cp's OEIS Frontend

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

Showing 1-10 of 12 results. Next

A126952 a(0)=1, a(n+1) = 5*a(n)-4*A117641(n) for n>=0.

Original entry on oeis.org

1, 1, 5, 21, 93, 421, 1937, 9017, 42349, 200277, 952425, 4549953, 21818841, 104966889, 506372277, 2448641061, 11865563853, 57604036309, 280110716777, 1364092539041, 6651682319233, 32474171399649, 158714415664557
Offset: 0

Views

Author

Philippe Deléham, Mar 19 2007

Keywords

Comments

Hankel transform is 4^n=A000302(n).

Crossrefs

Programs

  • Mathematica
    Block[{$MaxExtraPrecision = 10^3, s = Rest@ CoefficientList[Series[(1 + 3 x - Sqrt[1 - 6 x + 5 x^2])/(2 x^2 + 6 x), {x, 0, 21}], x]}, Nest[Append[#, 5 #[[-1]] - 4 s[[Length@ # - 1]] ] &, {1, 1}, Length@ s]] (* Michael De Vlieger, Dec 15 2019, after Robert G. Wilson v at A117641 *)

Formula

a(n) = Sum_{k = 0..n} binomial(n,k)*b(k), where b(n) = Sum_{k = 0..n} binomial(n+k,k) * (-2)^(n-k). - Peter Bala, Jun 18 2025
From Vaclav Kotesovec, Jun 22 2025: (Start)
Recurrence: 5*(n-2)*a(n-3) + (9*n-26)*a(n-2) + (12-17*n)*a(n-1) + 3*n*a(n) = 0.
a(n) ~ 5^(n + 1/2) / (4*sqrt(Pi*n)). (End)

Extensions

a(11) and a(22) corrected by Michael De Vlieger, Dec 15 2019

A171224 Riordan array (f(x),x*f(x)) where f(x) is the g.f. of A117641.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 3, 2, 0, 1, 11, 6, 3, 0, 1, 42, 23, 9, 4, 0, 1, 167, 90, 36, 12, 5, 0, 1, 684, 365, 144, 50, 15, 6, 0, 1, 2867, 1518, 595, 204, 65, 18, 7, 0, 1, 12240, 6441, 2511, 858, 270, 81, 21, 8, 0, 1, 53043, 27774, 10782, 3672, 1155, 342, 98, 24, 9, 0, 1
Offset: 0

Views

Author

Philippe Deléham, Dec 05 2009

Keywords

Examples

			Triangle begins
    1;
    0,  1;
    1,  0,  1;
    3,  2,  0,  1;
   11,  6,  3,  0,  1;
   42, 23,  9,  4,  0,  1;
  167, 90, 36, 12,  5,  0,  1;
  ...
Production array begins
    0,  1;
    1,  0,  1;
    3,  1,  0,  1;
    9,  3,  1,  0,  1;
   27,  9,  3,  1,  0,  1;
   81, 27,  9,  3,  1,  0,  1;
  243, 81, 27,  9,  3,  1,  0,  1;
  ... - _Philippe Deléham_, Mar 04 2013
		

Crossrefs

Programs

  • Magma
    [[((k+1)/(n+1))*(&+[3^(n-k-2*j)*Binomial(n+1,j)*Binomial(n-k-j-1, n-k-2*j): j in [0..Floor((n-k)/2)]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Apr 04 2019
    
  • Mathematica
    T[n_, k_]:= (k+1)/(n+1)*Sum[3^(n-k-2*j)*Binomial[n+1,j]*Binomial[n-k-j-1, n-k-2*j], {j, 0, Floor[(n-k)/2]}]; Table[T[n, k], {n,0,10}, {k,0,n} ]//Flatten (* G. C. Greubel, Apr 04 2019 *)
  • Maxima
    T(n,k):=(k+1)/(n+1)*sum(3^(n-k-2*j)*binomial(n+1,j)*binomial(n-k-j-1,n-k-2*j),j,0,floor((n-k)/2)); /* Vladimir Kruchinin, Apr 04 2019 */
    
  • PARI
    {T(n,k) = ((k+1)/(n+1))*sum(j=0, floor((n-k)/2), 3^(n-k-2*j) *binomial(n+1,j)*binomial(n-k-j-1, n-k-2*j))}; \\ G. C. Greubel, Apr 04 2019
    
