) = Sum_{k = 0..n} C(n,k) * C(x+k,n), whose values [p_n(0), p_n(1), p_n(2), ... ] give the n-th row of the square array, is the Ehrhart polynomial of the n-dimensional cross polytope (the hyperoctahedron) [Bump et al. (2000), Theorem 6].
The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
The reciprocity law p_n(m) = p_m(n) reflects the symmetry of the table.
The polynomial p_n(x) is the unique polynomial solution of the difference equation (x+1)*f(x+1) - x*f(x-1) = (2*n+1)*f(x), normalized so that f(0) = 1.
These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_n(x-1), n = 1,2,3,..., satisfy a Riemann hypothesis [Bump et al. (2000), Theorem 4]. The o.g.f. for the p_n(x) is (1 + t)^x/(1 - t)^(x + 1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ... .
The square array of Delannoy numbers has a close connection with the constant log(2). The entries in the n-th row of the array occur in the series acceleration formula log(2) = (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n) + (-1)^n * Sum_{k>=1} (-1)^(k+1)/(k*D(n,k-1)*D(n,k)). [T(n,k) was replaced with D(n,k) in the formula to agree with the beginning of the paragraph. - Petros Hadjicostas, Aug 05 2020]
For example, the fourth row of the table (n = 3) gives the series log(2) = 1 - 1/2 + 1/3 - 1/(1*1*7) + 1/(2*7*25) - 1/(3*25*63) + 1/(4*63*129) - ... . See A142979 for further details.
Also the main diagonal entries (the central Delannoy numbers) give the series acceleration formula Sum_{n>=1} 1/(n*D(n-1,n-1)*D(n,n)) = (1/2)*log(2), a result due to Burnside. [T(n,n) was replaced here with D(n,n) to agree with the previous paragraphs. - Petros Hadjicostas, Aug 05 2020]
Similar relations hold between log(2) and the crystal ball sequences of the C_n lattices A142992. For corresponding results for the constants zeta(2) and zeta(3), involving the crystal ball sequences for root lattices of type A_n and A_n x A_n, see A108625 and A143007 respectively. (End)
From Peter Bala, Oct 28 2008: (Start)
Hilbert transform of Pascal's triangle A007318 (see A145905 for the definition of this term).
D(n+a,n) = P_n(a,0;3) for all integer a such that a >= -n, where P_n(a,0;x) is the Jacobi polynomial with parameters (a,0) [Hetyei]. The related formula A(n,k) = P_k(0,n-k;3) defines the table of asymmetric Delannoy numbers, essentially A049600. (End)
Seen as a triangle read by rows: T(n, k) = Hyper2F1([k-n, -k], [1], 2). - Peter Luschny, Aug 02 2014, Oct 13 2024.
From Peter Bala, Jun 25 2015: (Start)
O.g.f. for triangle T(n,k): A(z,t) = 1/(1 - (1 + t)*z - t*z^2) = 1 + (1 + t)*z + (1 + 3*t + t^2)*z^2 + (1 + 5*t + 5*t^2 + t^3)*z^3 + ....
1 + z*d/dz(A(z,t))/A(z,t) is the o.g.f. for A102413. (End)
E.g.f. for the n-th subdiagonal of T(n,k), n >= 0, equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(2*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 4*x + 4*x^2/2) = 1 + 5*x + 13*x^2/2! + 25*x^3/3! + 41*x^4/4! + 61*x^5/5! + .... - Peter Bala, Mar 05 2017 [The n-th subdiagonal of triangle T(n,k) is the n-th row of array D(n,k).]
Let a_i(n) be multiplicative with a_i(p^e) = D(i, e), p prime and e >= 0, then Sum_{n > 0} a_i(n)/n^s = (zeta(s))^(2*i+1)/(zeta(2*s))^i for i >= 0. - Werner Schulte, Feb 14 2018
Seen as a triangle read by rows: T(n,k) = Sum_{i=0..k} binomial(n-i, i) * binomial(n-2*i, k-i) for 0 <= k <= n. - Werner Schulte, Jan 09 2019
Univariate generating function: Sum_{k >= 0} D(n,k)*z^k = (1 + z)^n/(1 - z)^(n+1). [Dziemianczuk (2013), Eq. 5.3] - Matt Majic, Nov 24 2019
(n+1)*D(n+1,k) = (2*k+1)*D(n,k) + n*D(n-1,k). [Majic (2019), Eq. 22] - Matt Majic, Nov 24 2019
For i, j >= 1, D(i,j) = D(i,j-1) + 2*Sum_{k=0..i-1} D(k,j-1), or, because D(i,j) = D(j,i), D(i,j) = D(i-1,j) + 2*Sum_{k=0..j-1} D(i-1,k). - Shel Kaphan, Jan 01 2023
Sum_{k=0..n} T(n,k)^2 = A026933(n). - R. J. Mathar, Nov 07 2023
Let S(x) = (1 - x - (1 - 6*x + x^2)^(1/2))/(2*x) denote the g.f. of the sequence of large Schröder numbers A006318. Read as a lower triangular array, the signed n-th row polynomial R(n, -x) = 1/sqrt(1 - 6*x + x^2) *( 1/S(x)^(n+1) + (x*S(x))^(n+1) ). For example, R(4, -x) = 1 - 7*x + 13*x^2 - 7*x^3 + x^4 = 1/sqrt(1 - 6*x + x^2) * ( 1/S(x)^5 + (x*S(x))^5 ). Cf. A102413. - Peter Bala, Aug 01 2024

Extensions

Expanded description from Clark Kimberling, Jun 15 1997
Additional references from Sylviane R. Schwer (schwer(AT)lipn.univ-paris13.fr), Nov 28 2001
Changed the notation to make the formulas more precise. - N. J. A. Sloane, Jul 01 2002

A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 14, 28, 20, 7, 1, 42, 90, 75, 35, 9, 1, 132, 297, 275, 154, 54, 11, 1, 429, 1001, 1001, 637, 273, 77, 13, 1, 1430, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 4862, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005
The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005
Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005
Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006
Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007
The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007
Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007
Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007
From Philippe Deléham, Mar 30 2007: (Start)
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):
(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970
(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;
(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;
(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;
(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;
(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)
The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007
Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007
Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007
Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007
Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009
The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011
The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011
Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):
rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.
For the odd powers of rho(n) see A039598. (End)
Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016
The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

Examples

			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)

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
  • T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

Links

Crossrefs

Columns give: A000108, A000245, A000344, A000588, A001392, A000589, A000590, A000012, A005408, A014107(n>1).
Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).
Cf. A008313, A039598, A084938, A000007, A067311, A321620.

Programs

  • Magma
    /* 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
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    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
  • Sage
    # 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

Formula

T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.
From Emeric Deutsch, May 06 2006: (Start)
T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).
G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
The following formulas were added by Philippe Deléham during 2003 to 2009: (Start)
Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.
T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
T(0, 0) = 1, T(n, k) = 0 if n<0 or n=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).
a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.
T(n, k) = A050165(n, n-k).
Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).
Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.
Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n.
T(n, k)*(-2)^(n-k) = A114193(n, k).
Sum_{k>=h} T(n,k) = binomial(2n,n-h).
Sum_{k=0..n} T(n,k)*5^k = A127628(n).
Sum_{k=0..n} T(n,k)*7^k = A115970(n).
T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).
Sum_{k=0..n} T(n,k)*6^k = A126694(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).
Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).
Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).
Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).
T(2*n,n) = A126596(n).
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.
Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).
T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).
T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k).
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).
T(n,k) = A053121(2*n,2*k).
Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).
T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).
Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).
Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).
Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.
Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).
Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).
Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).
Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).
Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.
Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.
Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).
Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).
T(n,k) + T(n,k+1) = A039598(n,k).
T(n,k) = A128899(n,k)+A128899(n,k+1).
Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.
Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.
Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).
Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).
Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).
Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).
Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).
Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.
T(n,2*k)+T(n,2*k+1) = A118919(n,k).
Sum_{k=0..j} T(n,k) = A050157(n,j).
Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).
Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).
Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).
Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).
Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).
Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).
Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).
Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).
(End)
T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - Paul Barry, Feb 17 2011
Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - Werner Schulte, Jul 22 2015
Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - Werner Schulte, Jul 22 2015
Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - Werner Schulte, Dec 03 2015
T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - Peter Luschny, May 13 2016
T(n,n-2) = A014107(n). - R. J. Mathar, Jan 30 2019
T(n,n-3) = n*(2*n-1)*(2*n-5)/3. - R. J. Mathar, Jan 30 2019
T(n,n-4) = n*(n-1)*(2*n-1)*(2*n-7)/6. - R. J. Mathar, Jan 30 2019
T(n,n-5) = n*(n-1)*(2*n-1)*(2*n-3)*(2*n-9)/30. - R. J. Mathar, Jan 30 2019

