cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A339019 Square table read by upwards antidiagonals: T(m,n) = A103438(2*m-1,n)/A103438(1,n) for m>=1, n>=1.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 11, 6, 1, 1, 43, 46, 10, 1, 1, 171, 386, 130, 15, 1, 1, 683, 3366, 1870, 295, 21, 1, 1, 2731, 29866, 28234, 6455, 581, 28, 1, 1, 10923, 267086, 437350, 149031, 17941, 1036, 36, 1
Offset: 1

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Author

Franz Vrabec, Dec 24 2020

Keywords

Examples

			T(3,4) = A103438(2*3-1,4)/A103438(1,4) = 1300/10 = 130.
By formula: a(2,3) = 4*15*1*1*B(4) = -2 and a(3,3) = (-8)*15*4*(2/4)*B(4) = 8 yields T(3,n) = (-N+4*N^2)/3. Since N = 4*5/2 = 10, T(3,4) = (4*10^2-10)/3 = 130.
Table begins:
m\n| 1    2      3       4        5         6          7
---+-----------------------------------------------------
1  | 1    1      1       1        1         1          1
2  | 1    3      6      10       15        21         28
3  | 1   11     46     130      295       581       1036
4  | 1   43    386    1870     6455     17941      42868
5  | 1  171   3366   28234   149031    586341    1880956
6  | 1  683  29866  437350  3546775  19809461   85475908
7  | 1 2731 267086 6871138 85960967 683338501 3972825676
		

Crossrefs

Cf. A103438.

Formula

Let a(i,m) = ((-2)^i)*Sum_{j=0..i} C(2*m,i-j)*C(i+j,j)*((i-j)/(i+j))*B(2*m-i+j), B(s) = A027641(s)/A027642(s) the Bernoulli numbers and N = n*(n+1)/2, then T(m,n) = (1/(2*m))*Sum_{i=2..m} a(i,m)*N^(i-1).

A000051 a(n) = 2^n + 1.

Original entry on oeis.org

2, 3, 5, 9, 17, 33, 65, 129, 257, 513, 1025, 2049, 4097, 8193, 16385, 32769, 65537, 131073, 262145, 524289, 1048577, 2097153, 4194305, 8388609, 16777217, 33554433, 67108865, 134217729, 268435457, 536870913, 1073741825, 2147483649, 4294967297, 8589934593
Offset: 0

Views

Author

Keywords

Comments

Same as Pisot sequence L(2,3).
Length of the continued fraction for Sum_{k=0..n} 1/3^(2^k). - Benoit Cloitre, Nov 12 2003
See also A004119 for a(n) = 2a(n-1)-1 with first term = 1. - Philippe Deléham, Feb 20 2004
From the second term on (n>=1), in base 2, these numbers present the pattern 1000...0001 (with n-1 zeros), which is the "opposite" of the binary 2^n-2: (0)111...1110 (cf. A000918). - Alexandre Wajnberg, May 31 2005
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)* charpoly(A,3). - Milan Janjic, Jan 27 2010
First differences of A006127. - Reinhard Zumkeller, Apr 14 2011
The odd prime numbers in this sequence form A019434, the Fermat primes. - David W. Wilson, Nov 16 2011
Pisano period lengths: 1, 1, 2, 1, 4, 2, 3, 1, 6, 4, 10, 2, 12, 3, 4, 1, 8, 6, 18, 4, ... . - R. J. Mathar, Aug 10 2012
Is the mentioned Pisano period lengths (see above) the same as A007733? - Omar E. Pol, Aug 10 2012
Only positive integers that are not 1 mod (2k+1) for any k>1. - Jon Perry, Oct 16 2012
For n >= 1, a(n) is the total length of the segments of the Hilbert curve after n iterations. - Kival Ngaokrajang, Mar 30 2014
Frénicle de Bessy (1657) proved that a(3) = 9 is the only square in this sequence. - Charles R Greathouse IV, May 13 2014
a(n) is the number of distinct possible sums made with at most two elements in {1,...,a(n-1)} for n > 0. - Derek Orr, Dec 13 2014
For n > 0, given any set of a(n) lattice points in R^n, there exist 2 distinct members in this set whose midpoint is also a lattice point. - Melvin Peralta, Jan 28 2017
Also the number of independent vertex sets, irredundant sets, and vertex covers in the (n+1)-star graph. - Eric W. Weisstein, Aug 04 and Sep 21 2017
Also the number of maximum matchings in the 2(n-1)-crossed prism graph. - Eric W. Weisstein, Dec 31 2017
Conjecture: For any integer n >= 0, a(n) is the permanent of the (n+1) X (n+1) matrix with M(j, k) = -floor((j - k - 1)/(n + 1)). This conjecture is inspired by the conjecture of Zhi-Wei Sun in A036968. - Peter Luschny, Sep 07 2021

References

  • Paul Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 75.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 60, 244.
  • 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, page 141.

Crossrefs

Apart from the initial 1, identical to A094373.
See A008776 for definitions of Pisot sequences.
Column 2 of array A103438.
Cf. A007583 (a((n-1)/2)/3 for odd n).

Programs

  • Haskell
    a000051 = (+ 1) . a000079
    a000051_list = iterate ((subtract 1) . (* 2)) 2
    -- Reinhard Zumkeller, May 03 2012
    
  • Magma
    [2^n+1: n in [0..40]]; // G. C. Greubel, Jan 18 2025
  • Maple
    A000051:=-(-2+3*z)/(2*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation
    a := n -> add(binomial(n,k)*bernoulli(n-k,1)*2^(k+1)/(k+1),k=0..n); # Peter Luschny, Apr 20 2009
  • Mathematica
    Table[2^n + 1, {n,0,40}]
    2^Range[0,40] + 1 (* Eric W. Weisstein, Jul 17 2017 *)
    LinearRecurrence[{3, -2}, {2, 3}, 40] (* Eric W. Weisstein, Sep 21 2017 *)
  • PARI
    a(n)=2^n+1
    
  • PARI
    first(n) = Vec((2 - 3*x)/((1 - x)*(1 - 2*x)) + O(x^n)) \\ Iain Fox, Dec 31 2017
    
  • Python
    def A000051(n): return (1<Chai Wah Wu, Dec 21 2022
    

Formula

a(n) = 2*a(n-1) - 1 = 3*a(n-1) - 2*a(n-2).
G.f.: (2-3*x)/((1-x)*(1-2*x)).
First differences of A052944. - Emeric Deutsch, Mar 04 2004
a(0) = 1, then a(n) = (Sum_{i=0..n-1} a(i)) - (n-2). - Gerald McGarvey, Jul 10 2004
Inverse binomial transform of A007689. Also, V sequence in Lucas sequence L(3, 2). - Ross La Haye, Feb 07 2005
a(n) = A127904(n+1) for n>0. - Reinhard Zumkeller, Feb 05 2007
Equals binomial transform of [2, 1, 1, 1, ...]. - Gary W. Adamson, Apr 23 2008
a(n) = A000079(n)+1. - Omar E. Pol, May 18 2008
E.g.f.: exp(x) + exp(2*x). - Mohammad K. Azarian, Jan 02 2009
a(n) = A024036(n)/A000225(n). - Reinhard Zumkeller, Feb 14 2009
From Peter Luschny, Apr 20 2009: (Start)
A weighted binomial sum of the Bernoulli numbers A027641/A027642 with A027641(1)=1 (which amounts to the definition B_{n} = B_{n}(1)).
a(n) = Sum_{k=0..n} C(n,k)*B_{n-k}*2^(k+1)/(k+1). (See also A052584.) (End)
a(n) is the a(n-1)-th odd number for n >= 1. - Jaroslav Krizek, Apr 25 2009
From Reinhard Zumkeller, Feb 28 2010: (Start)
a(n)*A000225(n) = A000225(2*n).
a(n) = A173786(n,0). (End)
If p[i]=Fibonacci(i-4) and if A is the Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise, then, for n>=1, a(n-1)= det A. - Milan Janjic, May 08 2010
a(n+2) = a(n) + a(n+1) + A000225(n). - Ivan N. Ianakiev, Jun 24 2012
a(A006521(n)) mod A006521(n) = 0. - Reinhard Zumkeller, Jul 17 2014
a(n) = 3*A007583((n-1)/2) for n odd. - Eric W. Weisstein, Jul 17 2017
Sum_{n>=0} 1/a(n) = A323482. - Amiram Eldar, Nov 11 2020

A000330 Square pyramidal numbers: a(n) = 0^2 + 1^2 + 2^2 + ... + n^2 = n*(n+1)*(2*n+1)/6.

