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 1-8 of 8 results.

A132816 Triangle read by rows: A007318^(-1) * A132812.

Original entry on oeis.org

1, 1, 2, 0, 5, 3, 0, 3, 15, 4, 0, 0, 22, 34, 5, 0, 0, 10, 90, 65, 6, 0, 0, 0, 95, 270, 111, 7, 0, 0, 0, 35, 490, 665, 175, 8, 0, 0, 0, 0, 406, 1820, 1428, 260, 9, 0, 0, 0, 0, 126, 2520, 5460, 2772, 369, 10, 0, 0, 0, 0, 0, 1722, 11130, 14070, 4980, 505, 11, 0, 0, 0, 0, 0, 462, 12474, 39270, 32340, 8415, 671, 12
Offset: 0

Views

Author

Gary W. Adamson, Sep 01 2007

Keywords

Comments

Row sums = A025566 starting (1, 3, 8, 22, 61, 171, 483, ...).

Examples

			First few rows of the triangle:
  1;
  1, 2;
  0, 5,  3;
  0, 3, 15,  4;
  0, 0, 22, 34,   5;
  0, 0, 10, 90,  65,   6;
  0, 0,  0, 95, 270, 111, 7;
  ...
		

Crossrefs

Formula

Inverse binomial transform of A132812.
Matrix product A130595 * A132812. - Georg Fischer, Jun 01 2023

Extensions

a(52) corrected and more terms from Georg Fischer, Jun 01 2023

A001791 a(n) = binomial coefficient C(2n, n-1).

Original entry on oeis.org

0, 1, 4, 15, 56, 210, 792, 3003, 11440, 43758, 167960, 646646, 2496144, 9657700, 37442160, 145422675, 565722720, 2203961430, 8597496600, 33578000610, 131282408400, 513791607420, 2012616400080, 7890371113950, 30957699535776, 121548660036300, 477551179875952
Offset: 0