  • Sage
    [[((k+1)/(n+1))*sum(3^(n-k-2*j)*binomial(n+1,j)*binomial(n-k-j-1, n-k-2*j) for j in (0..floor((n-k)/2))) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Apr 04 2019

Formula

Sum_{k=0..n} T(n,k)*x^k = A117641(n), A033321(n), A007317(n+1), A002212(n+1), A026378(n+1) for x = 0, 1, 2, 3, 4 respectively.
Triangle equals B*A065600*B^(-1) = B^2*A097609*B^(-2) = B^3*A053121*B^(-3), product considered as infinite lower triangular arrays and B = A007318. - Philippe Deléham, Dec 08 2009
T(n,k) = T(n-1,k-1) + Sum_{i>=0} T(n-1,k+1+i)*3^i, T(0,0) = 1. - Philippe Deléham, Feb 23 2012
T(n,k) = ((k+1)/(n+1))*Sum_{j=0..floor((n-k)/2)} 3^(n-k-2*j)*C(n+1,j)*C(n-k-j-1,n-k-2*j). - Vladimir Kruchinin, Apr 04 2019

Extensions

Terms a(55) onward added by G. C. Greubel, Apr 04 2019

A171243 Riordan array (f(x), x*g(x)), f(x) is the g.f. of A126952, g(x) is the g.f. of A117641.

Original entry on oeis.org

1, 1, 1, 5, 1, 1, 21, 6, 1, 1, 93, 25, 7, 1, 1, 421, 112, 29, 8, 1, 1, 1937, 510, 132, 33, 9, 1, 1, 9017, 2357, 606, 153, 37, 10, 1, 1, 42349, 11009, 2819, 709, 175, 41, 11, 1, 1, 200277, 51840, 13233, 3324, 819, 198, 45, 12, 1, 1
Offset: 0

Views

Author

Philippe Deléham, Dec 06 2009

Keywords

Comments

Expansion of row sums of T_(x,3), T_(x,y) defined in A039599.
Matrix product P^3 * Q * P^(-3), where P denotes Pascal's triangle A007318 and Q denotes A061554 (formed from P by sorting the rows into descending order). Cf. A158793 and A158815. - Peter Bala, Jul 13 2021

Examples

			Triangle begins:
    1;
    1,   1;
    5,   1,  1;
   21,   6,  1, 1;
   93,  25,  7, 1, 1;
  421, 112, 29, 8, 1, 1;
  ...
		

Crossrefs

Formula

Sum_{k=0..n} T(n,k)*x^k = A126952(n), A126568(n), A026375(n), A026378(n+1), A000351(n) for x = 0,1,2,3,4 respectively.

A035263 Trajectory of 1 under the morphism 0 -> 11, 1 -> 10; parity of 2-adic valuation of 2n: a(n) = A000035(A001511(n)).