Original entry on oeis.org

0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201, 6930, 7714, 8555, 9455, 10416, 11440, 12529, 13685, 14910, 16206, 17575, 19019, 20540, 22140, 23821, 25585, 27434, 29370
Offset: 0

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Author

Keywords

Comments

The sequence contains exactly one square greater than 1, namely 4900 (according to Gardner). - Jud McCranie, Mar 19 2001, Mar 22 2007 [This is a result from Watson. - Charles R Greathouse IV, Jun 21 2013] [See A351830 for further related comments and references.]
Number of rhombi in an n X n rhombus. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Number of acute triangles made from the vertices of a regular n-polygon when n is odd (cf. A007290). - Sen-Peng Eu, Apr 05 2001
Gives number of squares with sides parallel to the axes formed from an n X n square. In a 1 X 1 square, one is formed. In a 2 X 2 square, five squares are formed. In a 3 X 3 square, 14 squares are formed and so on. - Kristie Smith (kristie10spud(AT)hotmail.com), Apr 16 2002; edited by Eric W. Weisstein, Mar 05 2025
a(n-1) = B_3(n)/3, where B_3(x) = x(x-1)(x-1/2) is the third Bernoulli polynomial. - Michael Somos, Mar 13 2004
Number of permutations avoiding 13-2 that contain the pattern 32-1 exactly once.
Since 3*r = (r+1) + r + (r-1) = T(r+1) - T(r-2), where T(r) = r-th triangular number r*(r+1)/2, we have 3*r^2 = r*(T(r+1) - T(r-2)) = f(r+1) - f(r-1) ... (i), where f(r) = (r-1)*T(r) = (r+1)*T(r-1). Summing over n, the right hand side of relation (i) telescopes to f(n+1) + f(n) = T(n)*((n+2) + (n-1)), whence the result Sum_{r=1..n} r^2 = n*(n+1)*(2*n+1)/6 immediately follows. - Lekraj Beedassy, Aug 06 2004
Also as a(n) = (1/6)*(2*n^3 + 3*n^2 + n), n > 0: structured trigonal diamond numbers (vertex structure 5) (cf. A006003 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of triples of integers from {1, 2, ..., n} whose last component is greater than or equal to the others.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 12 2005
Sum of the first n positive squares. - Cino Hilliard, Jun 18 2007
Maximal number of cubes of side 1 in a right pyramid with a square base of side n and height n. - Pasquale CUTOLO (p.cutolo(AT)inwind.it), Jul 09 2007
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-3) is the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
We also have the identity 1 + (1+4) + (1+4+9) + ... + (1+4+9+16+ ... + n^2) = n(n+1)(n+2)(n+(n+1)+(n+2))/36; ... and in general the k-fold nested sum of squares can be expressed as n(n+1)...(n+k)(n+(n+1)+...+(n+k))/((k+2)!(k+1)/2). - Alexander R. Povolotsky, Nov 21 2007
The terms of this sequence are coefficients of the Engel expansion of the following converging sum: 1/(1^2) + (1/1^2)*(1/(1^2+2^2)) + (1/1^2)*(1/(1^2+2^2))*(1/(1^2+2^2+3^2)) + ... - Alexander R. Povolotsky, Dec 10 2007
Convolution of A000290 with A000012. - Sergio Falcon, Feb 05 2008
Hankel transform of binomial(2*n-3, n-1) is -a(n). - Paul Barry, Feb 12 2008
Starting (1, 5, 14, 30, ...) = binomial transform of [1, 4, 5, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jun 13 2008
Starting (1,5,14,30,...) = second partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+2,i+2)*b(i), where b(i)=1,2,0,0,0,... - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Convolution of A001477 with A005408: a(n) = Sum_{k=0..n} (2*k+1)*(n-k). - Reinhard Zumkeller, Mar 07 2009
Sequence of the absolute values of the z^1 coefficients of the polynomials in the GF1 denominators of A156921. See A157702 for background information. - Johannes W. Meijer, Mar 07 2009
The sequence is related to A000217 by a(n) = n*A000217(n) - Sum_{i=0..n-1} A000217(i) and this is the case d = 1 in the identity n^2*(d*n-d+2)/2 - Sum_{i=0..n-1} i*(d*i-d+2)/2 = n*(n+1)(2*d*n-2*d+3)/6, or also the case d = 0 in n^2*(n+2*d+1)/2 - Sum_{i=0..n-1} i*(i+2*d+1)/2 = n*(n+1)*(2*n+3*d+1)/6. - Bruno Berselli, Apr 21 2010, Apr 03 2012
a(n)/n = k^2 (k = integer) for n = 337; a(337) = 12814425, a(n)/n = 38025, k = 195, i.e., the number k = 195 is the quadratic mean (root mean square) of the first 337 positive integers. There are other such numbers -- see A084231 and A084232. - Jaroslav Krizek, May 23 2010
Also the number of moves to solve the "alternate coins game": given 2n+1 coins (n+1 Black, n White) set alternately in a row (BWBW...BWB) translate (not rotate) a pair of adjacent coins at a time (1 B and 1 W) so that at the end the arrangement shall be BBBBB..BW...WWWWW (Blacks separated by Whites). Isolated coins cannot be moved. - Carmine Suriano, Sep 10 2010
From J. M. Bergot, Aug 23 2011: (Start)
Using four consecutive numbers n, n+1, n+2, and n+3 take all possible pairs (n, n+1), (n, n+2), (n, n+3), (n+1, n+2), (n+1, n+3), (n+2, n+3) to create unreduced Pythagorean triangles. The sum of all six areas is 60*a(n+1).
Using three consecutive odd numbers j, k, m, (j+k+m)^3 - (j^3 + k^3 + m^3) equals 576*a(n) = 24^2*a(n) where n = (j+1)/2. (End)
From Ant King, Oct 17 2012: (Start)
For n > 0, the digital roots of this sequence A010888(a(n)) form the purely periodic 27-cycle {1, 5, 5, 3, 1, 1, 5, 6, 6, 7, 2, 2, 9, 7, 7, 2, 3, 3, 4, 8, 8, 6, 4, 4, 8, 9, 9}.
For n > 0, the units' digits of this sequence A010879(a(n)) form the purely periodic 20-cycle {1, 5, 4, 0, 5, 1, 0, 4, 5, 5, 6, 0, 9, 5, 0, 6, 5, 9, 0, 0}. (End)
Length of the Pisano period of this sequence mod n, n>=1: 1, 4, 9, 8, 5, 36, 7, 16, 27, 20, 11, 72, 13, 28, 45, 32, 17, 108, 19, 40, ... . - R. J. Mathar, Oct 17 2012
Sum of entries of n X n square matrix with elements min(i,j). - Enrique Pérez Herrero, Jan 16 2013
The number of intersections of diagonals in the interior of regular n-gon for odd n > 1 divided by n is a square pyramidal number; that is, A006561(2*n+1)/(2*n+1) = A000330(n-1) = (1/6)*n*(n-1)*(2*n-1). - Martin Renner, Mar 06 2013
For n > 1, a(n)/(2n+1) = A024702(m), for n such that 2n+1 = prime, which results in 2n+1 = A000040(m). For example, for n = 8, 2n+1 = 17 = A000040(7), a(8) = 204, 204/17 = 12 = A024702(7). - Richard R. Forberg, Aug 20 2013
A formula for the r-th successive summation of k^2, for k = 1 to n, is (2*n+r)*(n+r)!/((r+2)!*(n-1)!) (H. W. Gould). - Gary Detlefs, Jan 02 2014
The n-th square pyramidal number = the n-th triangular dipyramidal number (Johnson 12), which is the sum of the n-th + (n-1)-st tetrahedral numbers. E.g., the 3rd tetrahedral number is 10 = 1+3+6, the 2nd is 4 = 1+3. In triangular "dipyramidal form" these numbers can be written as 1+3+6+3+1 = 14. For "square pyramidal form", rebracket as 1+(1+3)+(3+6) = 14. - John F. Richardson, Mar 27 2014
Beukers and Top prove that no square pyramidal number > 1 equals a tetrahedral number A000292. - Jonathan Sondow, Jun 21 2014
Odd numbered entries are related to dissections of polygons through A100157. - Tom Copeland, Oct 05 2014
From Bui Quang Tuan, Apr 03 2015: (Start)
We construct a number triangle from the integers 1, 2, 3, ..., n as follows. The first column contains 2*n-1 integers 1. The second column contains 2*n-3 integers 2, ... The last column contains only one integer n. The sum of all the numbers in the triangle is a(n).
Here is an example with n = 5:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
(End)
The Catalan number series A000108(n+3), offset 0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset 0 (empirical observation). - Tony Foster III, Sep 05 2016; see Dougherty et al. link p. 2. - Andrey Zabolotskiy, Oct 13 2016
Number of floating point additions in the factorization of an (n+1) X (n+1) real matrix by Gaussian elimination as e.g. implemented in LINPACK subroutines sgefa.f or dgefa.f. The number of multiplications is given by A007290. - Hugo Pfoertner, Mar 28 2018
The Jacobi polynomial P(n-1,-n+2,2,3) or equivalently the sum of dot products of vectors from the first n rows of Pascal's triangle (A007318) with the up-diagonal Chebyshev T coefficient vector (1,3,2,0,...) (A053120) or down-diagonal vector (1,-7,32,-120,400,...) (A001794). a(5) = 1 + (1,1).(1,3) + (1,2,1).(1,3,2) + (1,3,3,1).(1,3,2,0) + (1,4,6,4,1).(1,3,2,0,0) = (1 + (1,1).(1,-7) + (1,2,1).(1,-7,32) + (1,3,3,1).(1,-7,32,-120) + (1,4,6,4,1).(1,-7,32,-120,400))*(-1)^(n-1) = 55. - Richard Turk, Jul 03 2018
Coefficients in the terminating series identity 1 - 5*n/(n + 4) + 14*n*(n - 1)/((n + 4)*(n + 5)) - 30*n*(n - 1)*(n - 2)/((n + 4)*(n + 5)*(n + 6)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A108674. - Peter Bala, Feb 12 2019
n divides a(n) iff n == +- 1 (mod 6) (see A007310). (See De Koninck reference.) Examples: a(11) = 506 = 11 * 46, and a(13) = 819 = 13 * 63. - Bernard Schott, Jan 10 2020
For n > 0, a(n) is the number of ternary words of length n+2 having 3 letters equal to 2 and 0 only occurring as the last letter. For example, for n=2, the length 4 words are 2221,2212,2122,1222,2220. - Milan Janjic, Jan 28 2020
Conjecture: Every integer can be represented as a sum of three generalized square pyramidal numbers. A related conjecture is given in A336205 corresponding to pentagonal case. A stronger version of these conjectures is that every integer can be expressed as a sum of three generalized r-gonal pyramidal numbers for all r >= 3. In here "generalized" means negative indices are included. - Altug Alkan, Jul 30 2020
The natural number y is a term if and only if y = a(floor((3 * y)^(1/3))). - Robert Israel, Dec 04 2024
Also the number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by a rotation of the board. Reflections are ignored. Equivalently, number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by an axial reflection of the board (horizontal or vertical). Rotations and diagonal reflections are ignored. - Hilko Koning, Aug 22 2025

