A080930 -id:A080930 - cp's OEIS frontend

A001789 a(n) = binomial(n,3)*2^(n-3).

1, 8, 40, 160, 560, 1792, 5376, 15360, 42240, 112640, 292864, 745472, 1863680, 4587520, 11141120, 26738688, 63504384, 149422080, 348651520, 807403520, 1857028096, 4244635648, 9646899200, 21810380800, 49073356800, 109924319232

Offset: 3

N. J. A. Sloane

Number of 3-dimensional cubes in n-dimensional hypercube. - Henry Bottomley, Apr 14 2000
With three leading zeros, this is the second binomial transform of (0,0,0,1,0,0,0,0,...). - Paul Barry, Mar 07 2003
With 3 leading zeros, binomial transform of C(n,3). - Paul Barry, Apr 10 2003
Let M=[1,0,i;0,1,0;i,0,1], i=sqrt(-1). Then 1/det(I-xM)=1/(1-2x)^4. - Paul Barry, Apr 27 2005
If X_1,X_2,...,X_n is a partition of a 2n-set X into 2-blocks then, for n>2, a(n+1) is equal to the number of (n+3)-subsets of X intersecting each X_i (i=1,2,...,n). - Milan Janjic, Jul 21 2007
With offset 0, a(n) is the number of ways to separate [n] into four non-overlapping intervals (allowed to be empty) and then choose a subset from each interval. - Geoffrey Critzer, Feb 07 2009

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009, page 282.
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).

T. D. Noe, Table of n, a(n) for n = 3..500
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.
Herbert Izbicki, Über Unterbaüme eines Baumes, Monatshefte für Mathematik, Vol. 74 (1970), pp. 56-62.
Milan Janjic, Two Enumerative Functions.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
C. W. Jones, J. C. P. Miller, J. F. C. Conn and R. C. Pankhurst, Tables of Chebyshev polynomials Proc. Roy. Soc. Edinburgh. Sect. A., Vol. 62, No. 2 (1946), pp. 187-203.
Dusko Letic, Nenad Cakic, Branko Davidovic and Ivana Berkovic, Orthogonal and diagonal dimension fluxes of hyperspherical function, Advances in Difference Equations 2012, 2012:22. - From _N. J. A. Sloane_, Sep 04 2012
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Eric Weisstein's World of Mathematics, Hypercube.
Alina F. Y. Zhao, Pattern Popularity in Multiply Restricted Permutations, Journal of Integer Sequences, Vol. 17 (2014), Article 14.10.3.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-24,32,-16).

Cf. A000079, A001787, A001788, A003472, A007318, A082138, A080930.

a(n) = A038207(n,3).

List([3..30], n-> Binomial(n,3)*2^(n-3)); # G. C. Greubel, Aug 27 2019

a001789 n = a007318 n 3 * 2 ^ (n - 3)
a001789_list = 1 : zipWith (+) (map (* 2) a001789_list) (drop 2 a001788_list)
-- Reinhard Zumkeller, Jul 12 2014

[Binomial(n,3)*2^(n-3): n in [3..30]]; // G. C. Greubel, Aug 27 2019

A001789:=1/(2*z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation
seq(binomial(n+3,3)*2^n,n=0..25); # Zerinvary Lajos, Jun 03 2008

Table[Binomial[n, 3]*2^(n-3), {n,3,30}] (* Stefan Steinerberger, Apr 18 2006 *)
LinearRecurrence[{8,-24,32,-16},{1,8,40,160},30] (* Harvey P. Dale, Feb 10 2016 *)