Original entry on oeis.org

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

Views

Author

Keywords

Comments

First Feigenbaum symbolic (or period-doubling) sequence, corresponding to the accumulation point of the 2^{k} cycles through successive bifurcations.
To construct the sequence: start with 1 and concatenate: 1,1, then change the last term (1->0; 0->1) gives: 1,0. Concatenate those 2 terms: 1,0,1,0, change the last term: 1,0,1,1. Concatenate those 4 terms: 1,0,1,1,1,0,1,1 change the last term: 1,0,1,1,1,0,1,0, etc. - Benoit Cloitre, Dec 17 2002
Let T denote the present sequence. Here is another way to construct T. Start with the sequence S = 1,0,1,,1,0,1,,1,0,1,,1,0,1,,... and fill in the successive holes with the successive terms of the sequence T (from paper by Allouche et al.). - Emeric Deutsch, Jan 08 2003 [Note that if we fill in the holes with the terms of S itself, we get A141260. - N. J. A. Sloane, Jan 14 2009]
From N. J. A. Sloane, Feb 27 2009: (Start)
In more detail: define S to be 1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1,0,1___...
If we fill the holes with S we get A141260:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........0.......1.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.0.1.... = A141260.
But instead, if we define T recursively by filling the holes in S with the terms of T itself, we get A035263:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........1.......0.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.1.1.0.1.0.1..0..1.1.1..0..1.0.1.. = A035263. (End)
Characteristic function of A003159, i.e., A035263(n)=1 if n is in A003159 and A035263(n)=0 otherwise (from paper by Allouche et al.). - Emeric Deutsch, Jan 15 2003
This is the sequence of R (=1), L (=0) moves in the Towers of Hanoi puzzle: R, L, R, R, R, L, R, L, R, L, R, R, R, ... - Gary W. Adamson, Sep 21 2003
Manfred Schroeder, p. 279 states, "... the kneading sequences for unimodal maps in the binary notation, 0, 1, 0, 1, 1, 1, 0, 1..., are obtained from the Morse-Thue sequence by taking sums mod 2 of adjacent elements." On p. 278, in the chapter "Self-Similarity in the Logistic Parabola", he writes, "Is there a closer connection between the Morse-Thue sequence and the symbolic dynamics of the superstable orbits? There is indeed. To see this, let us replace R by 1 and C and L by 0." - Gary W. Adamson, Sep 21 2003
Partial sums modulo 2 of the sequence 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), a(5), a(6), a(6), ... . - Philippe Deléham, Jan 02 2004
Parity of A007913, A065882 and A065883. - Philippe Deléham, Mar 28 2004
The length of n-th run of 1's in this sequence is A080426(n). - Philippe Deléham, Apr 19 2004
Also parity of A005043, A005773, A026378, A104455, A117641. - Philippe Deléham, Apr 28 2007
Equals parity of the Towers of Hanoi, or ruler sequence (A001511), where the Towers of Hanoi sequence (1, 2, 1, 3, 1, 2, 1, 4, ...) denotes the disc moved, labeled (1, 2, 3, ...) starting from the top; and the parity of (1, 2, 1, 3, ...) denotes the direction of the move, CW or CCW. The frequency of CW moves converges to 2/3. - Gary W. Adamson, May 11 2007
A conjectured identity relating to the partition sequence, A000041: p(x) = A(x) * A(x^2) when A(x) = the Euler transform of A035263 = polcoeff A174065: (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...). - Gary W. Adamson, Mar 21 2010
a(n) is 1 if the number of trailing zeros in the binary representation of n is even. - Ralf Stephan, Aug 22 2013
From Gary W. Adamson, Mar 25 2015: (Start)
A conjectured identity relating to the partition sequence, A000041 as polcoeff p(x); A003159, and its characteristic function A035263: (1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...); and A036554 indicating n-th terms with zeros in A035263: (2, 6, 8, 10, 14, 18, 22, ...).
The conjecture states that p(x) = A(x) = A(x^2) when A(x) = polcoeffA174065 = the Euler transform of A035263 = 1/(1-x)*(1-x^3)*(1-x^4)*(1-x^5)*... = (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...) and the aerated variant = the Euler transform of the complement of A035263: 1/(1-x^2)*(1-x^6)*(1-x^8)*... = (1 + x^2 + x^4 + 2x^6 + 3x^8 + 4x^10 + ...).
(End)
The conjecture above was proved by Jean-Paul Allouche on Dec 21 2013.
Regarded as a column vector, this sequence is the product of A047999 (Sierpinski's gasket) regarded as an infinite lower triangular matrix and A036497 (the Fredholm-Rueppel sequence) where the 1's have alternating signs, 1, -1, 0, 1, 0, 0, 0, -1, .... - Gary W. Adamson, Jun 02 2021
The numbers of 1's through n (A050292) can be determined by starting with the binary (say for 19 = 1 0 0 1 1) and writing: next term is twice current term if 0, otherwise twice plus 1. The result is 1, 2, 4, 9, 19. Take the difference row, = 1, 1, 2, 5, 10; and add the odd-indexed terms from the right: 5, 4, 3, 2, 1 = 10 + 2 + 1 = 13. The algorithm is the basis for determining the disc configurations in the tower of Hanoi game, as shown in the Jul 24 2021 comment of A060572. - Gary W. Adamson, Jul 28 2021

References

  • Karamanos, Kostas. "From symbolic dynamics to a digital approach." International Journal of Bifurcation and Chaos 11.06 (2001): 1683-1694. (Full version. See p. 1685)
  • Karamanos, K. (2000). From symbolic dynamics to a digital approach: chaos and transcendence. In Michel Planat (Ed.), Noise, Oscillators and Algebraic Randomness (Lecture Notes in Physics, pp. 357-371). Springer, Berlin, Heidelberg. (Short version. See p. 359)
  • Manfred R. Schroeder, "Fractals, Chaos, Power Laws", W. H. Freeman, 1991
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 892, column 2, Note on p. 84, part (a).

