A183134 -id:A183134 - cp's OEIS frontend

A039598 Triangle formed from odd-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.

1, 2, 1, 5, 4, 1, 14, 14, 6, 1, 42, 48, 27, 8, 1, 132, 165, 110, 44, 10, 1, 429, 572, 429, 208, 65, 12, 1, 1430, 2002, 1638, 910, 350, 90, 14, 1, 4862, 7072, 6188, 3808, 1700, 544, 119, 16, 1, 16796, 25194, 23256, 15504, 7752, 2907, 798, 152, 18, 1

Offset: 0

N. J. A. Sloane

T(n,k) is the number of leaves at level k+1 in all ordered trees with n+1 edges. - Emeric Deutsch, Jan 15 2005
Riordan array ((1-2x-sqrt(1-4x))/(2x^2),(1-2x-sqrt(1-4x))/(2x)). Inverse array is A053122. - Paul Barry, Mar 17 2005
T(n,k) is the number of walks of n steps, each in direction N, S, W, or E, starting at the origin, remaining in the upper half-plane and ending at height k (see the R. K. Guy reference, p. 5). Example: T(3,2)=6 because we have ENN, WNN, NEN, NWN, NNE and NNW. - Emeric Deutsch, Apr 15 2005
Triangle T(n,k), 0<=k<=n, read by rows given by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = 2*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 30 2007
Number of (2n+1)-step walks from (0,0) to (2n+1,2k+1) and consisting of steps u=(1,1) and d=(1,-1) in which the path stays in the nonnegative quadrant. Examples: T(2,0)=5 because we have uuudd, uudud, uuddu, uduud, ududu; T(2,1)=4 because we have uuuud, uuudu, uuduu, uduuu; T(2,2)=1 because we have uuuuu. - Philippe Deléham, Apr 16 2007, Apr 18 2007
Triangle read by rows: T(n,k)=number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and two types of steps H=(1,0); example: T(3,1)=14 because we have UDU, UUD, 4 HHU paths, 4 HUH paths and 4 UHH paths. - Philippe Deléham, Sep 25 2007
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. - Philippe Deléham, Sep 25 2007
With offset [1,1] this is the (ordinary) convolution triangle a(n,m) with o.g.f. of column m given by (c(x)-1)^m, where c(x) is the o.g.f. for Catalan numbers A000108. See the Riordan comment by Paul Barry.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) with exactly k fixed points. - Abdullahi Umar, Oct 02 2008
T(n,k)/2^(2n+1) = coefficients of the maximally flat lowpass digital differentiator of the order N=2n+3. - Pavel Holoborodko (pavel(AT)holoborodko.com), Dec 19 2008
The signed triangle S(n,k) := (-1)^(n-k)*T(n,k) provides the transformation matrix between f(n,l) := L(2*l)*5^n*F(2*l)^(2*n+1) (F=Fibonacci numbers A000045, L=Lucas numbers A000032) and F(4*l*(k+1)), k = 0, ..., n, for each l>=0: f(n,l) = Sum_{k=0..n} S(n,k)*F(4*l*(k+1)), n>=0, l>=0. Proof: the o.g.f. of the l.h.s., G(l;x) := Sum_{n>=0} f(n,l)*x^n = F(4*l)/(1 - 5*F(2*l)^2*x) is shown to match the o.g.f. of the r.h.s.: after an interchange of the n- and k-summation, the Riordan property of S = (C(x)/x,C(x)) (compare with the above comments by Paul Barry), with C(x) := 1 - c(-x), with the o.g.f. c(x) of A000108 (Catalan numbers), is used, to obtain, after an index shift, first Sum_{k>=0} F(4*l*(k))*GS(k;x), with the o.g.f of column k of triangle S which is GS(k;x) := Sum_{n>=k} S(n,k)*x^n = C(x)^(k+1)/x. The result is GF(l;C(x))/x with the o.g.f. GF(l,x) := Sum_{k>=0} F(4*l*k)*x^k = x*F(4*l)/(1-L(4*l)*x+x^2) (see a comment on A049670, and A028412). If one uses then the identity L(4*n) - 5*F(2*n)^2 = 2 (in Koshy's book [reference under A065563] this is No. 15, p. 88, attributed to Lucas, 1876), the proof that one recovers the o.g.f. of the l.h.s. from above boils down to a trivial identity on the Catalan o.g.f., namely 1/c^2(-x) = 1 + 2*x - (x*c(-x))^2. - Wolfdieter Lang, Aug 27 2012
O.g.f. for row polynomials R(x) := Sum_{k=0..n} a(n,k)*x^k:
  ((1+x) - C(z))/(x - (1+x)^2*z) with C the o.g.f. of A000108 (Catalan numbers). From Riordan ((C(x)-1)/x,C(x)-1), compare with a Paul Barry comment above. This coincides with the o.g.f. given by Emeric Deutsch in the formula section. - Wolfdieter Lang, Nov 13 2012
The A-sequence for this Riordan triangle is [1,2,1] and the Z-sequence is [2,1]. See a W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 13 2012
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n+1, 2*k+1). T(n, k) appears in the formula for the (2*n+1)-th power of the algebraic number rho(N) := 2*cos(Pi/N) = R(N, 2) in terms of the even-indexed diagonal/side length ratios R(N, 2*(k+1)) = S(2*k+1, 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+1) = 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 even powers of rho(n) see A039599. (End)
The tridiagonal Toeplitz production matrix P in the Example section corresponds to the unsigned Cartan matrix for the simple Lie algebra A_n as n tends to infinity (cf. Damianou ref. in A053122). -  Tom Copeland, Dec 11 2015 (revised Dec 28 2015)
T(n,k) is the number of pairs of non-intersecting walks of n steps, each in direction N or E, starting at the origin, and such that the end points of the two paths are separated by a horizontal distance of k. See Shapiro 1976. - Peter Bala, Apr 12 2017
Also the convolution triangle of the Catalan numbers A000108. - Peter Luschny, Oct 07 2022

