A001700 -id:A001700 - cp's OEIS frontend

A081721 Number of bracelets of n beads in up to n colors.

1, 3, 10, 55, 377, 4291, 60028, 1058058, 21552969, 500280022, 12969598086, 371514016094, 11649073935505, 396857785692525, 14596464294191704, 576460770691256356, 24330595997127372497, 1092955780817066765469, 52063675152021153895330, 2621440000054016000176044

Offset: 1

N. J. A. Sloane, based on information supplied by Gary W. Adamson, Apr 05 2003

nonn

T(n,n), T given in A081720.
From Olivier Gérard, Aug 01 2016: (Start)
Number of classes of functions of [n] to [n] under rotation and reversal.
.
Classes can be of size between 1 and 2n
depending on divisibility properties of n.
.
n   1   2   3   4    5         n     2n
----------------------------------------
1   1
2   2   1
3   3   0   6                         1
4   4   6   0                 30     15
5   5   0   0                120    252
6   6  15  30                725   3515
7   7   0   0               2394  57627
.
(End)

Alois P. Heinz, Table of n, a(n) for n = 1..200

Cf. A000312 All endofunctions
Cf. A000169 Classes under translation mod n
Cf. A001700 Classes under sort
Cf. A056665 Classes under rotation
Cf. A168658 Classes under complement to n+1
Cf. A130293 Classes under translation and rotation
Cf. A275549 Classes under reversal
Cf. A275550 Classes under reversal and complement
Cf. A275551 Classes under translation and reversal
Cf. A275552 Classes under translation and complement
Cf. A275553 Classes under translation, complement and reversal
Cf. A275554 Classes under translation, rotation and complement
Cf. A275555 Classes under translation, rotation and reversal
Cf. A275556 Classes under translation, rotation, complement and reversal
Cf. A275557 Classes under rotation and complement
Cf. A275558 Classes under rotation, complement and reversal
Row sums of partition array A213941.
Main diagonal of A321791.

Table[CycleIndex[DihedralGroup[n],s]/.Table[s[i]->n,{i,1,n}],{n,1,20}] (* Geoffrey Critzer, Jun 18 2013 *)
t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]); a[n_] := t[n, n]; Array[a, 20] (* Jean-François Alcover, Nov 02 2017, after Maple code for A081720 *)

a(n) ~ n^(n-1) / 2. - Vaclav Kotesovec, Mar 18 2017

Name changed by Olivier Gérard, Aug 05 2016

Name revised by Álvar Ibeas, Apr 20 2018

A224274 a(n) = binomial(4*n,n)/4.

Original entry on oeis.org

1, 7, 55, 455, 3876, 33649, 296010, 2629575, 23535820, 211915132, 1917334783, 17417133617, 158753389900, 1451182990950, 13298522298180, 122131734269895, 1123787895356412, 10358022441395860, 95615237915961100, 883829035553043580, 8179808679272664720, 75788358475481302185

Offset: 1

Gary Detlefs, Apr 02 2013

In general, binomial(k*n,n)/k = binomial(k*n-1,n-1).
Sequences in the OEIS related to this identity are:
. C(2n,n) = A000984, C(2n,n)/2 = A001700;
. C(3n,n) = A005809, C(3n,n)/3 = A025174;
. C(4n,n) = A005810, C(4n,n)/4 = a(n);
. C(5n,n) = A001449, C(5n,n)/5 = A163456;
. C(6n,n) = A004355, C(6n,n)/6 is not in the OEIS.
Conjecture: a(n) == 1 (mod n^3) iff n is an odd prime.
It is known that a(p) == 1(mod p^3) for prime p >= 3. See Mestrovic, Section 3. - Peter Bala, Oct 09 2015

			For n=2, binomial(4*n,n) = binomial(8,2) = 8*7/2 = 28, so a(2) = 28/4 = 7. - _Michael B. Porter_, Jul 12 2016

G. C. Greubel, Table of n, a(n) for n = 1..1000
Dmitry Kruchinin and Vladimir Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequence, Vol. 18 (2015), article 15.4.6.
Romeo Meštrović, Lucas’ theorem: its generalizations, extensions and applications (1878-2014), arXiv:1409.3820v1 [math.NT], 2014.
N. J. Wildberger and Dean Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly (2025). See section 12.

