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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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

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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

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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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

A100145 Structured great rhombicosidodecahedral numbers.

Original entry on oeis.org

1, 120, 579, 1600, 3405, 6216, 10255, 15744, 22905, 31960, 43131, 56640, 72709, 91560, 113415, 138496, 167025, 199224, 235315, 275520, 320061, 369160, 423039, 481920, 546025, 615576, 690795, 771904, 859125, 952680, 1052791, 1159680
Offset: 1

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Author

James A. Record (james.record(AT)gmail.com), Nov 07 2004

Keywords

Comments

Structured polyhedral numbers are a type of figurate polyhedral numbers. Structurate polyhedra differ from regular figurate polyhedra by having appropriate figurate polygonal faces at any iteration, i.e., a regular truncated octahedron, n=2, would have 7 points on its hexagonal faces, whereas a structured truncated octahedron, n=2, would have 6 points - just as a hexagon, n=2, would have. Like regular figurate polygons, structured polyhedra seem to originate at a vertex and since many polyhedra have different vertices (a pentagonal diamond has 2 "polar" vertices with 5 adjacent vertices and 5 "equatorial" vertices with 4 adjacent vertices), these polyhedra have multiple structured number sequences, dependent on the "vertex structures" which are each equal to the one vertex itself plus its adjacent vertices. For polystructurate polyhedra the notation, structured polyhedra (vertex structure x) is used to differentiate between alternate vertices, where VS stands for vertex structure.

Crossrefs

Cf. A051673, A100146 through A100156 - structured Archimedean solids; A100157 through A100175 - structured Catalan solids; A100147 - structured prisms; A000447 - structured diamonds; A100185 - structured anti-prisms; and A100188 - structured anti-diamonds.

Programs

Formula

a(n) = (1/6)*(222*n^3 - 312*n^2 + 96*n).
From Jaume Oliver Lafont, Sep 08 2009: (Start)
a(n) = (1+(n-1))*(1+22*(n-1)+37*(n-1)^2);
G.f.: x*(1+116*x+105*x^2)/(1-x)^4. (End)
E.g.f.: exp(x)*x*(1 + 59*x + 37*x^2). - Stefano Spezia, Jun 06 2025

Extensions

Corrected by T. D. Noe, Oct 25 2006

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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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.

A004466 a(n) = n*(5*n^2 - 2)/3.

Original entry on oeis.org

0, 1, 12, 43, 104, 205, 356, 567, 848, 1209, 1660, 2211, 2872, 3653, 4564, 5615, 6816, 8177, 9708, 11419, 13320, 15421, 17732, 20263, 23024, 26025, 29276, 32787, 36568, 40629, 44980, 49631, 54592
Offset: 0

Views

Author

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

Keywords

Comments

3-dimensional analog of centered polygonal numbers.
Also as a(n)=(1/6)*(10*n^3-4*n), n>0: structured pentagonal anti-diamond numbers (vertex structure 11) (Cf. A051673 = alternate vertex A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
a(n+1)-10*a(n) = (n+1)*(5*(n+1)^2-2)/3 - (10n(n+1)(n+2)/6) = n. The unit digits are 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,... . - Eric Desbiaux, Aug 18 2008

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.

Crossrefs

Cf. A062786 (first differences), A264853 (partial sums).
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+8*x+x^2)/(1-x)^4. - Colin Barker, Jan 08 2012
E.g.f.: (x/3)*(3 + 15*x + 5*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A063521 a(n) = n*(7*n^2-4)/3.

Original entry on oeis.org

0, 1, 16, 59, 144, 285, 496, 791, 1184, 1689, 2320, 3091, 4016, 5109, 6384, 7855, 9536, 11441, 13584, 15979, 18640, 21581, 24816, 28359, 32224, 36425, 40976, 45891, 51184, 56869, 62960, 69471, 76416, 83809, 91664, 99995, 108816, 118141, 127984, 138359, 149280, 160761
Offset: 0

Views

Author

N. J. A. Sloane, Aug 02 2001

Keywords

Comments

Also as a(n)=(1/6)*(14*n^3-8*n), n>0: structured heptagonal anti-diamond numbers (vertex structure 15) (Cf. A100186 = alternate vertex; A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004

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

G.f.: x*(1+12*x+x^2)/(1-x)^4. - Colin Barker, Jan 10 2012
E.g.f.: (x/3)*(3 + 21*x + 7*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A006484 a(n) = n*(n + 1)*(n^2 - 3*n + 5)/6.