			Triangle T(n,k) starts:
n\k     0      1      2      3      4     5    6    7   8  9 10
0:      1
1:      2      1
2:      5      4      1
3:     14     14      6      1
4:     42     48     27      8      1
5:    132    165    110     44     10     1
6:    429    572    429    208     65    12    1
7:   1430   2002   1638    910    350    90   14    1
8:   4862   7072   6188   3808   1700   544  119   16   1
9:  16796  25194  23256  15504   7752  2907  798  152  18  1
10: 58786  90440  87210  62016  33915 14364 4655 1120 189 20  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 13 2012.
Production matrix begins:
2, 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
0, 0, 0, 0, 0, 0, 1, 2, 1
- _Philippe Deléham_, Nov 07 2011
From _Wolfdieter Lang_, Nov 13 2012: (Start)
Recurrence: T(5,1) = 165 = 1*42 + 2*48 +1*27. The Riordan A-sequence is [1,2,1].
Recurrence from Riordan Z-sequence [2,1]: T(5,0) = 132 = 2*42 + 1*48. (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
  Example for rho(N) = 2*cos(Pi/N) powers:
  n=2: rho(N)^5 = 5*R(N, 2) + 4*R(N, 4) + 1*R(N, 6) = 5*S(1, rho(N)) + 4*S(3, rho(N)) + 1*S(5, rho(N)), identical in N >= 1. For N=5 (the pentagon with only one distinct diagonal) the degree delta(5) = 2, hence R(5, 4) and R(5, 6) can be reduced, namely to R(5, 1) = 1 and R(5, 6) = -R(5,1) = -1, respectively. Thus rho(5)^5 = 5*R(N, 2) + 4*1  + 1*(-1) = 3 + 5*R(N, 2) = 3 + 5*rho(5), with the golden section rho(5). (End)

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Mirror image of A050166. Row sums are A001700.

Cf. A000108, A008313, A039599, A183134, A094527, A033184, A035324, A053122, A172417.

/* As triangle: */ [[Binomial(2*n,n-k) - Binomial(2*n,n-k-2): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2015

T:=(n,k)->binomial(2*n, n-k) - binomial(2*n, n-k-2); # N. J. A. Sloane, Aug 26 2013
# Uses function PMatrix from A357368. Adds row and column above and to the left.
PMatrix(10, n -> binomial(2*n, n) / (n + 1)); # Peter Luschny, Oct 07 2022

Flatten[Table[Binomial[2n, n-k] - Binomial[2n, n-k-2], {n,0,9}, {k,0,n}]] (* Jean-François Alcover, May 03 2011 *)

T(n,k)=binomial(2*n,n-k) - binomial(2*n,n-k-2) \\ Charles R Greathouse IV, Nov 07 2016

# Algorithm of L. Seidel (1877)
# Prints the first n rows of the triangle.
def A039598_triangle(n) :
    D = [0]*(n+2); D[1] = 1
    b = True; h = 1
    for i in range(2*n) :
        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]
        b = not b
        if b : print([D[z] for z in (1..h-1) ])
A039598_triangle(10)  # Peter Luschny, May 01 2012