Cf. A001700, A001764, A025174, A163456, A002293, A069271, A004331, A005810, A052203, A257633, A262977, A227726.

[Binomial(4*n,n) div 4: n in [1..25]]; // Vincenzo Librandi, Jun 03 2015

Maple
```
seq(binomial(4*n,n)/4, n=1..17);
```

Table[Binomial[4 n, n]/4, {n, 30}] (* Vincenzo Librandi, Jun 03 2015 *)

a(n) = binomial(4*n,n)/4; /* Joerg Arndt, Apr 02 2013 */

a(n) = binomial(4*n,n)/4 = A005810(n)/4.
a(n) = binomial(4*n-1,n-1).
G.f.: A(x) = B'(x)/B(x), where B(x) = 1 + x*B(x)^4 is g.f. of A002293. - Vladimir Kruchinin, Aug 13 2015
From Peter Bala, Oct 08 2015: (Start)
a(n) = 1/2*[x^n] (C(x)^2)^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A163456.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 4*x^2 + 22*x^3 + ... is the o.g.f. for A002293.
exp( 2*Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*x + 9*x^2 + 52*x^3 + ... is the o.g.f. for A069271. (End)
From Peter Bala, Nov 04 2015: (Start)
With an offset of 1, the o.g.f. equals f(x)*g(x)^3, where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A262977 (k = -1), A005810 (k = 0), A052203 (k = 1), A257633 (k = 2) and A004331 (k = 4). (End)
a(n) = 1/5*[x^n] (1 + x)/(1 - x)^(3*n + 1) = 1/5*[x^n]( 1/C(-x) )^(5*n), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A227726. - Peter Bala, Jul 12 2016
a(n) ~ 2^(8*n-3/2)*3^(-3*n-1/2)*n^(-1/2)/sqrt(Pi). - Ilya Gutkovskiy, Jul 12 2016
O.g.f.: A(x) = f(x)/(1 - 3*f(x)), where f(x) = series reversion (x/(1 + x)^4) = x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + ... is the o.g.f. of A002293 with the initial term omitted. Cf. A025174. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/3)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+3)*n+k-1,k) = C(4*n,n)/4 and (1/4)*Sum_{k = 0..n} (-1)^k*C(x*n,n-k)*C((x-4)*n+k-1,k) = C(4*n,n)/4, both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
Right-hand side of the identity (1/3)*Sum_{k = 0..2*n} (-1)^k*binomial(5*n-k-1,2*n-k)*binomial(3*n+k-1,k) = binomial(4*n,n)/4. - Peter Bala, Mar 09 2022
a(n) = [x^n] G(x)^n, where G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... is the g.f. of A001764. - Peter Bala, Oct 17 2024

A003516 Binomial coefficients C(2n+1, n-2).

Original entry on oeis.org

1, 7, 36, 165, 715, 3003, 12376, 50388, 203490, 817190, 3268760, 13037895, 51895935, 206253075, 818809200, 3247943160, 12875774670, 51021117810, 202112640600, 800472431850, 3169870830126, 12551759587422

Offset: 2

N. J. A. Sloane

a(n) is the number of royal paths (A006318) from (0,0) to (n,n) with exactly one diagonal step off the line y=x. - David Callan, Mar 25 2004

a(n) is the total number of DDUU's in all Dyck (n+2)-paths. - David Scambler, May 03 2013

			For n=4, C(2*4+1,4-2) = C(9,2) = 9*8/2 = 36, so a(4) = 36. - _Michael B. Porter_, Sep 10 2016

