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 1-8 of 8 results.

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

A001793 a(n) = n*(n+3)*2^(n-3).

Original entry on oeis.org

1, 5, 18, 56, 160, 432, 1120, 2816, 6912, 16640, 39424, 92160, 212992, 487424, 1105920, 2490368, 5570560, 12386304, 27394048, 60293120, 132120576, 288358400, 627048448, 1358954496, 2936012800, 6325010432, 13589544960, 29125246976
Offset: 1

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Keywords

Comments

Coefficients of Chebyshev T polynomials: the subdiagonal A053120(n+3, n-1), for n > = 1. [rewritten by Wolfdieter Lang, Nov 25 2019]
Number of 132-avoiding permutations of [n+3] containing exactly two 123 patterns. - Emeric Deutsch, Jul 13 2001
Number of Dyck paths of semilength n+2 having pyramid weight n+1 (for pyramid weight see Denise and Simion). Example: a(2)=5 because the Dyck paths of semilength 4 having pyramid weight 3 are: (ud)u(ud)(ud)d, u(ud)(ud)d(ud), u(ud)(ud)(ud)d, u(ud)(uudd)d and u(uudd)(ud)d [here u=(1,1), d=(1,-1) and the maximal pyramids, of total length 3, are shown between parentheses]. - Emeric Deutsch, Mar 10 2004
a(n) is the number of dissections of a regular (n+3)-gon using n-1 noncrossing diagonals such that every piece of the dissection contains at least one non-base side of the (n+3)-gon. (One side of the (n+3)-gon is designated the base.) - David Callan, Mar 23 2004
If X_1,X_2,...,X_n are 2-blocks of a (2n+1)-set X then a(n) is the number of (n+2)-subsets of X intersecting each X_i, (i=1..n). - Milan Janjic, Nov 18 2007
The second corrector line for transforming 2^n offset 0 with a leading 1 into the Fibonacci sequence. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
Sum of all nodes of all integer compositions of n, see example. - Olivier Gérard, Oct 22 2011
Number of compositions of 2n with exactly two odd summands (see example). - Mamuka Jibladze, Sep 04 2013
4*a(n) is the number of North-East paths from (0,0) to (n+2,n+2) with exactly two east steps below y = x-1 or above y = x+1. It is related to paired pattern P_1 and P_6 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
From Paul Weisenhorn, Oct 18 2019: (Start)
The polynomials S(n,x)= Sum_(k>=1) b(n,k)*x^k has the recurrence relation S(n+2,x)=2*S(n+1,x))-x*S(n) with S(1,x)=1, S(2,x)=2-x and are generated by the coefficients b(n,k). b(n,k) is defined by b(n,k)=Sum_(j=1..k) binomials(k+1,j)*b(n-j,k) or by b(n,k)=((n-2+k)!*(n-1+2k)*2^n)/(4*(n-1)!*k!). b(n,1)=A001792, b(n,2)=A001793, b(n,3)=A001794, b(n,4)=A006974, b(n,5)=A006975, b(n,6)=A006976, b(n,7)=A209404.
The general formula for the sequences with k>=1: a(n)=((n-2+k)!*(n-1+2k)*2^n)/(4*(n-1)!*k!) with n >= 1. (End) [See a comment in A053120 on subdiagonal sequences. - Wolfdieter Lang, Jan 03 2020]

Examples

			a(2)=5 since 32415, 32451, 34125, 42135 and 52134 are the only 132-avoiding permutations of 12345 containing exactly two increasing subsequences of length 3.
a(4)=56: the compositions of 4 are 4, 3+1, 1+3, 2+2, 2+1+1, 1+2+1, 1+1+2, 1+1+1+1, the corresponding nodes (partial sums) are {0, 4}, {0, 3, 4}, {0, 1, 4}, {0, 2, 4}, {0, 2, 3, 4}, {0, 1, 3, 4}, {0, 1, 2, 4}, {0, 1, 2, 3, 4}, with individual sums {4, 7, 5, 6, 9, 8, 7, 10} and total 56. - _Olivier Gérard_, Oct 22 2011
The a(3)=18 compositions of 2*3=6 with two odd summands are 5+1, 1+5, 3+3, 4+1+1, 1+4+1, 1+1+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2. - _Mamuka Jibladze_, Sep 04 2013
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
  • 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

a(n) = A039991(n+3, 4) = A055252(n, 1).
Cf. A053120.

