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 93 results. Next

A329635 Numbers that are sums of consecutive centered octahedral numbers (A001845).

Original entry on oeis.org

1, 7, 8, 25, 32, 33, 63, 88, 95, 96, 129, 192, 217, 224, 225, 231, 360, 377, 423, 448, 455, 456, 575, 608, 737, 800, 825, 832, 833, 952, 1159, 1183, 1312, 1375, 1400, 1407, 1408, 1561, 1785, 1992, 2016, 2047, 2145, 2208, 2233, 2240, 2241, 2567, 2625, 2720, 2944
Offset: 1

Views

Author

Ilya Gutkovskiy, Nov 18 2019

Keywords

Crossrefs

A188050 a(n) = A016755(n) - A001845(n).

Original entry on oeis.org

0, 20, 100, 280, 600, 1100, 1820, 2800, 4080, 5700, 7700, 10120, 13000, 16380, 20300, 24800, 29920, 35700, 42180, 49400, 57400, 66220, 75900, 86480, 98000, 110500, 124020, 138600, 154280, 171100, 189100, 208320, 228800, 250580, 273700
Offset: 0

Views

Author

Damien Pras, Mar 19 2011

Keywords

Comments

A016755 are odd cubes and A001845 are centered octahedral numbers, so the sequence might be regarded as odd cubes without their octahedral content.
A000330 are square pyramidal numbers.

Crossrefs

Programs

  • Magma
    A016755:=func< n | (2*n+1)^3 >; A001845:=func< n | (2*n+1)*(2*n^2+2*n+3)/3 >; [ A016755(n)-A001845(n): n in [0..40] ]; // Klaus Brockhaus, Mar 20 2011
  • Maple
    (10/3)*n*(n+1)*(2*n+1)
  • Mathematica
    10n(n+1)(2n+1)/3
    LinearRecurrence[{4,-6,4,-1},{0,20,100,280},40] (* Harvey P. Dale, Jul 18 2016 *)

Formula

a(n) = (10/3)*n*(n + 1)*(2*n + 1).
a(n) = 20 * A000330(n).
G.f.: 20*x*(1+x)/(1-x)^4. - Klaus Brockhaus, Mar 20 2011

A342353 Number of ways to choose 3 points that form an equilateral triangle from the A001845(n) points in a regular octahedral grid of size n.

Original entry on oeis.org

0, 8, 80, 416, 1512, 4216, 10000, 21256
Offset: 0

Views

Author

Peter Kagey, Mar 08 2021

Keywords

Comments

The octahedral grid of side length n is the set of points (x,y,z) in Z^3 such that |x|+|y|+|z| <= n. The number of points in this grid is given by A001845(n).

Examples

			For n = 1, the a(1) = 8 equilateral triangles are given by the convex hulls of {(+-1,0,0),(0,+-1,0),(0,0,+-1)}.
		

Crossrefs

Cf. A001845.
Cf. A102698 and A334581 are analogous for the cubic grid and tetrahedral grid respectively.

A005897 a(n) = 6*n^2 + 2 for n > 0, a(0)=1.

Original entry on oeis.org

1, 8, 26, 56, 98, 152, 218, 296, 386, 488, 602, 728, 866, 1016, 1178, 1352, 1538, 1736, 1946, 2168, 2402, 2648, 2906, 3176, 3458, 3752, 4058, 4376, 4706, 5048, 5402, 5768, 6146, 6536, 6938, 7352, 7778, 8216, 8666, 9128, 9602, 10088, 10586
Offset: 0

Views

Author

Keywords

Comments

Number of points on surface of 3-dimensional cube in which each face has a square grid of dots drawn on it (with n+1 points along each edge, including the corners).
Coordination sequence for b.c.c. lattice.
Also coordination sequence for 3D uniform tiling with tile an equilateral triangular prism. - N. J. A. Sloane, Feb 06 2018
Binomial transform of [1, 7, 11, 1, -1, 1, -1, 1, ...]. - Gary W. Adamson, Oct 22 2007
First differences of A005898. - Jonathan Vos Post, Feb 06 2011
Apart from the first term, numbers of the form (r^2+2*s^2)*n^2+2 = (r*n)^2+(s*n-1)^2+(s*n+1)^2: in this case is r=2, s=1. After 8, all terms are in A000408. - Bruno Berselli, Feb 07 2012
For n > 0, the sequence of last digits (i.e., a(n) mod 10) is (8, 6, 6, 8, 2) repeating forever. - M. F. Hasler, Apr 05 2016
Number of cubes of edge length 1 required to make a hollow cube of edge length n+1. - Peter M. Chema, Apr 01 2017
a(n) is the number of pieces on the outside of a (n+1) X (n+1) X (n+1) Rubik's cube. For n > 0: corners = 8, edges = 12*(n-1), center pieces = 6*(n-1)^2. - Demilade Runsewe, Jan 08 2025

Examples

			For n = 1 we get the 8 corners of the cube; for n = 2 each face has 9 points, for a total of 8 + 12 + 6 = 26.
		

References

  • H. S. M. Coxeter, "Polyhedral numbers," in R. S. Cohen et al., editors, For Dirk Struik. Reidel, Dordrecht, 1974, pp. 25-35.
  • Gmelin Handbook of Inorg. and Organomet. Chem., 8th Ed., 1994, TYPIX search code (194) hP4
  • B. Grünbaum, Uniform tilings of 3-space, Geombinatorics, 4 (1994), 49-56. See tiling #11.
  • R. W. Marks and R. B. Fuller, The Dymaxion World of Buckminster Fuller. Anchor, NY, 1973, p. 46.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

See A005898 for partial sums.
The 28 uniform 3D tilings: cab: A299266, A299267; crs: A299268, A299269; fcu: A005901, A005902; fee: A299259, A299265; flu-e: A299272, A299273; fst: A299258, A299264; hal: A299274, A299275; hcp: A007899, A007202; hex: A005897, A005898; kag: A299256, A299262; lta: A008137, A299276; pcu: A005899, A001845; pcu-i: A299277, A299278; reo: A299279, A299280; reo-e: A299281, A299282; rho: A008137, A299276; sod: A005893, A005894; sve: A299255, A299261; svh: A299283, A299284; svj: A299254, A299260; svk: A010001, A063489; tca: A299285, A299286; tcd: A299287, A299288; tfs: A005899, A001845; tsi: A299289, A299290; ttw: A299257, A299263; ubt: A299291, A299292; bnn: A007899, A007202. See the Proserpio link in A299266 for overview.

Programs

  • Haskell
    a005897 n = if n == 0 then 1 else 6 * n ^ 2 + 2 -- Reinhard Zumkeller, Apr 27 2014
  • Magma
    [1] cat [6*n^2 + 2: n in [1..50]]; // Vincenzo Librandi, Oct 26 2011
    
  • Maple
    A005897:=-(z+1)*(z**2+4*z+1)/(z-1)**3; # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
  • Mathematica
    Join[{1},6Range[50]^2+2] (* or *) Join[{1},LinearRecurrence[{3,-3,1},{8,26,56},50]] (* Harvey P. Dale, Oct 25 2011 *)
  • PARI
    a(n)=if(n,6*n^2+2,1) \\ Charles R Greathouse IV, Mar 06 2014
    
  • PARI
    x='x+O('x^30); Vec(serlaplace(2*(1 + 3*x + 3*x^2)*exp(x) - 1)) \\ G. C. Greubel, Dec 01 2017
    

Formula

G.f.: (1+x)*(1+4*x+x^2)/(1-x)^3. - Simon Plouffe
a(0) = 1, a(n) = (n+1)^3 - (n-1)^3. - Ilya Nikulshin (ilyanik(AT)gmail.com), Aug 11 2009
a(0)=1, a(1)=8, a(2)=26, a(3)=56; for n>3, a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - Harvey P. Dale, Oct 25 2011
a(n) = A033581(n) + 2. - Reinhard Zumkeller, Apr 27 2014
E.g.f.: 2*(1 + 3*x + 3*x^2)*exp(x) - 1. - G. C. Greubel, Dec 01 2017
a(n) = A000567(n+1) + A045944(n-1), for n>0. See illustration. - John Elias, Mar 12 2022
a(n) = 2*A056107(n), n>0. - R. J. Mathar, May 30 2022
Sum_{n>=0} 1/a(n) = 3/4+ Pi*sqrt(3)*coth(Pi/sqrt 3)/12 = 1.2282133.. - R. J. Mathar, Apr 27 2024
a(n) = 8 + 12*(n-1) + 6*(n-1)^2 for n > 0. - Demilade Runsewe, Jan 08 2025

A001844 Centered square numbers: a(n) = 2*n*(n+1)+1. Sums of two consecutive squares. Also, consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z; then sequence gives Z values.