Crossrefs

Parity of A001511. Anti-parity of A007814.
Absolute values of first differences of A010060. Apart from signs, same as A029883. Essentially the same as A056832.
Swapping 0 and 1 gives A096268.
Cf. A033485, A050292 (partial sums), A089608, A088172, A019300, A039982, A073675, A121701, A141260, A000041, A174065, A220466, A154269 (Mobius transform).
Limit of A317957(n) for large n.

Programs

  • Haskell
    import Data.Bits (xor)
    a035263 n = a035263_list !! (n-1)
    a035263_list = zipWith xor a010060_list $ tail a010060_list
    -- Reinhard Zumkeller, Mar 01 2012
    
  • Maple
    nmax:=105: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := (p+1) mod 2 od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 07 2013
    A035263 := n -> 1 - padic[ordp](n, 2) mod 2:
    seq(A035263(n), n=1..105); # Peter Luschny, Oct 02 2018
  • Mathematica
    a[n_] := a[n] = If[ EvenQ[n], 1 - a[n/2], 1]; Table[ a[n], {n, 1, 105}] (* Or *)
    Rest[ CoefficientList[ Series[ Sum[ x^(2^k)/(1 + (-1)^k*x^(2^k)), {k, 0, 20}], {x, 0, 105}], x]]
    f[1] := True; f[x_] := Xor[f[x - 1], f[Floor[x/2]]]; a[x_] := Boole[f[x]] (* Ben Branman, Oct 04 2010 *)
    a[n_] := If[n == 0, 0, 1 - Mod[ IntegerExponent[n, 2], 2]]; (* Jean-François Alcover, Jul 19 2013, after Michael Somos *)
    Nest[ Flatten[# /. {0 -> {1, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Robert G. Wilson v, Jul 23 2014 *)
    SubstitutionSystem[{0->{1,1},1->{1,0}},1,{7}][[1]] (* Harvey P. Dale, Jun 06 2022 *)
  • PARI
    {a(n) = if( n==0, 0, 1 - valuation(n, 2)%2)}; /* Michael Somos, Sep 04 2006 */
    
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); subst( Pol(binary(n)) - Pol(binary(n-1)), x, 1)%2)}; /* Michael Somos, Sep 04 2006 */
    
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); direuler(p=2, n, 1 / (1 - X^((p<3) + 1)))[n])}; /* Michael Somos, Sep 04 2006 */
    
  • Python
    def A035263(n): return (n&-n).bit_length()&1 # Chai Wah Wu, Jan 09 2023
  • Scheme
    (define (A035263 n) (let loop ((n n) (i 1)) (cond ((odd? n) (modulo i 2)) (else (loop (/ n 2) (+ 1 i)))))) ;; (Use mod instead of modulo in R6RS) Antti Karttunen, Sep 11 2017
    