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

G. C. Greubel, Table of n, a(n) for n = 2..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Heidi Goodson, An Identity for Vertically Aligned Entries in Pascal's Triangle, arXiv:1901.08653 [math.CO], 2019.
Milan Janjic, Two Enumerative Functions.
Toufik Mansour and Mark Shattuck, Counting occurrences of subword patterns in non-crossing partitions, Art Disc. Appl. Math. (2022).
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Vol. 15 (2012), Article 12.3.3. - From _N. J. A. Sloane_, Sep 16 2012
Daniel W. Stasiuk, An Enumeration Problem for Sequences of n-ary Trees Arising from Algebraic Operads, Master's Thesis, University of Saskatchewan-Saskatoon (2018).

Diagonal 6 of triangle A100257.
Third unsigned column (s=2) of A113187. - Wolfdieter Lang, Oct 18 2012
Cf. triangle A114492 - Dyck paths with k DDUU's.
Cf. A001622, A115139.
Cf. binomial(2*n+m, n): A000984 (m = 0), A001700 (m = 1), A001791 (m = 2), A002054 (m = 3), A002694 (m = 4), A002696 (m = 6), A030053 - A030056, A004310 - A004318.

List([2..25], n-> Binomial(2*n+1, n-2)); # G. C. Greubel, Mar 21 2019

[Binomial(2*n+1,n-2): n in [2..25]]; // Vincenzo Librandi, Apr 13 2011

CoefficientList[ Series[ 32/(((Sqrt[1 - 4 x] + 1)^5)*Sqrt[1 - 4 x]), {x, 0, 25}], x] (* Robert G. Wilson v, Aug 08 2011 *)
Table[Binomial[2*n +1,n-2], {n,2,25}] (* G. C. Greubel, Jan 23 2017 *)

{a(n) = binomial(2*n+1, n-2)}; \\ G. C. Greubel, Mar 21 2019

[binomial(2*n+1, n-2) for n in (2..25)] # G. C. Greubel, Mar 21 2019

G.f.: 32*x^2/(sqrt(1-4*x)*(sqrt(1-4*x)+1)^5). - Marco A. Cisneros Guevara, Jul 18 2011
a(n) = Sum_{k=0..n-2} binomial(n+k+2,k). - Arkadiusz Wesolowski, Apr 02 2012
D-finite with recurrence (n+3)*(n-2)*a(n) = 2*n*(2*n+1)*a(n-1). - R. J. Mathar, Oct 13 2012
G.f.: x^2*c(x)^5/sqrt(1-4*x) = ((-1 + 2*x) + (1 - 3*x + x^2) * c(x))/(x^2*sqrt(1-4*x)), with c(x) the o.g.f. of the Catalan numbers A000108. See the W. Lang link under A115139 for powers of c. - Wolfdieter Lang, Sep 10 2016
a(n) ~ 2^(2*n+1)/sqrt(Pi*n). - Ilya Gutkovskiy, Sep 10 2016
From Amiram Eldar, Jan 24 2022: (Start)
Sum_{n>=2} 1/a(n) = 4 - 14*Pi/(9*sqrt(3)).
Sum_{n>=2} (-1)^n/a(n) = 228*log(phi)/(5*sqrt(5)) - 134/15, where phi is the golden ratio (A001622). (End)
G.f.: 2F1([7/2,3],[6],4*x). - Karol A. Penson, Apr 24 2024
a(n) = Integral_{x = 0..4} x^n * w(x) dx, where the weight function w(x) = 1/(2*Pi) * sqrt(x)*(x^2 - 5*x + 5)/sqrt(4 - x). - Peter Bala, Oct 13 2024

A005566 Number of walks of length n on square lattice, starting at origin, staying in first quadrant.