Extensions

Corrected by Philippe Deléham, Nov 26 2009, Dec 14 2009

A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

Original entry on oeis.org

0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, 1576239, 4126648, 10803704, 28284465, 74049690, 193864606, 507544127, 1328767776, 3478759200, 9107509825, 23843770274, 62423800998, 163427632719, 427859097160, 1120149658760
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n)/A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg (roberthappelberg(AT)yahoo.com), Jul 25 2005
Let phi be the golden ratio (cf. A001622). Then 1/phi = phi - 1 = Sum_{n>=1} (-1)^(n-1)/a(n), an alternating infinite series consisting solely of unit fractions. - Franz Vrabec, Sep 14 2005
a(n+2) is the Hankel transform of A005807 aerated. - Paul Barry, Nov 04 2008
A more exact name would be: Golden convergents to rectangle numbers. These rectangles are not actually golden (ratio of sides is not phi) but are golden convergents (sides are numerator and denominator of convergents in the continued fraction expansion of phi, whence ratio of sides converges to phi). - Daniel Forgues, Nov 29 2009
The Kn4 sums (see A180662 for definition) of the "Races with Ties" triangle A035317 lead to this sequence. - Johannes W. Meijer, Jul 20 2011
Numbers m such that m(5m+2)+1 or m(5m-2)+1 is a square. - Bruno Berselli, Oct 22 2012
In pairs, these numbers are important in finding binomial coefficients that appear in at least six places in Pascal's triangle. For instance, the pair (m,n) = (40, 104) finds the numbers binomial(n-1,m) = binomial(n,m-1). Two additional numbers are found on the other side of the triangle. The final two numbers appear in row binomial(n-1,m). See A003015. - T. D. Noe, Mar 13 2013
For n>1, a(n) is one-half the area of the trapezoid created by the four points (F(n),L(n)), (L(n),F(n)), (F(n+1), L(n+1)), (L(n+1), F(n+1)) where F(n) = A000045(n) and L(n) = A000032(n). - J. M. Bergot, May 14 2014
[Note on how to calculate: take the two points (a,b) and (c,d) with a
a(n) = A067962(n-1) / A067962(n-2), n > 1. - Reinhard Zumkeller, Sep 24 2015
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Fibonacci numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017

Examples

			G.f. = x + 2*x^2 + 6*x^3 + 15*x^4 + 40*x^5 + 104*x^6 + 273*x^7 + 714*x^8 + ...

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9.
  • 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).

Links

Crossrefs

Cf. A001655, A001656, A001657, A001658, A010048, A067962.
Cf. A005968, A005969, A098531, A098532, A098533, A119283, A128697.
Cf. A000071, A079472, A080145, A239798, A290565.
Bisection of A006498, A070550, A080239.
First differences of A064831.
Partial sums of A007598.

Programs

  • Haskell
    a001654 n = a001654_list !! n
a001654_list = zipWith (*) (tail a000045_list) a000045_list
-- Reinhard Zumkeller, Jun 08 2013
  • Magma
    I:=[0,1,2]; [n le 3 select I[n] else 2*Self(n-1) + 2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 17 2018
  • Maple
    with(combinat): A001654:=n->fibonacci(n)*fibonacci(n+1):
seq(A001654(n), n=0..28); # Zerinvary Lajos, Oct 07 2007
  • Mathematica
    LinearRecurrence[{2,2,-1}, {0,1,2}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
Times@@@Partition[Fibonacci[Range[0,30]],2,1] (* Harvey P. Dale, Aug 18 2011 *)
Accumulate[Fibonacci[Range[0, 30]]^2] (* Paolo Xausa, May 31 2024 *)
  • PARI
    A001654(n)=fibonacci(n)*fibonacci(n+1);
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
vector(30, n, b(n-1, 2))  \\ Joerg Arndt, May 08 2016
  • Python
    from sympy import fibonacci as F
def a(n): return F(n)*F(n + 1)
[a(n) for n in range(101)] # Indranil Ghosh, Aug 03 2017
  • Python
    from math import prod
from gmpy2 import fib2
def A001654(n): return prod(fib2(n+1)) # Chai Wah Wu, May 19 2022

Formula

a(n) = A010048(n+1, 2) = Fibonomial(n+1, 2).
a(n) = A006498(2*n-1).
a(n) = a(n - 1) + A007598(n) = a(n - 1) + A000045(n)^2 = Sum_{j <= n} Fibonacci(j)^2. - Henry Bottomley, Feb 09 2001 [corrected by Ridouane Oudra, Apr 12 2025]
For n > 0, 1 - 1/a(n+1) = Sum_{k=1..n} 1/(F(k)*F(k+2)) where F(k) is the k-th Fibonacci number. - Benoit Cloitre, Aug 31 2002.
G.f.: x/(1-2*x-2*x^2+x^3) = x/((1+x)*(1-3*x+x^2)). (Simon Plouffe in his 1992 dissertation; see Comments to A055870),
a(n) = 3*a(n-1) - a(n-2) - (-1)^n = -a(-1-n).
Let M = the 3 X 3 matrix [1 2 1 / 1 1 0 / 1 0 0]; then a(n) = the center term in M^n *[1 0 0]. E.g., a(5) = 40 since M^5 * [1 0 0] = [64 40 25]. - Gary W. Adamson, Oct 10 2004
a(n) = Sum{k=0..n} Fibonacci(k)^2. The proof is easy. Start from a square (1*1). On the right side, draw another square (1*1). On the above side draw a square ((1+1)*(1+1)). On the left side, draw a square ((1+2)*(1+2)) and so on. You get a rectangle (F(n)*F(1+n)) which contains all the squares of side F(1), F(2), ..., F(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 19 2007
With phi = (1+sqrt(5))/2, a(n) = round((phi^(2*n+1))/5) = floor((1/2) + (phi^(2*n+1))/5), n >= 0. - Daniel Forgues, Nov 29 2009
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), a(1)=1, a(2)=2, a(3)=6. - Sture Sjöstedt, Feb 06 2010
a(n) = (A002878(n) - (-1)^n)/5. - R. J. Mathar, Jul 22 2010
a(n) = 1/|F(n+1)/F(n) - F(n)/F(n-1)| where F(n) = Fibonacci numbers A000045. b(n) = F(n+1)/F(n) - F(n)/F(n-1): 1/1, -1/2, 1/6, -1/15, 1/40, -1/104, ...; c(n) = 1/b(n) = a(n)*(-1)^(n+1): 1, -2, 6, -15, 40, -104, ... (n=1,2,...). - Thomas Ordowski, Nov 04 2010
a(n) = (Fibonacci(n+2)^2 - Fibonacci(n-1)^2)/4. - Gary Detlefs, Dec 03 2010
Let d(n) = n mod 2, a(0)=0 and a(1)=1. For n > 1, a(n) = d(n) + 2*a(n-1) + Sum_{k=0..n-2} a(k). - L. Edson Jeffery, Mar 20 2011
From Tim Monahan, Jul 11 2011: (Start)
a(n+1) = ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5.
a(n) = ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n-2*(-1)^n)/10. (End)
From Wolfdieter Lang, Jul 21 2012: (Start)
a(n) = (2*A059840(n+2) - A027941(n))/3, n >= 0, with A059840(n+2) = Sum_{k=0..n} F(k)*F(k+2) and A027941(n) = A001519(n+1) - 1, n >= 0, where A001519(n+1) = F(2*n+1). (End)
a(n) = (-1)^n * Sum_{k=0..n} (-1)^k*F(2*k), n >= 0. - Wolfdieter Lang, Aug 11 2012
a(-1-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
a(n) = (L(2*n+1) - (-1)^n)/5 with L(k) = A000032(k). - J. M. Bergot, Apr 15 2016
E.g.f.: ((3 + sqrt(5))*exp((5+sqrt(5))*x/2) - 2*exp((2*x)/(3+sqrt(5))+x) - 1 - sqrt(5))*exp(-x)/(5*(1 + sqrt(5))). - Ilya Gutkovskiy, Apr 15 2016
From Klaus Purath, Apr 24 2019: (Start)
a(n) = A061646(n) - Fibonacci(n-1)^2.
a(n) = (A061646(n+1) - A061646(n))/2. (End)
a(n) = A226205(n+1) + (-1)^(n+1). - Flávio V. Fernandes, Apr 23 2020
Sum_{n>=1} 1/a(n) = A290565. - Amiram Eldar, Oct 06 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi^2/2 (A239798). - Amiram Eldar, Dec 02 2024
G.f.: x * exp( Sum_{k>=1} F(3*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extensions

