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 1-10 of 17 results. Next

A160674 A bisection of A063522.

Original entry on oeis.org

1, 63, 305, 847, 1809, 3311, 5473, 8415, 12257, 17119, 23121, 30383, 39025, 49167, 60929, 74431, 89793, 107135, 126577, 148239, 172241, 198703, 227745, 259487, 294049, 331551, 372113, 415855, 462897, 513359, 567361, 625023, 686465, 751807, 821169, 894671
Offset: 0

Views

Author

N. J. A. Sloane, Nov 17 2009

Keywords

Programs

  • Mathematica
    Table[LegendreP[3,2n+1],{n,0,50}]
  • PARI
    a(n)=pollegendre(3,2*n+1) \\ Charles R Greathouse IV, Mar 19 2017

A160699 A bisection of A063522.

Original entry on oeis.org

0, 17, 154, 531, 1268, 2485, 4302, 6839, 10216, 14553, 19970, 26587, 34524, 43901, 54838, 67455, 81872, 98209, 116586, 137123, 159940, 185157, 212894, 243271, 276408, 312425, 351442, 393579, 438956, 487693, 539910, 595727, 655264, 718641, 785978, 857395
Offset: 0

Views

Author

N. J. A. Sloane, Nov 17 2009

Keywords

Programs

  • Magma
    I:=[0,17,154,531]; [n le 4 select I[n] else 4*Self(n-1) - 6*Self(n-2) +4*Self(n-3) -Self(n-4): n in [1..50]]; // G. C. Greubel, May 02 2018
  • Mathematica
    Table[LegendreP[3,2n],{n,0,50}]
    LinearRecurrence[{4,-6,4,-1},{0,17,154,531},151] (* Harvey P. Dale, May 07 2011 *)
  • PARI
    a(n)=pollegendre(3,2*n) \\ Charles R Greathouse IV, Mar 19 2017
    

Formula

a(0)=0, a(1)=17, a(2)=154, a(3)=531, a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4). - Harvey P. Dale, May 07 2011

A000578 The cubes: a(n) = n^3.

Original entry on oeis.org

0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507
Offset: 0

Views

Author

Keywords

Comments

a(n) is the sum of the next n odd numbers; i.e., group the odd numbers so that the n-th group contains n elements like this: (1), (3, 5), (7, 9, 11), (13, 15, 17, 19), (21, 23, 25, 27, 29), ...; then each group sum = n^3 = a(n). Also the median of each group = n^2 = mean. As the sum of first n odd numbers is n^2 this gives another proof of the fact that the n-th partial sum = (n(n + 1)/2)^2. - Amarnath Murthy, Sep 14 2002
Total number of triangles resulting from criss-crossing cevians within a triangle so that two of its sides are each n-partitioned. - Lekraj Beedassy, Jun 02 2004. See Propp and Propp-Gubin for a proof.
Also structured triakis tetrahedral numbers (vertex structure 7) (cf. A100175 = alternate vertex); structured tetragonal prism numbers (vertex structure 7) (cf. A100177 = structured prisms); structured hexagonal diamond numbers (vertex structure 7) (cf. A100178 = alternate vertex; A000447 = structured diamonds); and structured trigonal anti-diamond numbers (vertex structure 7) (cf. A100188 = structured anti-diamonds). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {4, 3}.
Least multiple of n such that every partial sum is a square. - Amarnath Murthy, Sep 09 2005
Draw a regular hexagon. Construct points on each side of the hexagon such that these points divide each side into equally sized segments (i.e., a midpoint on each side or two points on each side placed to divide each side into three equally sized segments or so on), do the same construction for every side of the hexagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least one side of the polygon. The result is a triangular tiling of the hexagon and the creation of a number of smaller regular hexagons. The equation gives the total number of regular hexagons found where n = the number of points drawn + 1. For example, if 1 point is drawn on each side then n = 1 + 1 = 2 and a(n) = 2^3 = 8 so there are 8 regular hexagons in total. If 2 points are drawn on each side then n = 2 + 1 = 3 and a(n) = 3^3 = 27 so there are 27 regular hexagons in total. - Noah Priluck (npriluck(AT)gmail.com), May 02 2007
The solutions of the Diophantine equation: (X/Y)^2 - X*Y = 0 are of the form: (n^3, n) with n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^2k - XY = 0 are of the form: (m*n^(2k + 1), m*n^(2k - 1)) with m >= 1, k >= 1 and n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^(2k + 1) - XY = 0 are of the form: (m*n^(k + 1), m*n^k) with m >= 1, k >= 1 and n >= 1. - Mohamed Bouhamida, Oct 04 2007
Except for the first two terms, the sequence corresponds to the Wiener indices of C_{2n} i.e., the cycle on 2n vertices (n > 1). - K.V.Iyer, Mar 16 2009
Totally multiplicative sequence with a(p) = p^3 for prime p. - Jaroslav Krizek, Nov 01 2009
Sums of rows of the triangle in A176271, n > 0. - Reinhard Zumkeller, Apr 13 2010
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
Numbers n for which order of torsion subgroup t of the elliptic curve y^2 = x^3 - n is t = 2. - Artur Jasinski, Jun 30 2010
The sequence with the lengths of the Pisano periods mod k is 1, 2, 3, 4, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 16, 17, 6, 19, 20, ... for k >= 1, apparently multiplicative and derived from A000027 by dividing every ninth term through 3. Cubic variant of A186646. - R. J. Mathar, Mar 10 2011
The number of atoms in a bcc (body-centered cubic) rhombic hexahedron with n atoms along one edge is n^3 (T. P. Martin, Shells of atoms, eq. (8)). - Brigitte Stepanov, Jul 02 2011
The inverse binomial transform yields the (finite) 0, 1, 6, 6 (third row in A019538 and A131689). - R. J. Mathar, Jan 16 2013
Twice the area of a triangle with vertices at (0, 0), (t(n - 1), t(n)), and (t(n), t(n - 1)), where t = A000217 are triangular numbers. - J. M. Bergot, Jun 25 2013
If n > 0 is not congruent to 5 (mod 6) then A010888(a(n)) divides a(n). - Ivan N. Ianakiev, Oct 16 2013
For n > 2, a(n) = twice the area of a triangle with vertices at points (binomial(n,3),binomial(n+2,3)), (binomial(n+1,3),binomial(n+1,3)), and (binomial(n+2,3),binomial(n,3)). - J. M. Bergot, Jun 14 2014
Determinants of the spiral knots S(4,k,(1,1,-1)). a(k) = det(S(4,k,(1,1,-1))). - Ryan Stees, Dec 14 2014
One of the oldest-known examples of this sequence is shown in the Senkereh tablet, BM 92698, which displays the first 32 terms in cuneiform. - Charles R Greathouse IV, Jan 21 2015
From Bui Quang Tuan, Mar 31 2015: (Start)
We construct a number triangle from the integers 1, 2, 3, ... 2*n-1 as follows. The first column contains all the integers 1, 2, 3, ... 2*n-1. Each succeeding column is the same as the previous column but without the first and last items. The last column contains only n. The sum of all the numbers in the triangle is n^3.
Here is the example for n = 4, where 1 + 2*2 + 3*3 + 4*4 + 3*5 + 2*6 + 7 = 64 = a(4):
1
2 2
3 3 3
4 4 4 4
5 5 5
6 6
7
(End)
For n > 0, a(n) is the number of compositions of n+11 into n parts avoiding parts 2 and 3. - Milan Janjic, Jan 07 2016
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
Number of inequivalent face colorings of the cube using at most n colors such that each color appears at least twice. - David Nacin, Feb 22 2017
Consider A = {a,b,c} a set with three distinct members. The number of subsets of A is 8, including {a,b,c} and the empty set. The number of subsets from each of those 8 subsets is 27. If the number of such iterations is n, then the total number of subsets is a(n-1). - Gregory L. Simay, Jul 27 2018
By Fermat's Last Theorem, these are the integers of the form x^k with the least possible value of k such that x^k = y^k + z^k never has a solution in positive integers x, y, z for that k. - Felix Fröhlich, Jul 27 2018