Original entry on oeis.org

1, 2, 6, 18, 60, 200, 700, 2450, 8820, 31752, 116424, 426888, 1585584, 5889312, 22084920, 82818450, 312869700, 1181952200, 4491418360, 17067389768, 65166397296, 248817153312, 953799087696, 3656229836168, 14062422446800, 54086240180000, 208618354980000

Offset: 0

N. J. A. Sloane

a(n) is the number of involutions of length 2n which are invariant under the reverse-complement map and have no decreasing subsequences of length 5. - Eric S. Egge, Oct 21 2008

			G.f. = 1 + 2*x + 6*x^2 + 18*x^3 + 60*x^4 + 200*x^5 + 700*x^6 + 2450*x^7 + ... - _Michael Somos_, Oct 17 2019

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

Alin Bostan, Computer Algebra for Lattice Path Combinatorics, Seminaire de Combinatoire Ph. Flajolet, March 28 2013.
Alin Bostan, Calcul Formel pour la Combinatoire des Marches [The text is in English], Habilitation à Diriger des Recherches, Laboratoire d'Informatique de Paris Nord, Université Paris 13, December 2017.
Alin Bostan, Frédéric Chyzak, Mark van Hoeij, Manuel Kauers, and Lucien Pech, Hypergeometric expressions for generating functions of walks with small steps in the quarter plane. Eur. J. Comb. 61, 242-275 (2017)
R. K. Guy, Letter to N. J. A. Sloane, May 1990
R. K. Guy, Catwalks, sandsteps and Pascal pyramids, J. Integer Sequences, Vol. 3 (2000), Article #00.1.6.

Cf. A001700, A060897-A060900.

a(2*n) = A000894(n), a(2*n+1) = 2*A060150(n+1).

[Binomial(n, Floor(n/2))*Binomial(n+1, Floor((n+1)/2)): n in [0..30]]; // Vincenzo Librandi, Feb 18 2015

f[n_] := Binomial[n, Floor[n/2]] Binomial[n + 1, Floor[(n + 1)/2]]; Array[f, 25, 0] (* Robert G. Wilson v *)

a(n) = binomial(n, floor(n/2))*binomial(n+1, floor((n+1)/2)).
E.g.f.: (BesselI(0, 2*x)+BesselI(1, 2*x))^2. - Vladeta Jovovic, Apr 28 2003
EXPCONV of A001405 with itself, i.e., a(n) = sum_{k=0}^n binomial(n,k)*A001405(k)*A001405(n-k). - Max Alekseyev, May 18 2006
G.f.: (16*x^2-1)*hypergeom([3/2, 3/2],[2],16*x^2) + (1/(2x)+2)*hypergeom([1/2, 1/2],[1],16*x^2) - 1/(2x). - Mark van Hoeij, Oct 13 2009
G.f.: (hypergeom([1/2,1/2],[1],16*x^2) - 1)/(2*x) + hypergeom([1/2,3/2],[2],16*x^2). - Mark van Hoeij, Aug 14 2014
a(n) = A241530(n)*2*floor(n/2)/(floor(n/2)+1). - Peter Luschny, Apr 25 2014
D-finite with recurrence (n+2)*(n+1)*a(n) +4*(-2*n-1)*a(n-1) -16*n*(n-1)*a(n-2)=0. - R. J. Mathar, Mar 07 2015
0 = a(n)*(+16*a(n+2) -6*a(n+3)) +a(n+1)*(-2*a(n+2) +a(n+3)) if n >= 0. - Michael Somos, Oct 17 2019
a(n) = binomial(floor(n + 1/2), floor(n/2)) * binomial(ceiling(n + 1/2), ceiling(n/2)). - Peter Luschny, Dec 14 2024

Additional comments from David W. Wilson, May 05 2001

a(25)-a(26) from Vincenzo Librandi, Feb 18 2015

A100257 Triangle of expansions of 2^(k-1)*x^k in terms of T(n,x), in descending degrees n of T, with T the Chebyshev polynomials.

