A101096 -id:A101096 - cp's OEIS frontend

A001844 Centered square numbers: a(n) = 2n(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.

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

N. J. A. Sloane

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

			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

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.

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Ahmed, J. De Loera and R. Hemmecke, Polyhedral Cones of Magic Cubes and Squares, arXiv:math/0201108 [math.CO], 2002.
U. Alfred, n and n+1 consecutive integers with equal sums of squares, Math. Mag., 35 (1962), 155-164.
Bela Bajnok, Additive Combinatorics: A Menu of Research Problems, arXiv:1705.07444 [math.NT], May 2017. See Section 2.3.
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Matthias Beck, Moshe Cohen, Jessica Cuomo, and Paul Gribelyuk, The number of "magic" squares and hypercubes, arXiv:math/0201013 [math.CO], 2002-2005.
Arthur T. Benjamin and Doron Zeilberger, Pythagorean Primes and Palindromic Continued Fractions, Electronic Journal of Combinatorial Number Theory, 5(1) 2005, #A30.
J. A. De Loera, D. C. Haws and M. Koppe, Ehrhart Polynomials of Matroid Polytopes and Polymatroids, arXiv:0710.4346 [math.CO], 2007; Discrete Comput. Geom., 42 (2009), 670-702.
FiveThirtyEight, "Riddler Express" paper cutting problem and solution, Jan 28 2022.
D. C. Haws, Matroids [Broken link, Oct 30 2017]
D. C. Haws, Matroids [Copy on website of Matthias Koeppe]
D. C. Haws, Matroids [Cached copy, pdf file only]
L. Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Max Parrish and Co, London, 1950, pp. 22 and 36.
Milan Janjic, Two Enumerative Functions. [Broken link; WayBackMachine archive.]
Milan Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
Clark Kimberling and John E. Brown, Partial Complements and Transposable Dispersions, J. Integer Seqs., Vol. 7, 2004.
Ron Knott, Pythagorean Triples and Online Calculators.
G. Kreweras, Sur les hiérarchies de segments, Cahiers Bureau Universitaire Recherche Opérationnelle, Cahier 20, Inst. Statistiques, Univ. Paris, 1973.
G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973). (Annotated scanned copy)
A. O. Munagi, Pairing conjugate partitions by residue classes, Discrete Math., 308 (2008), 2492-2501.
Mitchell Paukner, Lucy Pepin, Manda Riehl, and Jarred Wieser, Pattern Avoidance in Task-Precedence Posets, arXiv:1511.00080 [math.CO], 2015-2016.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
John A. Jr. Rochowicz, Harmonic Numbers: Insights, Approximations and Applications, Spreadsheets in Education (eJSiE), 2015, Vol. 8: Iss. 2, Article 4.
Amelia Carolina Sparavigna, Groupoid of OEIS A001844 Numbers (centered square numbers), Politecnico di Torino, Italy (2019).
R. G. Stanton and D. D. Cowan, Note on a "square" functional equation, SIAM Rev., 12 (1970), 277-279.
David James Sycamore, Triangular array
Leo Tavares, Illustration: Diamond Rows
B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.
Eric Weisstein's World of Mathematics, Centered Polygonal Number, Centered Square Number, Diamond, Pythagorean Triple, and von Neumann Neighborhood.
Index entries for sequences related to centered polygonal numbers
Index entries for two-way infinite sequences.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Nicolay Avilov, Graphical representation of the sequence members

X values are A005408; Y values are A046092.
Cf. A008586 (first differences), A005900 (partial sums), A254373 (digital roots).
Cf. A069894, A000217, A000290, A001263, A001788, A001845, A002061, A002144, A003215, A005448, A005891, A005917, A008844 (square terms), A027862 (prime terms), A048395, A051890, A056106, A101096, A127876, A128064, A132778, A147973, A153869, A240876, A251599, A000982, A080827.
Subsequence of A004431.
Right edge of A055096; main diagonal of A069480, A078475, A129312.
Row n=2 (or column k=2) of A008288.
Cf. A016754.

a001844 n = 2 * n * (n + 1) + 1
a001844_list = zipWith (+) a000290_list $ tail a000290_list
-- Reinhard Zumkeller, Dec 04 2012

