A060747 -id:A060747 - cp's OEIS frontend

A289020 Number of Dyck paths having exactly one peak in each of the levels 1,...,n and no other peaks.

1, 1, 2, 10, 92, 1348, 28808, 845800, 32664944, 1605553552, 97868465696, 7245440815264, 640359291096512, 66598657958731840, 8051483595083729024, 1119653568781387712128, 177465810459239319017216, 31804047327185301634148608, 6398867435594240638421950976

Offset: 0

Alois P. Heinz, Jun 22 2017

nonn

The semilengths of Dyck paths counted by a(n) are elements of the integer interval [2*n-1, n*(n+1)/2] = [A060747(n), A000217(n)] for n>0.

			. a(2) = 2:      /\    /\
.             /\/  \  /  \/\  .

Alois P. Heinz, Table of n, a(n) for n = 0..100
Wikipedia, Counting lattice paths

Column k=1 of A288972.

Cf. A000217, A060747, A281874, A287846.

b:= proc(n, j, v) option remember; `if`(n=j,
      `if`(v=1, 1, 0), `if`(v<2, 0, add(b(n-j, i, v-1)*
       i*binomial(j-1, i-2), i=1..min(j+1, n-j))))
    end:
a:= n-> `if`(n=0, 1, add(b(w, 1, n), w=2*n-1..n*(n+1)/2)):
seq(a(n), n=0..18);

b[n_, j_, v_]:=b[n, j, v]=If[n==j, If[v==1, 1, 0], If[v<2, 0, Sum[b[n - j, i, v - 1]*i*Binomial[j - 1, i - 2], {i, Min[j + 1, n - j]}]]]; a[n_]:=If[n==0, 1, Sum[b[w, 1, n], {w, 2*n - 1, n*(n + 1)/2}]]; Table[a[n], {n, 0, 18}] (* Indranil Ghosh, Jul 06 2017, after Maple code *)

A340804 Triangle read by rows: T(n, k) = 1 + k(n - 1) + (2k - n - 1)*(k mod 2) with 0 < k <= n.

Original entry on oeis.org

1, 1, 3, 1, 5, 9, 1, 7, 11, 13, 1, 9, 13, 17, 25, 1, 11, 15, 21, 29, 31, 1, 13, 17, 25, 33, 37, 49, 1, 15, 19, 29, 37, 43, 55, 57, 1, 17, 21, 33, 41, 49, 61, 65, 81, 1, 19, 23, 37, 45, 55, 67, 73, 89, 91, 1, 21, 25, 41, 49, 61, 73, 81, 97, 101, 121, 1, 23, 27, 45, 53, 67, 79, 89, 105, 111, 131, 133

Offset: 1

Stefano Spezia, Jan 22 2021

T(n, k) is the k-th diagonal element of an n X n square matrix M(n) formed by writing the numbers 1, ..., n^2 successively forward and backward along the rows in zig-zag pattern.

It includes exclusively all the odd numbers (A005408). Except the term 1, all the other odd numbers appear a finite number of times.

			1
1,  3
1,  5,  9,
1,  7, 11, 13
1,  9, 13, 17, 25
1, 11, 15, 21, 29, 31
1, 13, 17, 25, 33, 37, 49
...

Stefano Spezia, Table of n, a(n) for n = 1..11325 (first 150 rows of the triangle, flattened).

Cf. A005408, A317614 (row sums).

Cf. A000012 (1st column), A006010 (sum of the first n rows), A060747 (2nd column), A074147 (antidiagonals of M matrices), A241016 (row sums of M matrices), A317617 (column sums of M matrices), A322277 (permanent of M matrices), A323723 (subdiagonal sum of M matrices), A323724 (superdiagonal sum of M matrices).

Table[1+k(n-1)+(2k-n-1)Mod[k,2],{n,12},{k,n}]//Flatten

T(n, k) = 1 + k*(n - 1) + (2*k - n - 1)*(k % 2); \\ Michel Marcus, Jan 25 2021

O.g.f.: (1 + y - 3*y^2 + y^3 + x*(-1 - y + 5*y^2 + y^3))/((-1 + x)^2*(-1 + y)^2*(1+y)^2).

E.g.f.: exp(x - y)*(1 + x + 2*y + exp(2*y)*(1 + x*(-1 + 2*y)))/2.

A033569 a(n) = (2n - 1)(3*n + 1).