a(n)=binomial(n,3)<<(n-3) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 2*a(n-1) + A001788(n-2).
For n>0, a(n+3) = 2*A082138(n) = 8*A080930(n+1).
From Paul Barry, Apr 10 2003: (Start)
G.f. (with three leading zeros): x^3/(1-2*x)^4.
With three leading zeros, a(n) = 8*a(n-1) - 24*a(n-2) + 32*a(n-3) - 16*a(n-4), a(0)=a(1)=a(2)=0, a(3)=1.
E.g.f.: (x^3/3!)*exp(2*x) (with 3 leading zeros). (End)
a(n) = Sum_{i=3..n} binomial(i,3)*binomial(n,i). Example: for n=6, a(6) = 1*20 + 4*15 + 10*6 + 20*1 = 160. - Bruno Berselli, Mar 23 2018
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=3} 1/a(n) = 6*log(2) - 3.
Sum_{n>=3} (-1)^(n+1)/a(n) = 54*log(3/2) - 21. (End)

More terms from James Sellers, Apr 15 2000
More terms from Stefan Steinerberger, Apr 18 2006
Formula fixed by Reinhard Zumkeller, Jul 12 2014

A131941 Partial sums of ceiling(n^2/2) (A000982).

Original entry on oeis.org

0, 1, 3, 8, 16, 29, 47, 72, 104, 145, 195, 256, 328, 413, 511, 624, 752, 897, 1059, 1240, 1440, 1661, 1903, 2168, 2456, 2769, 3107, 3472, 3864, 4285, 4735, 5216, 5728, 6273, 6851, 7464, 8112, 8797, 9519, 10280, 11080, 11921, 12803, 13728, 14696, 15709

Offset: 0

Gary W. Adamson, Oct 25 2007

Binomial transform of [0, 1, 1, 2, -2, 4, -8, 16, -32, ...].
Starting with offset 1 = (1, 3, 5, 7, ...) convolved with (1, 0, 3, 0, 5, ...). - Gary W. Adamson, Feb 16 2009
Also the number of undirected bishop moves on an n X n chessboard, counted up to rotations of the board. Equivalently, number of undirected bishop moves on an n X n chessboard, counted up to horizontal and vertical reflections of the board. - Hilko Koning, Aug 21 2025

			a(3) = 8 = 0 + 1 + 2 + 5.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions, J. Integer Sequences, Vol. 14 (2011), Article 11.9.1, p. 11.
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Cf. A000982, A080930 (binomial transform without leading 0).

[Ceiling((2*n^3+3*n^2+4*n)/12): n in [0..60]]; // Vincenzo Librandi, Jun 25 2011

a(n):=round(1/(12)(2*n^(3)+3*n^(2)+4*n))  # Mircea Merca, Oct 10 2010

CoefficientList[Series[x (1 + x^2)/(1 + x)/(1 - x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Mar 26 2014 *)
a[n_Integer?Positive] := If[EvenQ[n], With[{m = n/2}, (4 m^3 - 3 m^2 + 2 m)/3],
With[{m = (n - 1)/2}, (4 m^3 + 3 m^2 + 2 m)/3]];
Table[a[n], {n, 1, 50}] (* Hilko Koning, Aug 21 2025 *)

a(n) = (n+[0,1][n%2+1]) * (2*n^2 +[3,1][n%2+1]*n +[4,3][n%2+1])/12 \\ Washington Bomfim, Jul 31 2008

For even n, a(n) = n*(2*n^2 +3*n +4)/12. For odd n, a(n) = (n+1)*(2*n^2 +n +3)/12. - Washington Bomfim, Jul 31 2008
From R. J. Mathar, Feb 24 2010: (Start)
G.f.: x*(1+x^2)/((1+x)*(1-x)^4).
a(n) = 3*a(n-1) - 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - a(n-5). (End)
From Mircea Merca, Oct 10 2010: (Start)
a(n) = round((2*n^3 + 3*n^2 + 4*n)/12) = round((2*n+1)*(2*n^2 + 3*n + 3)/24) = floor((n+1)*(2*n^2 + n + 3)/12) = ceiling((2*n^3 + 3*n^2 + 4*n)/12).
a(n) = a(n-2) + n^2 - n + 1, n > 1. (End)
a(n) = (2*n*(2*n^2 + 3*n + 4) - 3*(-1)^n + 3)/24. - Bruno Berselli, Dec 07 2010
E.g.f.: (x*(9 + 9*x + 2*x^2)*cosh(x) + (3 + 9*x + 9*x^2 + 2*x^3)*sinh(x))/12. - Stefano Spezia, Dec 21 2021

A167875 One third of product plus sum of three consecutive nonnegative integers; a(n)=(n+1)(n^2+2n+3)/3.