Examples

			G.f. = x + 5*x^2 + 14*x^3 + 30*x^4 + 55*x^5 + 91*x^6 + 140*x^7 + 204*x^8 + ...
		

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. 813.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover Publications, NY, 1964, p. 194.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 215,223.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 122, see #19 (3(1)), I(n); p. 155.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 47-49.
  • 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.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p.165).
  • J. M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 310, pp. 46-196, Ellipses, Paris, 2004.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
  • M. Gardner, Fractal Music, Hypercards and More, Freeman, NY, 1991, p. 293.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 293.
  • M. Holt, Math puzzles and games, Walker Publishing Company, 1977, p. 2 and p. 89.
  • Simon Singh, The Simpsons and Their Mathematical Secrets. London: Bloomsbury Publishing PLC (2013): 188.
  • 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).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 126.

Crossrefs

Sums of 2 consecutive terms give A005900.
Column 0 of triangle A094414.
Column 1 of triangle A008955.
Right side of triangle A082652.
Row 2 of array A103438.
Partial sums of A000290.
Cf. similar sequences listed in A237616 and A254142.
Cf. |A084930(n, 1)|.
Cf. A253903 (characteristic function).
Cf. A034705 (differences of any two terms).

Programs

  • GAP
    List([0..30], n-> n*(n+1)*(2*n+1)/6); # G. C. Greubel, Dec 31 2019
  • Haskell
    a000330 n = n * (n + 1) * (2 * n + 1) `div` 6
    a000330_list = scanl1 (+) a000290_list
    -- Reinhard Zumkeller, Nov 11 2012, Feb 03 2012
    
  • Magma
    [n*(n+1)*(2*n+1)/6: n in [0..50]]; // Wesley Ivan Hurt, Jun 28 2014
    
  • Magma
    [0] cat [((2*n+3)*Binomial(n+2,2))/3: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014
    
  • Maple
    A000330 := n -> n*(n+1)*(2*n+1)/6;
    a := n->(1/6)*n*(n+1)*(2*n+1): seq(a(n),n=0..53); # Emeric Deutsch
    with(combstruct): ZL:=[st, {st=Prod(left, right), left=Set(U, card=r), right=Set(U, card=r), U=Sequence(Z, card>=1)}, unlabeled]: subs(r=1, stack): seq(count(subs(r=2, ZL), size=m*2), m=1..45) ; # Zerinvary Lajos, Jan 02 2008
    nmax := 44; for n from 0 to nmax do fz(n) := product( (1-(2*m-1)*z)^(n+1-m) , m=1..n); c(n) := abs(coeff(fz(n),z,1)); end do: a := n-> c(n): seq(a(n), n=0..nmax); # Johannes W. Meijer, Mar 07 2009
  • Mathematica
    Table[Binomial[w+2, 3] + Binomial[w+1, 3], {w, 0, 30}]
    CoefficientList[Series[x(1+x)/(1-x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 30 2014 *)
    Accumulate[Range[0,50]^2] (* Harvey P. Dale, Sep 25 2014 *)
  • Maxima
    A000330(n):=binomial(n+2,3)+binomial(n+1,3)$
    makelist(A000330(n),n,0,20); /* Martin Ettl, Nov 12 2012 */
    
  • PARI
    {a(n) = n * (n+1) * (2*n+1) / 6};
    
  • PARI
    upto(n) = [x*(x+1)*(2*x+1)/6 | x<-[0..n]] \\ Cino Hilliard, Jun 18 2007, edited by M. F. Hasler, Jan 02 2024
    