Extended by Wolfdieter Lang, Jun 27 2000

A009766 Catalan's triangle T(n,k) (read by rows): each term is the sum of the entries above and to the left, i.e., T(n,k) = Sum_{j=0..k} T(n-1,j).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 5, 5, 1, 4, 9, 14, 14, 1, 5, 14, 28, 42, 42, 1, 6, 20, 48, 90, 132, 132, 1, 7, 27, 75, 165, 297, 429, 429, 1, 8, 35, 110, 275, 572, 1001, 1430, 1430, 1, 9, 44, 154, 429, 1001, 2002, 3432, 4862, 4862, 1, 10, 54, 208, 637, 1638, 3640, 7072, 11934
Offset: 0

Views

Author

Wouter Meeussen

Keywords

Comments

The entries in this triangle (in its many forms) are often called ballot numbers.
T(n,k) = number of standard tableaux of shape (n,k) (n > 0, 0 <= k <= n). Example: T(3,1) = 3 because we have 134/2, 124/3 and 123/4. - Emeric Deutsch, May 18 2004
T(n,k) is the number of full binary trees with n+1 internal nodes and jump-length k. In the preorder traversal of a full binary tree, any transition from a node at a deeper level to a node on a strictly higher level is called a jump; the positive difference of the levels is called the jump distance; the sum of the jump distances in a given ordered tree is called the jump-length. - Emeric Deutsch, Jan 18 2007
The k-th diagonal from the right (k=1, 2, ...) gives the sequence obtained by asking in how many ways can we toss a fair coin until we first get k more heads than tails. The k-th diagonal has formula k(2m+k-1)!/(m!(m+k)!) and g.f. (C(x))^k where C(x) is the generating function for the Catalan numbers, (1-sqrt(1-4*x))/(2*x). - Anthony C Robin, Jul 12 2007
T(n,k) is also the number of order-decreasing and order-preserving full transformations (of an n-element chain) of waist k (waist (alpha) = max(Im(alpha))). - Abdullahi Umar, Aug 25 2008
Formatted as an upper right triangle (see tables) a(c,r) is the number of different triangulated planar polygons with c+2 vertices, with triangle degree c-r+1 for the same vertex X (c=column number, r=row number, with c >= r >= 1). - Patrick Labarque, Jul 28 2010
The triangle sums, see A180662 for their definitions, link Catalan's triangle, its mirror is A033184, with several sequences, see crossrefs. - Johannes W. Meijer, Sep 22 2010
The m-th row of Catalan's triangle consists of the unique nonnegative differences of the form binomial(m+k,m)-binomial(m+k,m+1) with 0 <= k <= m (See Links). - R. J. Cano, Jul 22 2014
T(n,k) is also the number of nondecreasing parking functions of length n+1 whose maximum element is k+1. For example T(3,2) = 5 because we have (1,1,1,3), (1,1,2,3), (1,2,2,3), (1,1,3,3), (1,2,3,3). - Ran Pan, Nov 16 2015
T(n,k) is the number of Dyck paths from (0,0) to (n+2,n+2) which start with n-k+2 east steps and touch the diagonal y=x only on the last north step. - Felipe Rueda, Sep 18 2019
T(n-1,k) for k < n is number of well-formed strings of n parenthesis pairs with prefix of exactly n-k opening parenthesis; T(n,n) = T(n,n-1). - Hermann Stamm-Wilbrandt, May 02 2021

Examples

			Triangle begins in row n=0 with 0 <= k <= n:
  1;
  1, 1;
  1, 2,  2;
  1, 3,  5,   5;
  1, 4,  9,  14,  14;
  1, 5, 14,  28,  42,   42;
  1, 6, 20,  48,  90,  132,  132;
  1, 7, 27,  75, 165,  297,  429,  429;
  1, 8, 35, 110, 275,  572, 1001, 1430, 1430;
  1, 9, 44, 154, 429, 1001, 2002, 3432, 4862, 4862;

References

  • William Feller, Introduction to Probability Theory and its Applications, vol. I, ed. 2, chap. 3, sect. 1,2.
  • Ki Hang Kim, Douglas G. Rogers, and Fred W. Roush, Similarity relations and semiorders. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 577-594, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561081 (81i:05013).
  • D. E. Knuth, TAOCP, Vol. 4, Section 7.2.1.6, Eq. 22, p. 451.
  • C. Krishnamachary and M. Bheemasena Rao, Determinants whose elements are Eulerian, prepared Bernoullian and other numbers, J. Indian Math. Soc., 14 (1922), 55-62, 122-138 and 143-146.
  • M. Bellon, Query 5467, L'Intermédiaire des Mathématiciens, 4 (1925), 11; H. Ory, 4 (1925), 120. - N. J. A. Sloane, Mar 09 2022
  • Andrzej Proskurowski and Ekaputra Laiman, Fast enumeration, ranking, and unranking of binary trees. Proceedings of the thirteenth Southeastern conference on combinatorics, graph theory and computing (Boca Raton, Fla., 1982). Congr. Numer. 35 (1982), 401-413.MR0725898 (85a:68152).
  • M. Welsch, Note #371, L'Intermédiaire des Mathématiciens, 2 (1895), pp. 235-237. - N. J. A. Sloane, Mar 02 2022

Links

Crossrefs

The following are all versions of (essentially) the same Catalan triangle: A009766, A008315, A028364, A030237, A047072, A053121, A059365, A062103, A099039, A106566, A130020, A140344.
Cf. A062745, A214292.
Sums of the shallow diagonals give A001405, central binomial coefficients: 1=1, 1=1, 1+1=2, 1+2=3, 1+3+2=6, 1+4+5=10, 1+5+9+5=20, ...
Row sums as well as the sums of the squares of the shallow diagonals give Catalan numbers (A000108).
Reflected version of A033184.
Diagonals give A000108, A000108, A000245, A002057, A000344, A003517, A000588, A003518, A003519, A001392, ...
Triangle sums (see the comments): A000108 (Row1), A000957 (Row2), A001405 (Kn11), A014495 (Kn12), A194124 (Kn13), A030238 (Kn21), A000984 (Kn4), A000958 (Fi2), A165407 (Ca1), A026726 (Ca4), A081696 (Ze2).

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],m->Binomial(n+m,n)*(n-m+1)/(n+1)))); # Muniru A Asiru, Feb 18 2018
  • Haskell
    a009766 n k = a009766_tabl !! n !! k
a009766_row n = a009766_tabl !! n
a009766_tabl = iterate (\row -> scanl1 (+) (row ++ [0])) [1]
-- Reinhard Zumkeller, Jul 12 2012
  • Magma
    [[Binomial(n+k,n)*(n-k+1)/(n+1): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Mar 07 2019
  • Maple
    A009766 := proc(n,k) binomial(n+k,n)*(n-k+1)/(n+1); end proc:
seq(seq(A009766(n,k), k=0..n), n=0..10); # R. J. Mathar, Dec 03 2010
  • Mathematica
    Flatten[NestList[Append[Accumulate[#], Last[Accumulate[#]]] &, {1}, 9]] (* Birkas Gyorgy, May 19 2012 *)
T[n_, k_] := T[n, k] = Which[k == 0, 1, k>n, 0, True, T[n-1, k] + T[n, k-1] ]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 07 2016 *)
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, binomial(n+1+k, k) * (n+1-k) / (n+1+k) )}; /* Michael Somos, Oct 17 2006 */
  • PARI
    b009766=(n1=0,n2=100)->{my(q=if(!n1,0,binomial(n1+1,2)));for(w=n1,n2,for(k=0,w,write("b009766.txt",1*q" "1*(binomial(w+k,w)-binomial(w+k,w+1)));q++))} \\ R. J. Cano, Jul 22 2014
  • Python
    from math import comb, isqrt
def A009766(n): return comb((a:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)))+(b:=n-comb(a+1,2)),b)*(a-b+1)//(a+1) # Chai Wah Wu, Nov 12 2024
  • Sage
    @CachedFunction
def ballot(p,q):
    if p == 0 and q == 0: return 1
    if p < 0 or p > q: return 0
    S = ballot(p-2, q) + ballot(p, q-2)
    if q % 2 == 1: S += ballot(p-1, q-1)
    return S
A009766 = lambda n, k: ballot(2*k, 2*n)
for n in (0..7): [A009766(n, k) for k in (0..n)]  # Peter Luschny, Mar 05 2014
  • Sage
    [[binomial(n+k,n)*(n-k+1)/(n+1) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Mar 07 2019

