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.

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A002415 4-dimensional pyramidal numbers: a(n) = n^2*(n^2-1)/12.

Original entry on oeis.org

0, 0, 1, 6, 20, 50, 105, 196, 336, 540, 825, 1210, 1716, 2366, 3185, 4200, 5440, 6936, 8721, 10830, 13300, 16170, 19481, 23276, 27600, 32500, 38025, 44226, 51156, 58870, 67425, 76880, 87296, 98736, 111265, 124950, 139860, 156066, 173641, 192660, 213200, 235340
Offset: 0

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Author

Keywords

Comments

Also number of ways to legally insert two pairs of parentheses into a string of m := n-1 letters. (There are initially 2C(m+4,4) (A034827) ways to insert the parentheses, but we must subtract 2(m+1) for illegal clumps of 4 parentheses, 2m(m+1) for clumps of 3 parentheses, C(m+1,2) for 2 clumps of 2 parentheses and (m-1)C(m+1,2) for 1 clump of 2 parentheses, giving m(m+1)^2(m+2)/12 = n^2*(n^2-1)/12.) See also A000217.
E.g., for n=2 there are 6 ways: ((a))b, ((a)b), ((ab)), (a)(b), (a(b)), a((b)).
Let M_n denote the n X n matrix M_n(i,j)=(i+j); then the characteristic polynomial of M_n is x^(n-2) * (x^2-A002378(n)*x - a(n)). - Benoit Cloitre, Nov 09 2002
Let M_n denote the n X n matrix M_n(i,j)=(i-j); then the characteristic polynomial of M_n is x^n + a(n)x^(n-2). - Michael Somos, Nov 14 2002 [See A114327 for the infinite matrix M in triangular form. - Wolfdieter Lang, Feb 05 2018]
Number of permutations of [n] which avoid the pattern 132 and have exactly 2 descents. - Mike Zabrocki, Aug 26 2004
Number of tilings of a <2,n,2> hexagon.
a(n) is the number of squares of side length at least 1 having vertices at the points of an n X n unit grid of points (the vertices of an n-1 X n-1 chessboard). [For a proof, see Comments in A051602. - N. J. A. Sloane, Sep 29 2021] For example, on the 3 X 3 grid (the vertices of a 2 X 2 chessboard) there are four 1 X 1 squares, one (skew) sqrt(2) X sqrt(2) square, and one 3 X 3 square, so a(3)=6. On the 4 X 4 grid (the vertices of a 3 X 3 chessboard) there are 9 1 X 1 squares, 4 2 X 2 squares, 1 3 X 3 square, 4 sqrt(2) X sqrt(2) squares, and 2 sqrt(5) X sqrt(5) squares, so a(4) = 20. See also A024206, A108279. [Comment revised by N. J. A. Sloane, Feb 11 2015]
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 12 2005
Number of distinct components of the Riemann curvature tensor. - Gene Ward Smith, Apr 24 2006
a(n) is the number of 4 X 4 matrices (symmetrical about each diagonal) M = [a,b,c,d;b,e,f,c;c,f,e,b;d,c,b,a] with a+b+c+d=b+e+f+c=n+2; (a,b,c,d,e,f natural numbers). - Philippe Deléham, Apr 11 2007
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-3) is the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
a(n) is the number of Dyck (n+1)-paths with exactly n-1 peaks. - David Callan, Sep 20 2007
Starting (1,6,20,50,...) = third partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} C(n+3,i+3)*b(i), where b(i)=[1,2,0,0,0,...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
4-dimensional square numbers. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Equals row sums of triangle A177877; a(n), n > 1 = (n-1) terms in (1,2,3,...) dot (...,3,2,1) with additive carryovers. Example: a(4) = 20 = (1,2,3) dot (3,2,1) with carryovers = (1*3) + (2*2 + 3) + (3*1 + 7) = (3 + 7 + 10).
Convolution of the triangular numbers A000217 with the odd numbers A004273.
a(n+2) is the number of 4-tuples (w,x,y,z) with all terms in {0,...,n} and w-x=max{w,x,y,z}-min{w,x,y,z}. - Clark Kimberling, May 28 2012
The second level of finite differences is a(n+2) - 2*a(n+1) + a(n) = (n+1)^2, the squares. - J. M. Bergot, May 29 2012
Because the differences of this sequence give A000330, this is also the number of squares in an n+1 X n+1 grid whose sides are not parallel to the axes.
a(n+2) gives the number of 2*2 arrays that can be populated with 0..n such that rows and columns are nondecreasing. - Jon Perry, Mar 30 2013
For n consecutive numbers 1,2,3,...,n, the sum of all ways of adding the k-tuples of consecutive numbers for n=a(n+1). As an example, let n=4: (1)+(2)+(3)+(4)=10; (1+2)+(2+3)+(3+4)=15; (1+2+3)+(2+3+4)=15; (1+2+3+4)=10 and the sum of these is 50=a(4+1)=a(5). - J. M. Bergot, Apr 19 2013
If P(n,k) = n*(n+1)*(k*n-k+3)/6 is the n-th (k+2)-gonal pyramidal number, then a(n) = P(n,k)*P(n-1,k-1) - P(n-1,k)*P(n,k-1). - Bruno Berselli, Feb 18 2014
For n > 1, a(n) = 1/6 of the area of the trapezoid created by the points (n,n+1), (n+1,n), (1,n^2+n), (n^2+n,1). - J. M. Bergot, May 14 2014
For n > 3, a(n) is twice the area of a triangle with vertices at points (C(n,4),C(n+1,4)), (C(n+1,4),C(n+2,4)), and (C(n+2,4),C(n+3,4)). - J. M. Bergot, Jun 03 2014
a(n) is the dimension of the space of metric curvature tensors (those having the symmetries of the Riemann curvature tensor of a metric) on an n-dimensional real vector space. - Daniel J. F. Fox, Dec 15 2018
Coefficients in the terminating series identity 1 - 6*n/(n + 5) + 20*n*(n - 1)/((n + 5)*(n + 6)) - 50*n*(n - 1)*(n - 2)/((n + 5)*(n + 6)*(n + 7)) + ... = 0 for n = 1,2,3,.... Cf. A000330 and A005585. - Peter Bala, Feb 18 2019