Formula

Absolute values of first differences (A029883) of Thue-Morse sequence (A001285 or A010060). Self-similar under 10->1 and 11->0.
Series expansion: (1/x) * Sum_{i>=0} (-1)^(i+1)*x^(2^i)/(x^(2^i)-1). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 17 2003
a(n) = Sum_{k>=0} (-1)^k*(floor((n+1)/2^k)-floor(n/2^k)). - Benoit Cloitre, Jun 03 2003
Another g.f.: Sum_{k>=0} x^(2^k)/(1+(-1)^k*x^(2^k)). - Ralf Stephan, Jun 13 2003
a(2*n) = 1-a(n), a(2*n+1) = 1. - Ralf Stephan, Jun 13 2003
a(n) = parity of A033485(n). - Philippe Deléham, Aug 13 2003
Equals A088172 mod 2, where A088172 = 1, 2, 3, 7, 13, 26, 53, 106, 211, 422, 845, ... (first differences of A019300). - Gary W. Adamson, Sep 21 2003
a(n) = a(n-1) - (-1)^n*a(floor(n/2)). - Benoit Cloitre, Dec 02 2003
a(1) = 1 and a(n) = abs(a(n-1) - a(floor(n/2))). - Benoit Cloitre, Dec 02 2003
a(n) = 1 - A096268(n+1); A050292 gives partial sums. - Reinhard Zumkeller, Aug 16 2006
Multiplicative with a(2^k) = 1 - (k mod 2), a(p^k) = 1, p > 2. Dirichlet g.f.: Product_{n = 4 or an odd prime} (1/(1-1/n^s)). - Christian G. Bower, May 18 2005
a(-n) = a(n). a(0)=0. - Michael Somos, Sep 04 2006
Dirichlet g.f.: zeta(s)*2^s/(2^s+1). - Ralf Stephan, Jun 17 2007
a(n+1) = a(n) XOR a(ceiling(n/2)), a(1) = 1. - Reinhard Zumkeller, Jun 11 2009
Let D(x) be the generating function, then D(x) + D(x^2) == x/(1-x). - Joerg Arndt, May 11 2010
a(n) = A010060(n) XOR A010060(n+1); a(A079523(n)) = 0; a(A121539(n)) = 1. - Reinhard Zumkeller, Mar 01 2012
a((2*n-1)*2^p) = (p+1) mod 2, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 07 2013
a(n) = A000035(A001511(n)). - Omar E. Pol, Oct 29 2013
a(n) = 2-A056832(n) = (5-A089608(n))/4. - Antti Karttunen, Sep 11 2017, after Benoit Cloitre
For n >= 0, a(n+1) = M(2n) mod 2 where M(n) is the Motzkin number A001006 (see Deutsch and Sagan 2006 link). - David Callan, Oct 02 2018
a(n) = A038712(n) mod 3. - Kevin Ryde, Jul 11 2019
Given any n in the form (k * 2^m, k odd), extract k and m. Categorize the results into two outcomes of (k, m, even or odd). If (k, m) is (odd, even) substitute 1. If (odd, odd), denote the result 0. Example: 5 = (5 * 2^0), (odd, even, = 1). (6 = 3 * 2^1), (odd, odd, = 0). - Gary W. Adamson, Jun 23 2021

Extensions

Alternative description added to the name by Antti Karttunen, Sep 11 2017

A091965 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 three types of steps H=(1,0) (left factors of 3-Motzkin steps).

Original entry on oeis.org

1, 3, 1, 10, 6, 1, 36, 29, 9, 1, 137, 132, 57, 12, 1, 543, 590, 315, 94, 15, 1, 2219, 2628, 1629, 612, 140, 18, 1, 9285, 11732, 8127, 3605, 1050, 195, 21, 1, 39587, 52608, 39718, 19992, 6950, 1656, 259, 24, 1, 171369, 237129, 191754, 106644, 42498, 12177, 2457
Offset: 0

Views

Author

Emeric Deutsch, Mar 13 2004

Keywords

Comments

T(n,0) = A002212(n+1), T(n,1) = A045445(n+1); row sums give A026378.
The inverse is A207815. - Gary W. Adamson, Dec 17 2006 [corrected by Philippe Deléham, Feb 22 2012]
Reversal of A084536. - Philippe Deléham, Mar 23 2007
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) = 3*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 3*T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Mar 27 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
5^n = (n-th row terms) dot (first n+1 terms in (1,2,3,...)). Example for row 4: 5^4 = 625 = (137, 132, 57, 12, 1) dot (1, 2, 3, 4, 5) = (137 + 264 + 171 + 48 + 5) = 625. - Gary W. Adamson, Jun 15 2011
Riordan array ((1-3*x-sqrt(1-6*x+5*x^2))/(2*x^2), (1-3*x-sqrt(1-6*x+5*x^2))/(2*x)). - Philippe Deléham, Feb 19 2012

Examples

			Triangle begins:
     1;
     3,    1;
    10,    6,    1;
    36,   29,    9,    1;
   137,  132,   57,   12,    1;
   543,  590,  315,   94,   15,    1;
  2219, 2628, 1629,  612,  140,   18,    1;
T(3,1)=29 because we have UDU, UUD, 9 HHU paths, 9 HUH paths and 9 UHH paths.
Production matrix begins
  3, 1;
  1, 3, 1;
  0, 1, 3, 1;
  0, 0, 1, 3, 1;
  0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 1, 3, 1;
- _Philippe Deléham_, Nov 07 2011
		

References

  • A. Nkwanta, Lattice paths and RNA secondary structures, DIMACS Series in Discrete Math. and Theoretical Computer Science, 34, 1997, 137-147.