Original entry on oeis.org

1, 5, 13, 25, 41, 61, 85, 113, 145, 181, 221, 265, 313, 365, 421, 481, 545, 613, 685, 761, 841, 925, 1013, 1105, 1201, 1301, 1405, 1513, 1625, 1741, 1861, 1985, 2113, 2245, 2381, 2521, 2665, 2813, 2965, 3121, 3281, 3445, 3613, 3785, 3961, 4141, 4325, 4513
Offset: 0

Views

Author

Keywords

Comments

These are Hogben's central polygonal numbers denoted by
...2...
....P..
...4.n.
Numbers of the form (k^2+1)/2 for k odd.
(y(2x+1))^2 + (y(2x^2+2x))^2 = (y(2x^2+2x+1))^2. E.g., let y = 2, x = 1; (2(2+1))^2 + (2(2+2))^2 = (2(2+2+1))^2, (2(3))^2 + (2(4))^2 = (2(5))^2, 6^2 + 8^2 = 10^2, 36 + 64 = 100. - Glenn B. Cox (igloos_r_us(AT)canada.com), Apr 08 2002
a(n) is also the number of 3 X 3 magic squares with sum 3(n+1). - Sharon Sela (sharonsela(AT)hotmail.com), May 11 2002
For n > 0, a(n) is the smallest k such that zeta(2) - Sum_{i=1..k} 1/i^2 <= zeta(3) - Sum_{i=1..n} 1/i^3. - Benoit Cloitre, May 17 2002
Number of convex polyominoes with a 2 X (n+1) minimal bounding rectangle.
The prime terms are given by A027862. - Lekraj Beedassy, Jul 09 2004
First difference of a(n) is 4n = A008586(n). Any entry k of the sequence is followed by k + 2*(1 + sqrt(2k - 1)). - Lekraj Beedassy, Jun 04 2006
Integers of the form 1 + x + x^2/2 (generating polynomial is Schur's polynomial as in A127876). - Artur Jasinski, Feb 04 2007
If X is an n-set and Y and Z disjoint 2-subsets of X then a(n-4) is equal to the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Row sums of triangle A132778. - Gary W. Adamson, Sep 02 2007
Binomial transform of [1, 4, 4, 0, 0, 0, ...]; = inverse binomial transform of A001788: (1, 6, 24, 80, 240, ...). - Gary W. Adamson, Sep 02 2007
Narayana transform (A001263) of [1, 4, 0, 0, 0, ...]. Equals A128064 (unsigned) * [1, 2, 3, ...]. - Gary W. Adamson, Dec 29 2007
k such that the Diophantine equation x^3 - y^3 = x*y + k has a solution with y = x-1. If that solution is (x,y) = (m+1,m) then m^2 + (m+1)^2 = k. Note that this Diophantine equation is an elliptic curve and (m+1,m) is an integer point on it. - James R. Buddenhagen, Aug 12 2008
Numbers k such that (k, k, 2*k-2) are the sides of an isosceles triangle with integer area. Also, k such that 2*k-1 is a square. - James R. Buddenhagen, Oct 17 2008
a(n) is also the least weight of self-conjugate partitions having n+1 different odd parts. - Augustine O. Munagi, Dec 18 2008
Prefaced with a "1": (1, 1, 5, 13, 25, 41, ...) = A153869 * (1, 2, 3, ...). - Gary W. Adamson, Jan 03 2009
Prefaced with a "1": (1, 1, 5, 13, 25, 41, ...) where a(n) = 2n*(n-1)+1, all tuples of square numbers (X-Y, X, X+Y) are produced by ((m*(a(n)-2n))^2, (m*a(n))^2, (m*(a(n)+2n-2))^2) where m is a whole number. - Doug Bell, Feb 27 2009
Equals (1, 2, 3, ...) convolved with (1, 3, 4, 4, 4, ...). E.g., a(3) = 25 = (1, 2, 3, 4) dot (4, 4, 3, 1) = (4 + 8 + 9 + 4). - Gary W. Adamson, May 01 2009
The running sum of squares taken two at a time. - Al Hakanson (hawkuu(AT)gmail.com), May 18 2009
Equals the odd integers convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 25 2009
Equals the triangular numbers convolved with [1, 2, 1, 0, 0, 0, ...]. - Gary W. Adamson & Alexander R. Povolotsky, May 29 2009
When the positive integers are written in a square array by diagonals as in A038722, a(n) gives the numbers appearing on the main diagonal. - Joshua Zucker, Jul 07 2009
The finite continued fraction [n,1,1,n] = (2n+1)/(2n^2 + 2n + 1) = (2n+1)/a(n); and the squares of the first two denominators of the convergents = a(n). E.g., the convergents and value of [4,1,1,4] = 1/4, 1/5, 2/9, 9/41 where 4^2 + 5^2 = 41. - Gary W. Adamson, Jul 15 2010
From Keith Tyler, Aug 10 2010: (Start)
Running sum of A008574.
Square open pyramidal number; that is, the number of elements in a square pyramid of height (n) with only surface and no bottom nodes. (End)
For k>0, x^4 + x^2 + k factors over the integers iff sqrt(k) is in this sequence. - James R. Buddenhagen, Aug 15 2010
Create the simple continued fraction from Pythagorean triples to get [2n + 1; 2n^2 + 2n, 2n^2 + 2n + 1]; its value equals the rational number 2n + 1 + a(n) / (4n^4 + 8n^3 + 6n^2 + 2n + 1). - J. M. Bergot, Sep 30 2011
a(n), n >= 1, has in its prime number factorization only primes of the form 4*k+1, i.e., congruent to 1 (mod 4) (see A002144). This follows from the fact that a(n) is a primitive sum of two squares and odd. See Theorem 3.20, p. 164, in the given Niven-Zuckerman-Montgomery reference. E.g., a(3) = 25 = 5^2, a(6) = 85 = 5*17. - Wolfdieter Lang, Mar 08 2012
From Ant King, Jun 15 2012: (Start)
a(n) is congruent to 1 (mod 4) for all n.
The digital roots of the a(n) form a purely periodic palindromic 9-cycle 1, 5, 4, 7, 5, 7, 4, 5, 1.
The units' digits of the a(n) form a purely periodic palindromic 5-cycle 1, 5, 3, 5, 1.
(End)
Number of integer solutions (x,y) of |x| + |y| <= n. Geometrically: number of lattice points inside a square with vertices (n,0), (0,-n), (-n,0), (0,n). - César Eliud Lozada, Sep 18 2012
(a(n)-1)/a(n) = 2*x / (1+x^2) where x = n/(n+1). Note that in this form, this is the velocity-addition formula according to the special theory of relativity (two objects traveling at 1/(n+1) slower than c relative to each other appear to travel at 1/a(n) less than c to a stationary observer). - Christian N. K. Anderson, May 20 2013 [Corrected by Rémi Guillaume, May 22 2025]
A geometric curiosity: the envelope of the circles x^2 + (y-a(n)/2)^2 = ((2n+1)/2)^2 is the parabola y = x^2, the n=0 circle being the osculating circle at the parabola vertex. - Jean-François Alcover, Dec 02 2013
Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; a(n-1) gives the number of internal regions into which the plane is divided (cf. A051890, A046092); a(n-1) = A051890(n) - 1 = A046092(n-1) + 1. - Jaroslav Krizek, Dec 27 2013
a(n) is also, of course, the scalar product of the 2-vector (n, n+1) (or (n+1, n)) with itself. The unique inverse of (n, n+1) as vector in the Clifford algebra over the Euclidean 2-space is (1/a(n))(0, n, n+1, 0) (similarly for the other vector). In general the unique inverse of such a nonzero vector v (odd element in Cl_2) is v^(-1) = (1/|v|^2) v. Note that the inverse with respect to the scalar product is not unique for any nonzero vector. See the P. Lounesto reference, sects. 1.7 - 1.12, pp. 7-14. See also the Oct 15 2014 comment in A147973. - Wolfdieter Lang, Nov 06 2014
Subsequence of A004431, for n >= 1. - Bob Selcoe, Mar 23 2016
Numbers k such that 2k - 1 is a perfect square. - Juri-Stepan Gerasimov, Apr 06 2016
The number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 574", based on the 5-celled von Neumann neighborhood. - Robert Price, May 13 2016
a(n) is the first integer in a sum of (2*n + 1)^2 consecutive integers that equals (2*n + 1)^4. - Patrick J. McNab, Dec 24 2016
Central elements of odd-length rows of the triangular array of positive integers. a(n) is the mean of the numbers in the (2*n + 1)-th row of this triangle. - David James Sycamore, Aug 01 2018
Intersection of A000982 and A080827. - David James Sycamore, Aug 07 2018
An off-diagonal of the array of Delannoy numbers, A008288, (or a row/column when the array is shown as a square). As such, this is one of the crystal ball sequences. - Jack W Grahl, Feb 15 2021 and Shel Kaphan, Jan 18 2023
a(n) appears as a solution to a "Riddler Express" puzzle on the FiveThirtyEight website. The Jan 21 2022 issue (problem) and the Jan 28 2022 issue (solution) present the following puzzle and include a proof. - Fold a square piece of paper in half, obtaining a rectangle. Fold again to obtain a square with 1/4 the size of the original square. Then make n cuts through the folded paper. a(n) is the greatest number of pieces of the unfolded paper after the cutting. - Manfred Boergens, Feb 22 2022
a(n) is (1/6) times the number of 2 X 2 triangles in the n-th order hexagram with 12*n^2 cells. - Donghwi Park, Feb 06 2024
If k is a centered square number, its index in this sequence is n = (sqrt(2k-1)-1)/2. - Rémi Guillaume, Mar 30 2025.
Row sums of the symmetric triangle of odd numbers [1]; [1, 3, 1]; [1, 3, 5, 3, 1]; [1, 3, 5, 7, 5, 3, 1]; .... - Marco Zárate, Jun 15 2025