  • Python
    a=lambda n: (n*(n+1)*(2*n+1))//6 # Indranil Ghosh, Jan 04 2017
    
  • Sage
    [n*(n+1)*(2*n+1)/6 for n in (0..30)] # G. C. Greubel, Dec 31 2019
    

Formula

G.f.: x*(1+x)/(1-x)^4. - Simon Plouffe (in his 1992 dissertation: generating function for sequence starting at a(1))
E.g.f.: (x + 3*x^2/2 + x^3/3)*exp(x).
a(n) = n*(n+1)*(2*n+1)/6 = binomial(n+2, 3) + binomial(n+1, 3).
2*a(n) = A006331(n). - N. J. A. Sloane, Dec 11 1999
Can be extended to Z with a(n) = -a(-1-n) for all n in Z.
a(n) = A002492(n)/4. - Paul Barry, Jul 19 2003
a(n) = (((n+1)^4 - n^4) - ((n+1)^2 - n^2))/12. - Xavier Acloque, Oct 16 2003
From Alexander Adamchuk, Oct 26 2004: (Start)
a(n) = sqrt(A271535(n)).
a(n) = (Sum_{k=1..n} Sum_{j=1..n} Sum_{i=1..n} (i*j*k)^2)^(1/3). (End)
a(n) = Sum_{i=1..n} i*(2*n-2*i+1); sum of squares gives 1 + (1+3) + (1+3+5) + ... - Jon Perry, Dec 08 2004
a(n+1) = A000217(n+1) + 2*A000292(n). - Creighton Dement, Mar 10 2005
Sum_{n>=1} 1/a(n) = 6*(3-4*log(2)); Sum_{n>=1} (-1)^(n+1)*1/a(n) = 6*(Pi-3). - Philippe Deléham, May 31 2005
Sum of two consecutive tetrahedral (or pyramidal) numbers a(n) = A000292(n-1) + A000292(n). - Alexander Adamchuk, May 17 2006
Euler transform of length-2 sequence [ 5, -1 ]. - Michael Somos, Sep 04 2006
a(n) = a(n-1) + n^2. - Rolf Pleisch, Jul 22 2007
a(n) = A132121(n,0). - Reinhard Zumkeller, Aug 12 2007
a(n) = binomial(n, 2) + 2*binomial(n, 3). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009, corrected by M. F. Hasler, Jan 02 2024
a(n) = A168559(n) + 1 for n > 0. - Reinhard Zumkeller, Feb 03 2012
a(n) = Sum_{i=1..n} J_2(i)*floor(n/i), where J_2 is A007434. - Enrique Pérez Herrero, Feb 26 2012
a(n) = s(n+1, n)^2 - 2*s(n+1, n-1), where s(n, k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 03 2012
a(n) = A001477(n) + A000217(n) + A007290(n+2) + 1. - J. M. Bergot, May 31 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 2. - Ant King, Oct 17 2012
a(n) = Sum_{i = 1..n} Sum_{j = 1..n} min(i,j). - Enrique Pérez Herrero, Jan 15 2013
a(n) = A000217(n) + A007290(n+1). - Ivan N. Ianakiev, May 10 2013
a(n) = (A047486(n+2)^3 - A047486(n+2))/24. - Richard R. Forberg, Dec 25 2013
a(n) = Sum_{i=0..n-1} (n-i)*(2*i+1), with a(0) = 0. After 0, row sums of the triangle in A101447. - Bruno Berselli, Feb 10 2014
a(n) = n + 1 + Sum_{i=1..n+1} (i^2 - 2i). - Wesley Ivan Hurt, Feb 25 2014
a(n) = A000578(n+1) - A002412(n+1). - Wesley Ivan Hurt, Jun 28 2014
a(n) = Sum_{i = 1..n} Sum_{j = i..n} max(i,j). - Enrique Pérez Herrero, Dec 03 2014
a(n) = A055112(n)/6, see Singh (2013). - Alonso del Arte, Feb 20 2015
For n >= 2, a(n) = A028347(n+1) + A101986(n-2). - Bui Quang Tuan, Apr 03 2015
For n > 0: a(n) = A258708(n+3,n-1). - Reinhard Zumkeller, Jun 23 2015
a(n) = A175254(n) + A072481(n), n >= 1. - Omar E. Pol, Aug 12 2015
a(n) = A000332(n+3) - A000332(n+1). - Antal Pinter, Dec 27 2015
Dirichlet g.f.: zeta(s-3)/3 + zeta(s-2)/2 + zeta(s-1)/6. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A080851(2,n-1). - R. J. Mathar, Jul 28 2016
a(n) = (A005408(n) * A046092(n))/12 = (2*n+1)*(2*n*(n+1))/12. - Bruce J. Nicholson, May 18 2017
12*a(n) = (n+1)*A001105(n) + n*A001105(n+1). - Bruno Berselli, Jul 03 2017
a(n) = binomial(n-1, 1) + binomial(n-1, 2) + binomial(n, 3) + binomial(n+1, 2) + binomial(n+1, 3). - Tony Foster III, Aug 24 2018
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Nathan Fox, Dec 04 2019
Let T(n) = A000217(n), the n-th triangular number. Then a(n) = (T(n)+1)^2 + (T(n)+2)^2 + ... + (T(n)+n)^2 - (n+2)*T(n)^2. - Charlie Marion, Dec 31 2019
a(n) = 2*n - 1 - a(n-2) + 2*a(n-1). - Boštjan Gec, Nov 09 2023
a(n) = 2/(2*n)! * Sum_{j = 1..n} (-1)^(n+j) * j^(2*n+2) * binomial(2*n, n-j). Cf. A060493. - Peter Bala, Mar 31 2025

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A000537 Sum of first n cubes; or n-th triangular number squared.

Original entry on oeis.org

0, 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025, 4356, 6084, 8281, 11025, 14400, 18496, 23409, 29241, 36100, 44100, 53361, 64009, 76176, 90000, 105625, 123201, 142884, 164836, 189225, 216225, 246016, 278784, 314721, 354025, 396900, 443556, 494209, 549081
Offset: 0

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Author

Keywords

Comments

Number of parallelograms in an n X n rhombus. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Or, number of orthogonal rectangles in an n X n checkerboard, or rectangles in an n X n array of squares. - Jud McCranie, Feb 28 2003. Compare A085582.
Also number of 2-dimensional cage assemblies (cf. A059827, A059860).
The n-th triangular number T(n) = Sum_{r=1..n} r = n(n+1)/2 satisfies the relations: (i) T(n) + T(n-1) = n^2 and (ii) T(n) - T(n-1) = n by definition, so that n^2*n = n^3 = {T(n)}^2 - {T(n-1)}^2 and by summing on n we have Sum_{ r = 1..n } r^3 = {T(n)}^2 = (1+2+3+...+n)^2 = (n*(n+1)/2)^2. - Lekraj Beedassy, May 14 2004
Number of 4-tuples of integers from {0,1,...,n}, without repetition, whose last component is strictly bigger than the others. Number of 4-tuples of integers from {1,...,n}, with repetition, whose last component is greater than or equal to the others.
Number of ordered pairs of two-element subsets of {0,1,...,n} without repetition.
Number of ordered pairs of 2-element multisubsets of {1,...,n} with repetition.
1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2.
a(n) is the number of parameters needed in general to know the Riemannian metric g of an n-dimensional Riemannian manifold (M,g), by knowing all its second derivatives; even though to know the curvature tensor R requires (due to symmetries) (n^2)*(n^2-1)/12 parameters, a smaller number (and a 4-dimensional pyramidal number). - Jonathan Vos Post, May 05 2006
Also number of hexagons with vertices in an hexagonal grid with n points in each side. - Ignacio Larrosa Cañestro, Oct 15 2006
Number of permutations of n distinct letters (ABCD...) each of which appears twice with 4 and n-4 fixed points. - Zerinvary Lajos, Nov 09 2006
With offset 1 = binomial transform of [1, 8, 19, 18, 6, ...]. - Gary W. Adamson, Dec 03 2008
The sequence is related to A000330 by a(n) = n*A000330(n) - Sum_{i=0..n-1} A000330(i): this is the case d=1 in the identity n*(n*(d*n-d+2)/2) - Sum_{i=0..n-1} i*(d*i-d+2)/2 = n*(n+1)*(2*d*n-2*d+3)/6. - Bruno Berselli, Apr 26 2010, Mar 01 2012
From Wolfdieter Lang, Jan 11 2013: (Start)
For sums of powers of positive integers S(k,n) := Sum_{j=1..n}j^k one has the recurrence S(k,n) = (n+1)*S(k-1,n) - Sum_{l=1..n} S(k-1,l), n >= 1, k >= 1.
This was used for k=4 by Ibn al-Haytham in an attempt to compute the volume of the interior of a paraboloid. See the Strick reference where the trick he used is shown, and the W. Lang link.
This trick generalizes immediately to arbitrary powers k. For k=3: a(n) = (n+1)*A000330(n) - Sum_{l=1..n} A000330(l), which coincides with the formula given in the previous comment by Berselli. (End)
Regarding to the previous contribution, see also Matem@ticamente in Links field and comments on this recurrences in similar sequences (partial sums of n-th powers). - Bruno Berselli, Jun 24 2013
A rectangular prism with sides A000217(n), A000217(n+1), and A000217(n+2) has surface area 6*a(n+1). - J. M. Bergot, Aug 07 2013, edited with corrected indices by Antti Karttunen, Aug 09 2013
A formula for the r-th successive summation of k^3, for k = 1 to n, is (6*n^2+r*(6*n+r-1)*(n+r)!)/((r+3)!*(n-1)!), (H. W. Gould). - Gary Detlefs, Jan 02 2014
Note that this sequence and its formula were known to (and possibly discovered by) Nicomachus, predating Ibn al-Haytham by 800 years. - Charles R Greathouse IV, Apr 23 2014
a(n) is the number of ways to paint the sides of a nonsquare rectangle using at most n colors. Cf. A039623. - Geoffrey Critzer, Jun 18 2014
For n > 0: A256188(a(n)) = A000217(n) and A256188(m) != A000217(n) for m < a(n), i.e., positions of first occurrences of triangular numbers in A256188. - Reinhard Zumkeller, Mar 26 2015
There is no cube in this sequence except 0 and 1. - Altug Alkan, Jul 02 2016
Also the number of chordless cycles in the complete bipartite graph K_{n+1,n+1}. - Eric W. Weisstein, Jan 02 2018
a(n) is the sum of the elements in the multiplication table [0..n] X [0..n]. - Michel Marcus, May 06 2021