Examples

			a(7) = 6*21 - (6*0 + 4*1 + 2*3 + 0*6 - 2*10 - 4*15) = 196. - _Bruno Berselli_, Jun 22 2013
G.f. = x^2 + 6*x^3 + 20*x^4 + 50*x^5 + 105*x^6 + 196*x^7 + 336*x^8 + ...
		

References

  • O. D. Anderson, Find the next sequence, J. Rec. Math., 8 (No. 4, 1975-1976), 241.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 195.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p.165).
  • R. Euler and J. Sadek, "The Number of Squares on a Geoboard", Journal of Recreational Mathematics, 251-5 30(4) 1999-2000 Baywood Pub. NY
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 238.
  • 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

a(n) = ((-1)^n)*A053120(2*n, 4)/8 (one-eighth of fifth unsigned column of Chebyshev T-triangle, zeros omitted). Cf. A001296.
Second row of array A103905.
Third column of Narayana numbers A001263.
Partial sums of A000330.
The expression binomial(m+n-1,n)^2-binomial(m+n,n+1)*binomial(m+n-2,n-1) for the values m = 2 through 14 produces sequences A000012, A000217, A002415, A006542, A006857, A108679, A134288, A134289, A134290, A134291, A140925, A140935, A169937.
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers (A000027) with the k-gonal numbers.

Programs

  • GAP
    List([0..45],n->Binomial(n^2,2)/6); # Muniru A Asiru, Dec 15 2018
  • Magma
    [n^2*(n^2-1)/12: n in [0..50]]; // Wesley Ivan Hurt, May 14 2014
    
  • Maple
    A002415 := proc(n) binomial(n^2,2)/6 ; end proc: # Zerinvary Lajos, Jan 07 2008
  • Mathematica
    Table[(n^4 - n^2)/12, {n, 0, 40}] (* Zerinvary Lajos, Mar 21 2007 *)
    LinearRecurrence[{5,-10,10,-5,1},{0,0,1,6,20},40] (* Harvey P. Dale, Nov 29 2011 *)
  • PARI
    a(n) = n^2 * (n^2 - 1) / 12;
    
  • PARI
    x='x+O('x^200); concat([0, 0], Vec(x^2*(1+x)/(1-x)^5)) \\ Altug Alkan, Mar 23 2016
    

Formula

G.f.: x^2*(1+x)/(1-x)^5. - Simon Plouffe in his 1992 dissertation
a(n) = Sum_{i=0..n} (n-i)*i^2 = a(n-1) + A000330(n-1) = A000217(n)*A000292(n-2)/n = A000217(n)*A000217(n-1)/3 = A006011(n-1)/3, convolution of the natural numbers with the squares. - Henry Bottomley, Oct 19 2000
a(n)+1 = A079034(n). - Mario Catalani (mario.catalani(AT)unito.it), Feb 12 2003
a(n) = 2*C(n+2, 4) - C(n+1, 3). - Paul Barry, Mar 04 2003
a(n) = C(n+2, 4) + C(n+1, 4). - Paul Barry, Mar 13 2003
a(n) = Sum_{k=1..n} A000330(n-1). - Benoit Cloitre, Jun 15 2003
a(n) = n*C(n+1,3)/2 = C(n+1,3)*C(n+1,2)/(n+1). - Mitch Harris, Jul 06 2006
a(n) = A006011(n)/3 = A008911(n)/2 = A047928(n-1)/12 = A083374(n)/6. - Zerinvary Lajos, May 09 2007
a(n) = (1/2)*Sum_{1 <= x_1, x_2 <= n} (det V(x_1,x_2))^2 = (1/2)*Sum_{1 <= i,j <= n} (i-j)^2, where V(x_1,x_2) is the Vandermonde matrix of order 2. - Peter Bala, Sep 21 2007
a(n) = C(n+1,3) + 2*C(n+1,4). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
a(n) = (1/48)*sinh(2*arccosh(n))^2. - Artur Jasinski, Feb 10 2010
a(n) = n*A000292(n-1)/2. - Tom Copeland, Sep 13 2011
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), n > 4. - Harvey P. Dale, Nov 29 2011
a(n) = (n-1)*A000217(n-1) - Sum_{i=0..n-2} (n-1-2*i)*A000217(i) for n > 1. - Bruno Berselli, Jun 22 2013
a(n) = C(n,2)*C(n+1,3) - C(n,3)*C(n+1,2). - J. M. Bergot, Sep 17 2013
a(n) = Sum_{k=1..n} ( (2k-n)* k(k+1)/2 ). - Wesley Ivan Hurt, Sep 26 2013
a(n) = floor(n^2/3) + 3*Sum_{k=1..n} k^2*floor((n-k+1)/3). - Mircea Merca, Feb 06 2014
Euler transform of length 2 sequence [6, -1]. - Michael Somos, May 28 2014
G.f. x^2*2F1(3,4;2;x). - R. J. Mathar, Aug 09 2015
Sum_{n>=2} 1/a(n) = 21 - 2*Pi^2 = 1.260791197821282762331... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A080852(2,n-2). - R. J. Mathar, Jul 28 2016
a(n) = A046092(n) * A046092(n-1)/48 = A000217(n) * A000217(n-1)/3. - Bruce J. Nicholson, Jun 06 2017
E.g.f.: (1/12)*exp(x)*x^2*(6 + 6*x + x^2). - Stefano Spezia, Dec 07 2018
Sum_{n>=2} (-1)^n/a(n) = Pi^2 - 9 (See A002388). - Amiram Eldar, Jun 28 2020