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 0, 3, 0, 1, 0, 4, 0, 3, 1, 0, 5, 0, 10, 0, 1, 0, 6, 0, 15, 0, 10, 1, 0, 7, 0, 21, 0, 35, 0, 1, 0, 8, 0, 28, 0, 56, 0, 35, 1, 0, 9, 0, 36, 0, 84, 0, 126, 0, 1, 0, 10, 0, 45, 0, 120, 0, 210, 0, 126, 1, 0, 11, 0, 55, 0, 165, 0, 330, 0, 462, 0, 1, 0, 12, 0, 66, 0, 220, 0

Offset: 0

Ralf Stephan, Nov 13 2004

			x^0 = T(0,x)
x^1 = T(1,x) + 0T(0,x)
2x^2 = T(2,x) + 0T(1,x) + 1T(0,x)
4x^3 = T(3,x) + 0T(2,x) + 3T(1,x) + 0T(0,x)
8x^4 = T(4,x) + 0T(3,x) + 4T(2,x) + 0T(1,x) + 3T(0,x)
16x^5 = T(5,x) + 0T(4,x) + 5T(3,x) + 0T(2,x) + 10T(1,x) + 0T(0,x)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.

Vincenzo Librandi, Table of n, a(n) for n = 0..6104
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
H. J. Brothers, Pascal's Prism: Supplementary Material.
Daniel J. Greenhoe, Frames and Bases: Structure and Design, Version 0.20, Signal Processing ABCs series (2019) Vol. 4, see page 175.
Daniel J. Greenhoe, A Book Concerning Transforms, Version 0.10, Signal Processing ABCs series (2019) Vol. 5, see page 97.
Index entries for sequences related to Chebyshev polynomials.

Without zeros: A008311. Row sums are A011782. Cf. A092392.

Diagonals are (with interleaved zeros) twice A001700, A001791, A002054, A002694, A003516, A002696, A030053, A004310, A030054, A004311, A030055, A004312, A030056, A004313.

a[k_, n_] := If[k == 1, 1, If[EvenQ[n] || k < 0 || n > k, 0, If[n >= k - 1, Binomial[2*Floor[k/2], Floor[k/2]]/2, Binomial[k - 1, Floor[n/2]]]]];
Table[a[k, n], {k, 1, 13}, {n, 1, k}] // Flatten (* Jean-François Alcover, May 04 2017, translated from PARI *)

a(k,n)=if(k==1,1,if(n%2==0||k<0||n>k,0,if(n>=k-1,binomial(2*floor(k/2),floor(k/2))/2,binomial(k-1,floor(n/2)))))

A303639 Number of ways to write n as a^2 + b^2 + binomial(2c+1,c) + binomial(2d+1,d), where a,b,c,d are nonnegative integers with a <= b and c <= d.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 2, 1, 2, 3, 3, 3, 3, 4, 2, 2, 2, 3, 4, 4, 5, 2, 4, 1, 2, 3, 3, 5, 3, 5, 1, 3, 1, 1, 6, 3, 8, 3, 6, 2, 4, 4, 2, 7, 5, 6, 2, 5, 2, 4, 5, 4, 8, 4, 7, 2, 4, 1, 3, 6, 4, 7, 3, 5, 2, 4, 2, 4, 9, 5, 6, 2, 6, 4, 5, 4, 7, 5, 2

Offset: 1

Zhi-Wei Sun, Apr 27 2018

nonn

Conjecture: a(n) > 0 for all n > 1.
This is similar to the author's conjecture in A303540.
It has been verified that a(n) > 0 for all n = 2..6*10^8.

			a(9) = 1 with 9 = 1^2 + 2^2 + binomial(2*0+1,0) + binomial(2*1+1,1).
a(2530) = 1 with 2530 = 0^2 + 49^2 + binomial(2*1+1,1) + binomial(2*4+1,4).
a(3258) = 1 with 3258 = 22^2 + 52^2 + binomial(2*3+1,3) + binomial(2*3+1,3).
a(5300) = 1 with 5300 = 10^2 + 59^2 + binomial(2*1+1,1) + binomial(2*6+1,6).
a(13453) = 1 with 13453 = 51^2 + 104^2 + binomial(2*0+1,0) + binomial(2*3+1,3).
a(20964) = 1 with 20964 = 13^2 + 138^2 + binomial(2*3+1,3) + binomial(2*6+1,6).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