Examples

			G.f. = x + 9*x^2 + 36*x^3 + 100*x^4 + 225*x^5 + 441*x^6 + ... - _Michael Somos_, Aug 29 2022
		

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. 813.
  • Avner Ash and Robert Gross, Summing it up, Princeton University Press, 2016, p. 62, eq. (6.3) for k=3.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 110ff.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • John H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, pp. 36, 58.
  • Clifford Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Oxford University Press, 2001, p. 325.
  • 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).
  • H. K. Strick, Geschichten aus der Mathematik II, Spektrum Spezial 3/11, p. 13.
  • D. Wells, You Are A Mathematician, "Counting rectangles in a rectangle", Problem 8H, pp. 240; 254, Penguin Books 1995.

Crossrefs

Convolution of A000217 and A008458.
Row sums of triangles A094414 and A094415.
Second column of triangle A008459.
Row 3 of array A103438.
Cf. A236770 (see crossrefs).

Programs

  • GAP
    List([0..40],n->(n*(n+1)/2)^2); # Muniru A Asiru, Dec 05 2018
    
  • Haskell
    a000537 = a000290 . a000217  -- Reinhard Zumkeller, Mar 26 2015
    
  • Magma
    [(n*(n+1)/2)^2: n in [0..50]]; // Wesley Ivan Hurt, Jun 06 2014
    
  • Maple
    a:= n-> (n*(n+1)/2)^2:
    seq(a(n), n=0..40);
  • Mathematica
    Accumulate[Range[0, 50]^3] (* Harvey P. Dale, Mar 01 2011 *)
    f[n_] := n^2 (n + 1)^2/4; Array[f, 39, 0] (* Robert G. Wilson v, Nov 16 2012 *)
    Table[CycleIndex[{{1, 2, 3, 4}, {3, 2, 1, 4}, {1, 4, 3, 2}, {3, 4, 1, 2}}, s] /. Table[s[i] -> n, {i, 1, 2}], {n, 0, 30}] (* Geoffrey Critzer, Jun 18 2014 *)
    Accumulate @ Range[0, 50]^2 (* Waldemar Puszkarz, Jan 24 2015 *)
    Binomial[Range[20], 2]^2 (* Eric W. Weisstein, Jan 02 2018 *)
    LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 9, 36, 100}, 20] (* Eric W. Weisstein, Jan 02 2018 *)
    CoefficientList[Series[-((x (1 + 4 x + x^2))/(-1 + x)^5), {x, 0, 20}], x] (* Eric W. Weisstein, Jan 02 2018 *)
  • PARI
    a(n)=(n*(n+1)/2)^2
    
  • Python
    def A000537(n): return (n*(n+1)>>1)**2 # Chai Wah Wu, Oct 20 2023

Formula

a(n) = (n*(n+1)/2)^2 = A000217(n)^2 = Sum_{k=1..n} A000578(k), that is, 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2.
G.f.: (x+4*x^2+x^3)/(1-x)^5. - Simon Plouffe in his 1992 dissertation
a(n) = Sum ( Sum ( 1 + Sum (6*n) ) ), rephrasing the formula in A000578. - Xavier Acloque, Jan 21 2003
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j, row sums of A127777. - Alexander Adamchuk, Oct 24 2004
a(n) = A035287(n)/4. - Zerinvary Lajos, May 09 2007
This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=1. - Alexander R. Povolotsky, May 17 2008
G.f.: x*F(3,3;1;x). - Paul Barry, Sep 18 2008
Sum_{k > 0} 1/a(k) = (4/3)*(Pi^2-9). - Jaume Oliver Lafont, Sep 20 2009
a(n) = Sum_{1 <= k <= m <= n} A176271(m,k). - Reinhard Zumkeller, Apr 13 2010
a(n) = Sum_{i=1..n} J_3(i)*floor(n/i), where J_ 3 is A059376. - Enrique Pérez Herrero, Feb 26 2012
a(n) = Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} min(i,j,k). - Enrique Pérez Herrero, Feb 26 2013 [corrected by Ridouane Oudra, Mar 05 2025]
a(n) = 6*C(n+2,4) + C(n+1,2) = 6*A000332(n+2) + A000217(n), (Knuth). - Gary Detlefs, Jan 02 2014
a(n) = -Sum_{j=1..3} j*Stirling1(n+1,n+1-j)*Stirling2(n+3-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*(3-4*log(2)). - Vaclav Kotesovec, Feb 13 2015
a(n)*((s-2)*(s-3)/2) = P(3, P(s, n+1)) - P(s, P(3, n+1)), where P(s, m) = ((s-2)*m^2-(s-4)*m)/2 is the m-th s-gonal number. For s=7, 10*a(n) = A000217(A000566(n+1)) - A000566(A000217(n+1)). - Bruno Berselli, Aug 04 2015
From Ilya Gutkovskiy, Jul 03 2016: (Start)
E.g.f.: x*(4 + 14*x + 8*x^2 + x^3)*exp(x)/4.
Dirichlet g.f.: (zeta(s-4) + 2*zeta(s-3) + zeta(s-2))/4. (End)
a(n) = (Bernoulli(4, n+1) - Bernoulli(4, 1))/4, n >= 0, with the Bernoulli polynomial B(4, x) from row n=4 of A053382/A053383. See, e.g., the Ash-Gross reference, p. 62, eq. (6.3) for k=3. - Wolfdieter Lang, Mar 12 2017
a(n) = A000217((n+1)^2) - A000217(n+1)^2. - Bruno Berselli, Aug 31 2017
a(n) = n*binomial(n+2, 3) + binomial(n+2, 4) + binomial(n+1, 4). - Tony Foster III, Nov 14 2017
Another identity: ..., a(3) = (1/2)*(1*(2+4+6)+3*(4+6)+5*6) = 36, a(4) = (1/2)*(1*(2+4+6+8)+3*(4+6+8)+5*(6+8)+7*(8)) = 100, a(5) = (1/2)*(1*(2+4+6+8+10)+3*(4+6+8+10)+5*(6+8+10)+7*(8+10)+9*(10)) = 225, ... - J. M. Bergot, Aug 27 2022
Comment from Michael Somos, Aug 28 2022: (Start)
The previous comment expresses a(n) as the sum of all of the n X n multiplication table array entries.
For example, for n = 4:
1 2 3 4
2 4 6 8
3 6 9 12
4 8 12 16
This array sum can be split up as follows:
+---+---------------+
| 0 | 1 2 3 4 | (0+1)*(1+2+3+4)
| +---+-----------+
| 0 | 2 | 4 6 8 | (1+2)*(2+3+4)
| | +---+-------+
| 0 | 3 | 6 | 9 12 | (2+3)*(3+4)
| | | +---+---+
| 0 | 4 | 8 |12 |16 | (3+4)*(4)
+---+---+---+---+---+
This kind of row+column sums was used by Ramanujan and others for summing Lambert series. (End)
a(n) = 6*A000332(n+4) - 12*A000292(n+1) + 7*A000217(n+1) - n - 1. - Adam Mohamed, Sep 05 2024

Extensions

Edited by M. F. Hasler, May 02 2015

A001550 a(n) = 1^n + 2^n + 3^n.