Row n: C(2n, n-k) - C(2n, n-k-2).
a(n, k) = C(2n+1, n-k)*2*(k+1)/(n+k+2) = A050166(n, n-k) = a(n-1, k-1) + 2*a(n-1, k)+ a (n-1, k+1) [with a(0, 0) = 1 and a(n, k) = 0 if n<0 or nHenry Bottomley, Sep 24 2001
From Philippe Deléham, Feb 14 2004: (Start)
T(n, 0) = A000108(n+1), T(n, k) = 0 if n0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+2) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
Sum_{k>=0} T(m, k)*T(n, k) = A000108(m+n+1). (End)
T(n, k) = A009766(n+k+1, n-k) = A033184(n+k+2, 2k+2). - Philippe Deléham, Feb 14 2004
Sum_{j>=0} T(k, j)*A039599(n-k, j) = A028364(n, k). - Philippe Deléham, Mar 04 2004
Antidiagonal Sum_{k=0..n} T(n-k, k) = A000957(n+3). - Gerald McGarvey, Jun 05 2005
The triangle may also be generated from M^n * [1,0,0,0,...], where M = an infinite tridiagonal matrix with 1's in the super- and subdiagonals and [2,2,2,...] in the main diagonal. - Gary W. Adamson, Dec 17 2006
G.f.: G(t,x) = C^2/(1-txC^2), where C = (1-sqrt(1-4x))/(2x) is the Catalan function. From here G(-1,x)=C, i.e., the alternating row sums are the Catalan numbers (A000108). - Emeric Deutsch, Jan 20 2007
Sum_{k=0..n} T(n,k)*x^k = A000957(n+1), A000108(n), A000108(n+1), A001700(n), A049027(n+1), A076025(n+1), A076026(n+1) for x=-2,-1,0,1,2,3,4 respectively (see square array in A067345). - Philippe Deléham, Mar 21 2007, Nov 04 2011
Sum_{k=0..n} T(n,k)*(k+1) = 4^n. - Philippe Deléham, Mar 30 2007
Sum_{j>=0} T(n,j)*binomial(j,k) = A035324(n,k), A035324 with offset 0 (0 <= k <= n). - Philippe Deléham, Mar 30 2007
T(n,k) = A053121(2*n+1,2*k+1). - Philippe Deléham, Apr 16 2007, Apr 18 2007
T(n,k) = A039599(n,k) + A039599(n,k+1). - Philippe Deléham, Sep 11 2007
Sum_{k=0..n+1} T(n+1,k)*k^2 = A029760(n). - Philippe Deléham, Dec 16 2007
Sum_{k=0..n} T(n,k)*A059841(k) = A000984(n). - Philippe Deléham, Nov 12 2008
G.f.: 1/(1-xy-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-.... (continued fraction).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A001700(n), A194723(n+1), A194724(n+1), A194725(n+1), A194726(n+1), A194727(n+1), A194728(n+1), A194729(n+1), A194730(n+1) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Nov 03 2011
From Peter Bala, Dec 21 2014: (Start)
This triangle factorizes in the Riordan group as ( C(x), x*C(x) ) * ( 1/(1 - x), x/(1 - x) ) = A033184 * A007318, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
Let U denote the lower unit triangular array with 1's on or below the main diagonal and zeros elsewhere. For k = 0,1,2,... define U(k) to be the lower unit triangular block array
/I_k 0\
\ 0  U/ having the k X k identity matrix I_k as the upper left block; in particular, U(0) = U. Then this array equals the bi-infinite product (...*U(2)*U(1)*U(0))*(U(0)*U(1)*U(2)*...). (End)
From Peter Bala, Jul 21 2015: (Start)
O.g.f. G(x,t) = (1/x) * series reversion of ( x/f(x,t) ), where f(x,t) = ( 1 + (1 + t)*x )^2/( 1 + t*x ).
1 + x*d/dx(G(x,t))/G(x,t) = 1 + (2 + t)*x + (6 + 4*t + t^2)*x^2 + ... is the o.g.f for A094527. (End)
Conjecture: Sum_{k=0..n} T(n,k)/(k+1)^2 = H(n+1)*A000108(n)*(2*n+1)/(n+1), where H(n+1) = Sum_{k=0..n} 1/(k+1). - Werner Schulte, Jul 23 2015
From Werner Schulte, Jul 25 2015: (Start)
Sum_{k=0..n} T(n,k)*(k+1)^2 = (2*n+1)*binomial(2*n,n). (A002457)
Sum_{k=0..n} T(n,k)*(k+1)^3 = 4^n*(3*n+2)/2.
Sum_{k=0..n} T(n,k)*(k+1)^4 = (2*n+1)^2*binomial(2*n,n).
Sum_{k=0..n} T(n,k)*(k+1)^5 = 4^n*(15*n^2+15*n+4)/4. (End)
The o.g.f. G(x,t) is such that G(x,t+1) is the o.g.f. for A035324, but with an offset of 0, and G(x,t-1) is the o.g.f. for A033184, again with an offset of 0. - Peter Bala, Sep 20 2015
Denote this lower triangular array by L; then L * transpose(L) is the Cholesky factorization of the Hankel matrix ( 1/(i+j)*binomial(2*i+2*j-2, i+j-1) )A172417%20read%20as%20a%20square%20array.%20See%20Chamberland,%20p.%201669.%20-%20_Peter%20Bala">i,j >= 1 = A172417 read as a square array. See Chamberland, p. 1669. - _Peter Bala, Oct 15 2023

Typo in one entry corrected by Philippe Deléham, Dec 16 2007

A183135 Square array A(n,k) by antidiagonals. A(n,k) is the number of length 2n k-ary words (n,k>=0) that can be built by repeatedly inserting doublets into the initially empty word.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 6, 1, 0, 1, 4, 15, 20, 1, 0, 1, 5, 28, 87, 70, 1, 0, 1, 6, 45, 232, 543, 252, 1, 0, 1, 7, 66, 485, 2092, 3543, 924, 1, 0, 1, 8, 91, 876, 5725, 19864, 23823, 3432, 1, 0, 1, 9, 120, 1435, 12786, 71445, 195352, 163719, 12870, 1, 0