Original entry on oeis.org

1, 4, 11, 24, 45, 76, 119, 176, 249, 340, 451, 584, 741, 924, 1135, 1376, 1649, 1956, 2299, 2680, 3101, 3564, 4071, 4624, 5225, 5876, 6579, 7336, 8149, 9020, 9951, 10944, 12001, 13124, 14315, 15576, 16909, 18316, 19799, 21360, 23001, 24724, 26531

Offset: 0

Klaus Brockhaus, Nov 14 2009

a(n) = ((n*(n+1)*(n+2))+(n+(n+1)+(n+2)))/3, n >= 0.
Equals A006527 without initial term 0: a(n) = A006527(n+1).
Binomial transform of A167876.
Inverse binomial transform of A080930.
a(n) = A007290(n+2)+n+1.
a(n) = A014820(n)/(n+1) for n > 0.
a(n) = A116731(n+2)-1.
a(n) = A033547(n+1)-n.
a(n) = A054602(n)/3.
a(n) = A086514(n+3)-2.
a(n) = A002061(n+1)+a(n-1) for n > 0.
a(n) = A005894(n)-a(n-1) for n > 0.
First bisection is A057813.
Second differences are in A004277.
a(n) = A177342(n)*(-1)+a(n-1)*5 with n>0. For n=8, a(8)=-A177342(8)+a(7)*5=-631+176*5=249. - Bruno Berselli, May 18 2010

			a(0) = (0*1*2+0+1+2)/3 = (0+3)/3 = 1.
a(1) = (1*2*3+1+2+3)/3 = (6+6)/3 = 4.
a(6)-4*a(5)+6*a(4)-4*a(3)+a(2) = 119-4*76+6*45-4*24+11 = 0. - _Bruno Berselli_, May 26 2010

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

Cf. A001477 (nonnegative integers),
A006527 ((n^3+2*n)/3),
A167876 (1, 3, 4, 2, 0, 0, 0, 0, ...),
A080930,
A007290 (2*C(n, 3)),
A014820 ((1/3)*(n^2+2*n+3)*(n+1)^2),
A116731,
A033547 (n*(n^2+5)/3),
A054602 (Sum_{d|3} phi(d)*n^(3/d)),
A086514 ((n^3-6*n^2+14*n-6)/3),
A002061 (n^2-n+1),
A005894 (centered tetrahedral numbers),
A057813 ((2*n+1)*(4*n^2+4*n+3)/3),
A004277 (1 and the positive even numbers),
A028387 (n+(n+1)^2),
A166941, A166942, A166943.

[ (&*s + &+s)/3 where s is [n..n+2]: n in [0..42] ];

Select[Table[(n*(n+1)*(n+2)+n+(n+1)+(n+2))/3,{n,0,5!}],IntegerQ[#]&] (* Vladimir Joseph Stephan Orlovsky, Dec 04 2010 *)
(Times@@#+Total[#])/3&/@Partition[Range[0,65],3,1]  (* Harvey P. Dale, Mar 14 2011 *)

a(n)=(n+1)*(n^2+2*n+3)/3 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = (n^3+3*n^2+5*n+3)/3.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3)+2 for n > 3; a(0)=1, a(1)=4, a(2)=11, a(3)=24.
G.f.: (1+x^2)/(1-x)^4.
a(n) = SUM(A109613(k)*A005408(n-k): 0<=k<=n). - Reinhard Zumkeller, Dec 05 2009
a(n)-4*a(n-1)+6*a(n-2)-4*a(n-3)+a(n-4)=0 for n>3. - Bruno Berselli, May 26 2010

A080928 Triangle T(n,k) read by rows: T(n,k) = Sum_{i=0..n} C(n,2i)*C(2i,k).