Views

Author

Keywords

Comments

Number of peaks at even level in all Dyck paths of semilength n+1. Example: a(2)=4 because UDUDUD, UDUU*DD, UU*DDUD, UU*DU*DD, UUUDDD, where U=(1,1), D=(1,-1) and the peaks at even level are shown by *. - Emeric Deutsch, Dec 05 2003
Also number of long ascents (i.e., ascents of length at least two) in all Dyck paths of semilength n+1. Example: a(2)=4 because in the five Dyck paths of semilength 3, namely UDUDUD, UD(UU)DD, (UU)DDUD, (UU)DUDD and (UUU)DDD, we have four long ascents (shown between parentheses). Here U=(1,1) and D=(1,-1). Also number of branch nodes (i.e., vertices of outdegree at least two) in all ordered trees with n+1 edges. - Emeric Deutsch, Feb 22 2004
Number of lattice paths from (0,0) to (n,n) with steps E=(1,0) and N=(0,1) which touch or cross the line x-y=1. Example: For n=2 these are the paths EENN, ENEN, ENNE and NEEN. - Herbert Kociemba, May 23 2004
Narayana transform (A001263) of [1, 3, 5, 7, 9, ...] = (1, 4, 15, 56, 210, ...). Row sums of triangles A136534 and A136536. - Gary W. Adamson, Jan 04 2008
Starting with offset 1 = the Catalan sequence starting (1, 2, 5, 14, ...) convolved with A000984: (1, 2, 6, 20, ...). - Gary W. Adamson, May 17 2009
Also number of peaks plus number of valleys in all Dyck n-paths. - David Scambler, Oct 08 2012
Apparently counts UDDUD in all Dyck paths of semilength n+2. - David Scambler, Apr 22 2013
Apparently the number of peaks strictly left of the midpoint in all Dyck paths of semilength n+1. - David Scambler, Apr 30 2013
For n>0, a(n) is the number of compositions of n into at most n parts if zeros are allowed as parts (so-called "weak" compositions). - L. Edson Jeffery, Jul 24 2014
Number of paths in the half-plane x >= 0, from (0,0) to (2n,2), and consisting of steps U=(1,1) and D=(1,-1). For example, for n=2, we have the 4 paths: UUUD, UUDU, UDUU, DUUU. - José Luis Ramírez Ramírez, Apr 19 2015
For n>1, 1/a(n) is the probability that when a stick is broken up at n points independently and uniformly chosen at random along its length any triple of pieces of the n+1 pieces can form a triangle. The corresponding probability for the existence of at least one triple is A339392(n)/A339393(n). - Amiram Eldar, Dec 04 2020
a(n) is the number of lattice paths of 2n steps taken from the step set {U=(1,1), D=(1,-1)} that start at the origin, never go below the x-axis, and end strictly above the x-axis; more succinctly, proper left factors of Dyck paths. For example, a(2)=4 counts UUUU, UUUD, UUDU, UDUU. - David Callan and Emeric Deutsch, Jan 25 2021
From Gus Wiseman, Jul 21 2021: (Start)
Also the number of integer compositions of 2n+1 with alternating sum -1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(1) = 1 through a(3) = 15 compositions are:
(1,2) (2,3) (3,4)
(1,3,1) (1,4,2)
(1,1,1,2) (2,4,1)
(1,2,1,1) (1,1,2,3)
(1,2,2,2)
(1,3,2,1)
(2,1,1,3)
(2,2,1,2)
(2,3,1,1)
(1,1,1,3,1)
(1,2,1,2,1)
(1,3,1,1,1)
(1,1,1,1,1,2)
(1,1,1,2,1,1)
(1,2,1,1,1,1)
The following relate to these compositions.
- The unordered version is A000070.
- Allowing any negative alternating sum gives A000346.
- The opposite (positive 1) version is A000984.
- The version for reverse-alternating sum is also A001791 (this sequence).
- Taking alternating sum -2 instead of -1 gives A002054.
- The shifted version for alternating sum 0 is counted by A088218 and ranked by A344619.
- Ranked by A345910 (reverse: A345912).
Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 0 than 1's. For example, the a(2) = 4 binary numbers are: 10001, 10010, 10100, 11000.
(End)
The diagonal of a square n X n matrix where cells of the first row are the nonnegative integers and cells of subsequent rows are sums of cells of the previous row up to and including n. - Torlach Rush, Apr 24 2024
For n>=1, a(n) is the independence number of the odd graph O_{n+1}. - Miquel A. Fiol, Jul 07 2024

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. 828.
  • Cornelius Lanczos, Applied Analysis, Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 517.
  • R. C. Mullin, E. Nemeth and P. J. Schellenberg, The enumeration of almost cubic maps, pp. 281-295 in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and Computer Science. Vol. 1, edited R. C. Mullin et al., 1970.
  • 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).

Crossrefs

Diagonal 3 of triangle A100257.
First differences are in A076540.
A345197 counts compositions by length and alternating sum.

Programs

  • GAP
    List([0..30],n->Binomial(2*n,n-1)); # Muniru A Asiru, Aug 09 2018
  • Magma
    [Binomial(2*n, n-1): n in [0..30]]; // Vincenzo Librandi, Apr 20 2015
    
  • Mathematica
    Table[Binomial[2n,n-1],{n,0,30}] (* Harvey P. Dale, Jul 12 2012 *)
    CoefficientList[ Series[(1 - 2x - Sqrt[1 - 4x])/(2x*Sqrt[1 - 4x]), {x, 0, 26}], x] (* Robert G. Wilson v, Aug 10 2018 *)
  • Maxima
    A001791(n):=binomial(2*n,n-1)$
    makelist(A001791(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n)=if(n<1,0,(2*n)!/(n+1)!/(n-1)!)
    