Original entry on oeis.org

3, 6, 14, 36, 98, 276, 794, 2316, 6818, 20196, 60074, 179196, 535538, 1602516, 4799354, 14381676, 43112258, 129271236, 387682634, 1162785756, 3487832978, 10462450356, 31385253914, 94151567436, 282446313698, 847322163876
Offset: 0

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Author

Keywords

Comments

a(n)*(-1)^n, n>=0, gives the z-sequence for the Sheffer triangle A049458 ((signed) 3-restricted Stirling1 numbers), which is the inverse triangle of A143495 with offset [0,0] (3-restricted Stirling2 numbers). See the W. Lang link under A006232 for a- and z-sequences for Sheffer matrices. The a-sequence for each (signed) r-restricted Stirling1 Sheffer triangle is A027641/A027642 (Bernoulli numbers). - Wolfdieter Lang, Oct 10 2011

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. 813.
  • 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

Programs

  • Haskell
    a001550 n = sum $ map (^ n) [1..3]  -- Reinhard Zumkeller, Mar 01 2012
    
  • Magma
    [1^n + 2^n + 3^n : n in [0..30]]; // Wesley Ivan Hurt, Jun 25 2020
    
  • Maple
    A001550:=-(3-12*z+11*z^2)/(z-1)/(3*z-1)/(2*z-1); # Simon Plouffe in his 1992 dissertation.
  • Mathematica
    Table[1^n + 2^n + 3^n, {n, 0, 30}]
    CoefficientList[Series[(3-12x+11x^2)/(1-6x+11x^2-6x^3),{x,0,30}],x] (* or *) LinearRecurrence[{6,-11,6},{3,6,14},31] (* Harvey P. Dale, Apr 30 2011 *)
    Total[Range[3]^#]&/@Range[0,30] (* Harvey P. Dale, Sep 23 2019 *)
  • PARI
    a(n)=1+2^n+3^n \\ Charles R Greathouse IV, Jun 10 2011
    
  • Python
    def A001550(n): return 3**n+(1<Chai Wah Wu, Nov 01 2024

Formula

From Michael Somos: (Start)
G.f.: (3 -12*x +11*x^2)/(1 -6*x +11*x^2 -6*x^3).
a(n) = 5*a(n-1) - 6*a(n-2) + 2. (End)
E.g.f.: exp(x) + exp(2*x) + exp(3*x). - Mohammad K. Azarian, Dec 26 2008
a(0)=3, a(1)=6, a(2)=14, a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3). - Harvey P. Dale, Apr 30 2011
a(n) = A007689(n) + 1. - Reinhard Zumkeller, Mar 01 2012
From Kai Wang, May 18 2020: (Start)
a(n) = 3*A000392(n+3) - 12*A000392(n+2) + 11*A000392(n+1).
A000392(n) = (3*a(n+1) - 12*a(n) + 10*a(n-1))/2. (End)

Extensions

Additional terms from Michael Somos
Attribute "conjectured" removed from Simon Plouffe's g.f. by R. J. Mathar, Mar 11 2009

A031971 a(n) = Sum_{k=1..n} k^n.

Original entry on oeis.org

1, 5, 36, 354, 4425, 67171, 1200304, 24684612, 574304985, 14914341925, 427675990236, 13421957361110, 457593884876401, 16841089312342855, 665478473553144000, 28101527071305611528, 1262899292504270591313, 60182438244917445266889, 3031284048960901518840700
Offset: 1

Views

Author

Chris du Feu (chris(AT)beckingham0.demon.co.uk)

Keywords

Comments

From Alexander Adamchuk, Jul 21 2006: (Start)
p^(3k - 1) divides a(p^k) for prime p > 2 and k = 1, 2, 3, 4, ... or p^2 divides a(p) for prime p > 2. p^5 divides a(p^2) for prime p > 2. p^8 divides a(p^3) for prime p > 2. p^11 divides a(p^4) for prime p > 2.
p^2 divides a(2p) for prime p > 3. p^3 divides a(3p) for prime p > 2. p^2 divides a(4p) for prime p > 5. p^3 divides a(5p) for prime p > 3. p^2 divides a(6p) for prime p > 7.
p divides a(2p - 1) for all prime p. p^3 divides a(2p^2 - 1) for all prime p. p^5 divides a(2p^3 - 1) for all prime p.
p divides a((p - 1)/2) for p = 5, 13, 17, 29, 37, 41, 53, 61, ... = A002144 Pythagorean primes: primes of form 4n + 1.
(End)
If p prime then a(p-1) == -1 (mod p) [see De Koninck & Mercier reference]. Example: for p = 7, a(6) = 67171 = 7 * 9596 - 1. - Bernard Schott, Mar 06 2020

References

  • J.-M. De Koninck et A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 327 pp. 48-200, Ellipses, Paris (2004).
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 21.

Crossrefs

A diagonal of array A103438.
For a(n) mod n see A182398.

Programs

Formula

a(n) is asymptotic to (e/(e - 1))*n^n. - Benoit Cloitre, Dec 17 2003
a(n) = zeta(-n) - zeta(-n, n + 1), where zeta(s) is the Riemann zeta function and zeta(s, a) is the Hurwitz zeta function, a generalization of the Riemann zeta function. - Alexander Adamchuk, Jul 21 2006
a(n) == 1 (mod n) <==> n is in A014117 = 1, 2, 6, 42, 1806 (see the link "On the congruence ..."). - Jonathan Sondow, Oct 18 2013
a(A054377(n)) = A233045(n). - Jonathan Sondow, Dec 11 2013
a(n) = n! * [x^n] exp(x)*(exp(n*x) - 1)/(exp(x) - 1). - Ilya Gutkovskiy, Apr 07 2018
a(n) ~ ((e*n+1)/((e-1)*(n+1))) * n^n. - N. J. A. Sloane, Oct 13 2018, based on email from Claude F. Leibovici who claims this is slightly better than Cloitre's version when n is small.

A000538 Sum of fourth powers: 0^4 + 1^4 + ... + n^4.

Original entry on oeis.org

0, 1, 17, 98, 354, 979, 2275, 4676, 8772, 15333, 25333, 39974, 60710, 89271, 127687, 178312, 243848, 327369, 432345, 562666, 722666, 917147, 1151403, 1431244, 1763020, 2153645, 2610621, 3142062, 3756718, 4463999, 5273999, 6197520, 7246096, 8432017, 9768353
Offset: 0

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Keywords

Comments

This sequence is related to A000537 by the transform a(n) = n*A000537(n) - Sum_{i=0..n-1} A000537(i). - Bruno Berselli, Apr 26 2010
A formula for the r-th successive summation of k^4, for k = 1 to n, is ((12*n^2+(12*n-5)*r+r^2)*(2*n+r)*(n+r)!)/((r+4)!*(n-1)!), (H. W. Gould). - Gary Detlefs, Jan 02 2014
The number of four dimensional hypercubes in a 4D grid with side lengths n. This applies in general to k dimensions. That is, the number of k-dimensional hypercubes in a k-dimensional grid with side lengths n is equal to the sum of 1^k + 2^k + ... + n^k. - Alejandro Rodriguez, Oct 20 2020

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. 813.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 222.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1991, p. 275.
  • 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

Programs

  • Haskell
    a000538 n = (3 * n * (n + 1) - 1) * (2 * n + 1) * (n + 1) * n `div` 30
    -- Reinhard Zumkeller, Nov 11 2012
    