Formula

T(n, m) = binomial(n+m, n)*(n-m+1)/(n+1), 0 <= m <= n.
G.f. for column m: (x^m)*N(2; m-1, x)/(1-x)^(m+1), m >= 0, with the row polynomials from triangle A062991 and N(2; -1, x) := 1.
G.f.: C(t*x)/(1-x*C(t*x)) = 1 + (1+t)*x + (1+2*t+2*t^2)*x^2 + ..., where C(x) = (1-sqrt(1-4*x))/(2*x) is the Catalan function. - Emeric Deutsch, May 18 2004
Another version of triangle T(n, k) given by [1, 0, 0, 0, 0, 0, ...] DELTA [0, 1, 1, 1, 1, 1, 1, ...] = 1; 1, 0; 1, 1, 0; 1, 2, 2, 0; 1, 3, 5, 5, 0; 1, 4, 9, 14, 14, 0; ... where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 16 2005
O.g.f. (with offset 1) is the series reversion of x*(1+x*(1-t))/(1+x)^2 = x - x^2*(1+t) + x^3*(1+2*t) - x^4*(1+3*t) + ... . - Peter Bala, Jul 15 2012
Sum_{k=0..floor(n/2)} T(n-k+p-1, k+p-1) = A001405(n+2*p-2) - C(n+2*p-2, p-2), p >= 1. - Johannes W. Meijer, Oct 03 2013
Let A(x,t) denote the o.g.f. Then 1 + x*d/dx(A(x,t))/A(x,t) = 1 + (1 + t)*x + (1 + 2*t + 3*t^2)*x^2 + (1 + 3*t + 6*t^2 + 10*t^3)*x^3 + ... is the o.g.f. for A059481. - Peter Bala, Jul 21 2015
The n-th row polynomial equals the n-th degree Taylor polynomial of the function (1 - 2*x)/(1 - x)^(n+2) about 0. For example, for n = 4, (1 - 2*x)/(1 - x)^6 = 1 + 4*x + 9*x^2 + 14*x^3 + 14*x^4 + O(x^5). - Peter Bala, Feb 18 2018
T(n,k) = binomial(n + k + 1, k) - 2*binomial(n + k, k - 1), for 0 <= k <= n. - David Callan, Jun 15 2022

A006918 a(n) = binomial(n+3, 3)/4 for odd n, n*(n+2)*(n+4)/24 for even n.

Original entry on oeis.org

0, 1, 2, 5, 8, 14, 20, 30, 40, 55, 70, 91, 112, 140, 168, 204, 240, 285, 330, 385, 440, 506, 572, 650, 728, 819, 910, 1015, 1120, 1240, 1360, 1496, 1632, 1785, 1938, 2109, 2280, 2470, 2660, 2870, 3080, 3311, 3542, 3795, 4048, 4324, 4600, 4900, 5200, 5525, 5850, 6201, 6552, 6930
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Maximal number of inconsistent triples in a tournament on n+2 nodes [Kac]. - corrected by Leen Droogendijk, Nov 10 2014
a(n-4) is the number of aperiodic necklaces (Lyndon words) with 4 black beads and n-4 white beads.
a(n-3) is the maximum number of squares that can be formed from n lines, for n>=3. - Erich Friedman; corrected by Leen Droogendijk, Nov 10 2014
Number of trees with diameter 4 where at most 2 vertices 1 away from the graph center have degree > 2. - Jon Perry, Jul 11 2003
a(n+1) is the number of partitions of n into parts of two kinds, with at most two parts of each kind. Also a(n-3) is the number of partitions of n with Durfee square of size 2. - Franklin T. Adams-Watters, Jan 27 2006
Factoring the g.f. as x/(1-x)^2 times 1/(1-x^2)^2 we find that the sequence equals (1, 2, 3, 4, ...) convolved with (1, 0, 2, 0, 3, 0, 4, ...), A000027 convolved with its aerated variant. - Gary W. Adamson, May 01 2009
Starting with "1" = triangle A171238 * [1,2,3,...]. - Gary W. Adamson, Dec 05 2009
The Kn21, Kn22, Kn23, Fi2 and Ze2 triangle sums, see A180662 for their definitions, of the Connell-Pol triangle A159797 are linear sums of shifted versions of this sequence, e.g., Kn22(n) = a(n+1) + a(n) + 2*a(n-1) + a(n-2) and Fi2(n) = a(n) + 4*a(n-1) + a(n-2). - Johannes W. Meijer, May 20 2011
For n>3, a(n-4) is the number of (w,x,y,z) having all terms in {1,...,n} and w+x+y+z=|x-y|+|y-z|. - Clark Kimberling, May 23 2012
a(n) is the number of (w,x,y) having all terms in {0,...,n} and w+x+y < |w-x|+|x-y|. - Clark Kimberling, Jun 13 2012
For n>0 number of inequivalent (n-1) X 2 binary matrices, where equivalence means permutations of rows or columns or the symbol set. - Alois P. Heinz, Aug 17 2014
Number of partitions p of n+5 such that p[3] = 2. Examples: a(1)=1 because we have (2,2,2); a(2)=2 because we have (2,2,2,1) and (3,2,2); a(3)=5 because we have (2,2,2,1,1), (2,2,2,2), (3,2,2,1), (3,3,2), and (4,2,2). See the R. P. Stanley reference. - Emeric Deutsch, Oct 28 2014
Sum over each antidiagonal of A243866. - Christopher Hunt Gribble, Apr 02 2015
Number of nonisomorphic outer planar graphs of order n>=3, size n+2, and maximum degree 3. - Christian Barrientos and Sarah Minion, Feb 27 2018
a(n) is the number of 2413-avoiding odd Grassmannian permutations of size n+1. - Juan B. Gil, Mar 09 2023

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 8*x^4 + 14*x^5 + 20*x^6 + 30*x^7 + 40*x^8 + 55*x^9 + ...
From _Gus Wiseman_, Apr 06 2019: (Start)
The a(4 - 3) = 1 through a(8 - 3) = 14 integer partitions with Durfee square of length 2 are the following (see Franklin T. Adams-Watters's second comment). The Heinz numbers of these partitions are given by A325164.
  (22)  (32)   (33)    (43)     (44)
        (221)  (42)    (52)     (53)
               (222)   (322)    (62)
               (321)   (331)    (332)
               (2211)  (421)    (422)
                       (2221)   (431)
                       (3211)   (521)
                       (22111)  (2222)
                                (3221)
                                (3311)
                                (4211)
                                (22211)
                                (32111)
                                (221111)
The a(0 + 1) = 1 through a(4 + 1) = 14 integer partitions of n into parts of two kinds with at most two parts of each kind are the following (see Franklin T. Adams-Watters's first comment).
  ()()  ()(1)  ()(2)   ()(3)    ()(4)
        (1)()  (2)()   (3)()    (4)()
               ()(11)  (1)(2)   (1)(3)
               (1)(1)  ()(21)   ()(22)
               (11)()  (2)(1)   (2)(2)
                       (21)()   (22)()
                       (1)(11)  ()(31)
                       (11)(1)  (3)(1)
                                (31)()
                                (11)(2)
                                (1)(21)
                                (2)(11)
                                (21)(1)
                                (11)(11)
The a(6 - 5) = 1 through a(10 - 5) = 14 integer partitions whose third part is 2 are the following (see Emeric Deutsch's comment). The Heinz numbers of these partitions are given by A307373.
  (222)  (322)   (332)    (432)     (442)
         (2221)  (422)    (522)     (532)
                 (2222)   (3222)    (622)
                 (3221)   (3321)    (3322)
                 (22211)  (4221)    (4222)
                          (22221)   (4321)
                          (32211)   (5221)
                          (222111)  (22222)
                                    (32221)
                                    (33211)
                                    (42211)
                                    (222211)
                                    (322111)
                                    (2221111)
(End)

References

  • J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 147.
  • M. Kac, An example of "counting without counting", Philips Res. Reports, 30 (1975), 20*-22* [Special issue in honour of C. J. Bouwkamp].
  • E. V. McLaughlin, Numbers of factorizations in non-unique factorial domains, Senior Thesis, Allegeny College, Meadville, PA, 2004.
  • K. B. Reid and L. W. Beineke "Tournaments", pp. 169-204 in L. W. Beineke and R. J. Wilson, editors, Selected Topics in Graph Theory, Academic Press, NY, 1978, p. 186, Theorem 6.11.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 1, 2nd ed., 2012, Exercise 4.16, pp. 530, 552.
  • W. A. Whitworth, DCC Exercises in Choice and Chance, Stechert, NY, 1945, p. 33.