Formula

a(n) = n*A000108(n).
G.f.: x*(d/dx)c(x) where c(x) = Catalan g.f. - Wolfdieter Lang
Convolution of A001700 (central binomial of odd order) and A000108 (Catalan): a(n+1) = Sum_{k=0..n} C(k)*binomial(2*(n-k)+1, n-k), C(k): Catalan. - Wolfdieter Lang
E.g.f.: exp(2x) * I_1(2x), where I_1 is Bessel function. - Michael Somos, Sep 08 2002
a(n) = Sum_{k=0..n} C(n, k)*C(n, k+1). - Paul Barry, May 15 2003
a(n) = Sum_{i=1..n} binomial(i+n-1, n).
G.f.: (1-2x-sqrt(1-4x))/(2x*sqrt(1-4x)). - Emeric Deutsch, Dec 05 2003
a(n) = A092956/(n!). - Amarnath Murthy, Jun 16 2004
a(n) = binomial(2n,n) - A000108(n). - Paul Barry, Apr 21 2005
a(n) = (1/(2*Pi))*Integral_{x=0..4} (x^n*(x-2)/sqrt(x(4-x))) is the moment sequence representation. - Paul Barry, Jan 11 2007
Row sums of triangle A132812 starting (1, 4, 15, 56, 210, ...). - Gary W. Adamson, Sep 01 2007
Starting (1, 4, 15, 56, 210, ...) gives the binomial transform of A025566 starting (1, 3, 8, 22, 61, 171, ...). - Gary W. Adamson, Sep 01 2007
For n >= 1, a(2^n) = 2^(n+1)*A001795(2^(n-1)). - Vladimir Shevelev, Sep 05 2010
D-finite with recurrence: (n-1)*(n+1)*a(n) = 2*n*(2n-1)*a(n-1). - R. J. Mathar, Dec 17 2011
From Sergei N. Gladkovskii, Jul 07 2012: (Start)
G.f.: -1/(2*x) - G(0) where G(k) = 1 - 1/(2*x - 8*x^3*(2*k+1)/(4*x^2*(2*k+1)- (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step);
E.g.f.: BesselI(1,2*x)*exp(2*x) = x*G(0) where G(k) = 1 + 2*x*(4*k+3)/((2*k+1)*(2*k+3) - x*(2*k+1)*(2*k+3)*(4*k+5)/(x*(4*k+5) + 2*(k+1)*(k+2)/G(k+1))); (continued fraction, 3rd kind, 3-step).
(End)
G.f.: c(x)^3/(2-c(x)) where c(x) is the g.f. for A000108. - Cheyne Homberger, May 05 2014
G.f.: z*C(z)^2/(1-2*z*C(z)), where C(z) is the g.f. of Catalan numbers. - José Luis Ramírez Ramírez, Apr 19 2015
G.f.: x*2F1(3/2,2;3;4x). - R. J. Mathar, Aug 09 2015
a(n) = Sum_{i=1..n} binomial(2*i-2,i-1)*binomial(2*(n-i+1),n-i+2)/(n-i+1). - Vladimir Kruchinin, Sep 07 2015
L.g.f.: 1/(1 - x/(1 - x/(1 - x/(1 - x/(1 - x/(1 - ...)))))) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, May 10 2017
Sum_{n>=1} 1/a(n) = 1/3 + 5*Pi/(9*sqrt(3)). - Amiram Eldar, Dec 04 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/5 + 14*sqrt(5)*log(phi)/25, where log(phi) = A002390. - Amiram Eldar, Feb 20 2021
a(n) = Product_{i=1..(n - 1)} (((4*i + 6)*i + 2)/((i + 2)*i)), for n>=1. - Neven Sajko, Oct 10 2021
a(n) = 2^(2*n)*gamma(n + 1/2)/(sqrt(Pi)*gamma(n)*(n+1)) for n > 0, and a(0) = lim_{n->0} a(n). - Karol A. Penson, Apr 24 2025

A103371 Number triangle T(n,k) = C(n,n-k)*C(n+1,n-k).