Crossrefs

Programs

  • Mathematica
    nmax = 9; t[n_, k_] := ((k+1)*n!*Hypergeometric2F1[k+3/2, k-n, 2k+3, -4]) / ((k+1)!*(n-k)!); Flatten[ Table[ t[n, k], {n, 0, nmax}, {k, 0, n}]] (* Jean-François Alcover, Nov 14 2011, after Vladimir Kruchinin *)
    T[0, 0, x_, y_] := 1; T[n_, 0, x_, y_] := x*T[n - 1, 0, x, y] + T[n - 1, 1, x, y]; T[n_, k_, x_, y_] := T[n, k, x, y] = If[k < 0 || k > n, 0,
    T[n - 1, k - 1, x, y] + y*T[n - 1, k, x, y] + T[n - 1, k + 1, x, y]];
    Table[T[n, k, 3, 3], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, May 22 2017 *)
  • Maxima
    T(n,k):=(k+1)*sum((binomial(2*(m+1),m-k)*binomial(n,m))/(m+1),m,k,n); /* Vladimir Kruchinin, Oct 08 2011 */
    
  • Sage
    @CachedFunction
    def A091965(n,k):
        if n==0 and k==0: return 1
        if k<0 or k>n: return 0
        if k==0: return 3*A091965(n-1,0)+A091965(n-1,1)
        return A091965(n-1,k-1)+3*A091965(n-1,k)+A091965(n-1,k+1)
    for n in (0..7):
        [A091965(n,k) for k in (0..n)] # Peter Luschny, Nov 05 2012

Formula

G.f.: G = 2/(1 - 3*z - 2*t*z + sqrt(1-6*z+5*z^2)). Alternatively, G = M/(1 - t*z*M), where M = 1 + 3*z*M + z^2*M^2.
Sum_{k>=0} T(m, k)*T(n, k) = T(m+n, 0) = A002212(m+n+1). - Philippe Deléham, Sep 14 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 [3,3,3,...] in the main diagonal. - Gary W. Adamson, Dec 17 2006
Sum_{k=0..n} T(n,k)*(k+1) = 5^n. - Philippe Deléham, Mar 27 2007
Sum_{k=0..n} T(n,k)*x^k = A117641(n), A033321(n), A007317(n), A002212(n+1), A026378(n+1) for x = -3, -2, -1, 0, 1 respectively. - Philippe Deléham, Nov 28 2009
T(n,k) = (k+1)*Sum_{m=k..n} binomial(2*(m+1),m-k)*binomial(n,m)/(m+1). - Vladimir Kruchinin, Oct 08 2011
The n-th row polynomial R(n,x) equals the n-th degree Taylor polynomial of the function (1 - x^2)*(1 + 3*x + x^2)^n expanded about the point x = 0. - Peter Bala, Sep 06 2022

A126970 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) = T(n-1,1), T(n,k) = T(n-1,k-1) + 3*T(n-1,k) + T(n-1,k+1) for k >= 1.

Original entry on oeis.org

1, 0, 1, 1, 3, 1, 3, 11, 6, 1, 11, 42, 30, 9, 1, 42, 167, 141, 58, 12, 1, 167, 684, 648, 327, 95, 15, 1, 684, 2867, 2955, 1724, 627, 141, 18, 1, 2867, 12240, 13456, 8754, 3746, 1068, 196, 21, 1, 12240, 53043, 61362, 43464, 21060, 7146, 1677, 260, 24, 1
Offset: 0

Views

Author

Philippe Deléham, Mar 19 2007

Keywords

Comments

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 from 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