Offset: 0

Alois P. Heinz, Dec 26 2010

A(n,k) is also the number of rooted closed walks of length 2n on the infinite rooted k-ary tree. - Danny Rorabaugh, Oct 31 2017

A(n,2k) is the number of unreduced words of length 2n that reduce to the empty word in the free group with k generators. - Danny Rorabaugh, Nov 09 2017

			A(2,2) = 6, because 6 words of length 4 can be built over 2-letter alphabet {a, b} by repeatedly inserting doublets (words with two equal letters) into the initially empty word: aaaa, aabb, abba, baab, bbaa, bbbb.
Square array A(n,k) begins:
  1,  1,   1,    1,     1,     1,  ...
  0,  1,   2,    3,     4,     5,  ...
  0,  1,   6,   15,    28,    45,  ...
  0,  1,  20,   87,   232,   485,  ...
  0,  1,  70,  543,  2092,  5725,  ...
  0,  1, 252, 3543, 19864, 71445,  ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
Jason Bell, Marni Mishna, On the Complexity of the Cogrowth Sequence, arXiv:1805.08118 [math.CO], 2018.
Beth Bjorkman et al., k-foldability of words, arXiv preprint arXiv:1710.10616 [math.CO], 2017.

Columns k=0-10 give: A000007, A000012, A000984, A089022, A035610, A130976, A130977, A130978, A130979, A130980, A131521.
Rows n=0-3 give: A000012, A001477, A000384, A027849(k-1) for k>0.
Main diagonal gives A294491.
Coefficients of row polynomials in k, (k-1) are given by A157491, A039599.
Cf. A007318, A183134, A256116, A256117.

A:= proc(n, k) local j;
      if n=0 then 1
             else k/n *add(binomial(2*n,j) *(n-j) *(k-1)^j, j=0..n-1)
      fi
    end:
seq(seq(A(n, d-n), n=0..d), d=0..10);

A[, 1] = 1; A[n, k_] := If[n == 0, 1, k/n*Sum[Binomial[2*n, j]*(n - j)*(k - 1)^j, {j, 0, n - 1}]]; Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Dec 27 2013, translated from Maple *)

A(n,k) = k/n * Sum_{j=0..n-1} C(2*n,j)*(n-j)*(k-1)^j if n>0, A(0,k) = 1.
A(n,k) = A183134(n,k) if n=0 or k<2, A(n,k) = A183134(n,k)*k otherwise.
G.f. of column k: 1/(1-k*x) if k<2, 2*(k-1)/(k-2+k*sqrt(1-(4*k-4)*x)) otherwise.

A213027 Number A(n,k) of 3n-length k-ary words, either empty or beginning with the first letter of the alphabet, that can be built by repeatedly inserting triples of identical letters into the initially empty word; square array A(n,k), n>=0, k>=0, by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 4, 1, 0, 1, 1, 7, 19, 1, 0, 1, 1, 10, 61, 98, 1, 0, 1, 1, 13, 127, 591, 531, 1, 0, 1, 1, 16, 217, 1810, 6101, 2974, 1, 0, 1, 1, 19, 331, 4085, 27631, 65719, 17060, 1, 0, 1, 1, 22, 469, 7746, 82593, 441604, 729933, 99658, 1, 0

Offset: 0

Alois P. Heinz, Jun 03 2012

In general, column k > 1 is asymptotic to a(n) ~ 3^(3*n+1/2) * (k-1)^(n+1) / (sqrt(Pi) * (2*k-3)^2 * 4^n * n^(3/2)). - Vaclav Kotesovec, Aug 31 2014

			A(0,k) = 1: the empty word.
A(n,1) = 1: (aaa)^n.
A(2,2) = 4: there are 4 words of length 6 over alphabet {a,b}, either empty or beginning with the first letter of the alphabet, that can be built by repeatedly inserting triples of identical letters into the initially empty word: aaaaaa, aaabbb, aabbba, abbbaa.
A(2,3) = 7: aaaaaa, aaabbb, aaaccc, aabbba, aaccca, abbbaa, acccaa.
A(3,2) = 19: aaaaaaaaa, aaaaaabbb, aaaaabbba, aaaabbbaa, aaabaaabb, aaabbaaab, aaabbbaaa, aaabbbbbb, aabaaabba, aabbaaaba, aabbbaaaa, aabbbabbb, aabbbbbba, abaaabbaa, abbaaabaa, abbbaaaaa, abbbaabbb, abbbabbba, abbbbbbaa.
Square array A(n,k) begins:
  1, 1,    1,     1,      1,       1,       1, ...
  0, 1,    1,     1,      1,       1,       1, ...
  0, 1,    4,     7,     10,      13,      16, ...
  0, 1,   19,    61,    127,     217,     331, ...
  0, 1,   98,   591,   1810,    4085,    7746, ...
  0, 1,  531,  6101,  27631,   82593,  195011, ...
  0, 1, 2974, 65719, 441604, 1751197, 5153626, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0-10 give: A000007, A000012, A047099, A218473, A218474, A218475, A218476, A218477, A218478, A218479, A218480.
Rows n=0-3 give: A000012, A057427, A016777(k-1), A127854(k-1).
Main diagonal gives: A218472.
Cf. A183134, A183135, A213028.