Original entry on oeis.org

-1, 4, 21, 50, 91, 144, 209, 286, 375, 476, 589, 714, 851, 1000, 1161, 1334, 1519, 1716, 1925, 2146, 2379, 2624, 2881, 3150, 3431, 3724, 4029, 4346, 4675, 5016, 5369, 5734, 6111, 6500, 6901, 7314, 7739, 8176, 8625, 9086, 9559, 10044, 10541, 11050, 11571

Offset: 0

N. J. A. Sloane

For n>0, a(n) is the sum of the numbers from 2n+2 to 4n. The last digit of a(n) corresponds to the last digit of the squares mod 10 (A008959). Binomial Transform of a(n) starts: -1, 3, 28, 124, 432, 1328, 3776, 10176, 26368, ... . - Wesley Ivan Hurt, Dec 06 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A008959, A060747, A016777, A259758 (subsequence).

List([0..50], n-> (2*n-1)*(3*n+1)) # G. C. Greubel, Apr 02 2019

a033569 n = (2 * n - 1) * (3 * n + 1)
a033569_list = map a033569 [0..]
-- Reinhard Zumkeller, Jul 05 2015

[(2*n-1)*(3*n+1): n in [0..50]]; // Vincenzo Librandi, Jul 07 2012

A033569:=n->(2*n-1)*(3*n+1): seq(A033569(n), n=0..50); # Wesley Ivan Hurt, Dec 06 2014

CoefficientList[Series[(-1+7*x+6*x^2)/(1-x)^3,{x,0,50}],x] (* Vincenzo Librandi, Jul 07 2012 *)

a(n)=(2*n-1)*(3*n+1) \\ Charles R Greathouse IV, Jun 17 2017

[(2*n-1)*(3*n+1) for n in (0..50)] # G. C. Greubel, Apr 02 2019

G.f.: (-1+7*x+6*x^2)/(1-x)^3. - Vincenzo Librandi, Jul 07 2012
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - Vincenzo Librandi, Jul 07 2012
E.g.f.: (-1+5*x+6*x^2)*e^x. - Robert Israel, Dec 07 2014
a(n) = A060747(n) * A016777(n). - Reinhard Zumkeller, Jul 05 2015
Sum_{n>=0} 1/a(n) = 2/5*(log(2)-1) -sqrt(3)*Pi/30 -3*log(3)/10 = -0.6337047... - R. J. Mathar, Apr 22 2024

A084643 a(n) = 3^(n-1)(2n-3) + 2^(n+1).

Original entry on oeis.org

1, 3, 11, 43, 167, 631, 2315, 8275, 28943, 99439, 336659, 1126027, 3728279, 12239527, 39890843, 129205699, 416249375, 1334710495, 4262149667, 13560765691, 43005771431, 135988785943, 428882869931, 1349402340403

Offset: 0

Paul Barry, Jun 09 2003

Binomial transform of A048495. Second binomial transform of 1, 1, 3, 5, 7, ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-21,18).

Cf. A048495, A060747.

[3^(n-1)*(2*n-3)+2^(n+1) : n in [0..30]]; // Vincenzo Librandi, Sep 25 2011

LinearRecurrence[{8,-21,18},{1,3,11},30] (* Harvey P. Dale, Dec 12 2015 *)

Vec((1-5*x+8*x^2)/(1-2*x)/(1-3*x)^2+O(x^99)) \\ Charles R Greathouse IV, Mar 22 2012

[2^(n+1) +3^(n-1)*(2*n-3) for n in range(41)] # G. C. Greubel, Mar 22 2023

G.f.: (1 - 5*x + 8*x^2)/((1-2*x)*(1-3*x)^2). - Colin Barker, Mar 22 2012

E.g.f.: 2*exp(2*x) + (2*x-1)*exp(3*x). - G. C. Greubel, Mar 22 2023

A132774 A natural number operator.

Original entry on oeis.org

1, 2, 3, 0, 4, 5, 0, 0, 6, 7, 0, 0, 0, 8, 9, 0, 0, 0, 0, 10, 11, 0, 0, 0, 0, 0, 12, 13, 0, 0, 0, 0, 0, 0, 14, 15, 0, 0, 0, 0, 0, 0, 0, 16, 17, 0, 0, 0, 0, 0, 0, 0, 0, 18, 19, 0, 0, 0, 0, 0, 0, 0, 0, 0, 20, 21, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 22, 23

Offset: 1

Gary W. Adamson, Aug 28 2007

Row sums = A016813: (1, 5, 9, 13, ...).

A132774 * [1, 2, 3, ...] = A033951.

			First few rows of the triangle are:
  1;
  2,  3;
  0,  4,  5;
  0,  0,  6,  7;
  0,  0,  0,  8,  9;
  0,  0,  0,  0, 10, 11;
  ...