Examples

			Triangle begins:
    1;
    0,   1;
    1,   3,   1;
    3,  11,   6,   1;
   11,  42,  30,   9,  1;
   42, 167, 141,  58, 12,  1;
  167, 684, 648, 327, 95, 15, 1; ...
From _Philippe Deléham_, Nov 07 2011: (Start)
Production matrix begins:
  0, 1
  1, 3, 1
  0, 1, 3, 1
  0, 0, 1, 3, 1
  0, 0, 0, 1, 3, 1
  0, 0, 0, 0, 1, 3, 1
  0, 0, 0, 0, 0, 1, 3, 1
  0, 0, 0, 0, 0, 0, 1, 3, 1
  0, 0, 0, 0, 0, 0, 0, 1, 3, 1 (End)
		

Programs

  • Mathematica
    T[0, 0, x_, y_] := 1; T[n_, 0, x_, y_] := x*T[n - 1, 0, x, y] + T[n - 1, 1, x, y]; T[n_, k_, x_, y_] := T[n, k, x, y] = If[k < 0 || k > n, 0,  T[n - 1, k - 1, x, y] + y*T[n - 1, k, x, y] + T[n - 1, k + 1, x, y]]; Table[T[n, k, 0, 3], {n, 0, 49}, {k, 0, n}] // Flatten (* G. C. Greubel, Apr 21 2017 *)

Formula

Sum_{k=0..n} T(n,k) = A126952(n).
Sum_{k>=0} T(m,k)*T(n,k) = T(m+n,0) = A117641(m+n).
Sum_{k=0..n} T(n,k)*(4*k+1) = 5^n. - Philippe Deléham, Mar 22 2007

A128723 Number of skew Dyck paths of semilength n having no peaks at level 1.

Original entry on oeis.org

1, 0, 2, 6, 22, 84, 334, 1368, 5734, 24480, 106086, 465462, 2063658, 9231084, 41610162, 188820726, 861891478, 3954732384, 18230522422, 84390187986, 392120098258, 1828220666844, 8550445900442, 40103716079436
Offset: 0

Views

Author

Emeric Deutsch, Mar 31 2007

Keywords

Comments

A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of a path is defined to be the number of steps in it.

Examples

			a(3)=6 because we have UUDUDD, UUUDDD, UUUDLD, UUDUDL, UUUDDL and UUUDLL.
		

Crossrefs

Programs

  • Maple
    G:=(3-3*z-sqrt(1-6*z+5*z^2))/(1+3*z+sqrt(1-6*z+5*z^2)): Gser:=series(G,z=0,30): seq(coeff(Gser,z,n),n=0..27);
  • Mathematica
    CoefficientList[Series[(3-3*x-Sqrt[1-6*x+5*x^2])/(1+3*x+Sqrt[1-6*x+5*x^2]), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 20 2014 *)
  • PARI
    my(z='x+O('z^50)); Vec((3-3*z-sqrt(1-6*z+5*z^2)) /(1+3*z +sqrt(1-6*z+5*z^2))) \\ G. C. Greubel, Mar 19 2017

Formula

a(n) = A128722(n,0).
a(n) = 2*A117641(n) for n>=1.
G.f.: (3-3*z-sqrt(1-6*z+5*z^2))/(1+3*z+sqrt(1-6*z+5*z^2)).
a(n) ~ 5^(n+3/2)/(16*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 20 2014
Conjecture: +3*(n+1)*a(n) +(-17*n+10)*a(n-1) +9*(n-3)*a(n-2) +5*(n-2)*a(n-3)=0. - R. J. Mathar, Jun 17 2016

A171368 Another version of A126216.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 0, 5, 0, 1, 0, 0, 5, 0, 9, 0, 1, 0, 0, 0, 21, 0, 14, 0, 1, 0, 0, 14, 0, 56, 0, 20, 0, 1, 0, 0, 0, 84, 0, 120, 0, 27, 0, 1, 0, 0, 42, 0, 300, 0, 225, 0, 35, 0, 1, 0, 0, 0, 330, 0, 825, 0, 385, 0, 44, 0, 1, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Dec 06 2009

Keywords

Comments

Expansion of the first column of the triangle T_(0,x), T_(x,y) defined in A039599; T_(0,0)= A053121, T_(0,1)= A089942, T_(0,2)= A126093, T_(0,3)= A126970.
T(n,k) is the number of Riordan paths of length n with k horizontal steps. A Riordan path is a Motzkin path with no horizontal steps on the x-axis. - Emanuele Munarini, Oct 14 2023

Examples

			Triangle begins:
  1 ;
  0,0 ;
  1,0,0 ;
  0,1,0,0 ;
  2,0,1,0,0 ;
  0,5,0,1,0,0 ;
  5,0,9,0,1,0,0 ;
  ...
		