Links

Crossrefs

Cf. A000031, A001037, A028723, A051168. a(n) = T(n,4), array T as in A051168.
Cf. A000094.
Cf. A171238. - Gary W. Adamson, Dec 05 2009
Row sums of A173997. - Gary W. Adamson, Mar 05 2010
Column k=2 of A242093. Column k=2 of A115720 and A115994.
Cf. A117485, A257990, A307373, A325164, A325168.

Programs

  • Haskell
    a006918 n = a006918_list !! n
a006918_list = scanl (+) 0 a008805_list
-- Reinhard Zumkeller, Feb 01 2013
  • Magma
    [Floor(Binomial(n+4, 4)/(n+4))-Floor((n+2)/8)*(1+(-1)^n)/2: n in [0..60]]; // Vincenzo Librandi, Nov 10 2014
  • Maple
    with(combstruct):ZL:=[st,{st=Prod(left,right),left=Set(U,card=r),right=Set(U,card=r),U=Sequence(Z,card>=3)}, unlabeled]: subs(r=1,stack): seq(count(subs(r=2,ZL),size=m),m=11..58) ; # Zerinvary Lajos, Mar 09 2007
A006918 := proc(n)
    if type(n,'even') then
        n*(n+2)*(n+4)/24 ;
    else
        binomial(n+3,3)/4 ;
    fi ;
end proc: # R. J. Mathar, May 17 2016
  • Mathematica
    f[n_]:=If[EvenQ[n],(n(n+2)(n+4))/24,Binomial[n+3,3]/4]; Join[{0},Array[f,60]]  (* Harvey P. Dale, Apr 20 2011 *)
durf[ptn_]:=Length[Select[Range[Length[ptn]],ptn[[#]]>=#&]];
Table[Length[Select[IntegerPartitions[n],durf[#]==2&]],{n,0,30}] (* Gus Wiseman, Apr 06 2019 *)
  • PARI
    { parttrees(n)=local(pt,k,nk); if (n%2==0, pt=(n/2+1)^2, pt=ceil(n/2)*(ceil(n/2)+1)); pt+=floor(n/2); for (x=1,floor(n/2),pt+=floor(x/2)+floor((n-x)/2)); if (n%2==0 && n>2, pt-=floor(n/4)); k=1; while (3*k<=n, for (x=k,floor((n-k)/2), pt+=floor(k/2); if (x!=k, pt+=floor(x/2)); if ((n-x-k)!=k && (n-x-k)!=x, pt+=floor((n-x-k)/2))); k++); pt }
  • PARI
    {a(n) = n += 2; (n^3 - n * (2-n%2)^2) / 24}; /* Michael Somos, Aug 15 2009 */

Formula

G.f.: x/((1-x)^2*(1-x^2)^2) = x/((1+x)^2*(1-x)^4).
0, 0, 0, 1, 2, 5, 8, 14, ... has a(n) = (Sum_{k=0..n} floor(k(n-k)/2))/2. - Paul Barry, Sep 14 2003
0, 0, 0, 0, 0, 1, 2, 5, 8, 14, 20, 30, 40, 55, ... has a(n) = binomial(floor(1/2 n), 3) + binomial(floor(1/2 n + 1/2), 3) [Eke]. - N. J. A. Sloane, May 12 2012
a(0)=0, a(1)=1, a(n) = (2/(n-1))*a(n-1) + ((n+3)/(n-1))*a(n-2). - Benoit Cloitre, Jun 28 2004
a(n) = floor(binomial(n+4, 4)/(n+4)) - floor((n+2)/8)(1+(-1)^n)/2. - Paul Barry, Jan 01 2005
a(n+1) = a(n) + binomial(floor(n/2)+2,2), i.e., first differences are A008805. Convolution of A008619 with itself, then shifted right (or A004526 with itself, shifted left by 3). - Franklin T. Adams-Watters, Jan 27 2006
a(n+1) = (A027656(n) + A003451(n+5))/2 with a(1)=0. - Yosu Yurramendi, Sep 12 2008
Linear recurrence: a(n) = 2a(n-1) + a(n-2) - 4a(n-3) + a(n-4) + 2a(n-5) - a(n-6). - Jaume Oliver Lafont, Dec 05 2008
Euler transform of length 2 sequence [2, 2]. - Michael Somos, Aug 15 2009
a(n) = -a(-4-n) for all n in Z.
a(n+1) + a(n) = A002623(n). - Johannes W. Meijer, May 20 2011
a(n) = (n+2)*(2*n*(n+4)-3*(-1)^n+3)/48. - Bruno Berselli, May 21 2011
a(2n) = A007290(n+2). - Jon Perry, Nov 10 2014
G.f.: (1/(1-x)^4-1/(1-x^2)^2)/4. - Herbert Kociemba, Oct 23 2016
E.g.f.: (x*(18 + 9*x + x^2)*cosh(x) + (6 + 15*x + 9*x^2 + x^3)*sinh(x))/24. - Stefano Spezia, Dec 07 2021
From Amiram Eldar, Mar 20 2022: (Start)
Sum_{n>=1} 1/a(n) = 75/4 - 24*log(2).
Sum_{n>=1} (-1)^(n+1)/a(n) = 69/4 - 24*log(2). (End)

A007306 Denominators of Farey tree fractions (i.e., the Stern-Brocot subtree in the range [0,1]).