Examples

			For k=3, b(3) = 2 b(2) - b(1) = 4-1 = 3, so det(S(4,3,(1,1,-1))) = 3*3^2 = 27.
For n=3, a(3) = 3 + (3*0^2 + 3*0 + 3*1^2 + 3*1 + 3*2^2 + 3*2) = 27. - _Patrick J. McNab_, Mar 28 2016
		

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See p. 191.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 43, 64, 81.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 292.
  • T. Aaron Gulliver, "Sequences from cubes of integers", International Mathematical Journal, 4 (2003), no. 5, 439 - 445. See http://www.m-hikari.com/z2003.html for information about this journal. [I expanded the reference to make this easier to find. - N. J. A. Sloane, Feb 18 2019]
  • J. Propp and A. Propp-Gubin, "Counting Triangles in Triangles", Pi Mu Epsilon Journal (to appear).
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 6-7.
  • D. Wells, You Are A Mathematician, pp. 238-241, Penguin Books 1995.

Crossrefs

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
For sums of cubes, cf. A000537 (partial sums), A003072, A003325, A024166, A024670, A101102 (fifth partial sums).
Cf. A001158 (inverse Möbius transform), A007412 (complement), A030078(n) (cubes of primes), A048766, A058645 (binomial transform), A065876, A101094, A101097.
Subsequence of A145784.
Cf. A260260 (comment). - Bruno Berselli, Jul 22 2015
Cf. A000292 (tetrahedral numbers), A005900 (octahedral numbers), A006566 (dodecahedral numbers), A006564 (icosahedral numbers).
Cf. A098737 (main diagonal).

Programs

  • Haskell
    a000578 = (^ 3)
    a000578_list = 0 : 1 : 8 : zipWith (+)
       (map (+ 6) a000578_list)
       (map (* 3) $ tail $ zipWith (-) (tail a000578_list) a000578_list)
    -- Reinhard Zumkeller, Sep 05 2015, May 24 2012, Oct 22 2011
    
  • Magma
    [ n^3 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 14 2014
    
  • Magma
    I:=[0,1,8,27]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jul 05 2014
    
  • Maple
    A000578 := n->n^3;
    seq(A000578(n), n=0..50);
    isA000578 := proc(r)
        local p;
        if r = 0 or r =1 then
            true;
        else
            for p in ifactors(r)[2] do
                if op(2, p) mod 3 <> 0 then
                    return false;
                end if;
            end do:
            true ;
        end if;
    end proc: # R. J. Mathar, Oct 08 2013
  • Mathematica
    Table[n^3, {n, 0, 30}] (* Stefan Steinerberger, Apr 01 2006 *)
    CoefficientList[Series[x (1 + 4 x + x^2)/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jul 05 2014 *)
    Accumulate[Table[3n^2+3n+1,{n,0,20}]] (* or *) LinearRecurrence[{4,-6,4,-1},{1,8,27,64},20](* Harvey P. Dale, Aug 18 2018 *)
  • Maxima
    A000578(n):=n^3$
    makelist(A000578(n),n,0,30); /* Martin Ettl, Nov 03 2012 */
    
  • PARI
    A000578(n)=n^3 \\ M. F. Hasler, Apr 12 2008
    
  • PARI
    is(n)=ispower(n,3) \\ Charles R Greathouse IV, Feb 20 2012
    
  • Python
    A000578_list, m = [], [6, -6, 1, 0]
    for _ in range(10**2):
        A000578_list.append(m[-1])
        for i in range(3):
            m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015
    
  • Scheme
    (define (A000578 n) (* n n n)) ;; Antti Karttunen, Oct 06 2017

Formula

a(n) = Sum_{i=0..n-1} A003215(i).
Multiplicative with a(p^e) = p^(3e). - David W. Wilson, Aug 01 2001
G.f.: x*(1+4*x+x^2)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
Dirichlet generating function: zeta(s-3). - Franklin T. Adams-Watters, Sep 11 2005, Amarnath Murthy, Sep 09 2005
E.g.f.: (1+3*x+x^2)*x*exp(x). - Franklin T. Adams-Watters, Sep 11 2005 - Amarnath Murthy, Sep 09 2005
a(n) = Sum_{i=1..n} (Sum_{j=i..n+i-1} A002024(j,i)). - Reinhard Zumkeller, Jun 24 2007
a(n) = lcm(n, (n - 1)^2) - (n - 1)^2. E.g.: lcm(1, (1 - 1)^2) - (1 - 1)^2 = 0, lcm(2, (2 - 1)^2) - (2 - 1)^2 = 1, lcm(3, (3 - 1)^2) - (3 - 1)^2 = 8, ... - Mats Granvik, Sep 24 2007
Starting (1, 8, 27, 64, 125, ...), = binomial transform of [1, 7, 12, 6, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
a(n) = A007531(n) + A000567(n). - Reinhard Zumkeller, Sep 18 2009
a(n) = binomial(n+2,3) + 4*binomial(n+1,3) + binomial(n,3). [Worpitzky's identity for cubes. See. e.g., Graham et al., eq. (6.37). - Wolfdieter Lang, Jul 17 2019]
a(n) = n + 6*binomial(n+1,3) = binomial(n,1)+6*binomial(n+1,3). - Ron Knott, Jun 10 2019
A010057(a(n)) = 1. - Reinhard Zumkeller, Oct 22 2011
a(n) = A000537(n) - A000537(n-1), difference between 2 squares of consecutive triangular numbers. - Pierre CAMI, Feb 20 2012
a(n) = A048395(n) - 2*A006002(n). - J. M. Bergot, Nov 25 2012
a(n) = 1 + 7*(n-1) + 6*(n-1)*(n-2) + (n-1)*(n-2)*(n-3). - Antonio Alberto Olivares, Apr 03 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6. - Ant King Apr 29 2013
a(n) = A000330(n) + Sum_{i=1..n-1} A014105(i), n >= 1. - Ivan N. Ianakiev, Sep 20 2013
a(k) = det(S(4,k,(1,1,-1))) = k*b(k)^2, where b(1)=1, b(2)=2, b(k) = 2*b(k-1) - b(k-2) = b(2)*b(k-1) - b(k-2). - Ryan Stees, Dec 14 2014
For n >= 1, a(n) = A152618(n-1) + A033996(n-1). - Bui Quang Tuan, Apr 01 2015
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Jon Tavasanis, Feb 21 2016
a(n) = n + Sum_{j=0..n-1} Sum_{k=1..2} binomial(3,k)*j^(3-k). - Patrick J. McNab, Mar 28 2016
a(n) = A000292(n-1) * 6 + n. - Zhandos Mambetaliyev, Nov 24 2016
a(n) = n*binomial(n+1, 2) + 2*binomial(n+1, 3) + binomial(n,3). - Tony Foster III, Nov 14 2017
From Amiram Eldar, Jul 02 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(3) (A002117).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*zeta(3)/4 (A197070). (End)
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)/Pi.
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)/(3*Pi). (End)
a(n) = Sum_{d|n} sigma_3(d)*mu(n/d) = Sum_{d|n} A001158(d)*A008683(n/d). Moebius transform of sigma_3(n). - Ridouane Oudra, Apr 15 2021