Original entry on oeis.org

0, 1, 3, 10, 30, 75, 161, 308, 540, 885, 1375, 2046, 2938, 4095, 5565, 7400, 9656, 12393, 15675, 19570, 24150, 29491, 35673, 42780, 50900, 60125, 70551, 82278, 95410, 110055, 126325, 144336, 164208, 186065, 210035, 236250, 264846, 295963, 329745, 366340
Offset: 0

Views

Author

Dennis S. Kluk (mathemagician(AT)ameritech.net)

Keywords

Comments

Structured meta-pyramidal numbers, the n-th number from an n-gonal pyramidal number sequence. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The Gi4 triangle sums of A139600 are given by the terms of this sequence. For the definitions of the Gi4 and other triangle sums see A180662. - Johannes W. Meijer, Apr 29 2011

References

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

Crossrefs

Cf. other meta sequences: A100177: prism; A000447: "polar" diamond; A059722: "equatorial diamond"; A100185: anti-prism; A100188: "polar" anti-diamond; and A100189: "equatorial" anti-diamond. Cf. A100145 for more on structured numbers.
Cf. A000332.

Programs

Formula

a(n) = (1/6)*(n^4 - 2*n^3 + 2*n^2 + 5*n). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
a(n) = binomial(n+3,4) - 2*binomial(n+2,4) + 5*binomial(n+1,4). - Johannes W. Meijer, Apr 29 2011

A063523 a(n) = n*(8*n^2 - 5)/3.

Original entry on oeis.org

0, 1, 18, 67, 164, 325, 566, 903, 1352, 1929, 2650, 3531, 4588, 5837, 7294, 8975, 10896, 13073, 15522, 18259, 21300, 24661, 28358, 32407, 36824, 41625, 46826, 52443, 58492, 64989, 71950, 79391, 87328, 95777, 104754, 114275, 124356, 135013, 146262, 158119, 170600
Offset: 0

Views

Author

N. J. A. Sloane, Aug 02 2001

Keywords

Comments

Also as a(n)=(1/6)*(16*n^3-10*n), n>0: structured octagonal anti-diamond numbers (vertex structure 17) (Cf. A100187 = alternate vertex; A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004

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

  • Mathematica
    Table[n(8n^2-5)/3,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
    LinearRecurrence[{4,-6,4,-1},{0,1,18,67},81] (* or *) CoefficientList[ Series[ (x+14 x^2+x^3)/(x-1)^4,{x,0,80}],x] (* Harvey P. Dale, Jul 11 2011 *)
  • PARI
    a(n) = n*(8*n^2 - 5)/3 \\ Harry J. Smith, Aug 25 2009

Formula

a(0)=0, a(1)=1, a(2)=18, a(3)=67, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)- a(n-4). - Harvey P. Dale, Jul 11 2011
G.f.: (x+14*x^2+x^3)/(x-1)^4. - Harvey P. Dale, Jul 11 2011
E.g.f.: (x/3)*(3 + 24*x + 8*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A005915 Hexagonal prism numbers: a(n) = (n + 1)*(3*n^2 + 3*n + 1).

Original entry on oeis.org

1, 14, 57, 148, 305, 546, 889, 1352, 1953, 2710, 3641, 4764, 6097, 7658, 9465, 11536, 13889, 16542, 19513, 22820, 26481, 30514, 34937, 39768, 45025, 50726, 56889, 63532, 70673, 78330, 86521, 95264, 104577, 114478, 124985, 136116, 147889, 160322, 173433, 187240
Offset: 0

Views

Author

Keywords

Comments

Also as a(n) = (1/6)*(18*n^3 - 18*n^2 + 6*n), n>0: structured rhombic dodecahedral numbers (vertex structure 7) (A100157 = alternate vertex); structured tetrakis hexahedral numbers (vertex structure 7) (Cf. A100174 = alternate vertex); and structured hexagonal anti-diamond numbers (vertex structure 7) (Cf. A007588 = alternate vertex) (Cf. A100188 = structured anti-diamonds). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
a(n) is the number of 4-tuples (w,x,y,z) with all terms in {0,...,n} and w=x or x=y or y=z. - Clark Kimberling, May 31 2012

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), pp. 4545-4558.