Crossrefs

Formula

Sum_{k, 0<=k<=n} T(n,k)*x^k = A099323(n+1), A126120(n), A005043(n), A000957(n+1), A117641(n) for x = -1, 0, 1, 2, 3 respectively.

A257363 Number of 3-Motzkin paths with no level steps at height 1.

Original entry on oeis.org

1, 3, 10, 33, 110, 369, 1247, 4248, 14603, 50724, 178314, 635526, 2300829, 8477382, 31842897, 122103276, 478372886, 1915188093, 7831613468, 32674683984, 138871668314, 600140517762, 2631926843602, 11690520554421, 52498671870181, 237966449687118, 1087246253873875, 5001141997115010, 23137102115963262
Offset: 0

Views

Author

Keywords

Comments

For n=2 we have 10 paths: H(1)H(1), H(1)H(2), H(2)H(1), H(2)H(2), H(1)H(3), H(3)H(1), H(3)H(3), H(2)H(3), H(3)H(2), UD.

Crossrefs

Programs

  • Maple
    rec:= (95+95*n)*a(n)+(-180-9*n)*a(n+1)+(-329-197*n)*a(n+2)+(369+144*n)*a(n+3)+(-117-36*n)*a(4+n)+(12+3*n)*a(n+5):
    f:= gfun:-rectoproc({rec,a(0)=1,a(1)=3,a(2)=10,a(3)=33,a(4)=110},a(n),remember):
    seq(f(n),n=0..100); # Robert Israel, Apr 28 2015
  • Mathematica
    CoefficientList[Series[2*(3+x)/(6-17*x-9*x^2+x*Sqrt[1-6*x+5*x^2]), {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 21 2015 *)

Formula

G.f.: 1/(1-3*x-x*F(x)), where F(x) is the g.f. of the sequence A117641.
G.f.: 2*(3+x)/(6-17*x-9*x^2+x*sqrt(1-6*x+5*x^2)).
a(n) ~ 5^(n+3/2)/(98*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Apr 21 2015
From Robert Israel, Apr 28 2015 (Start):
G.f.: (6-x*sqrt(1-6*x+5*x^2)-17*x-9*x^2)/(6-36*x+42*x^2+38*x^3).
3*(-n+1)*a(n) +9*(4*n-7)*a(n-1) +9*(-16*n+39)*a(n-2) +(197*n-656)*a(n-3) +9*(n+15)*a(n-4) +95*(-n+4)*a(n-5)=0. (End)

A171486 Riordan array (f(x), x*f(x)) where f(x) is the g.f. of A033321.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 6, 5, 3, 1, 21, 16, 9, 4, 1, 79, 58, 31, 14, 5, 1, 311, 224, 117, 52, 20, 6, 1, 1265, 900, 465, 205, 80, 27, 7, 1, 5275, 3720, 1910, 840, 330, 116, 35, 8, 1, 22431, 15713, 8034, 3532, 1396, 501, 161, 44, 9, 1, 96900, 67522, 34419, 15136, 6015, 2190
Offset: 0

Views

Author

Philippe Deléham, Dec 09 2009

Keywords

Comments

Equal to B*A065600 = A171224*B where B = A007318 ; equal to B*A039598*B^(-2).

Examples

			Triangle begins :
1
1, 1
2, 2, 1
6, 5, 3, 1
21, 16, 9, 4, 1
79, 58, 31, 14, 5, 1
311, 224, 117, 52, 20, 6, 1
		

Crossrefs

Formula

Sum_{k, 0<=k<=n} T(n,k)*x^k = A117641(n), A033321(n), A007317(n+1), A002212(n+1), A026378(n+1) for x = -1, 0, 1, 2, 3 respectively.
T(n,k) = T(n-1,k-1) + T(n-1,k) + sum_{i, i>=0} T(n-1,k+1+i)*2^i. - Philippe Deléham, Feb 23 2012
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