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 18, 12, 1, 5, 40, 60, 20, 1, 6, 75, 200, 150, 30, 1, 7, 126, 525, 700, 315, 42, 1, 8, 196, 1176, 2450, 1960, 588, 56, 1, 9, 288, 2352, 7056, 8820, 4704, 1008, 72, 1, 10, 405, 4320, 17640, 31752, 26460, 10080, 1620, 90, 1, 11, 550, 7425, 39600, 97020
Offset: 0

Views

Author

Paul Barry, Feb 03 2005

Keywords

Comments

Columns include A000027, A002411, A004302, A108647, A134287. Row sums are C(2n+1,n+1) or A001700.
T(n-1,k-1) is the number of ways to put n identical objects into k of altogether n distinguishable boxes. See the partition array A035206 from which this triangle arises after summing over all entries related to partitions with fixed part number k.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) of height k (height(alpha) = |Im(alpha)|). - Abdullahi Umar, Oct 02 2008
The o.g.f. of the (n+1)-th diagonal is given by G(n, x) = (n+1)*Sum_{k=1..n} A001263(n, k)*x^(k-1) / (1 - x)^(2*n+1), for n >= 1 and for n = 0 it is G(0, x) = 1/(1-x). - Wolfdieter Lang, Jul 31 2017

Examples

			The triangle T(n, k) begins:
n\k  0   1    2     3     4     5     6    7  8 9 ...
0:   1
1:   2   1
2:   3   6    1
3:   4  18   12     1
4:   5  40   60    20     1
5:   6  75  200   150    30     1
6:   7 126  525   700   315    42     1
7:   8 196 1176  2450  1960   588    56    1
8:   9 288 2352  7056  8820  4704  1008   72  1
9:  10 405 4320 17640 31752 26460 10080 1620 90 1
...  reformatted. - _Wolfdieter Lang_, Jul 31 2017
From _R. J. Mathar_, Mar 29 2013: (Start)
The matrix inverse starts
       1;
      -2,       1;
       9,      -6,      1;
     -76,      54,    -12,      1;
    1055,    -760,    180,    -20,   1;
  -21906,   15825,  -3800,    450, -30,   1;
  636447, -460026, 110775, -13300, 945, -42, 1; (End)
O.g.f. of 4th diagonal [4, 40,200, ...] is G(3, x) = 4*(1 + 3*x + x^2)/(1 - x)^7, from the n = 3 row [1, 3, 1] of A001263. See a comment above. - _Wolfdieter Lang_, Jul 31 2017
		

Crossrefs

Cf. A007318, A000894 (central terms), A132813 (mirrored).

Programs

  • Haskell
    a103371 n k = a103371_tabl !! n !! k
    a103371_row n = a103371_tabl !! n
    a103371_tabl = map reverse a132813_tabl
    -- Reinhard Zumkeller, Apr 04 2014
    
  • Magma
    /* As triangle */ [[Binomial(n,n-k)*Binomial(n+1,n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 01 2017
    
  • Maple
    A103371 := (n,k) -> binomial(n,k)^2*(n+1)/(k+1);
    seq(print(seq(A103371(n, k), k=0..n)), n=0..7); # Peter Luschny, Oct 19 2011
  • Mathematica
    Flatten[Table[Binomial[n,n-k]Binomial[n+1,n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, May 26 2014 *)
    CoefficientList[Series[Series[E^(x(1+y))(BesselI[0,2*x*Sqrt[y]]+BesselI[1,2*x*Sqrt[y]]/Sqrt[y]),{x,0,8}],{y,0,8}],{x,y}]*Range[0,8]! (* Natalia L. Skirrow, Apr 14 2025 *)
  • Maxima
    create_list(binomial(n,k)*binomial(n+1,k+1),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    for(n=0,10, for(k=0,n, print1(binomial(n,k)*binomial(n+1,k+1), ", "))) \\ G. C. Greubel, Nov 09 2018