Extensions

Typo in link fixed by Matthew Vandermast, Nov 22 2010
Redundant comment deleted and more detail on relationship with A000330 added by Joshua Zucker, Jan 01 2013

A005585 5-dimensional pyramidal numbers: a(n) = n*(n+1)*(n+2)*(n+3)*(2n+3)/5!.

Original entry on oeis.org

1, 7, 27, 77, 182, 378, 714, 1254, 2079, 3289, 5005, 7371, 10556, 14756, 20196, 27132, 35853, 46683, 59983, 76153, 95634, 118910, 146510, 179010, 217035, 261261, 312417, 371287, 438712, 515592, 602888, 701624, 812889, 937839, 1077699, 1233765, 1407406
Offset: 1

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Author

Keywords

Comments

Convolution of triangular numbers (A000217) and squares (A000290) (n>=1). - Graeme McRae, Jun 07 2006
p^k divides a(p^k-3), a(p^k-2), a(p^k-1) and a(p^k) for prime p > 5 and integer k > 0. p^k divides a((p^k-3)/2) for prime p > 5 and integer k > 0. - Alexander Adamchuk, May 08 2007
If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-5) is the number of 6-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007
5-dimensional square numbers, fourth partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+4, i+4)*b(i), where b(i)=[1,2,0,0,0,...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Antidiagonal sums of the convolution array A213550. - Clark Kimberling, Jun 17 2012
Binomial transform of (1, 6, 14, 16, 9, 2, 0, 0, 0, ...). - Gary W. Adamson, Jul 28 2015
2*a(n) is number of ways to place 4 queens on an (n+3) X (n+3) chessboard so that they diagonally attack each other exactly 6 times. The maximal possible attack number, p=binomial(k,2)=6 for k=4 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form a corresponding complete graph. - Antal Pinter, Dec 27 2015
While adjusting for offsets, add A000389 to find the next in series A000389, A005585, A051836, A034263, A027800, A051843, A051877, A051878, A051879, A051880, A056118, A271567. (See Bruno Berselli's comments in A271567.) - Bruce J. Nicholson, Jun 21 2018
Coefficients in the terminating series identity 1 - 7*n/(n + 6) + 27*n*(n - 1)/((n + 6)*(n + 7)) - 77*n*(n - 1)*(n - 2)/((n + 6)*(n + 7)*(n + 8)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A040977. - Peter Bala, Feb 18 2019

Examples

			G.f. = x + 7*x^2 + 27*x^3 + 77*x^4 + 182*x^5 + 378*x^6 + 714*x^7 + 1254*x^8 + ... - _Michael Somos_, Jun 24 2018
		

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. 797.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = ((-1)^(n+1))*A053120(2*n+3, 5)/16, (1/16 of sixth unsigned column of Chebyshev T-triangle, zeros omitted).
Partial sums of A002415.
Cf. A006542, A040977, A047819, A111125 (third column).
Cf. a(n) = ((-1)^(n+1))*A084960(n+1, 2)/16 (compare with the first line). - Wolfdieter Lang, Aug 04 2014

Programs

  • Magma
    I:=[1, 7, 27, 77, 182, 378]; [n le 6 select I[n] else 6*Self(n-1)-15*Self(n-2)+20*Self(n-3)-15*Self(n-4)+6*Self(n-5)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Jun 09 2013
    