A:= (n, k)-> `if`(n=0, 1, `if`(k<2, k,
    1/n *add(binomial(3*n, j) *(n-j) *(k-1)^j, j=0..n-1))):
seq(seq(A(n, d-n), n=0..d), d=0..12);

a[0, ] = 1; a[, k_ /; k < 2] := k; a[n_, k_] := 1/n*Sum[Binomial[3*n, j]*(n-j)*(k-1)^j, {j, 0, n-1}]; Table[a[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 11 2013 *)

A(n,k) = 1/n * Sum_{j=0..n-1} C(3*n,j) * (n-j) * (k-1)^j if n>0, k>1; A(0,k) = 1; A(n,k) = k if n>0, k<2.

A(n,k) = 1/k * A213028(n,k) if n>0, k>1; else A(n,k) = A213028(n,k).

A194723 Number of ternary words either empty or beginning with the first character of the alphabet, that can be built by inserting n doublets into the initially empty word.

Original entry on oeis.org

1, 1, 5, 29, 181, 1181, 7941, 54573, 381333, 2699837, 19319845, 139480397, 1014536117, 7426790749, 54669443141, 404388938349, 3004139083221, 22402851226749, 167640057210981, 1258340276153229, 9471952718661621, 71481616200910749, 540715584181142661

Offset: 0

Alois P. Heinz, Sep 02 2011

nonn

			a(2) = 5: aaaa, aabb, aacc, abba, acca (with ternary alphabet {a,b,c}).

Alois P. Heinz, Table of n, a(n) for n = 0..1000
C. Kassel and C. Reutenauer, Algebraicity of the zeta function associated to a matrix over a free group algebra, arXiv preprint arXiv:1303.3481 [math.CO], 2013-2014.

Column k=3 of A183134.

Cf. A194726.

[1] cat [&+[(Binomial(2*n, k)*(n-k)*2^k)/n: k in [0..n]]: n in [1..25]]; // Vincenzo Librandi, Apr 08 2018

a:= n-> `if`(n=0, 1, add(binomial(2*n, j) *(n-j) *2^j, j=0..n-1)/n):
seq(a(n), n=0..25);

CoefficientList[Series[2/3+4/(3*(1+3*Sqrt[1-8*x])), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)
a[n_] := 2^(n+1) CatalanNumber[n] Hypergeometric2F1[2, 1-n, n+2, -2] - 3^(2n - 1);
Table[If[n == 0, 1, a[n]], {n, 0, 22}] (* Peter Luschny, Apr 08 2018 *)

a(n) = if (n==0, 1, sum(j=0, n-1, binomial(2*n,j)*(n-j)*2^j)/n); \\ Michel Marcus, Apr 07 2018

x='x+O('x^99); Vec(4/(3*(1+3*(1-8*x)^(1/2)))+2/3) \\ Altug Alkan, Apr 07 2018

G.f.: 2/3 + 4/(3*(1+3*sqrt(1-8*x))).
a(0) = 1, a(n) = 1/n * Sum_{j=0..n-1} C(2*n,j)*(n-j)*2^j for n>0.
D-finite with recurrence: n*a(n) = (17*n-12)*a(n-1) - 36*(2*n-3)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(3*n+1)/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012
G.f.:  2-4/( Q(0) + 3), where Q(k) = 1 + 8*x*(4*k+1)/( 4*k+2 - 8*x*(4*k+2)*(4*k+3)/( 8*x*(4*k+3) + 4*(k+1)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 20 2013
From Karol A. Penson, Jul 13 2015: (Start)
Special values of the hypergeometric function 2F1, in Maple notation:
a(n+1) = (16/9)*8^n*GAMMA(n+3/2)*hypergeom([1, n+3/2], [n+3],8/9)/(sqrt(Pi)*(n+2)!), n=0,1,... .
Integral representation as the n-th moment of a positive function W(x) = sqrt((8-x)*x)*(1/(9-x))/(2*Pi) on (0,8): a(n+1) = int(x^n*W(x),x=0..8), n=0,1,... . This representation is unique as W(x) is the solution of the Hausdorff moment problem. (End)
a(n) = 2^(n+1)*binomial(2*n,n)*hypergeom([2,1-n],[n+2],-2)/(n+1) - 3^(2*n-1) for n>=1. - Peter Luschny, Apr 07 2018

A194726 Number of 6-ary words either empty or beginning with the first character of the alphabet, that can be built by inserting n doublets into the initially empty word.

Original entry on oeis.org

1, 1, 11, 146, 2131, 32966, 530526, 8786436, 148733571, 2561439806, 44731364266, 790211926076, 14095578557486, 253519929631996, 4592415708939356, 83709533881191816, 1534227271236577251, 28256420350942562286, 522675506718404898546, 9706083027629177910156

Offset: 0

Alois P. Heinz, Sep 02 2011

nonn

			a(2) = 11: aaaa, aabb, aacc, aadd, aaee, aaff, abba, acca, adda, aeea, affa (with 6-ary alphabet {a,b,c,d,e,f}).