Examples

			G.f.: 1 + 5*x + 13*x^2 + 25*x^3 + 41*x^4 + 61*x^5 + 85*x^6 + 113*x^7 + 145*x^8 + ...
The first few triples are (1,0,1), (3,4,5), (5,12,13), (7,24,25), ...
The first four such partitions, corresponding to n = 0,1,2,3, i.e., to a(n) = 1,5,13,25, are 1, 3+1+1, 5+3+3+1+1, 7+5+5+3+3+1+1. - _Augustine O. Munagi_, Dec 18 2008
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
  • A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, p. 125, 1964.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 50.
  • Pertti Lounesto, Clifford Algebras and Spinors, second edition, Cambridge University Press, 2001.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 483.
  • Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.
  • 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).
  • Travers et al., The Mysterious Lost Proof, Using Advanced Algebra, (1976), pp. 27.

Crossrefs

X values are A005408; Y values are A046092.
Cf. A008586 (first differences), A005900 (partial sums), A254373 (digital roots).
Subsequence of A004431.
Right edge of A055096; main diagonal of A069480, A078475, A129312.
Row n=2 (or column k=2) of A008288.
Cf. A016754.

Programs

  • Haskell
    a001844 n = 2 * n * (n + 1) + 1
    a001844_list = zipWith (+) a000290_list $ tail a000290_list
    -- Reinhard Zumkeller, Dec 04 2012
    
  • Magma
    [2*n^2 + 2*n + 1: n in [0..50]]; // Vincenzo Librandi, Jan 19 2013
    
  • Magma
    [n: n in [0..4400] | IsSquare(2*n-1)]; // Juri-Stepan Gerasimov, Apr 06 2016
    
  • Maple
    A001844:=-(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation
  • Mathematica
    Table[2n(n + 1) + 1, {n, 0, 50}]
    FoldList[#1 + #2 &, 1, 4 Range@ 50] (* Robert G. Wilson v, Feb 02 2011 *)
    maxn := 47; Flatten[Table[SeriesCoefficient[Series[(n + (n - 1)*x)/(1 - x)^2, {x, 0, maxn}], k], {n, maxn}, {k, n - 1, n - 1}]] (* L. Edson Jeffery, Aug 24 2014 *)
    CoefficientList[ Series[-(x^2 + 2x + 1)/(x - 1)^3, {x, 0, 48}], x] (* or *)
    LinearRecurrence[{3, -3, 1}, {1, 5, 13}, 48] (* Robert G. Wilson v, Aug 01 2018 *)
    Total/@Partition[Range[0,50]^2,2,1] (* Harvey P. Dale, Dec 05 2020 *)
    Table[ j! Coefficient[Series[Exp[x]*(1 + 4*x + 2*x^2), {x, 0, 20}], x,
    j], {j, 0, 20}] (* Nikolaos Pantelidis, Feb 07 2023 *)
  • PARI
    {a(n) = 2*n*(n+1) + 1};
    
  • PARI
    x='x+O('x^200); Vec((1+x)^2/(1-x)^3) \\ Altug Alkan, Mar 23 2016
    
  • Python
    print([2*n*(n+1)+1 for n in range(48)]) # Michael S. Branicky, Jan 05 2021
  • Sage
    [i**2 + (i + 1)**2 for i in range(46)] # Zerinvary Lajos, Jun 27 2008
    

Formula

a(n) = 2*n^2 + 2*n + 1 = n^2 + (n+1)^2.
a(n) = 1 + 3 + 5 + ... + 2*n-1 + 2*n+1 + 2*n-1 + ... + 3 + 1. - Amarnath Murthy, May 28 2001
a(n) = 1/real(z(n+1)) where z(1)=i, (i^2=-1), z(k+1) = 1/(z(k)+2i). - Benoit Cloitre, Aug 06 2002
Nearest integer to 1/Sum_{k>n} 1/k^3. - Benoit Cloitre, Jun 12 2003
G.f.: (1+x)^2/(1-x)^3.
E.g.f.: exp(x)*(1+4x+2x^2).
a(n) = a(n-1) + 4n.
a(-n) = a(n-1).
a(n) = A064094(n+3, n) (fourth diagonal).
a(n) = 1 + Sum_{j=0..n} 4*j. - Xavier Acloque, Oct 08 2003
a(n) = A046092(n)+1 = (A016754(n)+1)/2. - Lekraj Beedassy, May 25 2004
a(n) = Sum_{k=0..n+1} (-1)^k*binomial(n, k)*Sum_{j=0..n-k+1} binomial(n-k+1, j)*j^2. - Paul Barry, Dec 22 2004
a(n) = ceiling((2n+1)^2/2). - Paul Barry, Jul 16 2006
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=1, a(1)=5, a(2)=13. - Jaume Oliver Lafont, Dec 02 2008
a(n)*a(n-1) = 4*n^4 + 1 for n > 0. - Reinhard Zumkeller, Feb 12 2009
Prefaced with a "1" (1, 1, 5, 13, 25, 41, ...): a(n) = 2*n*(n-1)+1. - Doug Bell, Feb 27 2009
a(n) = sqrt((A056220(n)^2 + A056220(n+1)^2) / 2). - Doug Bell, Mar 08 2009
a(n) = floor(2*(n+1)^3/(n+2)). - Gary Detlefs, May 20 2010
a(n) = A000330(n) - A000330(n-2). - Keith Tyler, Aug 10 2010
a(n) = A069894(n)/2. - J. M. Bergot, Jun 11 2012
a(n) = 2*a(n-1) - a(n-2) + 4. - Ant King, Jun 12 2012
Sum_{n>=0} 1/a(n) = (Pi/2)*tanh(Pi/2) = 1.4406595199775... = A228048. - Ant King, Jun 15 2012
a(n) = A209297(2*n+1,n+1). - Reinhard Zumkeller, Jan 19 2013
a(n)^3 = A048395(n)^2 + A048395(-n-1)^2. - Vincenzo Librandi, Jan 19 2013
a(n) = A000217(2n+1) - n. - Ivan N. Ianakiev, Nov 08 2013
a(n) = A251599(3*n+1). - Reinhard Zumkeller, Dec 13 2014
a(n) = A101321(4,n). - R. J. Mathar, Jul 28 2016
From Ilya Gutkovskiy, Jul 30 2016: (Start)
a(n) = Sum_{k=0..n} A008574(k).
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = exp(-1) = A068985. (End)
a(n) = 4 * A000217(n) + 1. - Bruce J. Nicholson, Jul 10 2017
a(n) = A002522(n) + A005563(n) = A002522(n+1) + A005563(n-1). - Bruce J. Nicholson, Aug 05 2017
Sum_{n>=0} a(n)/n! = 7*e. Sum_{n>=0} 1/a(n) = A228048. - Amiram Eldar, Jun 20 2020
a(n) = A000326(n+1) + A000217(n-1). - Charlie Marion, Nov 16 2020
a(n) = Integral_{x=0..2n+2} |1-x| dx. - Pedro Caceres, Dec 29 2020
From Amiram Eldar, Feb 17 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)*sech(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = Pi*csch(Pi)*sinh(Pi/2). (End)
a(n) = A001651(n+1) + 1 - A028242(n). - Charlie Marion, Apr 05 2022
a(n) = A016754(n) - A046092(n). - Leo Tavares, Sep 16 2022
For n>0, a(n) = A101096(n+2) / 30. - Andy Nicol, Feb 06 2025
From Rémi Guillaume, Apr 21 2025: (Start)
a(n) = (2*A003215(n)+1)/3.
a(n) = (4*A005448(n+1)-1)/3.
a(n) + a(n-1) = A001845(n) - A001845(n-1), for n >= 1.
a(n) = (A005917(n+1))/(2n+1). (End)