A006003 a(n) = n*(n^2 + 1)/2.

Original entry on oeis.org

0, 1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, 4010, 4641, 5335, 6095, 6924, 7825, 8801, 9855, 10990, 12209, 13515, 14911, 16400, 17985, 19669, 21455, 23346, 25345, 27455, 29679, 32020, 34481, 37065, 39775
Offset: 0

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Author

Keywords

Comments

Write the natural numbers in groups: 1; 2,3; 4,5,6; 7,8,9,10; ... and add the groups. In other words, "sum of the next n natural numbers". - Felice Russo
Number of rhombi in an n X n rhombus, if 'crossformed' rhombi are allowed. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Also the sum of the integers between T(n-1)+1 and T(n), the n-th triangular number (A000217). Sum of n-th row of A000027 regarded as a triangular array.
Unlike the cubes which have a similar definition, it is possible for 2 terms of this sequence to sum to a third. E.g., a(36) + a(37) = 23346 + 25345 = 48691 = a(46). Might be called 2nd-order triangular numbers, thus defining 3rd-order triangular numbers (A027441) as n(n^3+1)/2, etc. - Jon Perry, Jan 14 2004
Also as a(n)=(1/6)*(3*n^3+3*n), n > 0: structured trigonal diamond numbers (vertex structure 4) (cf. A000330 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The sequence M(n) of magic constants for n X n magic squares (numbered 1 through n^2) from n=3 begins M(n) = 15, 34, 65, 111, 175, 260, ... - Lekraj Beedassy, Apr 16 2005 [comment corrected by Colin Hall, Sep 11 2009]
The sequence Q(n) of magic constants for the n-queens problem in chess begins 0, 0, 0, 0, 34, 65, 111, 175, 260, ... - Paul Muljadi, Aug 23 2005
Alternate terms of A057587. - Jeremy Gardiner, Apr 10 2005
Also partial differences of A063488(n) = (2*n-1)*(n^2-n+2)/2. a(n) = A063488(n) - A063488(n-1) for n>1. - Alexander Adamchuk, Jun 03 2006
In an n X n grid of numbers from 1 to n^2, select -- in any manner -- one number from each row and column. Sum the selected numbers. The sum is independent of the choices and is equal to the n-th term of this sequence. - F.-J. Papp (fjpapp(AT)umich.edu), Jun 06 2006
Nonnegative X values of solutions to the equation (X-Y)^3 - (X+Y) = 0. To find Y values: b(n) = (n^3-n)/2. - Mohamed Bouhamida, May 16 2006
For the equation: m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 and m is an odd number the X values are given by the sequence defined by a(n) = (m*n^k+n)/2. The Y values are given by the sequence defined by b(n) = (m*n^k-n)/2. - Mohamed Bouhamida, May 16 2006
If X is an n-set and Y a fixed 3-subset of X then a(n-3) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
(m*(2n)^k+n, m*(2n)^k-n) solves the Diophantine equation: 2m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 where m is a positive integer. - Mohamed Bouhamida, Oct 02 2007
Also c^(1/2) in a^(1/2) + b^(1/2) = c^(1/2) such that a^2 + b = c. - Cino Hilliard, Feb 09 2008
a(n) = n*A000217(n) - Sum_{i=0..n-1} A001477(i). - Bruno Berselli, Apr 25 2010
a(n) is the number of triples (w,x,y) having all terms in {0,...,n} such that at least one of these inequalities fails: x+y < w, y+w < x, w+x < y. - Clark Kimberling, Jun 14 2012
Sum of n-th row of the triangle in A209297. - Reinhard Zumkeller, Jan 19 2013
The sequence starting with "1" is the third partial sum of (1, 2, 3, 3, 3, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the largest eigenvalue of the matrix returned by the MATLAB command magic(n) for n > 0. - Altug Alkan, Nov 10 2015
a(n) is the number of triples (x,y,z) having all terms in {1,...,n} such that all these triangle inequalities are satisfied: x+y > z, y+z > x, z+x > y. - Heinz Dabrock, Jun 03 2016
Shares its digital root with the stella octangula numbers (A007588). See A267017. - Peter M. Chema, Aug 28 2016
Can be proved to be the number of nonnegative solutions of a system of three linear Diophantine equations for n >= 0 even: 2*a_{11} + a_{12} + a_{13} = n, 2*a_{22} + a_{12} + a_{23} = n and 2*a_{33} + a_{13} + a_{23} = n. The number of solutions is f(n) = (1/16)*(n+2)*(n^2 + 4n + 8) and a(n) = n*(n^2 + 1)/2 is obtained by remapping n -> 2*n-2. - Kamil Bradler, Oct 11 2016
For n > 0, a(n) coincides with the trace of the matrix formed by writing the numbers 1...n^2 back and forth along the antidiagonals (proved, see A078475 for the examples of matrix). - Stefano Spezia, Aug 07 2018
The trace of an n X n square matrix where the elements are entered on the ascending antidiagonals. The determinant is A069480. - Robert G. Wilson v, Aug 07 2018
Bisections are A317297 and A005917. - Omar E. Pol, Sep 01 2018
Number of achiral colorings of the vertices (or faces) of a regular tetrahedron with n available colors. An achiral coloring is identical to its reflection. - Robert A. Russell, Jan 22 2020
a(n) is the n-th centered triangular pyramidal number. - Lechoslaw Ratajczak, Nov 02 2021
a(n) is the number of words of length n defined on 4 letters {b,c,d,e} that contain one or no b's, one c or two d's, and any number of e's. For example, a(3) = 15 since the words are (number of permutations in parentheses): bce (6), bdd (3), cee (3), and dde (3). - Enrique Navarrete, Jun 21 2025