Alois P. Heinz, Table of n, a(n) for n = 0..500

Column k=6 of A183134.

a:= n-> `if`(n=0, 1, add(binomial(2*n, j) *(n-j) *5^j, j=0..n-1) /n):
seq(a(n), n=0..25);

G.f.: 5/6 + 5/(3*(4+6*sqrt(1-20*x))).
a(0) = 1, a(n) = 1/n * Sum_{j=0..n-1} C(2*n,j)*(n-j)*5^j for n>0.
a(n) ~ 5*20^n/(16*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 13 2013
D-finite with recurrence: n*a(n) = 2*(28*n-15)*a(n-1) - 360*(2*n-3)*a(n-2). - Vaclav Kotesovec, Aug 13 2013
From Karol A. Penson, Jul 12 2015: (Start)
Special values of the hypergeometric function 2F1, in Maple notation:
a(n+1) = (25/9)*20^n*GAMMA(n+3/2)*hypergeom([1, n+3/2], [n+3],5/9)/(sqrt(Pi)*(n+2)!), n=0,1,... .
Integral representation as the n-th moment of a positive function W(x) = sqrt((20-x)*x)*(1/(36-x))/(2*Pi) on (0,20): a(n+1) = int(x^n*W(x),x=0..20), n=0,1,... . This representation is unique as W(x) is the solution of the Hausdorff moment problem. (End)

A079273 Octo numbers (a polygonal sequence): a(n) = 5n^2 - 6n + 2 = (n-1)^2 + (2*n-1)^2.

Original entry on oeis.org

1, 10, 29, 58, 97, 146, 205, 274, 353, 442, 541, 650, 769, 898, 1037, 1186, 1345, 1514, 1693, 1882, 2081, 2290, 2509, 2738, 2977, 3226, 3485, 3754, 4033, 4322, 4621, 4930, 5249, 5578, 5917, 6266, 6625, 6994, 7373, 7762, 8161, 8570, 8989, 9418, 9857, 10306

Offset: 1

Matthew Vandermast, Feb 06 2003

a(n+1) = a(n) + 10*n - 1, and n + a(n) is always congruent to 2 mod 10 (notice pattern of final digits). a(n) = the n-th hex number (3*n^2 - 3*n + 1) added to the (2n-2)-nd triangular number (2*n^2 - 3*n + 1). The formula for the n-th octo number can be written as (2n-1)^2 + (n-1)^2; compare to formula for n-th octagonal number, n*(3n-2) = (2n-1)^2 - (n-1)^2.
a(n+1) = 5*n^2 + 4*n + 1 is also the number of ways of realizing the amount 10n using only coins with values 1, 2 and 5. - Francois Brunault (brunault(AT)gmail.com), Nov 24 2009
a(n) is the number of length 6 n-ary words, beginning with the first character of the alphabet, that can be built by repeatedly inserting doublets into the initially empty word. - Alois P. Heinz, Sep 01 2011
For n > 1, a(n) is the Wiener index of the caterpillar of diameter 3 where each internal vertex has attached n - 2 pendent vertices. - Christian Barrientos, Mar 31 2023

			a(4) = 58 because 58 dots can be arranged into a simple octagonal pattern with 4 dots on each side, its rows from top to bottom containing 4,5,6,7,7,7,7,6,5 and 4 dots respectively. The pattern is similar to the pattern for hex numbers (see link), with the exception that while the n-th hex figure has only 1 row of length 2n-1 dots (the maximum length) in the center, the n-th octo figure has n such rows.
a(4) = 58:
     O O O O
    O O O O O
   O O O O O O
  O O O O O O O
  O O O O O O O
  O O O O O O O
  O O O O O O O
   O O O O O O
    O O O O O
     O O O O

Alois P. Heinz, Table of n, a(n) for n = 1..1000
Leo Tavares, Illustration: Compacted Hexagons
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217 (triangular numbers), A000567 (octagonal numbers), A003215 (hex numbers).
Row n=3 of A183134. - Alois P. Heinz, Aug 31 2011
Cf. A016873.

[n*(5*n-6) +2: n in [1..50]]; // G. C. Greubel, Apr 19 2023

Table[5n^2-6n+2,{n,50}] (* or *) LinearRecurrence[{3,-3,1}, {1,10,29}, 150] (* Harvey P. Dale, Apr 06 2011 & May 03 2011 *)

a(n)=5*n^2-6*n+2 \\ Charles R Greathouse IV, Oct 07 2015

[n*(5*n-6) +2 for n in range(1,51)] # G. C. Greubel, Apr 19 2023

a(n) = 10*n + a(n-1) - 11 for n > 1, a(1)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(1) = 1, a(2) = 10, a(3) = 29. - Harvey P. Dale, May 03 2011
G.f.: x*(1 + 7*x + 2*x^2)/(1 - x)^3. - Alois P. Heinz, Sep 01 2011
E.g.f.: -2 + (2 - x + 5*x^2)*exp(x). - G. C. Greubel, Apr 19 2023
5*a(n) = A016873(n-1)^2 + 1. - Charlie Marion, May 10 2024

A194728 Number of 8-ary words either empty or beginning with the first character of the alphabet, that can be built by inserting n doublets into the initially empty word.