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A046092 4 times triangular numbers: a(n) = 2*n*(n+1).

Original entry on oeis.org

0, 4, 12, 24, 40, 60, 84, 112, 144, 180, 220, 264, 312, 364, 420, 480, 544, 612, 684, 760, 840, 924, 1012, 1104, 1200, 1300, 1404, 1512, 1624, 1740, 1860, 1984, 2112, 2244, 2380, 2520, 2664, 2812, 2964, 3120, 3280, 3444, 3612, 3784, 3960, 4140, 4324
Offset: 0

Views

Author

Keywords

Comments

Consider all Pythagorean triples (X,Y,Z=Y+1) ordered by increasing Z; sequence gives Y values. X values are 1, 3, 5, 7, 9, ... (A005408), Z values are A001844.
In the triple (X, Y, Z) we have X^2=Y+Z. Actually, the triple is given by {x, (x^2 -+ 1)/2}, where x runs over the odd numbers (A005408) and x^2 over the odd squares (A016754). - Lekraj Beedassy, Jun 11 2004
a(n) is the number of edges in n X n square grid with all horizontal and vertical segments filled in. - Asher Auel, Jan 12 2000 [Corrected by Felix Huber, Apr 09 2024]
a(n) is the only number satisfying an inequality related to zeta(2) and zeta(3): Sum_{i>a(n)+1} 1/i^2 < Sum_{i>n} 1/i^3 < Sum_{i>a(n)} 1/i^2. - Benoit Cloitre, Nov 02 2001
Number of right triangles made from vertices of a regular n-gon when n is even. - Sen-Peng Eu, Apr 05 2001
Number of ways to change two non-identical letters in the word aabbccdd..., where there are n type of letters. - Zerinvary Lajos, Feb 15 2005
a(n) is the number of (n-1)-dimensional sides of an (n+1)-dimensional hypercube (e.g., squares have 4 corners, cubes have 12 edges, etc.). - Freek van Walderveen (freek_is(AT)vanwal.nl), Nov 11 2005
From Nikolaos Diamantis (nikos7am(AT)yahoo.com), May 23 2006: (Start)
Consider a triangle, a pentagon, a heptagon, ..., a k-gon where k is odd. We label a triangle with n=1, a pentagon with n=2, ..., a k-gon with n = floor(k/2). Imagine a player standing at each vertex of the k-gon.
Initially there are 2 frisbees, one held by each of two neighboring players. Every time they throw the frisbee to one of their two nearest neighbors with equal probability. Then a(n) gives the average number of steps needed so that the frisbees meet.
I verified this by simulating the processes with a computer program. For example, a(2) = 12 because in a pentagon that's the expected number of trials we need to perform. That is an exercise in Concrete Mathematics and it can be done using generating functions. (End)
A diagonal of A059056. - Zerinvary Lajos, Jun 18 2007
If X_1,...,X_n is a partition of a 2n-set X into 2-blocks then a(n-1) is equal to the number of 2-subsets of X containing none of X_i, (i=1,...,n). - Milan Janjic, Jul 16 2007
X values of solutions to the equation 2*X^3 + X^2 = Y^2. To find Y values: b(n) = 2n(n+1)(2n+1). - Mohamed Bouhamida, Nov 06 2007
Number of (n+1)-permutations of 3 objects u,v,w, with repetition allowed, containing n-1 u's. Example: a(1)=4 because we have vv, vw, wv and ww; a(2)=12 because we can place u in each of the previous four 2-permutations either in front, or in the middle, or at the end. - Zerinvary Lajos, Dec 27 2007
Sequence found by reading the line from 0, in the direction 0, 4, ... and the same line from 0, in the direction 0, 12, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, May 03 2008
a(n) is also the least weight of self-conjugate partitions having n different even parts. - Augustine O. Munagi, Dec 18 2008
From Peter Luschny, Jul 12 2009: (Start)
The general formula for alternating sums of powers of even integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,1)-(-1)^k P(n,2k+1))/2. Here n=2, thus
a(k) = |(P(2,1) - (-1)^k*P(2,2k+1))/2|. (End)
The sum of squares of n+1 consecutive numbers between a(n)-n and a(n) inclusive equals the sum of squares of n consecutive numbers following a(n). For example, for n = 2, a(2) = 12, and the corresponding equation is 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Tanya Khovanova, Jul 20 2009
Number of roots in the root system of type D_{n+1} (for n>2). - Tom Edgar, Nov 05 2013
Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; sequence gives number of intersections of these ellipses (cf. A051890, A001844); a(n) = A051890(n+1) - 2 = A001844(n) - 1. - Jaroslav Krizek, Dec 27 2013
a(n) appears also as the second member of the quartet [p0(n), a(n), p2(n), p3(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = -A147973(n+3), p2(n) = A054000(n+1) and p3(n) = A139570(n). See a comment on A147973, also with a reference. - Wolfdieter Lang, Oct 15 2014
a(n) appears also as the third and fourth member of the quartet [p0(n), p0(n), a(n), a(n)] of the square of [n, n, n+1, n+1] in the Clifford algebra Cl_2 for n >= 0. p0(n) = A001105(n). - Wolfdieter Lang, Oct 16 2014
Consider two equal rectangles composed of unit squares. Then surround the 1st rectangle with 1-unit-wide layers to build larger rectangles, and surround the 2nd rectangle just to hide the previous layers. If r(n) and h(n) are the number of unit squares needed for n layers in the 1st case and the 2nd case, then for all rectangles, we have a(n) = r(n) - h(n) for n>=1. - Michel Marcus, Sep 28 2015
When greater than 4, a(n) is the perimeter of a Pythagorean triangle with an even short leg 2*n. - Agola Kisira Odero, Apr 26 2016
Also the number of minimum connected dominating sets in the (n+1)-cocktail party graph. - Eric W. Weisstein, Jun 29 2017
a(n+1) is the harmonic mean of A000384(n+2) and A014105(n+1). - Bob Andriesse, Apr 27 2019
Consider a circular cake from which wedges of equal center angle c are cut out in clockwise succession and turned around so that the bottom comes to the top. This goes on until the cake shows its initial surface again. An interesting case occurs if 360°/c is not an integer. Then, with n = floor(360°/c), the number of wedges which have to be cut out and turned equals a(n). (For the number of cutting line segments see A005408.) - According to Peter Winkler's book "Mathematical Mind-Benders", which presents the problem and its solution (see Winkler, pp. 111, 115) the problem seems to be of French origin but little is known about its history. - Manfred Boergens, Apr 05 2022
a(n-3) is the maximum irregularity over all maximal 2-degenerate graphs with n vertices. The extremal graphs are 2-stars (K_2 joined to n-2 independent vertices). (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, May 29 2023
Number of ways of placing a domino on a (n+1)X(n+1) board of squares. - R. J. Mathar, Apr 24 2024
The sequence terms are the exponents in the expansion of (1/(1 + x)) * Sum_{n >= 0} x^n * Product_{k = 1..n} (1 - x^(2*k-1))/(1 + x^(2*k+1)) = 1 - x^4 + x^12 - x^24 + x^40 - x^60 + - ... (Andrews and Berndt, Entry 9.3.3, p. 229). Cf. A153140. - Peter Bala, Feb 15 2025
Number of edges in an (n+1)-dimensional orthoplex. 2D orthoplexes (diamonds) have 4 edges, 3D orthoplexes (octahedrons) have 12 edges, 4D orthoplexes (16-cell) have 24 edges, and so on. - Aaron Franke, Mar 23 2025