Formula

Number triangle T(n, k) = C(n, n-k)*C(n+1, n-k) = C(n, k)*C(n+1, k+1); Column k of this triangle has g.f. Sum_{j=0..k} (C(k, j)*C(k+1, j) * x^(k+j))/(1-x)^(2*k+2); coefficients of the numerators are the rows of the reverse triangle C(n, k)*C(n+1, k).
T(n,k) = C(n, k)*Sum_{j=0..(n-k)} C(n-j, k). - Paul Barry, Jan 12 2006
T(n,k) = (n+1-k)*N(n+1,k+1), with N(n,k):=A001263(n,k), the Narayana triangle (with offset [1,1]).
O.g.f.: ((1-(1-y)*x)/sqrt((1-(1+y)*x)^2-4*x^2*y) -1)/2, (from o.g.f. of A001263, Narayana triangle). - Wolfdieter Lang, Nov 13 2007
From Peter Bala, Jan 24 2008: (Start)
Matrix product of A007318 and A122899.
O.g.f. for row n: (1-x)^n*P(n,1,0,(1+x)/(1-x)) = 1/(2*x)*(1-x)^(n+1)*( Legendre_P(n+1,(1+x)/(1-x)) - Legendre_P(n,(1+x)/(1-x)) ), where P(n,a,b,x) denotes the Jacobi polynomial.
O.g.f. for column k: x^k/(1-x)^(k+2)*P(k,0,1,(1+x)/(1-x)). Compare with A008459. (End)
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,1). (Cf. A194595, A197653, A197654). - Peter Luschny, Oct 20 2011
T(n,k) = A003056(n+1,k+1)*C(n,k)^2/(k+1). - Peter Luschny, Oct 29 2011
T(n,k) = A007318(n, k)*A135278(n, k), n >= k >= 0. - Wolfdieter Lang, Jul 31 2017
From Natalia L. Skirrow, Apr 14 2025: (Start)
T(n,k) = A008459(n,k) + n*N(n,k+1).
E.g.f.: e^(x*(1+y))*(I_0(2*x*sqrt(y)) + I_1(2*x*sqrt(y))/sqrt(y)), where I_n is the modified Bessel function of the first kind. (The I_0 contributes A008459(n,k), the I_1 contributes n*N(n,k+1))
O.g.f. for row n: (n+1)*2F1(-n,-n;2;y) = (n+1)*2F1(2+n,2+n;2;y)*(1-y)^(2*(n+1)) (by Euler's hypergeometric transformation); (n+1)*2F1(2+n,2+n;2;y) is the o.g.f. for row n of (k+n+1)!^2/(k!*(k+1)!*n!*(n+1)!), which is column n+1 of A132812.
O.g.f. for column k: 2F1(1+k,2+k;1;x)*x^k = 2F1(-k,-1-k;1;x)*x^k/(1-x)^(2+2*k). 2F1(-k,-1-k;1;x) is the kth row of A132813, the reflection of the kth row of this triangle.
O.g.f. for diagonal d (beginning at a(d,0)): (d+1)*x^d*2F1(d+1,d+2;2;x*y). 2F1(d+1,d+2;2;x) = 2F1(1-d,-d;2;x)/(1-x)^(2*d+1), numerator being the o.g.f. of row d of the Narayana triangle.
These respectively yield:
T(n,k) = Sum_{i=0..n+k} C(2*(n+1),i)*(-1)^i*A132812(n+1+k-i,n+1),
T(d+k,k) = Sum_{i=0..k} C(d-i+1+2*k,d-i)*T(k,k-i),
T(d+k,k) = Sum_{i=0..d} C(k-i + 2*d,k-i)*N(d,i+1)*(d+1).
E.g.f. for column k: 1F1(2+k;1;x)*x^k/k!.
E.g.f. for diagonal d: (d+1)*x^d*1F1(d+2;2;x*y)/d!. (End)

A110320 Number of blocks in all RNA secondary structures with n nodes (an RNA secondary structure can be viewed as a restricted noncrossing partition).