  • Maple
    [seq(binomial(n+2,6)-binomial(n,6), n=4..45)]; # Zerinvary Lajos, Jul 21 2006
    A005585:=(1+z)/(z-1)**6; # Simon Plouffe in his 1992 dissertation
  • Mathematica
    With[{c=5!},Table[n(n+1)(n+2)(n+3)(2n+3)/c,{n,40}]] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{1,7,27,77,182,378},40] (* Harvey P. Dale, Oct 04 2011 *)
    CoefficientList[Series[(1 + x) / (1 - x)^6, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)
  • PARI
    a(n)=binomial(n+3,4)*(2*n+3)/5 \\ Charles R Greathouse IV, Jul 28 2015

Formula

G.f.: x*(1+x)/(1-x)^6.
a(n) = 2*C(n+4, 5) - C(n+3, 4). - Paul Barry, Mar 04 2003
a(n) = C(n+3, 5) + C(n+4, 5). - Paul Barry, Mar 17 2003
a(n) = C(n+2, 6) - C(n, 6), n >= 4. - Zerinvary Lajos, Jul 21 2006
a(n) = Sum_{k=1..n} T(k)*T(k+1)/3, where T(n) = n(n+1)/2 is a triangular number. - Alexander Adamchuk, May 08 2007
a(n-1) = (1/4)*Sum_{1 <= x_1, x_2 <= n} |x_1*x_2*det V(x_1,x_2)| = (1/4)*Sum_{1 <= i,j <= n} i*j*|i-j|, where V(x_1,x_2) is the Vandermonde matrix of order 2. First differences of A040977. - Peter Bala, Sep 21 2007
a(n) = C(n+4,4) + 2*C(n+4,5). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), a(1)=1, a(2)=7, a(3)=27, a(4)=77, a(5)=182, a(6)=378. - Harvey P. Dale, Oct 04 2011
a(n) = (1/6)*Sum_{i=1..n+1} (i*Sum_{k=1..i} (i-1)*k). - Wesley Ivan Hurt, Nov 19 2014
E.g.f.: x*(2*x^4 + 35*x^3 + 180*x^2 + 300*x + 120)*exp(x)/120. - Robert Israel, Nov 19 2014
a(n) = A000389(n+3) + A000389(n+4). - Bruce J. Nicholson, Jun 21 2018
a(n) = -a(-3-n) for all n in Z. - Michael Somos, Jun 24 2018
From Amiram Eldar, Jun 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 40*(16*log(2) - 11)/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 20*(8*Pi - 25)/3. (End)
a(n) = A004302(n+1) - A207361(n+1). - J. M. Bergot, May 20 2022
a(n) = Sum_{i=0..n+1} Sum_{j=i..n+1} i*j*(j-i)/2. - Darío Clavijo, Oct 11 2023
a(n) = (A000538(n+1) - A000330(n+1))/12. - Yasser Arath Chavez Reyes, Feb 21 2024

A006542 a(n) = binomial(n,3)*binomial(n-1,3)/4.

Original entry on oeis.org

1, 10, 50, 175, 490, 1176, 2520, 4950, 9075, 15730, 26026, 41405, 63700, 95200, 138720, 197676, 276165, 379050, 512050, 681835, 896126, 1163800, 1495000, 1901250, 2395575, 2992626, 3708810, 4562425, 5573800, 6765440, 8162176, 9791320, 11682825, 13869450
Offset: 4

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Author

Keywords

Comments

Number of permutations of n+4 that avoid the pattern 132 and have exactly 3 descents. - Mike Zabrocki, Aug 26 2004
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 20 2005
a(n) = number of Dyck n-paths with exactly 4 peaks. - David Callan, Jul 03 2006
Six-dimensional figurate numbers for a hyperpyramid with pentagonal base. This corresponds to the sum(sum(sum(sum(1+sum(5*n))))) interpretation, see the Munafo webpage. - Robert Munafo, Jun 18 2009

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 166, no. 1).
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 238.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The expression binomial(m+n-1,n)^2-binomial(m+n,n+1)*binomial(m+n-2,n-1) for the values m = 2 through 14 produces the sequences A000012, A000217, A002415, A006542, A006857, A108679, A134288, A134289, A134290, A134291, A140925, A140935, A169937.
Fourth column of the table of Narayana numbers A001263.
Apart from a scale factor, a column of A124428.

Programs

  • GAP
    List([4..40], n-> n*(n-1)^2*(n-2)^2*(n-3)/144); # G. C. Greubel, Feb 24 2019
  • Magma
    [ n*((n-1)*(n-2))^2*(n-3)/144 : n in [4..40] ]; // Wesley Ivan Hurt, Jun 17 2014
    
  • Maple
    A006542:=-(1+3*z+z**2)/(z-1)**7; # conjectured by Simon Plouffe in his 1992 dissertation
    A006542:=n->n*((n-1)*(n-2))^2*(n-3)/144; seq(A006542(n), n=4..40); # Wesley Ivan Hurt, Jun 17 2014
  • Mathematica
    Table[Binomial[n, 3]*Binomial[n-1, 3]/4, {n, 4, 40}]
  • PARI
    a(n)=n*((n-1)*(n-2))^2*(n-3)/144
    
  • Sage
    [n*(n-1)^2*(n-2)^2*(n-3)/144 for n in (4..40)] # G. C. Greubel, Feb 24 2019
    