Examples

			a(7)=112 because 112 = 2*7*(7+1).
The first few triples are (1,0,1), (3,4,5), (5,12,13), (7,24,25), ...
The first such partitions, corresponding to a(n)=1,2,3,4, are 2+2, 4+4+2+2, 6+6+4+4+2+2, 8+8+6+6+4+4+2+2. - _Augustine O. Munagi_, Dec 18 2008
		

References

  • George E. Andrews and Bruce C. Berndt, Ramanujan's Lost Notebook, Part I, Springer, 2005.
  • Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
  • Albert H. Beiler, Recreations in the Theory of Numbers. New York: Dover, p. 125, 1964.
  • Ronald L. Graham, D. E. Knuth and Oren Patashnik, Concrete Mathematics, Reading, Massachusetts: Addison-Wesley, 1994.
  • Peter Winkler, Mathematical Mind-Benders, Wellesley, Massachusetts: A K Peters, 2007.

Crossrefs

Main diagonal of array in A001477.
Equals A033996/2. Cf. A001844. - Augustine O. Munagi, Dec 18 2008
Cf. A078371, A141530 (see Librandi's comment in A078371).
Cf. similar sequences listed in A299645.
Cf. A005408.
Cf. A016754.
Cf. A002378, A046092, A028896 (irregularities of maximal k-degenerate graphs).

Programs

Formula

a(n) = A100345(n+1, n-1) for n>0.
a(n) = 2*A002378(n) = 4*A000217(n). - Lekraj Beedassy, May 25 2004
a(n) = C(2n, 2) - n = 4*C(n, 2). - Zerinvary Lajos, Feb 15 2005
From Lekraj Beedassy, Jun 04 2006: (Start)
a(n) - a(n-1)=4*n.
Let k=a(n). Then a(n+1) = k + 2*(1 + sqrt(2k + 1)). (End)
Array read by rows: row n gives A033586(n), A085250(n+1). - Omar E. Pol, May 03 2008
O.g.f.:4*x/(1-x)^3; e.g.f.: exp(x)*(2*x^2+4*x). - Geoffrey Critzer, May 17 2009
From Stephen Crowley, Jul 26 2009: (Start)
a(n) = 1/int(-(x*n+x-1)*(step((-1+x*n)/n)-1)*n*step((x*n+x-1)/(n+1)),x=0..1) where step(x)=piecewise(x<0,0,0<=x,1) is the Heaviside step function.
Sum_{n>=1} 1/a(n) = 1/2. (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=4, a(2)=12. - Harvey P. Dale, Jul 25 2011
For n > 0, a(n) = 1/(Integral_{x=0..Pi/2} (sin(x))^(2*n-1)*(cos(x))^3). - Francesco Daddi, Aug 02 2011
a(n) = A001844(n) - 1. - Omar E. Pol, Oct 03 2011
(a(n) - A000217(k))^2 = A000217(2n-k)*A000217(2n+1+k) - (A002378(n) - A000217(k)), for all k. See also A001105. - Charlie Marion, May 09 2013
From Ivan N. Ianakiev, Aug 30 2013: (Start)
a(n)*(2m+1)^2 + a(m) = a(n*(2m+1)+m), for any nonnegative integers n and m.
t(k)*a(n) + t(k-1)*a(n+1) = a((n+1)*(t(k)-t(k-1)-1)), where k>=2, n>=1, t(k)=A000217(k). (End)
a(n) = A245300(n,n). - Reinhard Zumkeller, Jul 17 2014
2*a(n)+1 = A016754(n) = A005408(n)^2, the odd squares. - M. F. Hasler, Oct 02 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) - 1/2 = A187832. - Ilya Gutkovskiy, Mar 16 2017
a(n) = lcm(2*n,2*n+2). - Enrique Navarrete, Aug 30 2017
a(n)*a(n+k) + k^2 = m^2 (a perfect square), n >= 1, k >= 0. - Ezhilarasu Velayutham, May 13 2019
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(Pi/2)/(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -2*cos(sqrt(3)*Pi/2)/Pi. (End)
a(n) = A016754(n) - A001844(n). - Leo Tavares, Sep 20 2022

A008288 Square array of Delannoy numbers D(i,j) (i >= 0, j >= 0) read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 13, 7, 1, 1, 9, 25, 25, 9, 1, 1, 11, 41, 63, 41, 11, 1, 1, 13, 61, 129, 129, 61, 13, 1, 1, 15, 85, 231, 321, 231, 85, 15, 1, 1, 17, 113, 377, 681, 681, 377, 113, 17, 1, 1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1, 1, 21, 181, 833, 2241, 3653, 3653
Offset: 0

Views

Author

Keywords

Comments

In the Formula section, some contributors use T(n,k) = D(n-k, k) (for 0 <= k <= n), which is the triangular version of the square array (D(n,k): n,k >= 0). Conversely, D(n,k) = T(n+k,k) for n,k >= 0. - Petros Hadjicostas, Aug 05 2020
Also called the tribonacci triangle [Alladi and Hoggatt (1977)]. - N. J. A. Sloane, Mar 23 2014
D(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0), (0,1), (1,1). - Joerg Arndt, Jul 01 2011 [Corrected by N. J. A. Sloane, May 30 2020]
Or, triangle read by rows of coefficients of polynomials P[n](x) defined by P[0] = 1, P[1] = x+1; for n >= 2, P[n] = (x+1)*P[n-1] + x*P[n-2].
D(n, k) is the number of k-matchings of a comb-like graph with n+k teeth. Example: D(1, 3) = 7 because the graph consisting of a horizontal path ABCD and the teeth Aa, Bb, Cc, Dd has seven 3-matchings: four triples of three teeth and the three triples {Aa, Bb, CD}, {Aa, Dd, BC}, {Cc, Dd, AB}. Also D(3, 1)=7, the 1-matchings of the same graph being the seven edges: {AB}, {BC}, {CD}, {Aa}, {Bb}, {Cc}, {Dd}. - Emeric Deutsch, Jul 01 2002
Sum of n-th antidiagonal of the array D is A000129(n+1). - Reinhard Zumkeller, Dec 03 2004 [Edited by Petros Hadjicostas, Aug 05 2020 so that the counting of antidiagonals of D starts at n = 0. That is, the sum of the terms in the n-th row of the triangles T is A000129(n+1).]
The A-sequence for this Riordan type triangle (see one of Paul Barry's comments under Formula) is A112478 and the Z-sequence the trivial: {1, 0, 0, 0, ...}. See the W. Lang link under A006232 for Sheffer a- and z-sequences where also Riordan A- and Z-sequences are explained. This leads to the recurrence for the triangle given below. - Wolfdieter Lang, Jan 21 2008
The triangle or chess sums, see A180662 for their definitions, link the Delannoy numbers with twelve different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 and Kn15 have been added. It is remarkable that all knight sums are related to the tribonacci numbers, that is, A000073 and A001590, but none of the others. - Johannes W. Meijer, Sep 22 2010
This sequence, A008288, is jointly generated with A035607 as an array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) and v(n,x) = 2*x*u(n-1,x) + v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 09 2012
Row n, for n > 0, of Roger L. Bagula's triangle in the Example section shows the coefficients of the polynomial u(n) = c(0) + c(1)*x + ... + c(n)*x^n which is the numerator of the n-th convergent of the continued fraction [k, k, k, ...], where k = sqrt(x) + 1/sqrt(x); see A230000. - Clark Kimberling, Nov 13 2013
In an n-dimensional hypercube lattice, D(n,k) gives the number of nodes situated at a Minkowski (Manhattan) distance of k from a given node. In cellular automata theory, the cells at Manhattan distance k are called the von Neumann neighborhood of radius k. For k=1, see A005843. - Dmitry Zaitsev, Dec 10 2015
These numbers appear as the coefficients of series relating spherical and bispherical harmonics, in the solutions of Laplace's equation in 3D. [Majic 2019, Eq. 22] - Matt Majic, Nov 24 2019
From Peter Bala, Feb 19 2020: (Start)
The following remarks assume an offset of 1 in the row and column indices of the triangle.
The sequence of row polynomials T(n,x), beginning with T(1,x) = x, T(2,x) = x + x^2, T(3,x) = x + 3*x^2 + x^3, ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd(T(n,x), T(m,x)) = T(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the sequence (T(n,x): n >= 1) is a divisibility sequence in the polynomial ring Z[x]; that is, if n divides m then T(n,x) divides T(m,x) in Z[x].
Let S(x) = 1 + 2*x + 6*x^2 + 22*x^3 + ... denote the o.g.f. for the large Schröder numbers A006318. The power series (x*S(x))^n, n = 2, 3, 4, ..., can be expressed as a linear combination with polynomial coefficients of S(x) and 1: (x*S(x))^n = T(n-1,-x) - T(n,-x)*S(x). The result can be extended to negative integer n if we define T(0,x) = 0 and T(-n,x) = (-1)^(n+1) * T(n,x)/x^n. Cf. A115139.
[In the previous two paragraphs, D(n,x) was replaced with T(n,x) because the contributor is referring to the rows of the triangle T(n,k), not the rows of the array D(n,k). - Petros Hadjicostas, Aug 05 2020] (End)
Named after the French amateur mathematician Henri-Auguste Delannoy (1833-1915). - Amiram Eldar, Apr 15 2021
D(i,j) = D(j,i). With this and Dmitry Zaitsev's Dec 10 2015 comment, D(i,j) can be considered the number of points at L1 distance <= i in Z^j or the number of points at L1 distance <= j in Z^i from any given point. The rows and columns of D(i,j) are the crystal ball sequences on cubic lattices. See the first example below. The n-th term in the k-th crystal ball sequence can be considered the number of points at distance <= n from any point in a k-dimensional cubic lattice, or the number of points at distance <= k from any point in an n-dimensional cubic lattice. - Shel Kaphan, Jan 01 2023 and Jan 07 2023
Dimensions of hom spaces Hom(R^{(i)}, R^{(j)}) in the Delannoy category attached to the oligomorphic group of order preserving self-bijections of the real line. - Noah Snyder, Mar 22 2023