Examples

			G.f. = x + 5*x^2 + 15*x^3 + 34*x^4 + 65*x^5 + 111*x^6 + 175*x^7 + 260*x^8 + ...
For a(2)=5, the five tetrahedra have faces AAAA, AAAB, AABB, ABBB, and BBBB with colors A and B. - _Robert A. Russell_, Jan 31 2020
		

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 15, p. 5, Ellipses, Paris 2008.
  • F.-J. Papp, Colloquium Talk, Department of Mathematics, University of Michigan-Dearborn, March 6, 2005.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000330, A000537, A066886, A057587, A027480, A002817 (partial sums).
Cf. A000578 (cubes).
(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, this sequence, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Antidiagonal sums of array in A000027. Row sums of the triangular view of A000027.
Cf. A063488 (sum of two consecutive terms), A005917 (bisection), A317297 (bisection).
Cf. A105374 / 8.
Tetrahedron colorings: A006008 (oriented), A000332(n+3) (unoriented), A000332 (chiral), A037270 (edges).
Other polyhedron colorings: A337898 (cube faces, octahedron vertices), A337897 (octahedron faces, cube vertices), A337962 (dodecahedron faces, icosahedron vertices), A337960 (icosahedron faces, dodecahedron vertices).
Row 3 of A325001 (simplex vertices and facets) and A337886 (simplex faces and peaks).

Programs

  • GAP
    a_n:=List([0..nmax], n->n*(n^2 + 1)/2); # Stefano Spezia, Aug 12 2018
    
  • Haskell
    a006003 n = n * (n ^ 2 + 1) `div` 2
    a006003_list = scanl (+) 0 a005448_list
    -- Reinhard Zumkeller, Jun 20 2013
    
  • MATLAB
    % Also works with FreeMat.
    for(n=0:nmax); tm=n*(n^2 + 1)/2; fprintf('%d\t%0.f\n', n, tm); end
    % Stefano Spezia, Aug 12 2018
    
  • Magma
    [n*(n^2 + 1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015
    
  • Magma
    [Binomial(n,3)+Binomial(n-1,3)+Binomial(n-2,3): n in [2..60]]; // Vincenzo Librandi, Sep 12 2015
    
  • Mathematica
    Table[ n(n^2 + 1)/2, {n, 0, 45}]
    LinearRecurrence[{4,-6,4,-1}, {0,1,5,15},50] (* Harvey P. Dale, May 16 2012 *)
    CoefficientList[Series[x (1 + x + x^2)/(x - 1)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)
    With[{n=50},Total/@TakeList[Range[(n(n^2+1))/2],Range[0,n]]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Nov 28 2017 *)
  • Maxima
    a(n):=n*(n^2 + 1)/2$ makelist(a(n), n, 0, nmax); /* Stefano Spezia, Aug 12 2018 */
    
  • PARI
    {a(n) = n * (n^2 + 1) / 2}; /* Michael Somos, Dec 24 2011 */
    
  • PARI
    concat(0, Vec(x*(1+x+x^2)/(x-1)^4 + O(x^20))) \\ Felix Fröhlich, Oct 11 2016
    
  • Python
    def A006003(n): return n*(n**2+1)>>1 # Chai Wah Wu, Mar 25 2024

Formula

a(n) = binomial(n+2, 3) + binomial(n+1, 3) + binomial(n, 3). [corrected by Michel Marcus, Jan 22 2020]
G.f.: x*(1+x+x^2)/(x-1)^4. - Floor van Lamoen, Feb 11 2002
Partial sums of A005448. - Jonathan Vos Post, Mar 16 2006
Binomial transform of [1, 4, 6, 3, 0, 0, 0, ...] = (1, 5, 15, 34, 65, ...). - Gary W. Adamson, Aug 10 2007
a(n) = -a(-n) for all n in Z. - Michael Somos, Dec 24 2011
a(n) = Sum_{k = 1..n} A(k-1, k-1-n) where A(i, j) = i^2 + i*j + j^2 + i + j + 1. - Michael Somos, Jan 02 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), with a(0)=0, a(1)=1, a(2)=5, a(3)=15. - Harvey P. Dale, May 16 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3. - Ant King, Jun 13 2012
a(n) = A000217(n) + n*A000217(n-1). - Bruno Berselli, Jun 07 2013
a(n) = A057145(n+3,n). - Luciano Ancora, Apr 10 2015
E.g.f.: (1/2)*(2*x + 3*x^2 + x^3)*exp(x). - G. C. Greubel, Dec 18 2015; corrected by Ilya Gutkovskiy, Oct 12 2016
a(n) = T(n) + T(n-1) + T(n-2), where T means the tetrahedral numbers, A000292. - Heinz Dabrock, Jun 03 2016
From Ilya Gutkovskiy, Oct 11 2016: (Start)
Convolution of A001477 and A008486.
Convolution of A000217 and A158799.
Sum_{n>=1} 1/a(n) = H(-i) + H(i) = 1.343731971048019675756781..., where H(k) is the harmonic number, i is the imaginary unit. (End)
a(n) = A000578(n) - A135503(n). - Miquel Cerda, Dec 25 2016
Euler transform of length 3 sequence [5, 0, -1]. - Michael Somos, Dec 25 2016
a(n) = A037270(n)/n for n > 0. - Kritsada Moomuang, Dec 15 2018
a(n) = 3*A000292(n-1) + n. - Bruce J. Nicholson, Nov 23 2019
a(n) = A011863(n) - A011863(n-2). - Bruce J. Nicholson, Dec 22 2019
From Robert A. Russell, Jan 22 2020: (Start)
a(n) = C(n,1) + 3*C(n,2) + 3*C(n,3), where the coefficient of C(n,k) is the number of tetrahedron colorings using exactly k colors.
a(n) = C(n+3,4) - C(n,4).
a(n) = 2*A000332(n+3) - A006008(n) = A006008(n) - 2*A000332(n) = A000332(n+3) - A000332(n).
a(n) = A325001(3,n). (End)
From Amiram Eldar, Aug 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 2 * (A248177 + A001620).
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi)/4.
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi). (End)

Extensions

Better description from Albert Rich (Albert_Rich(AT)msn.com), Mar 1997

A005900 Octahedral numbers: a(n) = n*(2*n^2 + 1)/3.