Stefano Spezia, First 150 rows of the triangle, flattened

Cf. A016813 (row sums), A033951, A060747 (main diagonal).

T[n_,k_]:=If[n==k,2n-1,If[n-k==1,2(n-1),0]]; Flatten[Table[T[n,k],{n,12},{k,n}]] (* Stefano Spezia, Dec 21 2021 *)
Join[{1},Flatten[{#,PadRight[{},#[[1]]/2,0]}&/@Partition[Range[2,30],2]]] (* Harvey P. Dale, Mar 24 2024 *)
Join[{1},Flatten[Table[Join[Range[2n,2n+1],PadRight[{},n,0]],{n,20}]]] (* Harvey P. Dale, Mar 25 2024 *)

As an infinite lower triangular matrix, (1, 3, 5, ...) in the main diagonal and (2, 4, 6, ...) in the subdiagonal; with the rest zeros.
From Stefano Spezia, Dec 21 2021: (Start)
T(n, k) = 2*n - 1 if n = k, T(n, k) = 2*(n - 1) if n - k = 1, otherwise T(n, k) = 0.
G.f.: x*y*(1 + x*(2 + y))/(1 - x*y)^2. (End)

More terms from Stefano Spezia, Dec 21 2021

A242412 a(n) = (2*n-1)^2 + 14.

Original entry on oeis.org

15, 23, 39, 63, 95, 135, 183, 239, 303, 375, 455, 543, 639, 743, 855, 975, 1103, 1239, 1383, 1535, 1695, 1863, 2039, 2223, 2415, 2615, 2823, 3039, 3263, 3495, 3735, 3983, 4239, 4503, 4775, 5055, 5343, 5639, 5943, 6255, 6575, 6903, 7239, 7583, 7935, 8295, 8663, 9039, 9423, 9815

Offset: 1

Aaron David Fairbanks, May 13 2014

The previous definition was "a(n) = normalized inverse radius of the inscribed circle that is tangent to the left circle of the symmetric arbelos and the n-th and (n-1)-st circles in the Pappus chain".
See links section for image of these circles, via Wolfram MathWorld (there an asymmetric arbelos is shown).
The Rothman-Fukagawa article has another picture of the circles, based on a Japanese 1788 sangaku problem. - N. J. A. Sloane, Jan 02 2020

			For n = 1, the radius of the outermost circle divided by the radius of a circle drawn tangent to all three of the initial inner circle, the opposite inner circle (the 0th circle in the chain), and the 1st circle in the chain is 15.
For n = 2, the radius of the outermost circle divided by the radius of a circle drawn tangent to all three of the initial inner circle, the 1st circle in the chain, and the 2nd circle in the chain is 23.

N. J. A. Sloane, Table of n, a(n) for n = 1..1000
Brady Haran and Simon Pampena, Epic Circles, Numberphile video (2014).
Tony Rothman and Hidetoshi Fukagawa, Japanese temple geometry, Scientific American, Vol. 278, No. 5, May 1998, pp. 85-91.
Eric Weisstein's World of Mathematics, Image of inscribed circles (in red).
Eric Weisstein's World of Mathematics, Pappus Chain.
Wikipedia, Pappus chain.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000012, A060747, A059100, A114949, A222465, A259555.

[4*n^2 - 4*n + 15: n in [1..50]]; // Wesley Ivan Hurt, May 13 2014

A242412:=n->4*n^2 - 4*n + 15; seq(A242412(n), n=1..50); # Wesley Ivan Hurt, May 13 2014

Table[4 n^2 - 4 n + 15, {n, 50}] (* Wesley Ivan Hurt, May 13 2014 *)
LinearRecurrence[{3,-3,1},{15,23,39},50] (* Harvey P. Dale, Feb 22 2023 *)

a(n) = 4*n^2 - 4*n + 15 \\ Charles R Greathouse IV, May 14 2014

a(n) = 4*n^2 - 4*n + 15.
From Colin Barker, May 14 2014: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: -x*(15*x^2 - 22*x + 15)/(x-1)^3. (End)
From Descartes three circle theorem:
a(n) = 2 + c(n) + c(n-1) + 2*sqrt(2*(c(n) + c(n-1)) + c(n)*c(n-1)), with c(n) = A059100(n) = n^2 + 2, n >= 1, which produces 4*n^2 - 4*n + 15. - Wolfdieter Lang, Jul 01 2015
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: exp(x)*(4*x^2 + 15) - 15.
a(n) = A060747(n)^2 + 14. (End)

More terms from Wesley Ivan Hurt, May 13 2014
More terms and links from Robert G. Wilson v, May 13 2014
Edited: Name reformulated (with consent of the author). - Wolfdieter Lang, Jul 01 2015
Edited by N. J. A. Sloane, Jan 02 2020, simplifying the definition and adding a reference to the fact that this sequence arose in a sangaku problem from 1788 in a temple in Tokyo Prefecture.