Original entry on oeis.org

1, 1, 0, 2, 2, 1, 4, 6, 3, 0, 8, 16, 12, 4, 1, 16, 40, 40, 20, 5, 0, 32, 96, 120, 80, 30, 6, 1, 64, 224, 336, 280, 140, 42, 7, 0, 128, 512, 896, 896, 560, 224, 56, 8, 1, 256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 0, 512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10

Offset: 0

Paul Barry, Feb 26 2003

Gives the general solution to a(n) = 2*a(n-1) + k(k+2)*a(n-2), a(0) = a(1) = 1. The value k=1 gives the row sums of the triangle, or 1,1,5,13,... This is A046717, the solution to a(n) = 2*a(n-1) + 3*a(n-2), a(0)=a(1)=1.

Product of A007318 and A007318 with every odd-indexed row set to zero. - Paul Barry, Nov 08 2005

			Triangle begins:
    1;
    1,    0;
    2,    2,    1;
    4,    6,    3,    0;
    8,   16,   12,    4,    1;
   16,   40,   40,   20,    5,    0;
   32,   96,  120,   80,   30,    6,   1;
   64,  224,  336,  280,  140,   42,   7,  0;
  128,  512,  896,  896,  560,  224,  56,  8, 1;
  256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 0; etc.

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identity 156.
J-L. Kim, Relation between weight distribution and combinatorial identities, Bulletin of the Institute of Combinatorics and its Applications, Canada, 31, 2001, pp. 69-79.

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150).
Saul Schleimer, Bert Wiest, On the conjugacy problem in braid groups: Garside theory and subsurfaces, arXiv:1807.01500 [math.GT], 2018.

Apart from k=n, T(n, k) equals (1/2)*A038207(n, k).

Columns include A011782, 2*A001792, A080929, 4*A080930. Row sums are in A046717.

Table[Sum[Binomial[n, 2 i] Binomial[2 i, k], {i, 0, n}], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Oct 11 2018 *)

T(n, n) = (n+1) mod 2, T(n, k) = C(n, k)*2^(n-k-1).
T(n, 0) = A011782(n), T(n, k)=0, k>n, T(2n, 2n)=1, T(2n-1, 2n-1)=0, T(n+1, n)=n+1. Otherwise T(n, k) = T(n-1, k-1) + 2T(n-1, k). Rows are the coefficients of the polynomials in the expansion of (1-x)/((1+k*x)*(1-(k+2)*x)). The main diagonal is 1, 0, 1, 0, 1, 0, ... with g.f. 1/(1-x^2). Subsequent subdiagonals are given by A011782(k)*C(n+k, k) with g.f. A011782(k)/(1-x)^k.
T(n, k) = Sum_{j=0..n} C(n, j)*C(j, k)*(1+(-1)^j)/2; T(n, k) = 2^(n-k-1)*(C(n, k) + (-1)^n*C(0, n-k)). - Paul Barry, Nov 08 2005

Edited by Ralf Stephan, Feb 04 2005

cp's OEIS Frontend

A001789 a(n) = binomial(n,3)*2^(n-3).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

Formula

Extensions

A131941 Partial sums of ceiling(n^2/2) (A000982).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A167875 One third of product plus sum of three consecutive nonnegative integers; a(n)=(n+1)(n^2+2n+3)/3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A080928 Triangle T(n,k) read by rows: T(n,k) = Sum_{i=0..n} C(n,2i)*C(2i,k).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A083929 Inverse function of N -> N injection A083930.

Views

Author

Keywords

Comments

Crossrefs

Formula