  • Magma
    [n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30: n in [0..35]]; // Vincenzo Librandi, Apr 04 2015
  • Maple
    A000538 := n-> n*(n+1)*(2*n+1)*(3*n^2+3*n-1)/30;
  • Mathematica
    Accumulate[Range[0,40]^4] (* Harvey P. Dale, Jan 13 2011 *)
    CoefficientList[Series[x (1 + 11 x + 11 x^2 + x^3)/(1 - x)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 07 2015 *)
    LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 17, 98, 354, 979}, 35] (* Jean-François Alcover, Feb 09 2016 *)
    Table[x^5/5+x^4/2+x^3/3-x/30,{x,40}] (* Harvey P. Dale, Jun 06 2021 *)
  • Maxima
    A000538(n):=n*(n+1)*(2*n+1)*(3*n^2+3*n-1)/30$
    makelist(A000538(n),n,0,30); /* Martin Ettl, Nov 12 2012 */
    
  • PARI
    a(n) = n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30 \\ Charles R Greathouse IV, Nov 20 2012
    
  • PARI
    concat(0, Vec(x*(1+11*x+11*x^2+x^3)/(1-x)^6 + O(x^100))) \\ Altug Alkan, Dec 07 2015
    
  • Python
    A000538_list, m = [0], [24, -36, 14, -1, 0, 0]
    for _ in range(10**2):
        for i in range(5):
            m[i+1] += m[i]
        A000538_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    
  • Python
    def A000538(n): return n*(n**2*(n*(6*n+15)+10)-1)//30 # Chai Wah Wu, Oct 03 2024
    
  • Sage
    [bernoulli_polynomial(n,5)/5 for n in range(1, 35)] # Zerinvary Lajos, May 17 2009
    

Formula

a(n) = n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30.
The preceding formula is due to al-Kachi (1394-1437). - Juri-Stepan Gerasimov, Jul 12 2009
G.f.: x*(x+1)*(1+10*x+x^2)/(1-x)^6. Simon Plouffe in his 1992 dissertation. More generally, the o.g.f. for Sum_{k=0..n} k^m is x*E(m, x)/(1-x)^(m+2), where E(m, x) is the Eulerian polynomial of degree m (cf. A008292). The e.g.f. for these o.g.f.s is: x/(1-x)^2*(exp(y/(1-x))-exp(x*y/(1-x)))/(exp(x*y/(1-x))-x*exp(y/(1-x))). - Vladeta Jovovic, May 08 2002
a(n) = Sum_{i = 1..n} J_4(i)*floor(n/i), where J_4 is A059377. - Enrique Pérez Herrero, Feb 26 2012
a(n) = 5*a(n-1) - 10* a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + 24. - Ant King, Sep 23 2013
a(n) = -Sum_{j=1..4} j*Stirling1(n+1,n+1-j)*Stirling2(n+4-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = -30*(4 + 3/cos(sqrt(7/3)*Pi/2))*Pi/7. - Vaclav Kotesovec, Feb 13 2015
a(n) = (n + 1)*(n + 1/2)*n*(n + 1/2 + sqrt(7/12))*(n + 1/2 - sqrt(7/12))/5, see the Graham et al. reference, p. 275. - Wolfdieter Lang, Apr 02 2015

Extensions

The general V. Jovovic formula has been slightly changed after his approval by Wolfdieter Lang, Nov 03 2011

A000539 Sum of 5th powers: 0^5 + 1^5 + 2^5 + ... + n^5.

Original entry on oeis.org

0, 1, 33, 276, 1300, 4425, 12201, 29008, 61776, 120825, 220825, 381876, 630708, 1002001, 1539825, 2299200, 3347776, 4767633, 6657201, 9133300, 12333300, 16417401, 21571033, 28007376, 35970000, 45735625, 57617001, 71965908, 89176276, 109687425, 133987425, 162616576
Offset: 0

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Comments

This sequence is related to A000538 by a(n) = n*A000538(n) - Sum_{i=0..n-1} A000538(i). - Bruno Berselli, Apr 26 2010
See comment in A008292 for a formula for r-th successive summation of Sum_{k=1..n} k^j. - Gary Detlefs, Jan 02 2014

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. 813.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1991, p. 275.
  • 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

Partial sums of A000584. Row 5 of array A103438.

Programs

  • Magma
    [n^2*(n+1)^2*(2*n^2+2*n-1)/12: n in [0..30]]; // Vincenzo Librandi, Apr 04 2015
    
  • Maple
    A000539:=-(1+26*z+66*z**2+26*z**3+z**4)/(z-1)**7; # Simon Plouffe in his 1992 dissertation
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^5 od: seq(a[n], n=0..30); # Zerinvary Lajos, Feb 22 2008
    a:=n->sum(j^5,j=0..n): seq(a(n), n=0..30); # Zerinvary Lajos, Jun 05 2008
  • Mathematica
    Accumulate[Range[0, 40]^5]
    LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {0, 1, 33, 276, 1300, 4425, 12201}, 41] (* Jean-François Alcover, Feb 09 2016 *)
  • Maxima
    A000539(n):=n^2*(n+1)^2*(2*n^2+2*n-1)/12$ makelist(A000539(n),n,0,30); /* Martin Ettl, Nov 12 2012 */
    
  • PARI
    a(n)=n^2*(n+1)^2*(2*n^2+2*n-1)/12 \\ Charles R Greathouse IV, Jul 15 2011
    
  • PARI
    concat(0, Vec(x*(1+26*x+66*x^2+26*x^3+x^4)/(1-x)^7 + O(x^100))) \\ Altug Alkan, Dec 07 2015
    
  • Python
    A000539_list, m = [0], [120, -240, 150, -30, 1, 0, 0]
    for _ in range(10**2):
        for i in range(6):
            m[i+1] += m[i]
        A000539_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    
  • Python
    def A000539(n): return n**2*(n**2*(n*(n+3<<1)+5)-1)//12 # Chai Wah Wu, Oct 03 2024

Formula

a(n) = n^2*(n+1)^2*(2*n^2+2*n-1)/12.
a(n) = sqrt(Sum_{j=1..n}Sum_{i=1..n}(i*j)^5). - Alexander Adamchuk, Oct 26 2004
a(n) = Sum_{i = 1..n} J_5(i)*floor(n/i), where J_5 is A059378. - Enrique Pérez Herrero, Feb 26 2012
a(n) = 6*a(n-1) - 15* a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) + 120. - Ant King, Sep 23 2013
a(n) = 120*C(n+3,6) + 30*C(n+2,4) + C(n+1,2) (Knuth). - Gary Detlefs, Jan 02 2014
a(n) = -Sum_{j=1..5} j*Stirling1(n+1,n+1-j)*Stirling2(n+5-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} 1/a(n) = 60 - 4*Pi^2 + 8*sqrt(3)*Pi * tan(sqrt(3)*Pi/2). - Vaclav Kotesovec, Feb 13 2015
a(n) = (n + 1)^2*n^2*(n + 1/2 + sqrt(3/4))*(n + 1/2 - sqrt(3/4))/6. See the Graham et al. reference, p. 275. - Wolfdieter Lang, Apr 02 2015
G.f.: x*(1+26*x+66*x^2+26*x^3+x^4)/(1-x)^7. - Robert Israel, Dec 07 2015
a(n) = (4/3)*A000217(n)^3 - (1/3)*A000217(n)^2. - Michael Raney, Feb 19 2016
a(n) = (binomial(n+1,4) + 6*binomial(n+2,4) + binomial(n+3,4))*(binomial(n+2,3) - binomial(n+1,3)). - Tony Foster III, Oct 21 2018
a(n) = 24*A006542(n+2) + A000537(n). - Yasser Arath Chavez Reyes, May 04 2024
E.g.f.: exp(x)*x*(12 + 186*x + 360*x^2 + 195*x^3 + 36*x^4 + 2*x^5)/12. - Stefano Spezia, May 04 2024

A000540 Sum of 6th powers: 0^6 + 1^6 + 2^6 + ... + n^6.

Original entry on oeis.org

0, 1, 65, 794, 4890, 20515, 67171, 184820, 446964, 978405, 1978405, 3749966, 6735950, 11562759, 19092295, 30482920, 47260136, 71397705, 105409929, 152455810, 216455810, 302221931, 415601835, 563637724, 754740700, 998881325, 1307797101, 1695217590
Offset: 0

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Comments

This sequence is related to A000539 by a(n) = n*A000539(n)-sum(A000539(i), i=0..n-1). - Bruno Berselli, Apr 26 2010

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. 813.
  • J. L. Bailey, Jr., A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, (2008), p. 289.
  • 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

Row 6 of array A103438.
Partial sums of A001014.