Formula

a(n) = C(n, 3)*C(n-1, 3)/4 = n*(n-1)^2*(n-2)^2*(n-3)/144.
a(n) = A000292(n-3)*A000292(n-2)/4.
E.g.f.: x^4*(6 + 6*x + x^2)*exp(x)/144. - Vladeta Jovovic, Jan 29 2003
a(n) = Sum(Sum(Sum(Sum(1 + Sum(5*n))))) = Sum (A006414). - Xavier Acloque, Oct 08 2003
a(n) = C(n, 6) + 3*C(n+1, 6) + C(n+2, 6). - Mike Zabrocki, Aug 26 2004
G.f.: x^4*(1 + 3*x + x^2)/(1-x)^7. - Emeric Deutsch, Jun 20 2005
a(n) = C(n-2, n-4)*C(n-1, n-3)*C(n, n-2)/18. - Zerinvary Lajos, Jul 29 2005
a(n) = C(n,4)*C(n,3)/n. - Mitch Harris, Jul 06 2006
a(n+2) = (1/4)*Sum_{1 <= x_1, x_2 <= n} x_1*x_2*(det V(x_1,x_2))^2 = (1/4)*Sum_{1 <= i,j <= n} i*j*(i-j)^2, where V(x_1,x_2) is the Vandermonde matrix of order 2. - Peter Bala, Sep 21 2007
a(n) = C(n-1,3)^2 - C(n-1,2)*C(n-1,4). - Gary Detlefs, Dec 05 2011
a(n) = A000292(A000217(n-1)) - A000217(A000292(n-1)). - Ivan N. Ianakiev, Jun 17 2014
a(n) = Product_{i=1..3} A002378(n-4+i)/A002378(i). - Bruno Berselli, Nov 12 2014 (Rewritten, Sep 01 2016.)
Sum_{n>=4} 1/a(n) = 238 - 24*Pi^2. - Jaume Oliver Lafont, Jul 10 2017
Sum_{n>=4} (-1)^n/a(n) = 134 - 192*log(2). - Amiram Eldar, Oct 19 2020
a(n) = A000332(n) + 5*A000579(n+1). - Yasser Arath Chavez Reyes, Aug 18 2024

Extensions

Zabroki and Lajos formulas offset corrected by Gary Detlefs, Dec 05 2011

A103905 Square array T(n,k) read by antidiagonals: number of tilings of an hexagon.

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 20, 20, 4, 1, 70, 175, 50, 5, 1, 252, 1764, 980, 105, 6, 1, 924, 19404, 24696, 4116, 196, 7, 1, 3432, 226512, 731808, 232848, 14112, 336, 8, 1, 12870, 2760615, 24293412, 16818516, 1646568, 41580, 540, 9, 1, 48620, 34763300
Offset: 1

Views

Author

Ralf Stephan, Feb 22 2005

Keywords

Comments

As a square array, T(n,k) = number of all k-watermelons without a wall of length n. - Steven Finch, Mar 30 2008

Examples

			Array begins:
  1,   2,     3,      4,        5,         6, ...
  1,   6,    20,     50,      105,       196, ...
  1,  20,   175,    980,     4116,     14112, ...
  1,  70,  1764,  24696,   232848,   1646568, ...
  1, 252, 19404, 731808, 16818516, 267227532, ...
  ...
		

Crossrefs

Rows include A002415, A047819, A047835, A047831.
Columns include A000984 and A000891.
Main diagonal is A008793.

Programs

  • Mathematica
    t[n_, k_] := Product[j!*(j + 2*n)!/(j + n)!^2, {j, 0, k - 1}]; Join[{1}, Flatten[ Table[ t[n - k , k], {n, 1, 10}, {k, 1, n}]]] (* Jean-François Alcover, May 16 2012, from 2nd formula *)

Formula

T(n, k) = [V(2n+k-1)V(k-1)V(n-1)^2]/[V(2n-1)V(n+k-1)^2], with V(n) the superfactorial numbers (A000178).
T(n, k) = Prod[j=0..k-1, j!(j+2n)!/(j+n)!^2 ].
T(n, k) = Prod[h=1..n, Prod[i=1..k, Prod[j=1..n, (h+i+j-1)/(h+i+j-2) ]]].
T(n, k) = Prod[i=1..k, Prod[j=n+1..2n+1, i+j]/Prod[j=0..n, i+j]]; - Paul Barry, Jun 13 2006
Conjectural formula as a sum of squares of Vandermonde determinants: T(n,k) = 1/((1!*2! ... *(n-1)!)^2*n!)* sum {1 <= x_1, ..., x_n <= k} (det V(x_1, ..., x_n))^2, where V(x_1, ..., x_n) is the Vandermonde matrix of order n. Compare with A133112. - Peter Bala, Sep 18 2007
For k >= 1, T(n,k)=det(binomial(2*n,n+i-j))1<=i,j<=k [Krattenhaller, Theorem 4].
Let H(n) = product {k = 1..n-1} k!. Then for a,b,c nonnegative integers (H(a)*H(b)*H(c)*H(a+b+c))/(H(a+b)*H(b+c)*H(c+a)) is an integer [MacMahon, Section 4.29 with x -> 1]. Setting a = b = n and c = k gives the entries for this table. - Peter Bala, Dec 22 2011

A107891 a(n) = (n+1)*(n+2)^2*(n+3)^2*(n+4)*(3n^2 + 15n + 20)/2880.