Cf. A000290, A001481, A001700, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303601.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
c[n_]:=c[n]=Binomial[2n+1,n];
f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
tab={};Do[r=0;k=0;Label[bb];If[c[k]>n,Goto[aa]];Do[If[QQ[n-c[k]-c[j]],Do[If[SQ[n-c[k]-c[j]-x^2],r=r+1],{x,0,Sqrt[(n-c[k]-c[j])/2]}]],{j,0,k}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,80}];Print[tab]

A275549 Number of classes of endofunctions of [n] under reversal.

Original entry on oeis.org

1, 1, 3, 18, 136, 1625, 23436, 412972, 8390656, 193739769, 5000050000, 142656721086, 4458051717120, 151437584670385, 5556003465485760, 218946946471875000, 9223372039002259456, 413620131002462320337, 19673204037747448432896, 989209827833222327690890

Offset: 0

Olivier Gérard, Aug 01 2016

f and g are in the same class if function g(i) = f(n+1-i) for all i.
Decomposition by class size
.
n    1      2
---------------
1    1      0
2    2      1
3    9      9
4    16     120
5    125    1500
6    216    23220
7    2401   410571
.
Demonstration for the formula: the classes are either of size 1 or 2.
The classes of size 1 is for functions invariant by reversal. They are specified by half their values, including one more if n is odd. Their number is n^(ceiling(n/2)).
So the number of classes under this symmetry is half (the number of functions + the number of classes of size 1).
a(n) is the number of unoriented length n strings with a maximum of n colors. - Andrew Howroyd, Sep 13 2019

Andrew Howroyd, Table of n, a(n) for n = 0..200

Main diagonal of A277504.
Cf. A000312 All endofunctions
Cf. A000169 Classes under translation mod n
Cf. A001700 Classes under sort
Cf. A056665 Classes under rotation
Cf. A168658 Classes under complement to n+1
Cf. A130293 Classes under translation and rotation
Cf. A081721 Classes under rotation and reversal
Cf. A275550 Classes under reversal and complement
Cf. A275551 Classes under translation and reversal
Cf. A275552 Classes under translation and complement
Cf. A275553 Classes under translation, complement and reversal
Cf. A275554 Classes under translation, rotation and complement
Cf. A275555 Classes under translation, rotation and reversal
Cf. A275556 Classes under translation, rotation, complement and reversal
Cf. A275557 Classes under rotation and complement
Cf. A275558 Classes under rotation, complement and reversal
Cf. A078707 Endofunctions symmetric around their middle (stable by reversal).

a(n) = {(n^n + n^((n+1)\2))/2} \\ Andrew Howroyd, Sep 13 2019

a(n) = (n^n+n^ceiling(n/2))/2.

A050186 Triangular array T read by rows: T(h,k) = number of binary words of k 1's and h-k 0's which are not a juxtaposition of 2 or more identical subwords.

Original entry on oeis.org

1, 1, 1, 0, 2, 0, 0, 3, 3, 0, 0, 4, 4, 4, 0, 0, 5, 10, 10, 5, 0, 0, 6, 12, 18, 12, 6, 0, 0, 7, 21, 35, 35, 21, 7, 0, 0, 8, 24, 56, 64, 56, 24, 8, 0, 0, 9, 36, 81, 126, 126, 81, 36, 9, 0, 0, 10, 40, 120, 200, 250, 200, 120, 40, 10, 0, 0, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11

Offset: 0

Clark Kimberling

			For example, T(4,2) counts 1100,1001,0011,0110; T(2,1) counts 10, 01 (hence also counts 1010, 0101).
Rows:
  1;
  1,  1;
  0,  2,  0;
  0,  3,  3,  0;
  0,  4,  4,  4,  0;
  0,  5, 10, 10,  5,  0;

M. F. Hasler, A050186, rows 0..50, Sep 27 2018
M. F. Hasler, A050186, rows 0..100, Sep 27 2018
M. F. Hasler, A050186, rows 0..200, Sep 27 2018
N. J. A. Sloane, Transforms
Index entries for triangles and arrays related to Pascal's triangle

Same triangle as A053727 except this one includes column 0.