Examples

			The square array D(i,j) (i >= 0, j >= 0) begins:
  1, 1,  1,   1,   1,   1,    1,    1,    1,    1, ... = A000012
  1, 3,  5,   7,   9,  11,   13,   15,   17,   19, ... = A005408
  1, 5, 13,  25,  41,  61,   85,  113,  145,  181, ... = A001844
  1, 7, 25,  63, 129, 231,  377,  575,  833, 1159, ... = A001845
  1, 9, 41, 129, 321, 681, 1289, 2241, 3649, 5641, ... = A001846
  ...
For D(2,5) = 61, which is seen above in the row labeled A001844, we calculate the sum (9 + 11 + 41) of the 3 nearest terms above and/or to the left. - _Peter Munn_, Jan 01 2023
D(2,5) = 61 can also be obtained from the row labeled A005408 using a recurrence mentioned in the formula section:  D(2,5) = D(1,5) + 2*Sum_{k=0..4} D(1,k), so D(2,5) = 11 + 2*(1+3+5+7+9) = 11 + 2*25. - _Shel Kaphan_, Jan 01 2023
As a triangular array (on its side) this begins:
   0,   0,   0,   0,   1,   0,  11,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   1,   0,   3,   0,  13,   0,  63,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   0,   0,   1,   0,  11,   0, ...
   [Edited by _Shel Kaphan_, Jan 01 2023]
From _Roger L. Bagula_, Dec 09 2008: (Start)
As a triangle T(n,k) (with rows n >= 0 and columns k = 0..n), this begins:
   1;
   1,  1;
   1,  3,   1;
   1,  5,   5,   1;
   1,  7,  13,   7,    1;
   1,  9,  25,  25,    9,    1;
   1, 11,  41,  63,   41,   11,    1;
   1, 13,  61, 129,  129,   61,   13,   1;
   1, 15,  85, 231,  321,  231,   85,  15,   1;
   1, 17, 113, 377,  681,  681,  377, 113,  17,  1;
   1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1;
   ... (End)
Triangle T(n,k) recurrence: 63 = T(6,3) = 25 + 13 + 25 = T(5,2) + T(4,2) + T(5,3).
Triangle T(n,k) recurrence with A-sequence A112478: 63 = T(6,3) = 1*25 + 2*25 - 2*9 + 6*1 (T entries from row n = 5 only). [Here the formula T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j) is used with n = 6 and k = 3; i.e., T(6,3) = Sum_{j=0..3} A111478(j) * T(5, 2+j). - _Petros Hadjicostas_, Aug 05 2020]
From _Philippe Deléham_, Mar 29 2012: (Start)
Subtriangle of the triangle given by (1, 0, 1, -1, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   1,  0;
   1,  1,  0;
   1,  3,  1,  0;
   1,  5,  5,  1,  0;
   1,  7, 13,  7,  1,  0;
   1,  9, 25, 25,  9,  1, 0;
   1, 11, 41, 63, 41, 11, 1, 0;
   ...
Subtriangle of the triangle given by (0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   0, 1;
   0, 1,  1;
   0, 1,  3,  1;
   0, 1,  5,  5,  1;
   0, 1,  7, 13,  7,  1;
   0, 1,  9, 25, 25,  9,  1;
   0, 1, 11, 41, 63, 41, 11, 1;
   ... (End)
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 593.
  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
  • L. Moser and W. Zayachkowski, Lattice paths with diagonal steps, Scripta Mathematica, 26 (1963), 223-229.
  • G. Picou, Note #2235, L'Intermédiaire des Mathématiciens, 8 (1901), page 281. - N. J. A. Sloane, Mar 02 2022
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 28.

Crossrefs

Sums of antidiagonals: A000129 (Pell numbers).
Main diagonal: A001850 (central Delannoy numbers), which has further information and references.
A002002, A026002, and A190666 are +-k-diagonals for k=1, 2, 3 resp. - Shel Kaphan, Jan 01 2023
See also A027618.
Cf. A059446.
Has same main diagonal as A064861. Different from A100936.
Read mod small primes: A211312, A211313, A211314, A211315.
Triangle sums (see the comments): A000129 (Row1); A056594 (Row2); A000073 (Kn11 & Kn21); A089068 (Kn12 & Kn22); A180668 (Kn13 & Kn23); A180669 (Kn14 & Kn24); A180670 (Kn15 & Kn25); A099463 (Kn3 & Kn4); A116404 (Fi1 & Fi2); A006498 (Ca1 & Ca2); A006498(3*n) (Ca3 & Ca4); A079972 (Gi1 & Gi2); A079972(4*n) (Gi3 & Gi4); A079973(3*n) (Ze1 & Ze2); A079973(2*n) (Ze3 & Ze4).
Cf. A102413, A128966. (D(n,1)) = A005843. Cf. A115139.