Original entry on oeis.org

1, 2, 5, 13, 32, 80, 201, 505, 1273, 3217, 8146, 20668, 52531, 133726, 340909, 870213, 2223958, 5689807, 14571335, 37350585, 95821071, 246015677, 632088930, 1625119218, 4180845277, 10762096850, 27718352411, 71426753423, 184146711578
Offset: 1

Views

Author

Emeric Deutsch, Jul 19 2005

Keywords

Comments

Antidiagonal sums of A132812. - Philippe Deléham, Jun 08 2013

Examples

			a(4)=13 because the 4 (=A004148(4)) RNA secondary structures of size 4, namely 1/2/3/4, 13/2/4, 14/2/3 and 1/24/3, have altogether 4+3+3+3=13 blocks.
		

Crossrefs

Programs

  • Maple
    G:=1/2*(1-z-z^2)/z^2/(1-2*z-z^2-2*z^3+z^4)^(1/2)-1/2*1/(z^2): Gser:=series(G,z=0,37): seq(coeff(Gser,z^n),n=1..33);
  • Mathematica
    Table[Sum[Binomial[n-j+1,j]Binomial[n-j+1,j-1],{j, 0, n}],{n,1,25}] (* Benedict W. J. Irwin, Sep 24 2016 *)

Formula

G.f.: (1-z-z^2)/(2*z^2*sqrt(1-2*z-z^2-2*z^3+z^4))-1/(2*z^2).
a(n) = Sum_{k=1..n} k*A110319(n,k).
Conjecture: a(n) = (A051292(n+2)-A051286(n+1))/2. - Gerald McGarvey, Jan 14 2007
a(n) = (A051286(n+2)-A051286(n+1)-A051286(n))/2. - Benedict W. J. Irwin, Sep 24 2016
a(n) ~ sqrt(4 + 9/sqrt(5)) * (3+sqrt(5))^n / (sqrt(Pi*n) * 2^(n+1)). - Vaclav Kotesovec, Sep 25 2016, equivalently, a(n) ~ phi^(2*n + 3) / (2 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 06 2021
D-finite with recurrence (n+2)*a(n) +3*(-n-1)*a(n-1) +(n-7)*a(n-3) +2*(2*n-3)*a(n-4) +(n-5)*a(n-5) +(-n+4)*a(n-6)=0. - R. J. Mathar, Feb 21 2020

A209805 Triangle read by rows: T(n,k) is the number of k-block noncrossing partitions of n-set up to rotations.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 4, 2, 1, 1, 3, 10, 10, 3, 1, 1, 3, 15, 25, 15, 3, 1, 1, 4, 26, 64, 64, 26, 4, 1, 1, 4, 38, 132, 196, 132, 38, 4, 1, 1, 5, 56, 256, 536, 536, 256, 56, 5, 1, 1, 5, 75, 450, 1260, 1764, 1260, 450, 75, 5, 1
Offset: 1

Views

Author

Tilman Piesk, Mar 13 2012

Keywords

Comments

Like the Narayana triangle A001263 (and unlike A152175) this triangle is symmetric.
The diagonal entries are 1, 1, 4, 25, 196, 1764, ... which is probably sequence A001246 - the squares of the Catalan numbers.
The above conjecture about the diagonal entries T(2*n-1, n) is true since gcd(2*n-1, n) = gcd(2*n-1, n-1) = 1 and then T(2*n-1, n) simplifies to A001246(n-1) using the formula given below. - Andrew Howroyd, Nov 15 2017

Examples

			Triangle begins:
  1;
  1,   1;
  1,   1,   1;
  1,   2,   2,   1;
  1,   2,   4,   2,   1;
  1,   3,  10,  10,   3,   1;
  1,   3,  15,  25,  15,   3,   1;
  1,   4,  26,  64,  64,  26,   4,   1;
  1,   4,  38, 132, 196, 132,  38,   4,   1;
  1,   5,  56, 256, 536, 536, 256,  56,   5,   1;
		

Crossrefs

Cf. A054357 (row sums), A001246 (square Catalan numbers).