Original entry on oeis.org

0, 1, 6, 19, 44, 85, 146, 231, 344, 489, 670, 891, 1156, 1469, 1834, 2255, 2736, 3281, 3894, 4579, 5340, 6181, 7106, 8119, 9224, 10425, 11726, 13131, 14644, 16269, 18010, 19871, 21856, 23969, 26214, 28595, 31116, 33781, 36594, 39559, 42680
Offset: 0

Views

Author

Keywords

Comments

Series reversion of g.f.: A(x) is Sum_{n>0} - A066357(n)(-x)^n.
Partial sums of centered square numbers A001844. - Paul Barry, Jun 26 2003
Also as a(n) = (1/6)*(4n^3 + 2n), n>0: structured tetragonal diamond numbers (vertex structure 5) (cf. A000447 - structured diamonds); and structured trigonal anti-prism numbers (vertex structure 5) (cf. A100185 - structured anti-prisms). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {3,4}.
If X is an n-set and Y and Z are disjoint 2-subsets of X then a(n-4) is equal to the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Starting with 1 = binomial transform of [1, 5, 8, 4, 0, 0, 0, ...] where (1, 5, 8, 4) = row 3 of the Chebyshev triangle A081277. - Gary W. Adamson, Jul 19 2008
a(n) = largest coefficient of (1 + ... + x^(n-1))^4. - R. H. Hardin, Jul 23 2009
Convolution square root of (1 + 6x + 19x^3 + ...) = (1 + 3x + 5x^2 + 7x^3 + ...) = A005408(x). - Gary W. Adamson, Jul 27 2009
Starting with offset 1 = the triangular series convolved with [1, 3, 4, 4, 4, ...]. - Gary W. Adamson, Jul 28 2009
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral, and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
Let b be any product of four different primes. Then the divisor lattice of b^n is of width a(n+1). - Jean Drabbe, Oct 13 2010
Arises in Bezdek's proof on contact numbers for congruent sphere packings (see preprint). - Jonathan Vos Post, Feb 08 2011
Euler transform of length 2 sequence [6, -2]. - Michael Somos, Mar 27 2011
a(n+1) is the number of 2 X 2 matrices with all terms in {0,1,...,n} and (sum of terms) = 2n. - Clark Kimberling, Mar 19 2012
a(n) is the number of semistandard Young tableaux over all partitions of 3 with maximal element <= n. - Alois P. Heinz, Mar 22 2012
Self convolution of the odd numbers. - Reinhard Zumkeller, Apr 04 2012
a(n) is the number of (w,x,y,z) with all terms in {1,...,n} and w+x=y+z; also the number of (w,x,y,z) with all terms in {0,...,n} and |w-x|<=y. - Clark Kimberling, Jun 02 2012
The sequence is the third partial sum of (0, 1, 3, 4, 4, 4, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the number of join-irreducible elements in the Weyl group of type B_n with respect to the strong Bruhat order. - Rafael Mrden, Aug 26 2020
Number of unit octahedra contained in an n-scale octahedron composed of a tetrahedral-octahedral honeycomb. The number of unit tetrahedra in it is 8*A000292(n-1) = 4*(n^3 - n)/3. Also, the number of unit tetrahedra and unit octahedra contained in an n-scale tetrahedron composed of a tetrahedral-octahedral honeycomb is respectively A006527(n) = (n^3 + 2*n)/3 and A000292(n-1) = (n^3 - n)/6. - Jianing Song, Feb 24 2025

Examples

			G.f. = x + 6*x^2 + 19*x^3 + 44*x^4 + 85*x^5 + 146*x^6 + 231*x^7 + ...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 50.
  • H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Sums of 2 consecutive terms give A001845. Cf. A001844.
1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A022521.
Cf. A081277.
Row n=3 of A210391. - Alois P. Heinz, Mar 22 2012
Cf. A005408.
Cf. A002061.
Cf. A000292 (tetrahedral numbers), A000578 (cubes), A006566 (dodecahedral numbers), A006564 (icosahedral numbers).
Similar sequence: A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193(m=7), A099195 (m=8), A099196 (m=9), A099197 (m=10).

Programs

  • Haskell
    a005900 n = sum $ zipWith (*) odds $ reverse odds
                where odds = take n a005408_list
    a005900_list = scanl (+) 0 a001844_list
    -- Reinhard Zumkeller, Jun 16 2013, Apr 04 2012
    
  • Magma
    [n*(2*n^2+1)/3: n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015
    
  • Magma
    I:=[0,1,6,19]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Sep 12 2015
    
  • Maple
    al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n), r=0..d-1); end; [seq(be(3,n), n=0..100)];
    A005900:=(z+1)**2/(z-1)**4; # Simon Plouffe in his 1992 dissertation
    with(combinat): seq(fibonacci(4,2*n)/12, n=0..40); # Zerinvary Lajos, Apr 21 2008
  • Mathematica
    Table[(2n^3+n)/3, {n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,1,6,19},50] (* Harvey P. Dale, Oct 10 2013 *)
    CoefficientList[Series[x (1 + x)^2/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)
  • Maxima
    makelist(n*(2*n^2+1)/3, n, 0, 20); /* Martin Ettl, Jan 07 2013 */
    
  • PARI
    {a(n) = n*(2*n^2+1)/3};
    
  • PARI
    concat([0],Vec(x*(1 + x)^2/(1 - x)^4 + O(x^50))) \\ Indranil Ghosh, Mar 16 2017
    
  • Python
    def a(n): return n*(2*n*n + 1)//3
    print([a(n) for n in range(41)]) # Michael S. Branicky, Sep 03 2021

Formula

a(n) = 1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2. - Amarnath Murthy, May 28 2001
G.f.: x * (1 + x)^2 / (1 - x)^4. a(n) = -a(-n) = (2*n^3 + n) / 3.
a(n) = ( ((n+1)^5-n^5) - (n^5-(n-1)^5) )/30. - Xavier Acloque, Oct 17 2003
a(n) is the sum of the products pq, where p and q are both positive and odd and p + q = 2n, e.g., a(4) = 7*1 + 5*3 + 3*5 + 1*7 = 44. - Jon Perry, May 17 2005
a(n) = 4*binomial(n,3) + 4*binomial(n,2) + binomial(n,1). - Mitch Harris, Jul 06 2006
a(n) = binomial(n+2,3) + 2*binomial(n+1,3) + binomial(n,3), (this pair generalizes; see A014820, the 4-cross polytope numbers).
Sum_{n>=1} 1/a(n) = 3*gamma + 3*Psi((I*(1/2))*sqrt(2)) - (1/2)*(3*I)*Pi*coth((1/2)*Pi*sqrt(2)) - (1/2)*(3*I)*sqrt(2) = A175577, where I=sqrt(-1). - Stephen Crowley, Jul 14 2009
a(n) = A035597(n)/2. - J. M. Bergot, Jun 11 2012
a(n) = A000578(n) - 2*A000292(n-1) for n>0. - J. M. Bergot, Apr 05 2014
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Wesley Ivan Hurt, Sep 11 2015
E.g.f.: (1/3)*x*(3 + 6*x + 2*x^2)*exp(x). - Ilya Gutkovskiy, Mar 16 2017
a(n) = (A002061(A002061(n+1)) - A002061(A002061(n)))/6. - Daniel Poveda Parrilla, Jun 10 2017
a(n) = 6*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 6*log(2) - 4 = 1/(6 + 2/(6 + 6/(6 + ... + n*(n-1)/(6 + ...)))). See A142983. - Peter Bala, Mar 06 2024

A000447 a(n) = 1^2 + 3^2 + 5^2 + 7^2 + ... + (2*n-1)^2 = n*(4*n^2 - 1)/3.