Programs

  • Haskell
    a008288 n k = a008288_tabl !! n !! k
    a008288_row n = a008288_tabl !! n
    a008288_tabl = map fst $ iterate
        (\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
                           zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
    -- Reinhard Zumkeller, Jul 21 2013
    
  • Maple
    A008288 := proc(n, k) option remember; if k = 0 then 1 elif n=k then 1 else procname(n-1, k-1) + procname(n-2, k-1) + procname(n-1, k) end if; end proc: seq(seq(A008288(n,k),k=0..n), n=0..10); # triangular indices n and k
    P[0]:=1; P[1]:=x+1; for n from 2 to 12 do P[n]:=expand((x+1)*P[n-1]+x*P[n-2]); lprint(P[n]); lprint(seriestolist(series(P[n],x,200))); end do:
  • Mathematica
    (* Next, A008288 jointly generated with A035607 *)
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
    v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x];
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]    (* A008288 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]    (* A035607 *)
    (* Clark Kimberling, Mar 09 2012 *)
    d[n_, k_] := Binomial[n+k, k]*Hypergeometric2F1[-k, -n, -n-k, -1]; A008288 = Flatten[Table[d[n-k, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Apr 05 2012, after 3rd formula *)
  • Python
    from functools import cache
    @cache
    def delannoy_row(n: int) -> list[int]:
        if n == 0: return [1]
        if n == 1: return [1, 1]
        rov = delannoy_row(n - 2)
        row = delannoy_row(n - 1) + [1]
        for k in range(n - 1, 0, -1):
            row[k] += row[k - 1] + rov[k - 1]
        return row
    for n in range(10): print(delannoy_row(n))  # Peter Luschny, Jul 30 2023
  • Sage
    for k in range(8):  # seen as an array, read row by row
        a = lambda n: hypergeometric([-n, -k], [1], 2)
        print([simplify(a(n)) for n in range(11)]) # Peter Luschny, Nov 19 2014
    

Formula

D(n, 0) = 1 = D(0, n) for n >= 0; D(n, k) = D(n, k-1) + D(n-1, k-1) + D(n-1, k).
Bivariate o.g.f.: Sum_{n >= 0, k >= 0} D(n, k)*x^n*y^k = 1/(1 - x - y - x*y).
D(n, k) = Sum_{d = 0..min(n,k)} binomial(k, d)*binomial(n+k-d, k) = Sum_{d=0..min(n,k)} 2^d*binomial(n, d)*binomial(k, d). [Edited by Petros Hadjicostas, Aug 05 2020]
Seen as a triangle read by rows: T(n, 0) = T(n, n) = 1 for n >= 0 and T(n, k) = T(n-1, k-1) + T(n-2, k-1) + T(n-1, k), 0 < k < n and n > 1. - Reinhard Zumkeller, Dec 03 2004
Read as a number triangle, this is the Riordan array (1/(1-x), x(1+x)/(1-x)) with T(n, k) = Sum_{j=0..n-k} C(n-k, j) * C(k, j) * 2^j. - Paul Barry, Jul 18 2005
T(n,k) = Sum_{j=0..n-k} C(k,j)*C(n-j,k). - Paul Barry, May 21 2006
Let y^k(n) be the number of Khalimsky-continuous functions f from [0,n-1] to Z such that f(0) = 0 and f(n-1) = k. Then y^k(n) = D(i,j) for i = (1/2)*(n-1-k) and j = (1/2)*(n-1+k) where n-1+k belongs to 2Z. - Shiva Samieinia (shiva(AT)math.su.se), Oct 08 2007
Recurrence for triangle from A-sequence (see the Wolfdieter Lang comment above): T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j), n >= 1, k >= 1. [For k > n, the sum is empty, in which case T(n,k) = 0.]
From Peter Bala, Jul 17 2008: (Start)
The n-th row of the square array is the crystal ball sequence for the product lattice A_1 x ... x A_1 (n copies). A035607 is the table of the associated coordination sequences for these lattices.
The polynomial p_n(x) := Sum {k = 0..n} 2^k * C(n,k) * C(x,k) = Sum_{k = 0..n} C(n,k) * C(x+k,n), whose values [p_n(0), p_n(1), p_n(2), ... ] give the n-th row of the square array, is the Ehrhart polynomial of the n-dimensional cross polytope (the hyperoctahedron) [Bump et al. (2000), Theorem 6].
The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
The reciprocity law p_n(m) = p_m(n) reflects the symmetry of the table.
The polynomial p_n(x) is the unique polynomial solution of the difference equation (x+1)*f(x+1) - x*f(x-1) = (2*n+1)*f(x), normalized so that f(0) = 1.
These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_n(x-1), n = 1,2,3,..., satisfy a Riemann hypothesis [Bump et al. (2000), Theorem 4]. The o.g.f. for the p_n(x) is (1 + t)^x/(1 - t)^(x + 1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ... .
The square array of Delannoy numbers has a close connection with the constant log(2). The entries in the n-th row of the array occur in the series acceleration formula log(2) = (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n) + (-1)^n * Sum_{k>=1} (-1)^(k+1)/(k*D(n,k-1)*D(n,k)). [T(n,k) was replaced with D(n,k) in the formula to agree with the beginning of the paragraph. - Petros Hadjicostas, Aug 05 2020]
For example, the fourth row of the table (n = 3) gives the series log(2) = 1 - 1/2 + 1/3 - 1/(1*1*7) + 1/(2*7*25) - 1/(3*25*63) + 1/(4*63*129) - ... . See A142979 for further details.
Also the main diagonal entries (the central Delannoy numbers) give the series acceleration formula Sum_{n>=1} 1/(n*D(n-1,n-1)*D(n,n)) = (1/2)*log(2), a result due to Burnside. [T(n,n) was replaced here with D(n,n) to agree with the previous paragraphs. - Petros Hadjicostas, Aug 05 2020]
Similar relations hold between log(2) and the crystal ball sequences of the C_n lattices A142992. For corresponding results for the constants zeta(2) and zeta(3), involving the crystal ball sequences for root lattices of type A_n and A_n x A_n, see A108625 and A143007 respectively. (End)
From Peter Bala, Oct 28 2008: (Start)
Hilbert transform of Pascal's triangle A007318 (see A145905 for the definition of this term).
D(n+a,n) = P_n(a,0;3) for all integer a such that a >= -n, where P_n(a,0;x) is the Jacobi polynomial with parameters (a,0) [Hetyei]. The related formula A(n,k) = P_k(0,n-k;3) defines the table of asymmetric Delannoy numbers, essentially A049600. (End)
Seen as a triangle read by rows: T(n, k) = Hyper2F1([k-n, -k], [1], 2). - Peter Luschny, Aug 02 2014, Oct 13 2024.
From Peter Bala, Jun 25 2015: (Start)
O.g.f. for triangle T(n,k): A(z,t) = 1/(1 - (1 + t)*z - t*z^2) = 1 + (1 + t)*z + (1 + 3*t + t^2)*z^2 + (1 + 5*t + 5*t^2 + t^3)*z^3 + ....
1 + z*d/dz(A(z,t))/A(z,t) is the o.g.f. for A102413. (End)
E.g.f. for the n-th subdiagonal of T(n,k), n >= 0, equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(2*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 4*x + 4*x^2/2) = 1 + 5*x + 13*x^2/2! + 25*x^3/3! + 41*x^4/4! + 61*x^5/5! + .... - Peter Bala, Mar 05 2017 [The n-th subdiagonal of triangle T(n,k) is the n-th row of array D(n,k).]
Let a_i(n) be multiplicative with a_i(p^e) = D(i, e), p prime and e >= 0, then Sum_{n > 0} a_i(n)/n^s = (zeta(s))^(2*i+1)/(zeta(2*s))^i for i >= 0. - Werner Schulte, Feb 14 2018
Seen as a triangle read by rows: T(n,k) = Sum_{i=0..k} binomial(n-i, i) * binomial(n-2*i, k-i) for 0 <= k <= n. - Werner Schulte, Jan 09 2019
Univariate generating function: Sum_{k >= 0} D(n,k)*z^k = (1 + z)^n/(1 - z)^(n+1). [Dziemianczuk (2013), Eq. 5.3] - Matt Majic, Nov 24 2019
(n+1)*D(n+1,k) = (2*k+1)*D(n,k) + n*D(n-1,k). [Majic (2019), Eq. 22] - Matt Majic, Nov 24 2019
For i, j >= 1, D(i,j) = D(i,j-1) + 2*Sum_{k=0..i-1} D(k,j-1), or, because D(i,j) = D(j,i), D(i,j) = D(i-1,j) + 2*Sum_{k=0..j-1} D(i-1,k). - Shel Kaphan, Jan 01 2023
Sum_{k=0..n} T(n,k)^2 = A026933(n). - R. J. Mathar, Nov 07 2023
Let S(x) = (1 - x - (1 - 6*x + x^2)^(1/2))/(2*x) denote the g.f. of the sequence of large Schröder numbers A006318. Read as a lower triangular array, the signed n-th row polynomial R(n, -x) = 1/sqrt(1 - 6*x + x^2) *( 1/S(x)^(n+1) + (x*S(x))^(n+1) ). For example, R(4, -x) = 1 - 7*x + 13*x^2 - 7*x^3 + x^4 = 1/sqrt(1 - 6*x + x^2) * ( 1/S(x)^5 + (x*S(x))^5 ). Cf. A102413. - Peter Bala, Aug 01 2024

Extensions

Expanded description from Clark Kimberling, Jun 15 1997
Additional references from Sylviane R. Schwer (schwer(AT)lipn.univ-paris13.fr), Nov 28 2001
Changed the notation to make the formulas more precise. - N. J. A. Sloane, Jul 01 2002

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

A005898 Centered cube numbers: n^3 + (n+1)^3.