Programs

  • Mathematica
    b[n_, k_] := Binomial[n-1, n-k] Binomial[n, n-k];
    T[n_, k_] := (DivisorSum[GCD[n, k], EulerPhi[#] b[n/#, k/#]&] + DivisorSum[GCD[n, k - 1], EulerPhi[#] b[n/#, (n + 1 - k)/#]&] - k Binomial[n, k]^2/(n - k + 1))/n;
    Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 01 2018, after Andrew Howroyd *)
  • PARI
    b(n,k)=binomial(n-1,n-k)*binomial(n,n-k);
    T(n,k)=(sumdiv(gcd(n,k), d, eulerphi(d)*b(n/d,k/d)) + sumdiv(gcd(n,k-1), d, eulerphi(d)*b(n/d,(n+1-k)/d)) - k*binomial(n,k)^2/(n-k+1))/n; \\ Andrew Howroyd, Nov 15 2017

Formula

T(n,k) = (1/n)*((Sum_{d|gcd(n,k)} phi(d)*A103371(n/d-1,k/d-1)) + (Sum_{d|gcd(n,k-1)} phi(d)*A103371(n/d-1,(n+1-k)/d-1)) - A132812(n,k)). - Andrew Howroyd, Nov 15 2017

A202409 Triangle read by rows, n>=1, 1<=k<=n, T(n,k) = k*binomial(n,k)^3*(n^2+n-k*n-k+k^2)/((n-k+1)^2*n).

Original entry on oeis.org

1, 4, 4, 9, 36, 9, 16, 168, 168, 16, 25, 550, 1400, 550, 25, 36, 1440, 7500, 7500, 1440, 36, 49, 3234, 30135, 61250, 30135, 3234, 49, 64, 6496, 98784, 356720, 356720, 98784, 6496, 64, 81, 11988, 278208, 1629936, 2889432, 1629936, 278208, 11988, 81
Offset: 1

Views

Author

Peter Luschny and Susanne Wienand, Dec 19 2011

Keywords

Comments

Let a meander be defined as in the link and m = 3. Then T(n,k) counts the invertible meanders of length m(n+1) built from arcs with central angle 360/m whose binary representation have mk '1's.

Examples

			[1]                1
[2]               4, 4
[3]             9, 36, 9
[4]         16, 168, 168, 16
[5]      25, 550, 1400, 550, 25
[6]  36, 1440, 7500, 7500, 1440, 36
T(2,1) = 4 because the invertible meanders of length 9 and central angle 120 degree which have three '1's in their binary representation are {100100100, 100011000, 110001000, 111000000}.
		

Crossrefs

Row sums: A201640. Cf. A132812.

Programs

  • Maple
    A202409 := (n,k) -> k*binomial(n,k)^3*(n^2+n-k*n-k+k^2)/((n-k+1)^2*n);
    seq(print(seq(A202409(n,k),k=1..n)),n=1..6);
  • Mathematica
    t[n_, k_] := k*Binomial[n, k]^3*(n^2 + n - k*n - k + k^2)/((n - k + 1)^2*n); Table[t[n, k], {n, 1, 9}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 02 2013 *)

A241475 Triangle t(n,r) = s(n,r)*s(n,r+1), where s(n,r) = lcm(n,n-1,...,n-r+1)/lcm(1,2,...,r-1,r), n >= 1 and 0 <= r < n.