Programs

Formula

G.f.: x*(1-x)/(1-2*x)^3. Binomial transform of squares [1, 4, 9, ...].
a(n) = Sum_{k=0..floor((n+4)/2)} C(n+4, 2k)*C(k, 2). - Paul Barry, May 15 2003
With two leading zeros, binomial transform of quarter-squares A002620. - Paul Barry, May 27 2003
a(n) = Sum_{k=0..n+2} C(n+2, k) * floor(k^2/4). - Paul Barry, May 27 2003
a(n) = Sum_{i=0..j} binomial(i+1, 2)*binomial(j, i). - Jon Perry, Feb 26 2004
With one leading zero, binomial transform of triangular numbers A000217. - Philippe Deléham, Aug 02 2005
a(n) = Sum_{k=0..n+1} (-1)^(n-k+1)*C(k, n-k+1)*k*C(2k, k)/2. - Paul Barry, Oct 07 2005
Left-shifted sequence is binomial transform of left-shifted squares (A000290). - Franklin T. Adams-Watters, Nov 29 2006
Binomial transform of a(n) = n^2 offset 1. a(3)=18. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
a(n) = (1/n) * Sum_{k=0..n} binomial(n,k)*k^3. - Gary Detlefs, Nov 26 2011
For n > 1, a(n) = Sum_{k=0..n-1} Sum_{i=0..n} (k+2) * C(n-2,i). - Wesley Ivan Hurt, Sep 20 2017
a(n) = a(-3-n)*2^(2*n+3), a(n)*(n+3) = -A058645(-3-n)*2^(2*n+3) for all n in Z. - Michael Somos, Apr 19 2019
E.g.f.: (1/2)*exp(2*x)*x*(2 + x). - Stefano Spezia, Aug 17 2019
From Amiram Eldar, Jan 05 2022: (Start)
Sum_{n>=1} 1/a(n) = 128/9 - 56*log(2)/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 24*log(3/2) - 80/9. (End)

A014483 Expansion of (1+2*x) / (1-2*x)^4.

Original entry on oeis.org

1, 10, 56, 240, 880, 2912, 8960, 26112, 72960, 197120, 518144, 1331200, 3354624, 8314880, 20316160, 49020928, 116981760, 276430848, 647495680, 1504706560, 3471835136, 7958691840, 18136170496, 41104179200, 92694118400, 208071032832, 465064427520
Offset: 0

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Crossrefs

Programs

Formula

a(n) = 2^n * A000330(n+1). - R. J. Mathar, Oct 23 2008
From Colin Barker, Feb 13 2017: (Start)
a(n) = 8*a(n-1) - 24*a(n-2) + 32*a(n-3) - 16*a(n-4) for n>3.
a(n) = (2^(n-1)*(6 + 13*n + 9*n^2 + 2*n^3)) / 3. (End)
a(n) = (1/2) * Sum_{k=0..n+1} Sum_{i=0..n+1} (n-i+1)^2 * C(n+1,k). - Wesley Ivan Hurt, Sep 21 2017

Extensions

More terms from Colin Barker, Feb 13 2017

A056468 a(n) = Sum_{k=1..n} k^6*binomial(n,k).

Original entry on oeis.org

0, 1, 66, 924, 7400, 44040, 217392, 942592, 3714048, 13593600, 46914560, 154328064, 487778304, 1490384896, 4423372800, 12801146880, 36235378688, 100580917248, 274361352192, 736775372800, 1950815354880, 5099601002496, 13176144920576, 33682341494784
Offset: 0

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Author

Benoit Cloitre, Dec 06 2002

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[k^6*Binomial[n, k], {k, n}], {n, 0, 30}] (* T. D. Noe, Nov 22 2013 *)
  • PARI
    a(n) = sum(k = 1, n, k^6*binomial(n,k)); \\ Michel Marcus, Nov 20 2013

Formula

a(n) = 2^(n-6)*n*(n+1)*(n^4 + 14*n^3 + 31*n^2 - 46*n + 16).
G.f.: -x*(136*x^4-272*x^3+84*x^2+52*x+1)/(2*x-1)^7. [Colin Barker, Sep 20 2012]

A087076 Sums of the squares of the elements in the subsets of the integers 1 to n.

Original entry on oeis.org

0, 1, 10, 56, 240, 880, 2912, 8960, 26112, 72960, 197120, 518144, 1331200, 3354624, 8314880, 20316160, 49020928, 116981760, 276430848, 647495680, 1504706560, 3471835136, 7958691840, 18136170496, 41104179200, 92694118400, 208071032832
Offset: 0

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Author

Alan Sutcliffe (alansut(AT)ntlworld.com), Aug 08 2003

Keywords

Comments

A000076 gives the number of subsets of 1 to n. A001787 gives the number of elements in the subsets. A001788 gives the sum of the elements.

Examples

			a(3)=56 since the subsets of (1,2,3) are ( ) (1) (1,2) (1,3) (1,2,3) (2) (2,3) (3) and the sum of the squares of the elements in these subsets is 56.
		

Crossrefs

Cf. A058645 has the same then similar initial values.
Equals A014483 shifted right.