Original entry on oeis.org

1, 9, 35, 91, 189, 341, 559, 855, 1241, 1729, 2331, 3059, 3925, 4941, 6119, 7471, 9009, 10745, 12691, 14859, 17261, 19909, 22815, 25991, 29449, 33201, 37259, 41635, 46341, 51389, 56791, 62559, 68705, 75241, 82179, 89531, 97309, 105525, 114191, 123319, 132921
Offset: 0

Views

Author

Keywords

Comments

Write the natural numbers in groups: 1; 2,3,4; 5,6,7,8,9; 10,11,12,13,14,15,16; ..... and add the groups, i.e., a(n) = Sum_{j=n^2-2(n-1)..n^2} j. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Sep 05 2001
The numbers 1, 9, 35, 91, etc. are divisible by 1, 3, 5, 7, etc. Therefore there are no prime numbers in this list. 9 is divisible by 3 and every third number after 9 is also divisible by 3. 35 is divisible by 5 and 7 and every fifth number after 35 is also divisible by 5 and every seventh number after 35 is also divisible by 7. This pattern continues indefinitely. - Howard Berman (howard_berman(AT)hotmail.com), Nov 07 2008
n^3 + (n+1)^3 = (2n+1)*(n^2+n+1), hence all terms are composite. - Zak Seidov, Feb 08 2011
This is the order of an n-ball centered at a node in the Kronecker product (or direct product) of three cycles, each of whose lengths is at least 2n+2. - Pranava K. Jha, Oct 10 2011
Positive y values of 4*x^3 - 3*x^2 = y^2. - Bruno Berselli, Apr 28 2018

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 52.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

(1/12)*t*(2*n^3 - 3*n^2 + n) + 2*n - 1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.
Partial sums of A005897.
The 28 uniform 3D tilings: cab: A299266, A299267; crs: A299268, A299269; fcu: A005901, A005902; fee: A299259, A299265; flu-e: A299272, A299273; fst: A299258, A299264; hal: A299274, A299275; hcp: A007899, A007202; hex: A005897, A005898; kag: A299256, A299262; lta: A008137, A299276; pcu: A005899, A001845; pcu-i: A299277, A299278; reo: A299279, A299280; reo-e: A299281, A299282; rho: A008137, A299276; sod: A005893, A005894; sve: A299255, A299261; svh: A299283, A299284; svj: A299254, A299260; svk: A010001, A063489; tca: A299285, A299286; tcd: A299287, A299288; tfs: A005899, A001845; tsi: A299289, A299290; ttw: A299257, A299263; ubt: A299291, A299292; bnn: A007899, A007202. See the Proserpio link in A299266 for overview.

Programs

Formula

a(n) = Sum_{i=0..n} A005897(i), partial sums. - Jonathan Vos Post, Feb 06 2011
G.f.: (x^2+4*x+1)*(1+x)/(1-x)^3. - Simon Plouffe (see MAPLE section) and Colin Barker, Jan 02 2012; edited by N. J. A. Sloane, Feb 07 2018
a(n) = A037270(n+1) - A037270(n). - Ivan N. Ianakiev, May 13 2012
a(n) = A000217(n+1)^2 - A000217(n-1)^2. - Bob Selcoe, Mar 25 2016
a(n) = A005408(n) * A002061(n+1). - Miquel Cerda, Oct 05 2016
From Ilya Gutkovskiy, Oct 06 2016: (Start)
E.g.f.: (1 + 8*x + 9*x^2 + 2*x^3)*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)
a(n) = (A081435(n))^2 - (A081435(n) - 1)^2. - Sergey Pavlov, Mar 01 2017

A005902 Centered icosahedral (or cuboctahedral) numbers, also crystal ball sequence for f.c.c. lattice.

Original entry on oeis.org

1, 13, 55, 147, 309, 561, 923, 1415, 2057, 2869, 3871, 5083, 6525, 8217, 10179, 12431, 14993, 17885, 21127, 24739, 28741, 33153, 37995, 43287, 49049, 55301, 62063, 69355, 77197, 85609, 94611, 104223, 114465, 125357, 136919, 149171, 162133, 175825, 190267, 205479
Offset: 0

Views

Author

Keywords

Comments

Called "magic numbers" in some chemical contexts.
Partial sums of A005901(n). - Lekraj Beedassy, Oct 30 2003
Equals binomial transform of [1, 12, 30, 20, 0, 0, 0, ...]. - Gary W. Adamson, Aug 01 2008
Crystal ball sequence for A_3 lattice. - Michael Somos, Jun 03 2012

Examples

			a(4) = 147 = (1, 3, 3, 1) dot (1, 12, 30, 20) = (1 + 36 + 90 + 20). - _Gary W. Adamson_, Aug 01 2008
G.f. = 1 + 13*x + 55*x^2 + 147*x^3 + 309*x^4 + 561*x^5 + 923*x^6 + 1415*x^7 + ...
		

References

  • 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

(1/12)*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.
The 28 uniform 3D tilings: cab: A299266, A299267; crs: A299268, A299269; fcu: A005901, A005902; fee: A299259, A299265; flu-e: A299272, A299273; fst: A299258, A299264; hal: A299274, A299275; hcp: A007899, A007202; hex: A005897, A005898; kag: A299256, A299262; lta: A008137, A299276; pcu: A005899, A001845; pcu-i: A299277, A299278; reo: A299279, A299280; reo-e: A299281, A299282; rho: A008137, A299276; sod: A005893, A005894; sve: A299255, A299261; svh: A299283, A299284; svj: A299254, A299260; svk: A010001, A063489; tca: A299285, A299286; tcd: A299287, A299288; tfs: A005899, A001845; tsi: A299289, A299290; ttw: A299257, A299263; ubt: A299291, A299292; bnn: A007899, A007202. See the Proserpio link in A299266 for overview.

Programs

  • Magma
    [(2*n+1)*(5*n^2+5*n+3)/3: n in [0..30]]; // G. C. Greubel, Dec 01 2017
    
  • Maple
    A005902 := n -> (2*n+1)*(5*n^2+5*n+3)/3;
    A005902:=(z+1)*(z**2+8*z+1)/(z-1)**4; # Simon Plouffe in his 1992 dissertation
  • Mathematica
    f[n_] := (2n + 1)(5n^2 + 5n + 3)/3; Array[f, 36, 0] (* Robert G. Wilson v, Feb 02 2011 *)
    LinearRecurrence[{4,-6,4,-1},{1,13,55,147},50] (* Harvey P. Dale, Oct 08 2015 *)
    CoefficientList[Series[(x^3 + 9*x^2 + 9*x + 1)/(x - 1)^4, {x, 0, 50}], x] (* Indranil Ghosh, Apr 08 2017 *)
  • PARI
    {a(n) = (2*n + 1) * (5*n^2 + 5*n + 3) / 3}; /* Michael Somos, Jun 03 2012 */
    
  • PARI
    x='x+O('x^50); Vec((x^3 + 9*x^2 + 9*x + 1)/(x - 1)^4) \\ Indranil Ghosh, Apr 08 2017
    
  • Python
    def a(n): return (2*n+1)*(5*n**2+5*n+3)//3
    print([a(n) for n in range(40)]) # Michael S. Branicky, Jan 13 2021

Formula

a(n) = (2*n+1)*(5*n^2+5*n+3)/3.
For n > 0, n*a(n) = (Sum_{i=0..n-1} a(i)) + 2*A005891(n)*A000217(n). - Bruno Berselli, Feb 02 2011
a(-1 - n) = -a(n). - Michael Somos, Jun 03 2012
From Indranil Ghosh, Apr 08 2017: (Start)
G.f.: (x^3 + 9x^2 + 9x + 1)/(x - 1)^4.
E.g.f.: (1/3)*exp(x)*(10x^3 + 45x^2 + 36x + 3).
(End)
a(n) = A100171(n+1) - A008778(n-1) = A100174(n+1) - A000290(n) = A005917(n+1) - A006331(n) = A051673(n+1) + A000578(n). - Bruce J. Nicholson, Jul 05 2018
Showing 1-10 of 93 results. Next