Original entry on oeis.org

1, 2, 2, 3, 9, 3, 4, 24, 12, 2, 5, 50, 100, 50, 5, 6, 90, 150, 50, 5, 1, 7, 147, 735, 1225, 245, 49, 7, 8, 224, 784, 1960, 980, 196, 28, 2, 9, 324, 3024, 3528, 1764, 1764, 252, 18, 3, 10, 450, 2700, 12600, 8820, 1764, 252, 18, 3, 1, 11, 605, 9075, 54450, 152460, 213444, 30492, 2178, 363, 121, 11
Offset: 1

Views

Author

Jean-François Alcover, Apr 23 2014

Keywords

Comments

The first eight terms and the first two terms of every row are identical to those of A132812.

Examples

			Triangle begins:
  1;
  2,  2;
  3,  9,   3;
  4, 24,  12,  2;
  5, 50, 100, 50, 5;
  6, 90, 150, 50, 5, 1;
  ...
		

Crossrefs

Programs

  • Mathematica
    s[, 0] = 1; s[n, r_?NumericQ] := LCM @@ Table[n-k+1, {k, 1, r}] / LCM @@ Table[k, {k, 1, r}]; t[n_, r_] := s[n, r]*s[n, r+1]; Table[t[n, r] , {n, 1, 12}, {r, 0, n-1}] // Flatten

A361894 Triangle read by rows. T(n, k) is the number of Fibonacci meanders with a central angle of 360/m degrees that make m*k left turns and whose length is m*n, where m = 2.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 6, 2, 1, 5, 16, 6, 2, 1, 6, 35, 20, 6, 2, 1, 7, 66, 65, 20, 6, 2, 1, 8, 112, 186, 70, 20, 6, 2, 1, 9, 176, 462, 246, 70, 20, 6, 2, 1, 10, 261, 1016, 812, 252, 70, 20, 6, 2, 1, 11, 370, 2025, 2416, 917, 252, 70, 20, 6, 2, 1, 12, 506, 3730, 6435, 3256, 924, 252, 70, 20, 6, 2, 1
Offset: 1

Views

Author

Peter Luschny, Mar 31 2023

Keywords

Comments

For an overview of the terms used see A361574. A201631 gives the row sums of this triangle.
The corresponding sequence counting meanders without the requirement of being Fibonacci is A103371 (for which in turn A103327 is a termwise majorant counting permutations of the same type).
The diagonals, starting from the main diagonal, apparently converge to A000984.

Examples

			Triangle T(n, k) starts:
  [ 1]  1;
  [ 2]  2,   1;
  [ 3]  3,   2,    1;
  [ 4]  4,   6,    2,    1;
  [ 5]  5,  16,    6,    2,    1;
  [ 6]  6,  35,   20,    6,    2,   1;
  [ 7]  7,  66,   65,   20,    6,   2,   1;
  [ 8]  8, 112,  186,   70,   20,   6,   2,  1;
  [ 9]  9, 176,  462,  246,   70,  20,   6,  2,  1;
  [10] 10, 261, 1016,  812,  252,  70,  20,  6,  2, 1;
  [11] 11, 370, 2025, 2416,  917, 252,  70, 20,  6, 2, 1;
  [12] 12, 506, 3730, 6435, 3256, 924, 252, 70, 20, 6, 2, 1.
.
T(4, k) counts Fibonacci meanders with central angle 180 degrees and length 8 that make k left turns. Written as binary strings (L = 1, R = 0):
k = 1: 11000000, 10010000, 10000100, 10000001;
k = 2: 11110000, 11100100, 11100001, 11010010, 11001001, 10100101;
k = 3: 11111100, 11111001;
k = 4: 11111111.
		

Crossrefs

Cf. A201631 (row sums), A361681 (m=3), A132812, A361574, A103371, A000984.

Programs

  • SageMath
    # using function 'FibonacciMeandersByLeftTurns' from A361681.
    for n in range(1, 12):
        print(FibonacciMeandersByLeftTurns(2, n))
Showing 1-8 of 8 results.