Programs

  • PARI
    a(n) = (2^(n-2))*n*(n+1)*(2*n+1)/3 \\ Michel Marcus, Jul 12 2013

Formula

a(n) = 2^(n - 2)*n*(n + 1)*(2*n + 1)/3.
G.f.: x*(1 + 2*x)/(1 - 2*x)^4.
a(n) = 8*a(n-1) - 24*a(n-2) + 32*a(n-3) - 16*a(n-4). - Matthew House, Feb 13 2017
a(n) = (1/2) * Sum_{k=0..n} Sum_{i=0..n} i^2 * C(n,k). - Wesley Ivan Hurt, Sep 21 2017

A084641 Binomial transform of n^7.

Original entry on oeis.org

0, 1, 130, 2574, 25904, 183200, 1040112, 5076400, 22171648, 88915968, 333209600, 1181548544, 4001402880, 13033885696, 41061830656, 125666611200, 374947708928, 1093874155520, 3128047828992, 8785866391552, 24280799641600, 66124498599936, 177683966197760
Offset: 0

Views

Author

Paul Barry, Jun 08 2003

Keywords

Comments

The binomial transforms of n, n^2, n^3, n^4, n^5, n^6 are A001787, A001788, A058645, A058649, A059338, A056468 respectively.

Crossrefs

Programs

  • Magma
    [n^2*(n^5+21*n^4+105*n^3+35*n^2-210*n+112)*2^(n-7): n in [0..40]]; // G. C. Greubel, Mar 20 2023
    
  • Mathematica
    LinearRecurrence[{16,-112,448,-1120,1792,-1792,1024,-256}, {0,1,130, 2574,25904,183200,1040112,5076400}, 41] (* Amiram Eldar, Nov 26 2021 *)
  • SageMath
    [n^2*(n^5+21*n^4+105*n^3+35*n^2-210*n+112)*2^(n-7) for n in range(41)] # G. C. Greubel, Mar 20 2023

Formula

a(n) = n^2*(n^5 + 21*n^4 + 105*n^3 + 35*n^2 - 210*n + 112)*2^(n-7).
a(n) = Sum_{k=0..n} C(n, k)*k^7.
G.f.: x*(1+114*x+606*x^2-1168*x^3-96*x^4+816*x^5-272*x^6)/(1-2*x)^8. - Colin Barker, Sep 20 2012

A084903 Binomial transform of positive cubes.

Original entry on oeis.org

1, 9, 44, 170, 576, 1792, 5248, 14688, 39680, 104192, 267264, 672256, 1662976, 4055040, 9764864, 23257088, 54853632, 128253952, 297533440, 685375488, 1568669696, 3569352704, 8078229504, 18192793600, 40785412096, 91049951232
Offset: 0

Views

Author

Paul Barry, Jun 13 2003

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{8,-24,32,-16},{1,9,44,170},30] (* Harvey P. Dale, Jul 30 2023 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)*(1+k)^3); \\ Michel Marcus, Oct 13 2016

Formula

a(n) = 2^(n-3)*(n^3+9n^2+18n+8).
a(n) = Sum_{k=0..n} C(n, k)*(1+k)^3.
O.g.f.: (x-1)*(2*x^2-2*x-1)/(-1+2*x)^4. - R. J. Mathar, Apr 02 2008
a(n) = A058649(n+1)/n. [Gary Detlefs, Nov 26 2011]

A349706 Square array T(n,k) = Sum_{j=0..k} binomial(k,j) * j^n for n and k >= 0, read by ascending antidiagonals.

Original entry on oeis.org

1, 0, 2, 0, 1, 4, 0, 1, 4, 8, 0, 1, 6, 12, 16, 0, 1, 10, 24, 32, 32, 0, 1, 18, 54, 80, 80, 64, 0, 1, 34, 132, 224, 240, 192, 128, 0, 1, 66, 342, 680, 800, 672, 448, 256, 0, 1, 130, 924, 2192, 2880, 2592, 1792, 1024, 512, 0, 1, 258, 2574, 7400, 11000, 10752, 7840, 4608, 2304, 1024
Offset: 0

Views

Author

Michel Marcus, Nov 26 2021

Keywords

Examples

			Square array begins:
  1 2  4   8   16    32
  0 1  4  12   32    80
  0 1  6  24   80   240
  0 1 10  54  224   800
  0 1 18 132  680  2880
  0 1 34 342 2192 11000
		

Crossrefs

Main diagonal gives A072034.
Cf. A209849.

Programs

  • Mathematica
    T[n_, k_] := Sum[Binomial[k, j] * If[j == n == 0, 1, j^n], {j, 0, k}]; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Nov 26 2021 *)
  • PARI
    T(n,k) = sum(j=0, k, binomial(k,j)*j^n);
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