Original entry on oeis.org

1, 19, 155, 805, 3136, 9996, 27468, 67320, 150645, 313027, 611611, 1134497, 2012920, 3436720, 5673648, 9093096, 14194881, 21643755, 32310355, 47319349, 68105576, 96479020, 134699500, 185562000, 252493605, 339663051, 452103939
Offset: 0

Views

Author

Emeric Deutsch, Jun 12 2005

Keywords

Comments

Kekulé numbers for certain benzenoids.
Partial sums of A114239. First differences of A047819. - Peter Bala, Sep 21 2007

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see pp. 167, 187 and p. 105 eq. (iii) for k=2 and m=5).

Crossrefs

Programs

  • Maple
    a:=n->(1/2880)*(n+1)*(n+2)^2*(n+3)^2*(n+4)*(3*n^2+15*n+20): seq(a(n),n=0..32);
  • Mathematica
    Table[((1+n) (2+n)^2 (3+n)^2 (4+n) (20+3 n (5+n)))/2880,{n,0,40}] (* or *) LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,19,155,805,3136,9996,27468,67320,150645},40] (* Harvey P. Dale, Dec 10 2021 *)

Formula

a(n-2) = (1/8) * Sum_{1 <= x_1, x_2 <= n} (x_1*x_2)^2*(det V(x_1,x_2))^2 = 1/8*sum {1 <= i,j <= n} (i*j*(i-j))^2, where V(x_1,x_2) is the Vandermonde matrix of order 2. - Peter Bala, Sep 21 2007
G.f.: (1+10*x+20*x^2+10*x^3+x^4)/(1-x)^9. - Colin Barker, Feb 08 2012
a(n) = (A000330(n+2)*A000538(n+2) - (A000537(n+2))^2)/4. - J. M. Bergot, Sep 17 2013
Sum_{n>=0} 1/a(n) = 17095/4 - 240*Pi^2 - 162*sqrt(15)*Pi*tanh(sqrt(5/3)*Pi/2). - Amiram Eldar, May 29 2022

A107915 a(n) = binomial(n+4,4)*binomial(n+5,4)*binomial(n+6,4)/75.

Original entry on oeis.org

1, 35, 490, 4116, 24696, 116424, 457380, 1557270, 4723719, 13026013, 33157124, 78835120, 176729280, 376375104, 766192176, 1498581756, 2828205765, 5168991135, 9177226366, 15870391460, 26794167400, 44253495000, 71627692500, 113794603650, 177694650315
Offset: 0

Views

Author

Emeric Deutsch, Jun 12 2005

Keywords

Comments

Kekulé numbers for certain benzenoids.
Partial sums of A107917. - Peter Bala, Sep 21 2007

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 229).

Crossrefs

Programs

  • Maple
    a:=n->(1/75)*binomial(n+4,4)*binomial(n+5,4)*binomial(n+6,4): seq(a(n),n=0..27);
    seq(binomial(n,n-2)*binomial(n+1,n-3)*binomial(n+2,n-4)/(5*3!), n=4..22); # Zerinvary Lajos, May 29 2007
  • Mathematica
    a[n_] := Binomial[n + 4, 4] * Binomial[n + 5, 4] * Binomial[n + 6, 4]/75; Array[a, 25, 0] (* Amiram Eldar, May 29 2022 *)

Formula

a(n) = C(n,n-2)*C(n+1,n-3)*C(n+2,n-4)/(5*3!), n>=4. - Zerinvary Lajos, May 29 2007
a(n-3) = (1/144) * Sum_{1 <= x_1, x_2, x_3 <= n} x_1*x_2*x_3*(det V(x_1,x_2,x_3))^2 = 1/144*sum {1 <= i,j,k <= n} i*j*k*((i-j)(i-k)(j-k))^2, where V(x_1,x_2,x_3) is the Vandermonde matrix of order 3. - Peter Bala, Sep 21 2007
G.f.: -(x^6+22*x^5+113*x^4+190*x^3+113*x^2+22*x+1)/(x-1)^13. - Colin Barker, Jun 06 2012
From Amiram Eldar, May 29 2022: (Start)
Sum_{n>=0} 1/a(n) = 25200*Pi^2 - 248713.
Sum_{n>=0} (-1)^n/a(n) = 376003 - 430080*log(2) - 64800*zeta(3). (End)

A359363 Triangle read by rows. The coefficients of the Baxter polynomials p(0, x) = 1 and p(n, x) = x*hypergeom([-1 - n, -n, 1 - n], [2, 3], -x) for n >= 1.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 4, 1, 0, 1, 10, 10, 1, 0, 1, 20, 50, 20, 1, 0, 1, 35, 175, 175, 35, 1, 0, 1, 56, 490, 980, 490, 56, 1, 0, 1, 84, 1176, 4116, 4116, 1176, 84, 1, 0, 1, 120, 2520, 14112, 24696, 14112, 2520, 120, 1, 0, 1, 165, 4950, 41580, 116424, 116424, 41580, 4950, 165, 1
Offset: 0

Views

Author

Peter Luschny, Dec 28 2022

Keywords

Comments

This triangle is a member of a family of Pascal-like triangles. Let T(n, k, m) = sf(m)*F(n - 1) / (F(k - 1)*F(n - k)) if k > 0 and otherwise k^n, where F(n) = Product_{j=0..m} (n + j)! and sf(m) are the superfactorials A000178. The case m = 2 gives this triangle, some other cases are given in the crossreferences. See also A342889 for a related representation of generalized binomial coefficients.