Original entry on oeis.org

1, 1, 15, 274, 5531, 118686, 2654646, 61189668, 1443039123, 34648845862, 844131474530, 20813234394492, 518373091849502, 13021801045587244, 329543346098061516, 8393705745623980104, 215009056951891319811, 5535306699430995140214, 143144289829339089562986

Offset: 0

Alois P. Heinz, Sep 02 2011

nonn

			a(2) = 15: aaaa, aabb, aacc, aadd, aaee, aaff, aagg, aahh, abba, acca, adda, aeea, affa, agga, ahha (with 8-ary alphabet {a,b,c,d,e,f,g,h}).

Alois P. Heinz, Table of n, a(n) for n = 0..500

Column k=8 of A183134.

Cf. A194723, A194726.

a:= n-> `if`(n=0, 1, add(binomial(2*n, j) *(n-j) *7^j, j=0..n-1) /n):
seq(a(n), n=0..20);

G.f.: 7/8 + 7/(4*(6+8*sqrt(1-28*x))).
a(0) = 1, a(n) = 1/n * Sum_{j=0..n-1} C(2*n,j)*(n-j)*7^j for n>0.
a(n) ~ 7 * 28^n / (36 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Sep 07 2014
D-finite with recurrence n*a(n) +2*(-46*n+21)*a(n-1) +896*(2*n-3)*a(n-2)=0. - R. J. Mathar, Mar 14 2015
From Karol A. Penson, Jul 13 2015: (Start)
Special values of the hypergeometric function 2F1, in Maple notation:
a(n+1) = (7/4)^2*(28)^n*GAMMA(n+3/2)*hypergeom([1, n+3/2], [n+3],7/16)/(sqrt(Pi)*(n+2)!), n=0,1,... .
Integral representation as the n-th moment of a positive function W(x) = sqrt((28-x)*x)*(1/(64-x))/(2*Pi) on (0,28): a(n+1) = int(x^n*W(x), x=0..28), n=0,1,... . This representation is unique as W(x) is the solution of the Hausdorff moment problem. (End)

A194729 Number of 9-ary words either empty or beginning with the first character of the alphabet, that can be built by inserting n doublets into the initially empty word.

Original entry on oeis.org

1, 1, 17, 353, 8113, 198401, 5060433, 133071009, 3581326065, 98156060225, 2730108129937, 76862217117665, 2186096427128369, 62718004238927233, 1812849590253944273, 52742324721313632033, 1543272031837984426353, 45386639860532255882433, 1340844916965007902013713

Offset: 0

Alois P. Heinz, Sep 02 2011

nonn

			a(2) = 17: aaaa, aabb, aacc, aadd, aaee, aaff, aagg, aahh, aaii, abba, acca, adda, aeea, affa, agga, ahha, aiia (with 9-ary alphabet {a,b,c,d,e,f,g,h,i}).

Alois P. Heinz, Table of n, a(n) for n = 0..200

Column k=9 of A183134.

Cf. A194723, A194726, A174728.

a:= n-> `if`(n=0, 1, add(binomial(2*n, j) *(n-j) *8^j, j=0..n-1) /n):
seq(a(n), n=0..20);

CoefficientList[Series[8/9 + 16/(9 (7 + 9 Sqrt[1 - 32 x])), {x, 0, 33}], x] (* Vincenzo Librandi, Jul 16 2015 *)

G.f.: 8/9 + 16/(9*(7+9*sqrt(1-32*x))).
a(0) = 1, a(n) = 1/n * Sum_{j=0..n-1} C(2*n,j)*(n-j)*8^j for n>0.
a(n) ~ 8 * 32^n / (49 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Sep 07 2014
n*a(n) +(-113*n+48)*a(n-1) +1296*(2*n-3)*a(n-2)=0. - R. J. Mathar, Mar 14 2015
From Karol A. Penson, Jul 15 2015: (Start)
Special values of the hypergeometric function 2F1, in Maple notation:
a(n+1) = 2^8*32^n*GAMMA(n+3/2)*hypergeom([1,n+3/2],[n+3],32/81)/(81*sqrt(Pi)*(n+2)!), n=0,1,... .
Integral representation as the n-th moment of a positive function W(x) = sqrt(x*(32-x))/(2*Pi*(81-x)) on (0,32): a(n+1) = Integral_{x=0..32} x^n*W(x) dx, n=0,1,... . This representation is unique as W(x) is the solution of the Hausdorff moment problem. (End)

A194724 Number of quaternary words either empty or beginning with the first character of the alphabet, that can be built by inserting n doublets into the initially empty word.