Original entry on oeis.org

0, 1, 10, 35, 84, 165, 286, 455, 680, 969, 1330, 1771, 2300, 2925, 3654, 4495, 5456, 6545, 7770, 9139, 10660, 12341, 14190, 16215, 18424, 20825, 23426, 26235, 29260, 32509, 35990, 39711, 43680, 47905, 52394, 57155, 62196, 67525, 73150, 79079, 85320, 91881, 98770, 105995, 113564, 121485
Offset: 0

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Author

Keywords

Comments

4 times the variance of the area under an n-step random walk: e.g., with three steps, the area can be 9/2, 7/2, 3/2, 1/2, -1/2, -3/2, -7/2, or -9/2 each with probability 1/8, giving a variance of 35/4 or a(3)/4. - Henry Bottomley, Jul 14 2003
Number of standard tableaux of shape (2n-1,1,1,1) (n>=1). - Emeric Deutsch, May 30 2004
Also a(n) = (1/6)*(8*n^3-2*n), n>0: structured octagonal diamond numbers (vertex structure 9). Cf. A059722 = alternate vertex; A000447 = structured diamonds; and structured tetragonal anti-diamond numbers (vertex structure 9). Cf. A096000 = alternate vertex; A100188 = structured anti-diamonds. Cf. A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The n-th tetrahedral (or pyramidal) number is n(n+1)(n+2)/6. This sequence contains the tetrahedral numbers of A000292 obtained for n= 1,3,5,7,... (see A015219). - Valentin Bakoev, Mar 03 2009
Using three consecutive numbers u, v, w, (u+v+w)^3-(u^3+v^3+w^3) equals 18 times the numbers in this sequence. - J. M. Bergot, Aug 24 2011
This sequence is related to A070893 by A070893(2*n-1) = n*a(n)-sum(i=0..n-1, a(i)). - Bruno Berselli, Aug 26 2011
Number of integer solutions to 1-n <= x <= y <= z <= n-1. - Michael Somos, Dec 27 2011
Partial sums of A016754. - Reinhard Zumkeller, Apr 02 2012
Also the number of cubes in the n-th Haüy square pyramid. - Eric W. Weisstein, Sep 27 2017

Examples

			G.f. = x + 10*x^2 + 35*x^3 + 84*x^4 + 165*x^5 + 286*x^6 + 455*x^7 + 680*x^8 + ...
a(2) = 10 since (-1, -1, -1), (-1, -1, 0), (-1, -1, 1), (-1, 0, 0), (-1, 0, 1), (-1, 1, 1), (0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1) are the 10 solutions (x, y, z) of -1 <= x <= y <= z <= 1.
a(0) = 0, which corresponds to the empty sum.
		

References

  • G. Chrystal, Textbook of Algebra, Vol. 1, A. & C. Black, 1886, Chap. XX, Sect. 10, Example 2.
  • F. E. Croxton and D. J. Cowden, Applied General Statistics. 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1955, p. 742.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.
  • C. V. Durell, Advanced Algebra, Volume 1, G. Bell & Son, 1932, Exercise IIIe, No. 4.
  • L. B. W. Jolley, Summation of Series. 2nd ed., Dover, NY, 1961, p. 7.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
  • 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

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Column 1 in triangles A008956 and A008958.
A000447 is related to partitions of 2^n into powers of 2, as it is shown in the formula, example and cross-references of A002577. - Valentin Bakoev, Mar 03 2009

Programs

Formula

a(n) = binomial(2*n+1, 3) = A000292(2*n-1).
G.f.: x*(1+6*x+x^2)/(1-x)^4.
a(n) = -a(-n) for all n in Z.
a(n) = A000330(2*n)-4*A000330(n) = A000466(n)*n/3 = A000578(n)+A007290(n-2) = A000583(n)-2*A024196(n-1) = A035328(n)/3. - Henry Bottomley, Jul 14 2003
a(n+1) = (2*n+1)*(2*n+2)(2*n+3)/6. - Valentin Bakoev, Mar 03 2009
a(0)=0, a(1)=1, a(2)=10, a(3)=35, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, May 25 2012
a(n) = v(n,n-1), where v(n,k) is the central factorial numbers of the first kind with odd indices. - Mircea Merca, Jan 25 2014
a(n) = A005917(n+1) - A100157(n+1), where A005917 are the rhombic dodecahedral numbers and A100157 are the structured rhombic dodecahedral numbers (vertex structure 9). - Peter M. Chema, Jan 09 2016
For any nonnegative integers m and n, 8*(n^3)*a(m) + 2*m*a(n) = a(2*m*n). - Ivan N. Ianakiev, Mar 04 2017
E.g.f.: exp(x)*x*(1 + 4*x + (4/3)*x^2). - Wolfdieter Lang, Mar 11 2017
a(n) = A002412(n) + A016061(n-1), for n>0. - Bruce J. Nicholson, Nov 12 2017
From Amiram Eldar, Jan 04 2022: (Start)
Sum_{n>=1} 1/a(n) = 6*log(2) - 3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3 - 3*log(2). (End)

Extensions

Chrystal and Durell references from R. K. Guy, Apr 02 2004

A004006 a(n) = C(n,1) + C(n,2) + C(n,3), or n*(n^2 + 5)/6.