Examples

			Triangle T(n, k) starts:
[0] 1
[1] 0, 1
[2] 0, 1,   1
[3] 0, 1,   4,    1
[4] 0, 1,  10,   10,     1
[5] 0, 1,  20,   50,    20,     1
[6] 0, 1,  35,  175,   175,    35,     1
[7] 0, 1,  56,  490,   980,   490,    56,    1
[8] 0, 1,  84, 1176,  4116,  4116,  1176,   84,   1
[9] 0, 1, 120, 2520, 14112, 24696, 14112, 2520, 120, 1
.
Let p = (p1, p2,..., pn) denote a permutation of {1, 2,..., n}. The pair (p(i), p(i+1)) is a 'rise' if p(i) < p(i+1). Additionally a conventional rise is counted at the beginning of p.
T(n, k) is the number of Baxter permutations of {1,2,...,n} with k rises. For example for n = 4, [T(n, k) for k = 0..n] = [0, 1, 10, 10, 1]. The permutations, with preceding number of rises, are:
.
1 [4, 3, 2, 1],  3 [2, 3, 4, 1],  2 [3, 4, 2, 1],  3 [2, 3, 1, 4],
2 [3, 2, 4, 1],  3 [2, 1, 3, 4],  2 [3, 2, 1, 4],  3 [1, 3, 4, 2],
2 [2, 4, 3, 1],  3 [1, 3, 2, 4],  2 [4, 2, 3, 1],  3 [3, 4, 1, 2],
2 [2, 1, 4, 3],  3 [3, 1, 2, 4],  2 [4, 2, 1, 3],  3 [1, 2, 4, 3],
2 [1, 4, 3, 2],  3 [1, 4, 2, 3],  2 [4, 1, 3, 2],  3 [4, 1, 2, 3],
2 [4, 3, 1, 2],  4 [1, 2, 3, 4].
		

Crossrefs

Special cases of the general formula: A097805 (m = 0), (0,1)-Pascal triangle; A090181 (m = 1), triangle of Narayana; this triangle (m = 2); A056940 (m = 3), with 1,0,0...; A056941 (m = 4), with 1,0,0...; A142465 (m = 5), with 1,0,0....
Variant: A056939. Diagonals: A000292, A006542, A047819.

Programs

  • Maple
    p := (n, x) -> ifelse(n = 0, 1, x*hypergeom([-1-n, -n, 1-n], [2, 3], -x)):
    seq(seq(coeff(simplify(p(n, x)), x, k), k = 0..n), n = 0..10);
    # Alternative:
    T := proc(n, k) local F; F := n -> n!*(n+1)!*(n+2)!;
    ifelse(k = 0, k^n, 2*F(n-1)/(F(k-1)*F(n-k))) end:
    for n from 0 to 9 do seq(T(n, k), k = 0..n) od;
  • PARI
    C=binomial;
    T(n, k) = if(n==0 && k==0, 1, ( C(n+1,k-1) * C(n+1,k) * C(n+1,k+1) ) / ( C(n+1,1) * C(n+1,2) ) );
    for(n=0,10,for(k=0,n,print1(T(n,k),", "));print());
    \\ Joerg Arndt, Jan 04 2024
    
  • Python
    from functools import cache
    from math import factorial
    @cache
    def A359363Row(n: int) -> list[int]:
        @cache
        def F(n: int): return factorial(n) ** 3 * ((n+1) * (n+1) * (n+2))
        if n == 0: return [1]
        return [0] + [(2*F(n-1))//(F(k-1) * F(n-k)) for k in range(1, n+1)]
    for n in range(0, 10): print(A359363Row(n))
    # Peter Luschny, Jan 04 2024
  • SageMath
    def A359363(n):
        if n == 0: return SR(1)
        h = x*hypergeometric([-1 - n, -n, 1 - n], [2, 3], -x)
        return h.series(x, n + 1).polynomial(SR)
    for n in range(10): print(A359363(n).list())
    def PolyA359363(n, t): return Integer(A359363(n)(x=t).n())
    # Peter Luschny, Jan 04 2024
    

Formula

T(n, k) = [x^k] p(n, x).
T(n, k) = 2*F(n-1)/(F(k-1)*F(n-k)) for k > 0 where F(n) = n!*(n+1)!*(n+2)!.
p(n, 1) = A001181(n), i.e. the Baxter numbers are the values of the Baxter polynomials at x = 1.
(-1)^(n + 1)*p(2*n + 1, -1) = A217800(n) .