Original entry on oeis.org

1, 1, 7, 58, 523, 4966, 48838, 492724, 5068915, 52955950, 560198962, 5987822380, 64563867454, 701383563388, 7668869344108, 84326618668648, 931894610845123, 10344218506421758, 115280448164645818, 1289346114476360188, 14467472108268263818, 162816535672067515828

Offset: 0

Alois P. Heinz, Sep 02 2011

nonn

			a(2) = 7: aaaa, aabb, aacc, aadd, abba, acca, adda (with quaternary alphabet {a,b,c,d}).

Alois P. Heinz, Table of n, a(n) for n = 0..500
C. Kassel and C. Reutenauer, Algebraicity of the zeta function associated to a matrix over a free group algebra, arXiv preprint arXiv:1303.3481, 2013

Column k=4 of A183134.

a:= n-> `if`(n=0, 1, add(binomial(2*n, j) *(n-j) *3^j, j=0..n-1)/n):
seq(a(n), n=0..25);
# second Maple program:
a:= proc(n) option remember; `if`(n<3, [1, 1, 7][n+1],
      ((28*n-18)*a(n-1) -(192*n-288)*a(n-2))/n)
    end:
seq(a(n), n=0..30);

CoefficientList[Series[3/4+3/(2(2+4Sqrt[1-12x])),{x,0,30}],x] (* Harvey P. Dale, Sep 30 2012 *)

G.f.: 3/4 + 3/(2*(2+4*sqrt(1-12*x))).
a(0) = 1, a(n) = 1/n * Sum_{j=0..n-1} C(2*n,j)*(n-j)*3^j for n>0.
a(n) ~ 3 * 12^n / (4 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Sep 07 2014
Conjecture: n*a(n) +2*(-14*n+9)*a(n-1) +96*(2*n-3)*a(n-2)=0. - R. J. Mathar, Mar 14 2015

A194725 Number of 5-ary words either empty or beginning with the first character of the alphabet, that can be built by inserting n doublets into the initially empty word.

Original entry on oeis.org

1, 1, 9, 97, 1145, 14289, 185193, 2467137, 33563481, 464221105, 6507351113, 92236247841, 1319640776249, 19031570387857, 276368559434025, 4037555902072065, 59299855337012505, 875056238174271345, 12967283824008178185, 192889769468751321825, 2879117809973276680185

Offset: 0

Alois P. Heinz, Sep 02 2011

nonn

			a(2) = 9: aaaa, aabb, aacc, aadd, aaee, abba, acca, adda, aeea (with 5-ary alphabet {a,b,c,d,e}).

Alois P. Heinz, Table of n, a(n) for n = 0..200
C. Kassel and C. Reutenauer, Algebraicity of the zeta function associated to a matrix over a free group algebra, arXiv preprint arXiv:1303.3481, 2013

Column k=5 of A183134.

a:= n-> `if`(n=0, 1, add(binomial(2*n, j) *(n-j) *4^j, j=0..n-1) /n):
seq(a(n), n=0..20);
# second Maple program
a:= proc(n) a(n):= `if`(n<3, [1, 1, 9][n+1],
       ((41*n-24)*a(n-1) +(600-400*n)*a(n-2))/n)
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, Oct 30 2012

FullSimplify[Flatten[{1,Table[4^(2*n+1)*(1/2 (2*n-1))! Hypergeometric2F1[1,1/2+n,2+n,16/25]/(25*Sqrt[Pi]*(n+1)!),{n,1,20}]}]] (* Vaclav Kotesovec, Aug 13 2013 *)

G.f.: 4/5 + 8/(5*(3+5*sqrt(1-16*x))).
a(0) = 1, a(n) = 1/n * Sum_{j=0..n-1} C(2*n,j)*(n-j)*4^j for n>0.
a(n) ~ 2^(4*n+2)/(9*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 13 2013
Recurrence: n*a(n) = (41*n-24)*a(n-1) - 200*(2*n-3)*a(n-2). - Vaclav Kotesovec, Aug 13 2013

cp's OEIS Frontend

A039598 Triangle formed from odd-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.

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A183135 Square array A(n,k) by antidiagonals. A(n,k) is the number of length 2n k-ary words (n,k>=0) that can be built by repeatedly inserting doublets into the initially empty word.

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A213027 Number A(n,k) of 3n-length k-ary words, either empty or beginning with the first letter of the alphabet, that can be built by repeatedly inserting triples of identical letters into the initially empty word; square array A(n,k), n>=0, k>=0, by antidiagonals.

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Maple

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A194723 Number of ternary words either empty or beginning with the first character of the alphabet, that can be built by inserting n doublets into the initially empty word.

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Magma

Maple

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PARI

PARI

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A194726 Number of 6-ary words either empty or beginning with the first character of the alphabet, that can be built by inserting n doublets into the initially empty word.

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Maple

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A079273 Octo numbers (a polygonal sequence): a(n) = 5*n^2 - 6*n + 2 = (n-1)^2 + (2*n-1)^2.

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Magma

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SageMath

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A079273 Octo numbers (a polygonal sequence): a(n) = 5n^2 - 6n + 2 = (n-1)^2 + (2*n-1)^2.