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 1-5 of 5 results.

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

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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

A053676 Let Oc(n) = A005900(n) = n-th octahedral number. Consider all integer triples (i,j,k), j >= k > 0, with Oc(i) = Oc(j)+Oc(k), ordered by increasing i; sequence gives i values.

Original entry on oeis.org

7, 41, 465, 2732, 3005, 20648, 48125, 94396, 129299, 282931, 789281, 835050, 1241217, 1292143, 1521647, 1603655, 2756953, 4847702, 5128447, 6242598
Offset: 1

Views

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Feb 16 2000

Keywords

Comments

a(21) > 10^7. - Donovan Johnson, Sep 29 2010

Examples

			Oc(7) = 231 = Oc(6) + Oc(5); Oc(41) = 45961 = Oc(40) + Oc(17); Oc(465) = 67029905 = Oc(454) + Oc(191)
		

References

  • Pollock, F. "On the Extension of the Principle of Fermat's Theorem of the Polygonal Numbers to the Higher Orders of Series Whose Ultimate Differences Are Constant. With a New Theorem Proposed, Applicable to All the Orders." Abs. Papers Commun. Roy. Soc. London 5, 922-924, 1843-1850.
  • Dickson, L. E., History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Dover, 2005, cites the Pollock reference.

Crossrefs

Cf. A005900, A053677 (j values), A053678 (k values).

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), May 07 2001
a(13)-a(16) from Donovan Johnson, Jun 21 2010
a(17)-a(20) from Donovan Johnson, Sep 29 2010

A053677 Let Oc(n) = A005900(n) = n-th octahedral number. Consider all integer triples (i,j,k), j >= k > 0, with Oc(i) = Oc(j)+Oc(k), ordered by increasing i; sequence gives j values.

Original entry on oeis.org

6, 40, 454, 2600, 2586, 20175, 48103, 93097, 105805, 265195, 755100, 803007, 1211000, 1111974, 1493421, 1499160, 2622000, 4309280, 5127195, 5574139
Offset: 1

Views

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Feb 16 2000

Keywords

Examples

			Oc(7) = 231 = Oc(6) + Oc(5); Oc(41) = 45961 = Oc(40) + Oc(17); Oc(465) = 67029905 = Oc(454) + Oc(191)
		

Crossrefs

i values are A053676 and k values are A053678.

Programs

  • Mathematica
    (* This is just a check of j-values, given i-values *) A053676 = {7, 41, 465, 2732, 3005, 20648, 48125, 94396, 129299, 282931, 789281, 835050, 1241217, 1292143, 1521647, 1603655, 2756953, 4847702, 5128447, 6242598}; r[i_] := Reduce[0 < k <= j && 2*i^3 + i == 2*j^3 + j + 2*k^3 + k, {j, k}, Integers]; A053677 = Table[res = {i, j, k} /. ToRules[r[i]]; Print[res]; res, {i, A053676}][[All, 2]] (* Jean-François Alcover, Dec 07 2012 *)

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), May 07 2001
a(13)-a(16) from Donovan Johnson, Jun 21 2010
a(17)-a(20) from Donovan Johnson, Sep 29 2010

A133330 Sums of exactly three positive octahedral numbers A005900.

Original entry on oeis.org

3, 8, 13, 18, 21, 26, 31, 39, 44, 46, 51, 56, 57, 64, 69, 82, 87, 89, 92, 94, 97, 105, 107, 110, 123, 130, 132, 135, 148, 153, 158, 166, 171, 173, 176, 184, 189, 191, 196, 209, 214, 232, 233, 234, 237, 238, 243, 250, 251, 255, 256, 269, 275, 276, 281, 293, 294
Offset: 1

Views

Author

Jonathan Vos Post, Oct 18 2007

Keywords

References

  • Dickson, L. E. History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Dover, 2005, cites the Pollock reference.
  • Pollock, F. "On the Extension of the Principle of Fermat's Theorem of the Polygonal Numbers to the Higher Orders of Series Whose Ultimate Differences Are Constant. With a New Theorem Proposed, Applicable to All the Orders." Abs. Papers Commun. Roy. Soc. London 5, 922-924, 1843-1850.

Crossrefs

Programs

  • Mathematica
    lim = 300; oc[n_] := (2*n^3 + n)/3; nmax = Floor[Solve[oc[n] + oc[1] + oc[1] == lim, n][[1, 1, 2]]]; t = Table[ oc[n], {n, nmax}]; Select[ Union[ Flatten[ Outer[ Plus, t, t, t]]], # <= lim &] (* Jean-François Alcover, Sep 08 2011 *)

A131280 Sums of exactly 4 positive octahedral numbers A005900.

Original entry on oeis.org

4, 9, 14, 19, 22, 24, 27, 32, 37, 40, 45, 47, 50, 52, 57, 58, 62, 63, 65, 70, 75, 76, 83, 88, 90, 93, 95, 98, 100, 101, 103, 106, 108, 111, 113, 116, 124, 126, 129, 131, 133, 136, 138, 141, 142, 149, 151, 154, 159, 164, 167, 172, 174, 176, 177, 179, 182, 185, 190
Offset: 1

Views

Author

Jonathan Vos Post, Oct 21 2007

Keywords

Comments

Pollock (1850) conjectured that every number is the sum of at most 7 octahedral numbers. Which octahedral numbers are themselves the sum of exactly 4 positive octahedral numbers? To begin with, Oc(3) = Oc(2) + Oc(2) + Oc(2) + Oc(1) = 6 + 6 + 6 + 1 = 19.

References

  • Dickson, L. E. History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Dover, 2005, cites the Pollock reference.
  • Pollock, F. "On the Extension of the Principle of Fermat's Theorem of the Polygonal Numbers to the Higher Orders of Series Whose Ultimate Differences Are Constant. With a New Theorem Proposed, Applicable to All the Orders." Abs. Papers Commun. Roy. Soc. London 5, 922-924, 1843-1850.

Crossrefs

Programs

  • Mathematica
    With[{octs=Table[(2n^3+n)/3,{n,10}]},Take[Union[Total/@Tuples[octs,4]], 60]] (* Harvey P. Dale, Nov 26 2013 *)
Showing 1-5 of 5 results.