Original entry on oeis.org

0, 1, 3, 7, 14, 25, 41, 63, 92, 129, 175, 231, 298, 377, 469, 575, 696, 833, 987, 1159, 1350, 1561, 1793, 2047, 2324, 2625, 2951, 3303, 3682, 4089, 4525, 4991, 5488, 6017, 6579, 7175, 7806, 8473, 9177, 9919, 10700, 11521, 12383, 13287, 14234, 15225
Offset: 0

Views

Author

Albert D. Rich (Albert_Rich(AT)msn.com)

Keywords

Comments

3-dimensional analog of centered polygonal numbers.
The Burnside group B(3,n) has order 3^a(n).
Answer to the question: if you have a tall building and 3 plates and you need to find the highest story, a plate thrown from which does not break, what is the number of stories you can handle given n tries? - Leonid Broukhis, Oct 24 2000
Equals row sums of triangle A144329 starting with "1". - Gary W. Adamson, Sep 18 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=A[i,i]:=1, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=4, a(n-1)=-coeff(charpoly(A,x),x^(n-3)). - Milan Janjic, Jan 24 2010
From J. M. Bergot, Aug 03 2011: (Start)
If one formed the 3 X 3 square
| n | n+1 | n+2 |
| n+3 | n+4 | n+5 |
| n+6 | n+7 | n+8 |
and found the sum of the horizontal products n*(n + 1)*(n + 2) + (n + 3)*(n + 4)*(n + 5) + (n + 6)*(n + 7)*(n + 8) and added the sum of the vertical products n*(n + 3)*(n + 6) + (n + 1)*(n + 4)*(n + 7) + (n + 2)*(n + 5)(n + 8) one gets 6*n^3 + 72*n^2 + 318*n + 504. This will give 36 times the values of all the terms in this sequence. (End)
a(n) is divisible by n for n congruent to {1,5} mod 6. (see A007310). - Gary Detlefs, Dec 08 2011
From Beimar Naranjo, Feb 22 2024: (Start)
Number of compositions with at most three parts and sum at most n.
Also the number of compositions with at most one part distinct from 1 and with a sum at most n. (End)
a(n) is the number of strings of length n defined on {0, 1, 2, 3} that contain one 1 and any number of 0's, or two 2's and any number of 0's, or three 3's and any number of 0's. For example, a(6) = 41 since the strings are the 20 permutations of 333000, the 15 permutations of 220000 and the 6 permutations of 100000. - Enrique Navarrete, Jun 18 2025

Examples

			G.f. = x + 3*x^2 + 7*x^3 + 14*x^4 + 25*x^5 + 41*x^6 + 63*x^7 + 92*x^8 + ... - _Michael Somos_, Dec 29 2019
		

References

  • W. Magnus, A. Karrass and D. Solitar, Combinatorial Group Theory, Wiley, 1966, see p. 380.

Crossrefs

Cf. A051576, A055795, A006552. Differences give A000217 + 1 or A000124.
1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Programs

Formula

G.f.: x*(1-x+x^2)/(1-x)^4.
E.g.f.: x*(1 + x/2 + x^2/6) * exp(x).
a(-n) = -a(n).
a(n) = binomial(n+2,n-1) - binomial(n,n-2). - Zerinvary Lajos, May 11 2006
Euler transform of length 6 sequence [3, 1, 1, 0, 0, -1]. - Michael Somos, May 04 2007
Starting (1, 3, 7, 14, ...) = binomial transform of [1, 2, 2, 1, 0, 0, 0, ...]. - Gary W. Adamson, Apr 24 2008
a(0)=0, a(1)=1, a(2)=3, a(3)=7, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Aug 21 2011
a(n+1) = A000292(n) + n + 1. - Reinhard Zumkeller, Mar 31 2012
a(n) = 2*a(n-1) + (n-1) - a(n-2) with a(0) = 0, a(1) = 1. - Richard R. Forberg, Jan 23 2014
a(n) = Sum_{i=1..n} binomial(n-2i,2). - Wesley Ivan Hurt, Nov 18 2017
a(n) = n + Sum_{k=0..n} k*(n-k). - Gionata Neri, May 19 2018
a(n) = Sum_{k=0..n-1} A000124(k). - Torlach Rush, Aug 05 2018
G.f.: ((1 - x^5)/(1 - x)^5 - 1)/5. - Michael Somos, Dec 29 2019
G.f.: g(f(x)), where g is g.f. of A001477 and f is g.f. of A128834. - Oboifeng Dira, Jun 21 2020
Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i*sqrt(5)) + polygamma(0, 1+i*sqrt(5)))/5 = 1.6787729555834452106286261834348972248... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

A006527 a(n) = (n^3 + 2*n)/3.

Original entry on oeis.org

0, 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, 28424, 30405
Offset: 0

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Keywords

Comments

Number of ways to color vertices (or edges) of a triangle using <= n colors, allowing only rotations.
Also: dot_product (1,2,...,n)*(2,3,...,n,1), n >= 0. - Clark Kimberling
Start from triacid and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink below on chemistry. - Robert G. Wilson v, Aug 02 2002
Starting with offset 1 = row sums of triangle A158822 and binomial transform of (1, 3, 4, 2, 0, 0, 0, ...). - Gary W. Adamson, Mar 28 2009
One-ninth of sum of three consecutive cubes: a(n) = ((n-1)^3 + n^3 + (n+1)^3)/9. - Zak Seidov, Jul 22 2013
For n > 2, number of different cubes, formed after splitting a cube in color C_1, by parallel planes in the colors C_2, C_3, ..., C_n in three spatial dimensions (in the order of the colors from a fixed vertex). Generally, in a large hypercube n^d is f(n,d) = C(n+d-1, d) + C(n, d) different small hypercubes. See below for my formula a(n) = f(n,3). - Thomas Ordowski, Jun 15 2014
a(n) is a square for n = 1, 2 & 24; and for no other values up to 10^7 (see M. Gardner). - Michel Marcus, Sep 06 2015
Number of unit tetrahedra contained in an n-scale tetrahedron composed of a tetrahedral-octahedral honeycomb. - Jason Pruski, Aug 23 2017

References

  • M. Gardner, New Mathematical Diversions from Scientific American. Simon and Schuster, NY, 1966, p. 246.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 483.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Column 1 of triangle A094414. Row 6 of the array in A107735.
Cf. A000292 (unoriented), A000292(n-2) (chiral), A000290 (achiral) triangle colorings.
Row 2 of A324999 (simplex vertices and facets) and A327083 (simplex edges and ridges).

Programs

  • Haskell
    a006527 n = n * (n ^ 2 + 2) `div` 3  -- Reinhard Zumkeller, Jan 06 2014
  • Magma
    [(n^3 + 2*n)/3: n in [0..50]]; // Vincenzo Librandi, May 15 2011
    
  • Maple
    A006527:=z*(1+z**2)/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation
    with(combinat):seq(lcm(fibonacci(4,n),fibonacci(2,n))/3,n=0..42); # Zerinvary Lajos, Apr 20 2008
  • Mathematica
    Table[ (n^3 + 2*n)/3, {n, 0, 45} ]
    LinearRecurrence[{4,-6,4,-1},{0,1,4,11},46] (* or *) CoefficientList[ Series[(x+x^3)/(x-1)^4,{x,0,49}],x] (* Harvey P. Dale, Jun 13 2011 *)
  • PARI
    a(n)=n*(n^2+2)/3 \\ Charles R Greathouse IV, Jul 25 2011
    