A047835 a(n) = Product_{i=1..n} ((i+4)*(i+5)*(i+6)*(i+7))/(i*(i+1)*(i+2)*(i+3)).

Original entry on oeis.org

1, 70, 1764, 24696, 232848, 1646568, 9343620, 44537922, 184225041, 677352676, 2254684432, 6892441920, 19571505408, 52101067968, 131018862096, 313203587004, 715536058545, 1569305708586, 3316911815140, 6778924352200, 13435361082000
Offset: 0

Views

Author

Keywords

Comments

Number of tilings of a <4,n,4> hexagon.
Partial sums of A133708. - Peter Bala, Sep 21 2007

References

  • O. D. Anderson, Find the next sequence, J. Rec. Math., 8 (No. 4, 1975-1976), 241.

Crossrefs

Fourth row of array A103905.

Programs

  • Maple
    seq(binomial(n,n-1)*binomial(n+1,n-2)*binomial(n+2,n-3)*binomial(n+3,n-4)/(10*4!), n=4..24); # Zerinvary Lajos, May 29 2007
  • Mathematica
    Table[Product[Times@@((i+Range[4,7])/(i+Range[0,3])),{i,n}],{n,0,20}] (* Harvey P. Dale, Nov 03 2011 *)

Formula

a(n) = C(n,n-1)*C(n+1,n-2)*C(n+2,n-3)*C(n+3,n-4)/(10*4!), n >= 4 . - Zerinvary Lajos, May 29 2007
a(n-4) = (1/3456)*Sum_{1 <= x_1, x_2, x_3, x_4 <= n} (det V(x_1,x_2,x_3,x_4))^2 = (1/3456)*Sum_{1 <= i,j,k,l <= n} ((i-j)(i-k)(i-l)(j-k)(j-l)(k-l))^2, where V(x_1,x_2,x_3,x_4) is the Vandermonde matrix of order 4. - Peter Bala, Sep 21 2007
Empirical g.f.: (x+1)*(x^8 + 52*x^7 + 658*x^6 + 2890*x^5 + 4810*x^4 + 2890*x^3 + 658*x^2 + 52*x + 1)/(1-x)^17. - Colin Barker, Jun 06 2012
Sum_{n>=0} 1/a(n) = 67200*Pi^4 + 5605600*Pi^2 - 185612833/3. - Amiram Eldar, May 29 2022

A133708 First differences of A047835.

Original entry on oeis.org

0, 0, 1, 69, 1694, 22932, 208152, 1413720, 7697052, 35194302, 139687119, 493127635, 1577331756, 4637757488, 12679063488, 32529562560, 78917794128, 182184724908, 402332471541, 853769650041, 1747606106554, 3462012537060, 6656436729800, 12452933493000
Offset: 1

Views

Author

Peter Bala, Sep 21 2007

Keywords

Crossrefs

Programs

  • Mathematica
    Join[{0,0,1},Differences[Table[Product[Times@@((i+Range[4,7])/(i+Range[0,3])),{i,n}],{n,0,30}]]] (* Harvey P. Dale, Aug 08 2015 *)

Formula

In terms of Vandermonde determinants, a(n) = 1/864*sum {1 <= x_1,x_2,x_3 <= n} (x_1*x_2*x_3)^2*(det V(x_1,x_2,x_3))^2 = 1/864 *sum {1 <= i,j,k <= n} (i*j*k*(i-j)(i-k)(j-k))^2, where V(x_1,x_2,x_3) is the Vandermonde matrix of order 3. a(n) = (n-2)*(n-1)^2*n^3*(n+1)^3*(n+2)^2*(n+3)*(n^2+n+3)*(2n+1)/108864000.
Empirical G.f.: x^3*(x+1)*(x^8+52*x^7+658*x^6+2890*x^5+4810*x^4+2890*x^3+658*x^2+52*x+1)/(x-1)^16. [Colin Barker, Jun 06 2012]

A114239 a(n) = (n+1)(n+2)^3*(n+3)(n^2 + 4n + 5)/120.

Original entry on oeis.org

1, 18, 136, 650, 2331, 6860, 17472, 39852, 83325, 162382, 298584, 522886, 878423, 1423800, 2236928, 3419448, 5101785, 7448874, 10666600, 15008994, 20786227, 28373444, 38220480, 50862500, 66931605, 87169446, 112440888, 143748766
Offset: 0

Views

Author

Emeric Deutsch, Nov 18 2005

Keywords

Comments

Kekulé numbers for certain benzenoids.
First differences of A107891. Partial sums of A083200. - Peter Bala, Sep 21 2007

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 167, Table 10.5/I/6).

Crossrefs

Programs

  • Maple
    a:=n->(n+1)*(n+2)^3*(n+3)*(n^2+4*n+5)/120: seq(a(n),n=0..33);
  • PARI
    a(n)=n-=2;(n^7-n^3)/120 \\ Charles R Greathouse IV, Feb 09 2012

Formula

a(n-2) = (n^7-n^3)/(2^7-2^3). - David Radcliffe, Dec 27 2008
G.f.: (1+10*x+20*x^2+10*x^3+x^4)/(1-x)^8. - Colin Barker, Feb 09 2012
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