Crossrefs

Cf. A143804.
Cf. A260260 (comment). - Bruno Berselli, Jul 22 2015

Programs

  • Magma
    [(n + 1)*(3*n^2 + 3*n + 1): n in [0..50]]; // Vincenzo Librandi, May 16 2011
    
  • Maple
    A005915:=(1+10*z+7*z**2)/(z-1)**4; # Conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    Table[(n+1)(3n^2+3n+1),{n,0,50}]  (* Harvey P. Dale, Mar 31 2011 *)
    LinearRecurrence[{4,-6,4,-1},{1,14,57,148},50] (* Harvey P. Dale, Jun 25 2011 *)
  • PARI
    a(n) = (n + 1)*(3*n^2 + 3*n + 1);

Formula

a(n) = (n+1)^3 + 6*(n*(n+1)*(2*n+1)/6). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=1, a(1)=14, a(2)=57, a(3)=148. - Harvey P. Dale, Jun 25 2011
G.f.: (1+10*x+7*x^2)/(1-x)^4. - Harvey P. Dale, Jun 25 2011
Equals row sums of triangle A143804 and binomial transform of [1, 13, 30, 18, 0, 0, 0, ...]. - Gary W. Adamson, Sep 01 2008
2*a(n+1) = A213829(n). - Clark Kimberling, Jul 04 2012
E.g.f.: exp(x)*(1 + x)*(1 + 12*x + 3*x^2). - Elmo R. Oliveira, Aug 04 2025

Extensions

More terms from James Sellers, Dec 24 1999

A051673 Cubic star numbers: a(n) = n^3 + 4*Sum_{i=0..n-1} i^2.

Original entry on oeis.org

0, 1, 12, 47, 120, 245, 436, 707, 1072, 1545, 2140, 2871, 3752, 4797, 6020, 7435, 9056, 10897, 12972, 15295, 17880, 20741, 23892, 27347, 31120, 35225, 39676, 44487, 49672, 55245, 61220, 67611, 74432, 81697, 89420, 97615, 106296, 115477, 125172
Offset: 0

Views

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)

Keywords

Comments

Also as a(n) = (1/6)*(14*n^3 - 12*n^2 + 4*n), n>0: structured cubeoctahedral numbers (vertex structure 7); and structured pentagonal anti-diamond numbers (vertex structure 7) (cf. A004466 = alternate vertex) (cf. A100188 = structured anti-diamonds). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Starting with offset 1 = binomial transform of [1, 11, 24, 14, 0, 0, 0, ...]. - Gary W. Adamson, Aug 05 2009
This is prime for a(3) = 47. The subsequence of semiprimes begins: 707, 7435, 10897, 20741, 115477, 341797, 825091, 897097, no more through a(100). - Jonathan Vos Post, May 27 2010

Examples

			a(51) = 51*(51*(7*51-6)+2)/3 = 304351 = 17 * 17903 is semiprime. - _Jonathan Vos Post_, May 27 2010
		

References

  • T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Crossrefs

Programs

Formula

a(n) = n*(n*(7*n-6) + 2)/3.
G.f.: x*(1+8*x+5*x^2)/(1-x)^4. - Bruno Berselli, May 12 2011
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=0, a(1)=1, a(2)=12, a(3)=47. - Harvey P. Dale, Jul 22 2011
From Reinhard Zumkeller, Jul 25 2012: (Start)
a(n) = A214659(n) - A002378(n).
a(n) = Sum_{k=1..n} A214661(n, k), for n > 0 (row sums). (End)
E.g.f.: (x/3)*(3 + 15*x + 7*x^2)*exp(x). - G. C. Greubel, Mar 10 2024

Extensions

Corrected by T. D. Noe, Nov 01 2006, Nov 08 2006
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