Programs

  • Haskell
    a000540 n = a000540_list !! n
    a000540_list = scanl1 (+) a001014_list -- Reinhard Zumkeller, Dec 04 2011
    
  • Magma
    [n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42: n in [0..30]]; // Vincenzo Librandi, Apr 04 2015
  • Maple
    a:=n->sum (j^6,j=0..n): seq(a(n),n=0..27); # Zerinvary Lajos, Jun 27 2007
    A000540:=(z+1)*(z**4+56*z**3+246*z**2+56*z+1)/(z-1)**8; # g.f. by Simon Plouffe in his 1992 dissertation, without the leading 0.
    A000540 := proc(n) n^7/7+n^6/2+n^5/2-n^3/6+n/42 ; end proc: # R. J. Mathar
  • Mathematica
    Accumulate[Range[0,30]^6] (* Harvey P. Dale, Jul 30 2009 *)
    LinearRecurrence[{8, -28, 56, -70, 56, -28, 8, -1}, {0, 1, 65, 794, 4890, 20515, 67171, 184820}, 31] (* Jean-François Alcover, Feb 09 2016 *)
  • PARI
    a(n)=n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42 \\ Edward Jiang, Sep 10 2014
    
  • PARI
    a(n)=sum(i=1, n, i^6); \\ Michel Marcus, Sep 11 2014
    
  • Python
    A000540_list, m = [0], [720, -1800, 1560, -540, 62, -1, 0, 0]
    for _ in range(10**2):
        for i in range(7):
            m[i+1] += m[i]
        A000540_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    
  • Sage
    [bernoulli_polynomial(n,7)/7 for n in range(1, 29)]# Zerinvary Lajos, May 17 2009
    

Formula

a(n) = n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42.
a(n) = sqrt(Sum_{j=1..n} Sum_{i=1..n} (i*j)^6). - Alexander Adamchuk, Oct 26 2004
G.f.: A(x) = 3*x/7*G(0); with G(k) = 1 + 2/(k+1+(k+1)/(2*k^2 + 4*k + 1 + 2*(k+1)^2/(3*k + 2 - 9*x*(k+1)*(k+2)^4*(k+3)*(2*k+5)/(3*x*(k+2)^4*(k+3)*(2*k+5)+(k+1)*(2*k+3)/G(k+1))))); (continued fraction). - Sergei N. Gladkovskii, Dec 03 2011
G.f.: x*(1+x)*(x^4 + 56*x^3 + 246*x^2 + 56*x + 1) / (x-1)^8 . - R. J. Mathar, Aug 07 2012
a(n) = Sum_{i=1..n} J_6(i)*floor(n/i), where J_6 is A069091. - Enrique Pérez Herrero, Mar 09 2013
a(n) = 7*a(n-1) - 21* a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) + 720. - Ant King, Sep 24 2013
a(n) = -Sum_{j=1..6} j*Stirling1(n+1,n+1-j)*Stirling2(n+6-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = 84*Pi*(8*cos(sqrt((sqrt(93) + 9)/6)*Pi) + 15*cos(sqrt((sqrt(93) + 9)/6)*Pi/2) * cosh(sqrt((sqrt(93) - 9)/6)*Pi/2) + 8*cosh(sqrt((sqrt(93) - 9)/6)*Pi) - 7*sqrt(3)*sin(sqrt((sqrt(93) + 9)/6)*Pi/2) * sinh(sqrt((sqrt(93) - 9)/6)*Pi/2)) / (31*(cos(sqrt((sqrt(93) + 9)/6)*Pi) + cosh(sqrt((sqrt(93) - 9)/6)*Pi))) = 0.985708051237101247832970793342271511... . - Vaclav Kotesovec, Feb 13 2015
a(n) = (n + 1)*(n + 1/2)*n*(n + 1/2 + z)*(n + 1/2 - z)*(n + 1/2 + zbar)*(n + 1/2 - zbar)/7, with I^2 = -1 and z = 2^(-3/2)*3^(-1/4)*(sqrt(sqrt(31) + 3*sqrt(3)) + I*sqrt(sqrt(31) - 3*sqrt(3))), and zbar is the complex conjugate of z. See the Graham et al. reference, eq. (6.98), pp. 288-289 (with n -> n+1). (There was a typo in the first edition, which was corrected in the second edition.) - Wolfdieter Lang, Apr 03 2015
a(n+2) = 36*A086020(n+1) + 24*A005585(n+1) + A000330(n+2). - Yasser Arath Chavez Reyes, Apr 16 2024

A000541 Sum of 7th powers: 1^7 + 2^7 + ... + n^7.

Original entry on oeis.org

0, 1, 129, 2316, 18700, 96825, 376761, 1200304, 3297456, 8080425, 18080425, 37567596, 73399404, 136147921, 241561425, 412420800, 680856256, 1091194929, 1703414961, 2597286700, 3877286700, 5678375241, 8172733129, 11577558576, 16164030000, 22267545625
Offset: 0

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Author

Keywords

Comments

a(n) is divisible by A000537(n) if and only n is congruent to 1 mod 3 (see A016777) - Artur Jasinski, Oct 10 2007
This sequence is related to A000540 by a(n) = n*A000540(n) - Sum_{i=0..n-1} A000540(i). - Bruno Berselli, Apr 26 2010

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. 815.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • 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

Row 7 of array A103438.

Programs

  • Magma
    [n^2*(n+1)^2*(3*n^4+6*n^3-n^2-4*n+2)/24: n in [0..30]]; // Vincenzo Librandi, Feb 20 2016
  • Maple
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^7 od: seq(a[n], n=0..25); # Zerinvary Lajos, Feb 22 2008
  • Mathematica
    Table[Sum[k^7, {k, 1, n}], {n, 0, 100}] (* Artur Jasinski, Oct 10 2007 *)
    s = 0; lst = {s}; Do[s += n^7; AppendTo[lst, s], {n, 1, 30, 1}]; lst (* Zerinvary Lajos, Jul 12 2009 *)
    LinearRecurrence[{9, -36, 84, -126, 126, -84, 36, -9, 1}, {0, 1, 129, 2316, 18700, 96825, 376761, 1200304, 3297456}, 35] (* Vincenzo Librandi, Feb 20 2016 *)
  • PARI
    a(n)=n^2*(n+1)^2*(3*n^4+6*n^3-n^2-4*n+2)/24 \\ Edward Jiang, Sep 10 2014
    
  • PARI
    a(n) = sum(i=1, n, i^7); \\ Michel Marcus, Sep 11 2014
    
  • Python
    A000541_list, m = [0], [5040, -15120, 16800, -8400, 1806, -126, 1, 0, 0]
    for _ in range(10**2):
        for i in range(8):
            m[i+1] += m[i]
        A000541_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    

Formula

a(n) = n^2*(n+1)^2*(3*n^4 + 6*n^3 - n^2 - 4*n + 2)/24.
a(n) = sqrt(Sum_{j=1..n} Sum_{i=1..n} (i*j)^7). - Alexander Adamchuk, Oct 26 2004
Jacobi formula: a(n) = 2(A000217(n))^4 - A000539(n). - Artur Jasinski, Oct 10 2007
G.f.: x*(1 + 120*x + 1191*x^2 + 2416*x^3 + 1191*x^4 + 120*x^5 + x^6)/(1-x)^9. - Colin Barker, May 25 2012
a(n) = 8*a(n-1) - 28* a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) + 5040. - Ant King, Sep 24 2013
a(n) = -Sum_{j=1..7} j*Stirling1(n+1,n+1-j)*Stirling2(n+7-j,n). - Mircea Merca, Jan 25 2014
a(n) = 2*A000217(n)^4 - (4/3)*A000217(n)^3 + (1/3)*A000217(n)^2. - Michael Raney, Feb 19 2016
a(n) = 72*A288876(n-2) + 48*A006542(n+2) + A000537(n). - Yasser Arath Chavez Reyes, Apr 27 2024
a(n) = Sum_{i=1..n} J_7(i)*floor(n/i), where J_7 is A069092. - Ridouane Oudra, Jul 17 2025
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