T(2n, n), T(2n+1, n) match A007727, A001700, respectively. Row sums match A027375.

T[n_, k_] := If[n == 0, 1, DivisorSum[GCD[k, n], MoebiusMu[#] Binomial[n/#, k/#]&]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 16 2022 *)

A050186(n,k)=sumdiv(gcd(n+!n,k),d,moebius(d)*binomial(n/d,k/d)) \\ M. F. Hasler, Sep 27 2018

MOEBIUS transform of A007318 Pascal's Triangle.

If rows n > 1 are divided by n, this yields the triangle A051168, which equals A245558 surrounded by 0's (except for initial terms). This differs from A011847 from row n = 9 on. - M. F. Hasler, Sep 29 2018

A245667 Number T(n,k) of sequences in {1,...,n}^n with longest increasing subsequence of length k; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 10, 16, 1, 0, 35, 175, 45, 1, 0, 126, 1771, 1131, 96, 1, 0, 462, 17906, 23611, 4501, 175, 1, 0, 1716, 184920, 461154, 161876, 13588, 288, 1, 0, 6435, 1958979, 8837823, 5179791, 759501, 34245, 441, 1, 0, 24310, 21253375, 169844455, 157279903, 36156355, 2785525, 75925, 640, 1

Offset: 0

Alois P. Heinz, Jul 28 2014

Sum_{k=0..1} T(n,k) = A088218(n).
Sum_{k=0..2} T(n,k) = A239295(n).
Sum_{k=0..3} T(n,k) = A239299(n).
Sum_{k=1..n} k * T(n,k) = A275576(n).

			T(3,1) = 10: [1,1,1], [2,1,1], [2,2,1], [2,2,2], [3,1,1], [3,2,1], [3,2,2], [3,3,1], [3,3,2], [3,3,3].
T(3,3) = 1: [1,2,3].
Triangle T(n,k) begins:
  1;
  0,    1;
  0,    3,      1;
  0,   10,     16,      1;
  0,   35,    175,     45,      1;
  0,  126,   1771,   1131,     96,     1;
  0,  462,  17906,  23611,   4501,   175,   1;
  0, 1716, 184920, 461154, 161876, 13588, 288,  1;
  ...

Alois P. Heinz, Rows n = 0..18, flattened

Columns k=0-10 give: A000007, A088218 or A001700(n-1) for n>0, A268869, A268870, A268871, A268872, A268873, A268874, A268875, A268876, A268877.
Main diagonal gives A000012.
T(n,n-1) gives A152618(n) for n>0.
T(n,n-2) gives A268936(n).
T(2n,n) gives A268949(n).
Row sums give A000312.
Cf. A047874, A239295, A239299, A275576.

b:= proc(n, l) option remember; `if`(n=0, 1, add(b(n-1, [seq(min(l[j],
      `if`(j=1 or l[j-1] `if`(k=0, `if`(n=0, 1, 0), b(n, [n$k])):
T:= (n, k)-> A(n, k) -`if`(k=0, 0, A(n, k-1)):
seq(seq(T(n, k), k=0..n), n=0..9);

b[n_, l_List] := b[n, l] = If[n == 0, 1, Sum[b[n-1, Table[Min[l[[j]], If[j == 1 || l[[j-1]]Jean-François Alcover, Feb 04 2015, after Alois P. Heinz *)

A005573 Number of walks on cubic lattice (starting from origin and not going below xy plane).