A033589 a(n) = (2n-1)(3n-1)(4*n-1).

Original entry on oeis.org

-1, 6, 105, 440, 1155, 2394, 4301, 7020, 10695, 15470, 21489, 28896, 37835, 48450, 60885, 75284, 91791, 110550, 131705, 155400, 181779, 210986, 243165, 278460, 317015, 358974, 404481, 453680, 506715

Offset: 0

N. J. A. Sloane

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A060747, A033568, A033590.

[(2*n-1)*(3*n-1)*(4*n-1): n in [0..30]]; // G. C. Greubel, Mar 05 2020

seq( mul(j*n-1, j=2..4), n=0..30); # G. C. Greubel, Mar 05 2020

Table[Times@@(n*Range[2,4]-1),{n,0,30}] (* or *) LinearRecurrence[{4,-6,4,-1},{-1,6,105,440},30] (* Harvey P. Dale, Sep 22 2014 *)

vector(31, n, my(m=n-1); prod(j=2,4, j*m-1) ) \\ G. C. Greubel, Mar 05 2020

[product(j*n-1 for j in (2..4)) for n in (0..30)] # G. C. Greubel, Mar 05 2020

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Sep 22 2014
G.f.: (-1 +10*x +75*x^2 +60*x^3)/(1-x)^4. - R. J. Mathar, Feb 06 2017
From G. C. Greubel, Mar 05 2020: (Start)
a(n) = n^3 * Pochhammer(2 - 1/n, 3) = Product_{j=2..4} (j*n-1).
E.g.f.: (-1 + 7*x + 46*x^2 + 24*x^3)*exp(x). (End)
Sum_{n>=1} 1/a(n) = (sqrt(3)/2-1)*Pi + 8*log(2) - 9*log(3)/2. - Amiram Eldar, Feb 22 2022

A033590 a(n) = (2n-1)(3n-1)(4n-1)(5*n-1).

Original entry on oeis.org

1, 24, 945, 6160, 21945, 57456, 124729, 238680, 417105, 680680, 1052961, 1560384, 2232265, 3100800, 4201065, 5571016, 7251489, 9286200, 11721745, 14607600, 17996121, 21942544, 26504985, 31744440

Offset: 0

N. J. A. Sloane

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A060747, A033568, A033589.

[(2*n-1)*(3*n-1)*(4*n-1)*(5*n-1): n in [0..40]]; // G. C. Greubel, Mar 05 2020

seq( mul(j*n-1, j=2..5), n=0..40); # G. C. Greubel, Mar 05 2020

Table[(2*n-1)*(3*n-1)*(4*n-1)*(5*n-1), {n,0,40}] (* G. C. Greubel, Mar 05 2020 *)

vector(41, n, my(m=n-1); prod(j=2,5, j*m-1) ) \\ G. C. Greubel, Mar 05 2020

[product(j*n-1 for j in (2..5)) for n in (0..40)] # G. C. Greubel, Mar 05 2020

G.f.: (1 + 19*x + 835*x^2 + 1665*x^3 + 360*x^4)/(1-x)^5. - R. J. Mathar, Feb 06 2017
From G. C. Greubel, Mar 05 2020: (Start)
a(n) = n^4 * Pochhammer(2 - 1/n, 4) = Product_{j=2..5} (j*n-1).
E.g.f.: (1 + 23*x + 449*x^2 + 566*x^3 + 120*x^4)*exp(x). (End)

A214955 Number of solid standard Young tableaux of shape [[n,n-1],[1]].

Original entry on oeis.org

1, 6, 25, 98, 378, 1452, 5577, 21450, 82654, 319124, 1234506, 4784276, 18572500, 72209880, 281150505, 1096087770, 4278278070, 16717354500, 65388738030, 256000696380, 1003116947820, 3933750236520, 15437682614250, 60625494924228, 238235373671148, 936735006679752

Offset: 1

Alois P. Heinz, Jul 30 2012

nonn

a(n) is odd if and only if n = 2^i-1 for i in {1, 2, 3, ...} = A000027.