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 5, 4, 5, 7, 8, 7, 7, 8, 7, 5, 6, 9, 11, 10, 11, 13, 12, 9, 9, 12, 13, 11, 10, 11, 9, 6, 7, 11, 14, 13, 15, 18, 17, 13, 14, 19, 21, 18, 17, 19, 16, 11, 11, 16, 19, 17, 18, 21, 19, 14, 13, 17, 18, 15, 13, 14, 11, 7, 8, 13, 17, 16, 19, 23, 22, 17, 19, 26, 29, 25, 24
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Also number of odd entries in n-th row of triangle of Stirling numbers of the second kind (A008277). - Benoit Cloitre, Feb 28 2004
Apparently (except for the first term) the number of odd entries in the alternated diagonals of Pascal's triangle at 45 degrees slope. - Javier Torres (adaycalledzero(AT)hotmail.com), Jul 26 2009
The Kn3 and Kn4 triangle sums, see A180662 for their definitions, of Sierpiński's triangle A047999 equal a(n+1). - Johannes W. Meijer, Jun 05 2011
From Yosu Yurramendi, Jun 23 2014: (Start)
If the terms (n>1) are written as an array:
2,
3, 3,
4, 5, 5, 4,
5, 7, 8, 7, 7, 8, 7, 5,
6, 9, 11, 10, 11, 13, 12, 9, 9, 12, 13, 11, 10, 11, 9, 6,
7, 11, 14, 13, 15, 18, 17, 13, 14, 19, 21, 18, 17, 19, 16, 11, 11, 16, 19,17,18,
then the sum of the k-th row is 2*3^(k-2), each column is an arithmetic progression. The differences of the arithmetic progressions give the sequence itself (a(2^(m+1)+1+k) - a(2^m+1+k) = a(k+1), m >= 1, 1 <= k <= 2^m), because a(n) = A002487(2*n-1) and A002487 has these properties. A071585 also has these properties. Each row is a palindrome: a(2^(m+1)+1-k) = a(2^m+k), m >= 0, 1 <= k <= 2^m.
If the terms (n>0) are written in this way:
1,
2, 3,
3, 4, 5, 5,
4, 5, 7, 8, 7, 7, 8, 7,
5, 6, 9, 11, 10, 11, 13, 12, 9, 9, 12, 13, 11, 10, 11, 9,
6, 7, 11, 14, 13, 15, 18, 17, 13, 14, 19, 21, 18, 17, 19, 16, 11, 11, 16, 19,
each column is an arithmetic progression and the steps also give the sequence itself (a(2^(m+1)+k) - a(2^m+k) = a(k), m >= 0, 0 <= k < 2^m). Moreover, by removing the first term of each column:
a(2^(m+1)+k) = A049448(2^m+k+1), m >= 0, 0 <= k < 2^m.
(End)
k > 1 occurs in this sequence phi(k) = A000010(k) times. - Franklin T. Adams-Watters, May 25 2015
Except for the initial 1, this is the odd bisection of A002487. The even bisection of A002487 is A002487 itself. - Franklin T. Adams-Watters, May 25 2015
For all m >= 0, max_{k=1..2^m} a(2^m+k) = A000045(m+3) (Fibonacci sequence). - Yosu Yurramendi, Jun 05 2016
For all n >= 2, max(m: a(2^m+k) = n, 1<=k<=2^m) = n-2. - Yosu Yurramendi, Jun 05 2016
a(2^m+1) = m+2, m >= 0; a(2^m+2) = 2m+1, m>=1; min_{m>=0, k=1..2^m} a(2^m+k) = m+2; min_{m>=2, k=2..2^m-1} a(2^m+k) = 2m+1. - Yosu Yurramendi, Jun 06 2016
a(2^(m+2) + 2^(m+1) - k) - a(2^(m+1) + 2^m-k) = 2*a(k+1), m >= 0, 0 <= k <= 2^m. - Yosu Yurramendi, Jun 09 2016
If the initial 1 is omitted, this is the number of nonzero entries in row n of the generalized Pascal triangle P_2, see A282714 [Leroy et al., 2017]. - N. J. A. Sloane, Mar 02 2017
Apparently, this sequence was introduced by Johann Gustav Hermes in 1894. His paper gives a strong connection between this sequence and the so-called "Gaussian brackets" ("Gauss'schen Klammer"). For an independent discussion about Gaussian brackets, see the relevant MathWorld article and the article by Herzberger (1943). Srinivasan (1958) gave another, more modern, explanation of the connection between this sequence and the Gaussian brackets. (Parenthetically, J. G. Hermes is the mathematician who completed or constructed the regular polygon with 65537 sides.) - Petros Hadjicostas, Sep 18 2019

Examples

			[ 0/1; 1/1; ] 1/2; 1/3, 2/3; 1/4, 2/5, 3/5, 3/4; 1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5; ...

References

  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 61.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 158.
  • J. C. Lagarias, Number Theory and Dynamical Systems, pp. 35-72 of S. A. Burr, ed., The Unreasonable Effectiveness of Number Theory, Proc. Sympos. Appl. Math., 46 (1992). Amer. Math. Soc.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

For numerators see A007305.
Cf. A001222, A002487, A006842, A006843, A047679, A054424, A065674-A065675, A065810, A260443, A277324, A277328, A283986, A283987, A283988, A284009, A284265, A284266, A284267, A284268, A284565, A284566, A285106, A285107, A285108, A287731, A287732.
See also A282714.

Programs

  • Magma
    [1] cat [&+[Binomial(n+k,2*k) mod 2: k in [0..n]]: n in [0..80]]; // Vincenzo Librandi, Jun 10 2019
  • Maple
    A007306 := proc(n): if n=0 then 1 else A002487(2*n-1) fi: end: A002487 := proc(m) option remember: local a, b, n; a := 1; b := 0; n := m; while n>0 do if type(n, odd) then b := a + b else a := a + b end if; n := floor(n/2); end do; b; end proc: seq(A007306(n),n=0..77); # Johannes W. Meijer, Jun 05 2011
  • Mathematica
    a[0] = 1; a[n_] := Sum[ Mod[ Binomial[n+k-1, 2k] , 2], {k, 0, n}]; Table[a[n], {n, 0, 77}] (* Jean-François Alcover, Dec 16 2011, after Paul Barry *)
a[0] = 0; a[1] = 1;
Flatten[{1,Table[a[2*n] = a[n]; a[2*n + 1] = a[n] + a[n + 1], {n, 0, 50}]}] (* Horst H. Manninger, Jun 09 2021 *)
  • PARI
    {a(n) = if( n<1, n==0, n--; sum( k=0, n, binomial( n+k, n-k)%2))};
  • PARI
    {a(n) = my(m); if( n<2, n>=0, m = 2^length( binary( n-1)); a(n - m/2) + a(m-n+1))}; /* Michael Somos, May 30 2005 */
  • Python
    from sympy import binomial
def a(n):
    return 1 if n<1 else sum(binomial(n + k - 1, 2*k) % 2 for k in range(n + 1))
print([a(n) for n in range(101)]) # Indranil Ghosh, Mar 22 2017
  • Python
    from functools import reduce
def A007306(n): return sum(reduce(lambda x,y:(x[0],sum(x)) if int(y) else (sum(x),x[1]),bin((n<<1)-1)[-1:2:-1],(1,0))) if n else 1 # Chai Wah Wu, May 18 2023
  • R
    maxrow <- 6 # by choice
a <- c(1,2)
for(m in 0:maxrow) for(k in 1:2^m){
  a[2^(m+1)+k  ] <- a[2^m+k] + a[k]
  a[2^(m+1)-k+1] <- a[2^m+k]
}
a
# Yosu Yurramendi, Jan 05 2015
  • R
    # Given n, compute directly a(n)
# by taking into account the binary representation of n-1
# aa <- function(n){
  b <- as.numeric(intToBits(n))
  l <- sum(b)
  m <- which(b == 1)-1
  d <- 1
  if(l > 1) for(j in 1:(l-1)) d[j] <- m[j+1]-m[j]+1
  f <- c(1,m[1]+2) # In A002487: f <- c(0,1)
  if(l > 1) for(j in 3:(l+1)) f[j] <- d[j-2]*f[j-1]-f[j-2]
  return(f[l+1])
}
# a(0) = 1, a(1) = 1, a(n) = aa(n-1)   n > 1
#
# Example
n <- 73
aa(n-1)
#
# Yosu Yurramendi, Dec 15 2016
  • Sage
    @CachedFunction
def a(n):
    return a((odd_part(n-1)+1)/2)+a((odd_part(n)+1)/2) if n>1 else 1
[a(n) for n in (0..77)] # after Alessandro De Luca, Peter Luschny, May 20 2014
  • Sage
    def A007306(n):
    if n == 0: return 1
    M = [1, 1]
    for b in (n-1).bits():
        M[b] = M[0] + M[1]
    return M[1]
print([A007306(n) for n in (0..77)]) # Peter Luschny, Nov 28 2017
  • Scheme
    (define (A007306 n) (if (zero? n) 1 (A002487 (+ n n -1)))) ;; Code for A002487 given in that entry. - Antti Karttunen, Mar 21 2017