Formula

a(0)=0, a(1)=1, a(2)=4, a(3)=11; for n > 3, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Jun 13 2011
From Paul Barry, Mar 13 2003: (Start)
a(n) = 2*binomial(n+1, 3) + binomial(n, 1).
G.f.: x*(1+x^2)/(1-x)^4. (End)
a(n) = A000292(n) + A000292(n-2). - Alexander Adamchuk, May 20 2006
a(n) = n*A059100(n)/3. - Lekraj Beedassy, Feb 06 2007
a(n) = A054602(n)/3. - Zerinvary Lajos, Apr 20 2008
a(n) = ( n + Sum_{i=1..n} A177342(i) )/(n+1), with n > 0. - Bruno Berselli, May 19 2010
a(n) = A002264(A000578(n) + A005843(n)). - Reinhard Zumkeller, Jun 16 2011
a(n) = binomial(n+2, 3) + binomial(n, 3). - Thomas Ordowski, Jun 15 2014
a(n) = A000292(n) - A000292(-n). - Bruno Berselli, Sep 22 2016
E.g.f.: (x/3)*(3 + 3*x + x^2)*exp(x). - G. C. Greubel, Sep 01 2017
From Robert A. Russell, Oct 20 2020: (Start)
a(n) = 1*C(n,1) + 2*C(n,2) + 2*C(n,3), where the coefficient of C(n,k) is the number of oriented triangle colorings using exactly k colors.
a(n) = 2*A000292(n) - A000290(n) = 2*A000292(n-2) + A000290(n). (End)
Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i*sqrt(2)) + polygamma(0, 1+i*sqrt(2)))/4 = 1.45245201414472469745354677573358867... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

Extensions

More terms from Alexander Adamchuk, May 20 2006
Corrected and replaced 5th formula from Harvey P. Dale, Jun 13 2011
Deleted an erroneous comment. - N. J. A. Sloane, Dec 10 2018

A007588 Stella octangula numbers: a(n) = n*(2*n^2 - 1).

Original entry on oeis.org

0, 1, 14, 51, 124, 245, 426, 679, 1016, 1449, 1990, 2651, 3444, 4381, 5474, 6735, 8176, 9809, 11646, 13699, 15980, 18501, 21274, 24311, 27624, 31225, 35126, 39339, 43876, 48749, 53970, 59551, 65504, 71841, 78574, 85715, 93276, 101269, 109706, 118599, 127960
Offset: 0

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Keywords

Comments

Also as a(n)=(1/6)*(12*n^3-6*n), n>0: structured hexagonal anti-diamond numbers (vertex structure 13) (Cf. A005915 = alternate vertex; A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The only known square stella octangula number for n>1 is a(169) = 169*(2*169^2 - 1) = 9653449 = 3107^2. - Alexander Adamchuk, Jun 02 2008
Ljunggren proved that 9653449 = (13*239)^2 is the only square stella octangula number for n>1. See A229384 and the Wikipedia link. - Jonathan Sondow, Sep 30 2013
4*A007588 = A144138(ChebyshevU[3,n]). - Vladimir Joseph Stephan Orlovsky, Jun 30 2011
If A016813 is regarded as a regular triangle (with leading terms listed in A001844), a(n) provides the row sums of this triangle: 1, 5+9=14, 13+17+21=51 and so on. - J. M. Bergot, Jul 05 2013
Shares its digital root, A267017, with n*(n^2 + 1)/2 ("sum of the next n natural numbers" see A006003). - Peter M. Chema, Aug 28 2016

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 51.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.
  • W. Ljunggren, Zur Theorie der Gleichung x^2 + 1 = Dy^4, Avh. Norske Vid. Akad. Oslo. I. 1942 (5): 27.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Backwards differences give star numbers A003154: A003154(n)=a(n)-a(n-1).
1/12*t*(n^3-n)+ n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A001653 = Numbers n such that 2*n^2 - 1 is a square.
a(169) = (A229384(3)*A229384(4))^2.

Programs

Formula

G.f.: x*(1+10*x+x^2)/(1-x)^4.
a(n) = n*A056220(n).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Harvey P. Dale, Sep 16 2011
From Ilya Gutkovskiy, Jul 02 2016: (Start)
E.g.f.: x*(1 + 6*x + 2*x^2)*exp(x).
Dirichlet g.f.: 2*zeta(s-3) - zeta(s-1). (End)
a(n) = A004188(n) + A135503(n). - Miquel Cerda, Dec 25 2016
a(n) = A061317(n) - A005843(n) = A062392(n) - A062392(n-1). - J.S. Seneschal, Jul 01 2025

Extensions

In the formula given in the 1995 Encyclopedia of Integer Sequences, the second 2 should be an exponent.

A004068 Number of atoms in a decahedron with n shells.

Original entry on oeis.org

0, 1, 7, 23, 54, 105, 181, 287, 428, 609, 835, 1111, 1442, 1833, 2289, 2815, 3416, 4097, 4863, 5719, 6670, 7721, 8877, 10143, 11524, 13025, 14651, 16407, 18298, 20329, 22505, 24831, 27312, 29953, 32759, 35735, 38886, 42217, 45733, 49439
Offset: 0

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Author

Albert D. Rich (Albert_Rich(AT)msn.com)

Keywords

Comments

Also as a(n)=(n/6)*(5*n^2+1), n>0: structured pentagonal diamond numbers (vertex structure 6) (cf. A081436 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of atoms in decahedron with n shells, number = 5/6*(n^3) + 1/6*(n) (T. P. Martin, Shells of atoms, eq.(3)). - Brigitte Stepanov, Jul 02 2011
a(n+1) is the number of triples (w,x,y) having all terms in {0,...,n} and x+y >= w. - Clark Kimberling, Jun 14 2012
a(n) = Sum_{k=1..n} A215630(n,k) for n > 0. - Reinhard Zumkeller, Nov 11 2012
a(n) - a(n-2) = A010001(n-1), for n>1. - K. G. Stier, Dec 21 2012
a(n) is also a figurate number representing a cube of side n with a vertex cut off by a tetrahedron of side n-1. As such, a(n) = A000578(n) - A000292(n-1), n > 0. - Jean M. Morales, Aug 11 2013
The sequence starting with 1 is the third partial sum of (1, 4, 5, 5, 5, ...) and the binomial transform of (1, 6, 10, 5, 0, 0, 0, ...). - Gary W. Adamson, Sep 27 2015

Crossrefs

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Programs

Formula

a(n) = 5*binomial(n + 1, 3) + binomial(n, 1).
a(n) = 5*n^3/6 + n/6.
a(n) = Sum_{i=0..n-1} A005891(i). - Xavier Acloque, Oct 08 2003
G.f.: x*(1+3*x+x^2) / (1-x)^4. - R. J. Mathar, Jun 05 2011
E.g.f.: (x/6)*(5x^2 + 15x + 6)*exp(x). - G. C. Greubel, Sep 27 2015
Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i/sqrt(5)) + polygamma(0, 1+i/sqrt(5))) = 1.233988011257952852492845364799197179252... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

Extensions

Typo in definition corrected by Jean M. Morales, Aug 11 2013
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