Form an array with m(1,n) = n*(n+1)/2, m(n,1) = n*(n-1)+1, and m(i,j) = m(i,j-1) + m(i-1,j); A000217 in the top row, A002061 in the first column, A086514 in the second column. Then on the diagonal m(n,n) = a(n). - J. M. Bergot, May 02 2013

Alois P. Heinz, Table of n, a(n) for n = 1..1000
S. B. Ekhad and D. Zeilberger, Computational and Theoretical Challenges on Counting Solid Standard Young Tableaux, arXiv:1202.6229 [math.CO], 2012.
Wikipedia, Young tableau.

Column k=1 of A214775.

Cf. A000027, A000108, A006752, A060747, A215002.

a:= proc(n) option remember;
      `if`(n<2, n, 2*(2*n-1)^2*a(n-1)/((n+1)*(2*n-3)))
    end:
seq(a(n), n=1..30);

a[n_]:= a[n] = If[n<2, n, 2*(2*n-1)^2*a[n-1]/((n+1)*(2*n-3))]; Array[a, 30] (* Jean-François Alcover, Aug 14 2017, translated from Maple *)

a(n) = (2*n-1) * binomial(2*n,n)/(n+1); \\ Michel Marcus, Mar 06 2022

a(n) = 2*(2*n-1)^2/((n+1)*(2*n-3)) * a(n-1) for n>1; a(1) = 1.
a(n) = (2*n-1) * C(2*n,n)/(n+1) = A060747(n) * A000108(n).
a(n) = [x^n] x*(1 + 2*x)/(1 - x)^(n+2). - Ilya Gutkovskiy, Oct 12 2017
Sum_{n>=1} 1/a(n) = 1/6 + G + 13*Pi/(36*sqrt(3)) - Pi*log(2+sqrt(3))/8, where G is Catalan's constant (A006752). - Amiram Eldar, Mar 06 2022
From Stefano Spezia, Mar 29 2023: (Start)
O.g.f.: 1 + (3 - 3*sqrt(1 - 4*x) - 8*x)/(2*x*sqrt(1 - 4*x)).
E.g.f.: 1 + exp(2*x)*(3*I_1(2*x) - I_0(2*x)), where I_n(x) is the modified Bessel function of the first kind.
a(n) ~ 2^(1+2*n)/sqrt(n*Pi). (End)

A343558 Irregular triangle read by rows: the n-th row gives the row indices of the consecutive elements of the spiral of the n X n matrix defined in A126224.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 1, 1, 2, 3, 3, 3, 2, 2, 1, 1, 1, 1, 2, 3, 4, 4, 4, 4, 3, 2, 2, 2, 3, 3, 1, 1, 1, 1, 1, 2, 3, 4, 5, 5, 5, 5, 5, 4, 3, 2, 2, 2, 2, 3, 4, 4, 4, 3, 3, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 6, 6, 6, 6, 6, 5, 4, 3, 2, 2, 2, 2, 2, 3, 4, 5, 5, 5, 5, 4, 3, 3, 3, 4, 4

Offset: 1

Stefano Spezia, Apr 19 2021

			The triangle begins
1
1   1   2   2
1   1   1   2   3   3   3   2   2
1   1   1   1   2   3   4   4   4   4   3   2   2   2   3   3
...

Stefano Spezia, First 30 rows of the triangle, flattened

Cf. A000290 (row length), A002265, A002411 (row sums), A010873, A060747, A126224, A343559 (column indices).

a:={};nmax:=6;For[n=1,n<=nmax,n++,For[s=1,s<=2n-1,s++,If[OddQ[s] &&Mod[s,4]==1,k=Ceiling[s/4];For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k]],If[EvenQ[s]&&Mod[s,4]==2,For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k+i]];k+=Ceiling[n-s/2],If[EvenQ[s]&&Mod[s,4]==0,For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k-i]];k=k-i+1,For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k]]]]]]];a

cp's OEIS Frontend

A289020 Number of Dyck paths having exactly one peak in each of the levels 1,...,n and no other peaks.

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A340804 Triangle read by rows: T(n, k) = 1 + k*(n - 1) + (2*k - n - 1)*(k mod 2) with 0 < k <= n.

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A033569 a(n) = (2*n - 1)*(3*n + 1).

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A084643 a(n) = 3^(n-1)*(2*n-3) + 2^(n+1).

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A132774 A natural number operator.

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A242412 a(n) = (2*n-1)^2 + 14.

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A340804 Triangle read by rows: T(n, k) = 1 + k(n - 1) + (2k - n - 1)*(k mod 2) with 0 < k <= n.

A033569 a(n) = (2n - 1)(3*n + 1).

A084643 a(n) = 3^(n-1)(2n-3) + 2^(n+1).

A033589 a(n) = (2n-1)(3n-1)(4*n-1).

A033590 a(n) = (2n-1)(3n-1)(4n-1)(5*n-1).