Formula

Recurrence: a(0) to a(8) are 1, 1, 2, 3, 3, 4, 5, 5, 4; thereafter a(n) = a(n-2^p) + a(2^(p+1)-n+1), where 2^p < n <= 2^(p+1). [J. Hermes, Math. Ann., 1894; quoted by Dickson, Vol. 1, p. 158] - N. J. A. Sloane, Mar 24 2019
a(4*n) = -a(n)+2*a(2*n); a(4*n+1) = -a(n)+a(2*n)+a(2*n+1); a(4*n+2)=a(n)-a(2*n)+2*a(2*n+1); a(4*n+3) = 4*a(n)-4*a(2*n)+3*a(2*n+1). Thus a(n) is a 2-regular sequence. - Jeffrey Shallit, Dec 26 2024
For n > 0, a(n) = A002487(n-1) + A002487(n) = A002487(2*n-1).
a(0) = 1; a(n) = Sum_{k=0..n-1} C(n-1+k, n-1-k) mod 2, n > 0. - Benoit Cloitre, Jun 20 2003
a(n+1) = Sum_{k=0..n} binomial(2*n-k, k) mod 2; a(n) = 0^n + Sum_{k=0..n-1} binomial(2(n-1)-k, k) mod 2. - Paul Barry, Dec 11 2004
a(n) = Sum_{k=0..n} C(n+k,2*k) mod 2. - Paul Barry, Jun 12 2006
a(0) = a(1) = 1; a(n) = a(A003602(n-1)) + a(A003602(n)), n > 1. - Alessandro De Luca, May 08 2014
a(n) = A007305(n+(2^m-1)), m=A029837(n), n=1,2,3,... . - Yosu Yurramendi, Jul 04 2014
a(n) = A007305(2^(m+1)-n) - A007305(2^m-n), m >= (A029837(n)+1), n=1,2,3,... - Yosu Yurramendi, Jul 05 2014
a(2^m) = m+1, a(2^m+1) = m+2 for m >= 0. - Yosu Yurramendi, Jan 01 2015
a(n+2) = A007305(n+2) + A047679(n) n >= 0. - Yosu Yurramendi, Mar 30 2016
a(2^m+2^r+k) = a(2^r+k)(m-r+1) - a(k), m >= 2, 0 <= r <= m-1, 0 <= k < 2^r. Example: a(73) = a(2^6+2^3+1) = a(2^3+1)*(6-3+1) - a(1) = 5*4 - 1 = 19 . - Yosu Yurramendi, Jul 19 2016
From Antti Karttunen, Mar 21 2017 & Apr 12 2017: (Start)
For n > 0, a(n) = A001222(A277324(n-1)) = A001222(A260443(n-1)*A260443(n)).
The following decompositions hold for all n > 0:
a(n) = A277328(n-1) + A284009(n-1).
a(n) = A283986(n) + A283988(n) = A283987(n) + 2*A283988(n).
a(n) = 2*A284265(n-1) + A284266(n-1).
a(n) = A284267(n-1) + A284268(n-1).
a(n) = A284565(n-1) + A284566(n-1).
a(n) = A285106(n-1) + A285108(n-1) = A285107(n-1) + 2*A285108(n-1). (End)
a(A059893(n)) = a(n+1) for n > 0. - Yosu Yurramendi, May 30 2017
a(n) = A287731(n) + A287732(n) for n > 0. - I. V. Serov, Jun 09 2017
a(n) = A287896(n) + A288002(n) for n > 1.
a(n) = A287896(n-1) + A002487(n-1) - A288002(n-1) for n > 1.
a(n) = a(n-1) + A002487(n-1) - 2*A288002(n-1) for n > 1. - I. V. Serov, Jun 14 2017
From Yosu Yurramendi, May 14 2019: (Start)
For m >= 0, M >= m, 0 <= k < 2^m,
a((2^(m+1) + A119608(2^m+k+1))*2^(M-m) - A000035(2^m+k)) =
a((2^(m+2) - A119608(2^m+k+1))*2^(M-m) - A000035(2^m+k)-1) =
a(2^(M+2) - (2^m+k)) = a(2^(M+1) + (2^m+k) + 1) =
a(2^m+k+1)*(M-m) + a(2^(m+1)+2^m+k+1). (End)
a(k) = sqrt(A007305(2^(m+1)+k)*A047679(2^(m+1)+k-2) - A007305(2^m+k)*A047679(2^m+k-2)), m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Jun 09 2019
G.f.: 1 + x * (1 + x) * Product_{k>=0} (1 + x^(2^k) + x^(2^(k+1))). - Ilya Gutkovskiy, Jul 19 2019
Conjecture: a(n) = a(n-1) + b(n-1) - 2*(a(n-1) mod b(n-1)) for n > 1 with a(0) = a(1) = 1 where b(n) = a(n) - b(n-1) for n > 1 with b(1) = 1. - Mikhail Kurkov, Mar 13 2022

Extensions

Formula fixed and extended by Franklin T. Adams-Watters, Jul 07 2009
Incorrect Maple program removed by Johannes W. Meijer, Jun 05 2011

A065941 T(n,k) = binomial(n-floor((k+1)/2), floor(k/2)). Triangle read by rows, for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 4, 3, 3, 1, 1, 1, 5, 4, 6, 3, 1, 1, 1, 6, 5, 10, 6, 4, 1, 1, 1, 7, 6, 15, 10, 10, 4, 1, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1, 1, 1, 11, 10, 45, 36, 84, 56, 70, 35, 21, 6, 1
Offset: 0

Views

Author

Len Smiley, Nov 29 2001

Keywords

Comments

Also the q-Stirling2 numbers at q = -1. - Peter Luschny, Mar 09 2020
Row sums give the Fibonacci sequence. So do the alternating row sums.
Triangle of coefficients of polynomials defined by p(-1,x) = p(0,x) = 1, p(n, x) = x*p(n-1, x) + p(n-2, x), for n >= 1. - Benoit Cloitre, May 08 2005 [rewritten with correct offset. - Wolfdieter Lang, Feb 18 2020]
Another version of triangle in A103631. - Philippe Deléham, Jan 01 2009
The T(n,k) coefficients appear in appendix 2 of Parks's remarkable article "A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov" if we assume that the b(n) coefficients are all equal to 1 and ignore the first column. The complete version of this triangle including the first column is A103631. - Johannes W. Meijer, Aug 11 2011
Signed ++--++..., the roots are chaotic using f(x) --> x^2 - 2 with cycle lengths shown in A003558 by n-th rows. Example: given row 3, x^3 + x^2 - 2x - 1; the roots are (a = 1.24697, ...; b = -0.445041, ...; c = -1.802937, ...). Then (say using seed b with x^2 - 2) we obtain the trajectory -0.445041, ... -> -1.80193, ... -> 1.24697, ...; matching the entry "3" in A003558(3). - Gary W. Adamson, Sep 06 2011
From Gary W. Adamson, Aug 25 2019: (Start)
Roots to the polynomials and terms in A003558 can all be obtained from the numbers below using a doubling series mod N procedure as follows: (more than one row may result). Any row ends when the trajectory produces a term already used. Then try the next higher odd term not used as the leftmost term, then repeat.
For example, for N = 11, we get: (1, 2, 4, 3, 5), showing that when confronted with two choices after the 4: (8 and -3), pick the smaller (abs) term, = 3. Then for the next row pick 7 (not used) and repeat the algorithm; succeeding only if the trajectory produces new terms. But 7 is also (-4) mod 11 and 4 was used. Therefore what I call the "r-t table" (for roots trajectory) has only one row: (1, 2, 4, 3, 5). Conjecture: The numbers of terms in the first row is equal to A003558 corresponding to N, i.e., 5 in this case with period 2.
Now for the roots to the polynomials. Pick N = 7. The polynomial is x^3 - x^2 - 2x + 1 = 0, with roots 1.8019..., -1.2469... and 0.445... corresponding to 2*cos(j*Pi/N), N = 7, and j = (1, 2, and 3). The terms (1, 2, 3) are the r-t terms for N = 7. For 11, the r-t terms are (1, 2, 4, 3, 5). This implies that given any roots of the corresponding polynomial, they are cyclic using f(x) --> x^2 - 2 with cycle lengths shown in A003558. The terms thus generated are 2*cos(j*Pi), with j = (1, 2, 4, 3, 5). Check: Begin with 2*j*Pi/N, with j = 1 (1.9189...). The other trajectory terms are: --> 1.6825..., --> 0.83083..., -1.3097...; 545...; (a 5 period and cyclic since we can begin with any of the constants). The r-t table for odd N begins as follows:
3...............1
5...............1, 2
7...............1, 2, 3
9...............1, 2, 4
...............3 (singleton terms reduce to "1") (9 has two rows)
11...............1, 2, 4, 3, 5
13...............1, 2, 4, 5, 3, 6
15...............1, 2, 4, 7
................3, 6 (dividing through by the gcd gives (1, 2))
................5. (singleton terms reduce to "1")
The result is that 15 has 3 factors (since 3 rows), and the values of those factors are the previous terms "N", corresponding to the r-t terms in each row. Thus, the first row is new, the second (1, 2), corresponds to N = 5, and the "1" in row 3 corresponds to N = 3. The factors are those values apart from 15 and 1. Note that all of the unreduced r-t terms in all rows for N form a complete set of the terms 1 through (N-1)/2 without duplication. (End)
From Gary W. Adamson, Sep 30 2019: (Start)
The 3 factors of the 7th degree polynomial for 15: (x^7 - x^6 - 6x^5 + 5x^4 + 10x^3 - 6x^2 - 4x + 1) can be determined by getting the roots for 2*cos(j*Pi/1), j = (1, 2, 4, 7) and finding the corresponding polynomial, which is x^4 + x^3 - 4x^2 - 4x + 1. This is the minimal polynomial for N = 15 as shown in Table 2, p. 46 of (Lang). The degree of this polynomial is 4, corresponding to the entry in A003558 for 15, = 4. The trajectories (3, 6) and (5) are j values for 2*cos(j*Pi/15) which are roots to x^2 - x - 1 (relating to the pentagon), and (x - 1), relating to the triangle. (End)
From Gary W. Adamson, Aug 21 2019: (Start)
Matrices M of the form: (1's in the main diagonal, -1's in the subdiagonal, and the rest zeros) are chaotic if we replace (f(x) --> x^2 - 2) with f(x) --> M^2 - 2I, where I is the Identity matrix [1, 0, 0; 0, 1, 0; 0, 0, 1]. For example, with the 3 X 3 matrix M: [0, 0, 1; 0, 1, -1; 1, -1, 0]; the f(x) trajectory is:
....M^2 - 2I: [-1, -1, 0; -1, 0, -1; 0, -1, 0], then for the latter,
....M^2 - 2I: [0, 1, 1; 1, 0, 0; 1, 0, -1]. The cycle ends with period 3 since the next matrix is (-1) * the seed matrix. As in the case with f(x) --> x^2 - 2, the eigenvalues of the 3 chaotic matrices are (abs) 1.24697, 0.44504... and 1.80193, ... Also, the characteristic equations of the 3 matrices are the same as or variants of row 4 of the triangle below: (x^3 + x - 2x - 1) with different signs. (End)
Received from Herb Conn, Jan 2004: (Start)
Let x = 2*cos(2A) (A = Angle); then
sin(A)/sin A = 1
sin(3A)/sin A = x + 1
sin(5A)/sin A = x^2 + x - 1
sin(7A)/sin A = x^3 + x - 2x - 1
sin(9A)/sin A = x^4 + x^3 - 3x^2 - 2x + 1
... (signed ++--++...). (End)
Or Pascal's triangle (A007318) with duplicated diagonals. Also triangle of coefficients of polynomials defined by P_0(x) = 1 and for n>=1, P_n(x) = F_n(x) + F_(n+1)(x), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 12 2012
The matrix inverse is given by
1;
1, 1;
0, -1, 1;
0, 1, -2, 1;
0, 0, 1, -2, 1;
0, 0, -1, 3, -3, 1;
0, 0, 0, -1, 3, -3, 1;
0, 0, 0, 1, -4, 6, -4, 1;
0, 0, 0, 0, 1, -4, 6, -4, 1;
... apart from signs the same as A124645. - R. J. Mathar, Mar 12 2013