Original entry on oeis.org

1, 5, 26, 139, 758, 4194, 23460, 132339, 751526, 4290838, 24607628, 141648830, 817952188, 4736107172, 27487711752, 159864676803, 931448227590, 5435879858958, 31769632683132, 185918669183370, 1089302293140564

Offset: 0

N. J. A. Sloane

Binomial transform of A026378, second binomial transform of A001700. - Philippe Deléham, Jan 28 2007

The Hankel transform of [1,1,5,26,139,758,...] is [1,4,15,56,209,...](see A001353). - Philippe Deléham, Apr 13 2007

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

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
Emeric Deutsch and Jim Brawner, Problem 10795: Three-Dimensional Lattice Walks in the Upper Half-Space, Amer. Math. Monthly, 108 (Dec. 2001), 980.
Rigoberto Flórez, Leandro Junes, and José L. Ramírez, Further Results on Paths in an n-Dimensional Cubic Lattice, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.2.
R. K. Guy, Letter to N. J. A. Sloane, May 1990
R. K. Guy, Catwalks, Sandsteps and Pascal Pyramids, J. Integer Seqs., Vol. 3 (2000), #00.1.6.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 97.

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (Sqrt((1-2*x)/(1-6*x)) -1)/(2*x) )); // G. C. Greubel, May 02 2019

CoefficientList[Series[(Sqrt[(1-2x)/(1-6x)]-1)/(2x),{x,0,20}],x] (* Harvey P. Dale, Jun 24 2011 *)
a[n_] := 6^n Hypergeometric2F1[1/2, -n, 2, 2/3]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Apr 11 2017 *)

my(x='x+O('x^30)); Vec((sqrt((1-2*x)/(1-6*x)) -1)/(2*x)) \\ G. C. Greubel, May 02 2019

((sqrt((1-2*x)/(1-6*x)) -1)/(2*x)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 02 2019

From Emeric Deutsch, Jan 09 2003; corrected by Roland Bacher: (Start)
a(n) = Sum_{i=0..n} (-1)^i*6^(n-i)*binomial(n, i)*binomial(2*i, i)/(i+1);
g.f. A(x) satisfies: x(1-6x)A^2 + (1-6x)A - 1 = 0. (End)
From Henry Bottomley, Aug 23 2001: (Start)
a(n) = 6*a(n-1) - A005572(n-1).
a(n) = Sum_{j=0..n} 4^(n-j)*binomial(n, floor(n/2))*binomial(n, j). (End)
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(2*k+1, k)*2^(n-k).
a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*Catalan(k)*6^(n-k).
D-finite with recurrence (n+1)*a(n) = (8*n+2)*a(n-1)-(12*n-12)*a(n-2). - Vladeta Jovovic, Jul 16 2004
a(n) = Sum_{k=0..n} A052179(n,k). - Philippe Deléham, Jan 28 2007
Conjecture: a(n)= 6^n * hypergeom([1/2,-n],[2], 2/3). - Benjamin Phillabaum, Feb 20 2011
From Paul Barry, Apr 21 2009: (Start)
G.f.: (sqrt((1-2*x)/(1-6*x)) - 1)/(2*x).
G.f.: 1/(1-5*x-x^2/(1-4*x-x^2/(1-4*x-x^2/(1-4*x-x^2/(1-... (continued fraction). (End)
G.f.: 1/(1 - 4*x - x*(1 - 2*x)/(1 - 2*x - x*(1 - 2*x)/(1 - 2*x - x*(1 - 2*x)/(1 - 2*x - x*(1 - 2*x)/(1...(continued fraction). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Jul 02 2010
a(n) ~ 6^(n+1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 05 2012
G.f.: G(0)/(2*x) - 1/(2*x), where G(k)= 1 + 4*x*(4*k+1)/( (4*k+2)*(1-2*x) - 2*x*(1-2*x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1-2*x)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 24 2013
a(n) = 2^n*hypergeom([-n, 3/2], [2], -2). - Peter Luschny, Apr 26 2016
E.g.f.: exp(4*x)*(BesselI(0,2*x) + BesselI(1,2*x)). - Ilya Gutkovskiy, Sep 20 2017

More terms from Henry Bottomley, Aug 23 2001

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