Examples

			Triangle T(n, k) begins:
n\k 0  1  2  3   4   5  6   7  8  9 ...
---------------------------------------
[0] 1,
[1] 1, 1,
[2] 1, 1, 1,
[3] 1, 1, 2, 1,
[4] 1, 1, 3, 2,  1,
[5] 1, 1, 4, 3,  3,  1,
[6] 1, 1, 5, 4,  6,  3,  1,
[7] 1, 1, 6, 5, 10,  6,  4,  1,
[8] 1, 1, 7, 6, 15, 10, 10,  4,  1,
[9] 1, 1, 8, 7, 21, 15, 20, 10,  5, 1,
---------------------------------------
From _Gary W. Adamson_, Oct 23 2019: (Start)
Consider the roots of the polynomials corresponding to odd N such that for N=7 the polynomial is (x^3 + x^2 - 2x - 1) and the roots (a, b, c) are (-1.8019377..., 1.247697..., and -0.445041...). The discriminant of a polynomial derived from the roots is the square of the product of successive differences: ((a-b), (b-c), (c-a))^2 in this case, resulting in 49, matching the method derived from the coefficients of a cubic. For our purposes we use the product of the differences, not the square, resulting in (3.048...) * (1.69202...) * (1.35689...) = 7.0. Conjecture: for all polynomials in the set, the product of the differences of the roots = the corresponding N. For N = 7, we get x^3 - 7x + 7. It appears that for all prime N's, these resulting companion polynomials are monic (left coefficient is 1), and all other coefficients are N or multiples thereof, with the rightmost term = N. The companion polynomials for the first few primes are:
  N =  5:  x^2 - 5;
  N =  7:  x^3 - 7x + 7;
  N = 11:  x^5 - 11x^3 + 11x^2 + 11x - 11;
  N = 13:  x^6 - 13x^4 + 13x^3 + 26x^2 - 39x + 13;
  N = 17:  x^8 - 17x^6 + 17x^5 + 68x^4 - 119x^3 + 17x^2 + 51x - 17;
  N = 19:  x^9 - 19x^7 + 19x^6 + 95x^5 - 171x^4 - 19x^3 + 190x^2 - 114x + 19. (End)

Links

Crossrefs

Cf. A065942 (central stalk sequence), A000045 (row sums), A108299.
Reflected version of A046854.
Some triangle sums (see A180662): A000045 (Fi1), A016116 (Kn21), A000295 (Kn23), A094967 (Fi2), A000931 (Ca2), A001519 (Gi3), A000930 (Ze3).
Cf. A003558, A182579, A059841, A333143.

Programs

  • Haskell
    a065941 n k = a065941_tabl !! n !! k
a065941_row n = a065941_tabl !! n
a065941_tabl = iterate (\row ->
   zipWith (+) ([0] ++ row) (zipWith (*) (row ++ [0]) a059841_list)) [1]
-- Reinhard Zumkeller, May 07 2012
  • Magma
    [Binomial(n - Floor((k+1)/2), Floor(k/2)): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 10 2019
  • Maple
    A065941 := proc(n,k): binomial(n-floor((k+1)/2),floor(k/2)) end: seq(seq(A065941(n,k), k=0..n), n=0..15); # Johannes W. Meijer, Aug 11 2011
A065941 := proc(n,k) option remember: local j: if k=0 then 1 elif k=1 then 1: elif k>=2 then add(procname(j,k-2), j=k-2..n-2) fi: end: seq(seq(A065941(n,k), k=0..n), n=0..15);  # Johannes W. Meijer, Aug 11 2011
# The function qStirling2 is defined in A333143.
seq(print(seq(qStirling2(n, k, -1), k=0..n)), n=0..9);
# Peter Luschny, Mar 09 2020
  • Mathematica
    Flatten[Table[Binomial[n-Floor[(k+1)/2],Floor[k/2]],{n,0,15},{k,0,n}]] (* Harvey P. Dale, Dec 11 2011 *)
  • PARI
    T065941(n, k) = binomial(n-(k+1)\2, k\2); \\ Michel Marcus, Apr 28 2014
  • Sage
    [[binomial(n - floor((k+1)/2), floor(k/2)) for k in (0..n)] for n in (0..15)] # G. C. Greubel, Jul 10 2019

Formula

T(n, k) = binomial(n-floor((k+1)/2), floor(k/2)).
As a square array read by antidiagonals, this is given by T1(n, k) = binomial(floor(n/2) + k, k). - Paul Barry, Mar 11 2003
Triangle is a reflection of that in A066170 (absolute values). - Gary W. Adamson, Feb 16 2004
Recurrences: T(k, 0) = 1, T(k, n) = T(k-1, n) + T(k-2, n-2), or T(k, n) = T(k-1, n) + T(k-1, n-1) if n even, T(k-1, n-1) if n odd. - Ralf Stephan, May 17 2004
G.f.: sum[n, sum[k, T(k, n)x^ky^n]] = (1+xy)/(1-y-x^2y^2). sum[n>=0, T(k, n)y^n] = y^k/(1-y)^[k/2]. - Ralf Stephan, May 17 2004
T(n, k) = A108299(n, k)*A087960(k) = abs(A108299(n, k)). - Reinhard Zumkeller, Jun 01 2005
From Johannes W. Meijer, Aug 11 2011: (Start)
T(n,k) = A046854(n, n-k) = abs(A066170(n, n-k)).
T(n+k, n-k) = A109223(n,k).
T(n, k) = sum(T(j, k-2), j=k-2..n-2), 2 <= k <= n, n>=2;
T(n, 0) =1, T(n+1, 1) = 1, n >= 0. (End)
For n > 1: T(n, k) = T(n-2, k) + T(n-1, k), 1 < k < n. - Reinhard Zumkeller, Apr 24 2013
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