[2*n^2 + 2*n + 1: n in [0..50]]; // Vincenzo Librandi, Jan 19 2013

[n: n in [0..4400] | IsSquare(2*n-1)]; // Juri-Stepan Gerasimov, Apr 06 2016

A001844:=-(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation

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};
```

x='x+O('x^200); Vec((1+x)^2/(1-x)^3) \\ Altug Alkan, Mar 23 2016

print([2*n*(n+1)+1 for n in range(48)]) # Michael S. Branicky, Jan 05 2021

[i**2 + (i + 1)**2 for i in range(46)] # Zerinvary Lajos, Jun 27 2008

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)

Partially edited by Joerg Arndt, Mar 11 2010

A101102 Fifth partial sums of cubes (A000578).

Original entry on oeis.org

1, 13, 82, 354, 1200, 3432, 8646, 19734, 41613, 82225, 153868, 274924, 472056, 782952, 1259700, 1972884, 3016497, 4513773, 6624046, 9550750, 13550680, 18944640, 26129610, 35592570, 47926125, 63846081, 84211128, 110044792, 142559824, 183185200, 233595912

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Colin Barker, Table of n, a(n) for n = 1..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 15.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Archive link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Partial sums of A101097.

Cf. A000537, A024166, A101094.

[Binomial(n+5,6)*(3*n^2+15*n+10)/28: n in [1..30]]; // G. C. Greubel, Dec 01 2018

Table[Binomial[n+5,6]*(3*n^2+15*n+10)/28, {n,1,30}] (* G. C. Greubel, Dec 01 2018 *)
Nest[Accumulate,Range[40]^3,5] (* Harvey P. Dale, Feb 06 2023 *)

a(n)=sum(t=1,n,sum(s=1,t,sum(l=1,s,sum(j=1,l, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2,(2*i-1))))))) \\ Alexander R. Povolotsky, May 17 2008

Vec(-x*(x^2+4*x+1)/(x-1)^9 + O(x^100)) \\ Colin Barker, Apr 23 2015

a(n) = binomial(n+5,6)*(3*n^2+15*n+10)/28 \\ Charles R Greathouse IV, Apr 23 2015

[binomial(n+5,6)*(3*n^2+15*n+10)/28 for n in  (1..30)] # G. C. Greubel, Dec 01 2018

a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)*(10 + 3*n*(n+5))/20160.
This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=5. - Alexander R. Povolotsky, May 17 2008
G.f.: x*(x^2+4*x+1) / (1-x)^9. - Colin Barker, Apr 23 2015
Sum_{n>=1} 1/a(n) = -162*sqrt(21/5)*Pi*tan(sqrt(35/3)*Pi/2) - 136269/100. - Amiram Eldar, Jan 26 2022

Edited by Ralf Stephan, Dec 16 2004

A101099 Third partial sums of fifth powers (A000584).

Original entry on oeis.org

1, 35, 345, 1955, 7990, 26226, 73470, 182490, 412335, 863005, 1695551, 3158805, 5624060, 9629140, 15933420, 25585476, 40005165, 61082055, 91292245, 133835735, 192796626, 273328550, 381867850, 526377150, 716622075, 964484001, 1284311835

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

G. C. Greubel, Table of n, a(n) for n = 1..5000
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Dead link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A000584.

[n*(1+n)*(2+n)*(3+n)*(-1+n*(2+n))*(2+n*(4+n))/336: n in [1..30]]; // G. C. Greubel, Dec 01 2018

Nest[Accumulate[#]&,Range[30]^5,3] (* or *) LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,35,345,1955,7990,26226,73470,182490,412335},30] (* Harvey P. Dale, Feb 20 2015 *)

vector(30, n, n*(1+n)*(2+n)*(3+n)*(-1+n*(2+n))*(2+n*(4+n))/336) \\ G. C. Greubel, Dec 01 2018

[n*(1+n)*(2+n)*(3+n)*(-1+n*(2+n))*(2+n*(4+n))/336 for n in (1..30)] # G. C. Greubel, Dec 01 2018

a(n) = n*(1 + n)*(2 + n)*(3 + n)*(-1 + n*(2 + n))*(2 + n*(4 + n))/336.
G.f.: x*(1 + 26*x + 66*x^2 + 26*x^3 + x^4)/(1-x)^9. - Colin Barker, Apr 16 2012
a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9). - Harvey P. Dale, Feb 20 2015
E.g.f.: x*(336 + 5544*x + 13608*x^2 + 10934*x^3 + 3696*x^4 + 574*x^5 + 40*x^6 + x^7)*exp(x)/336. - G. C. Greubel, Dec 01 2018
Sum_{n>=1} 1/a(n) = 224/3 - 60*sqrt(2)*Pi*cot(sqrt(2)*Pi). - Amiram Eldar, Jan 27 2022

Edited by Ralf Stephan, Dec 16 2004

A101091 Fourth partial sums of fourth powers (A000583).

Original entry on oeis.org

1, 20, 155, 760, 2814, 8592, 22770, 54120, 117975, 239668, 459173, 837200, 1463020, 2464320, 4019412, 6372144, 9849885, 14884980, 22040095, 32037896, 45795530, 64464400, 89475750, 122592600, 165968595, 222214356, 294471945, 386498080

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004

Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Dead link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A000583, A101090.

Nest[Accumulate,Range[30]^4,4] (* or *) LinearRecurrence[ {9,-36,84,-126,126,-84,36,-9,1},{1,20,155,760,2814,8592,22770,54120,117975},30] (* Harvey P. Dale, Dec 30 2011 *)

a(n) = n*(1 + n)*(2 + n)^2*(3 + n)*(4 + n)*(-1 + 3*n*(4 + n))/5040.
a(1)=1, a(2)=20, a(3)=155, a(4)=760, a(5)=2814, a(6)=8592, a(7)=22770, a(8)=54120, a(9)=117975, a(n)=9*a(n-1)-36*a(n-2)+84*a(n-3)- 126*a(n-4)+ 126*a(n-5)-84*a(n-6)+36*a(n-7)-9*a(n-8)+a(n-9). - Harvey P. Dale, Dec 30 2011
G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^9. - Colin Barker, Apr 04 2012
Sum_{n>=1} 1/a(n) = 3934693/3380 - 210*Pi^2/13 - (2268/13)*sqrt(3/13)*Pi*cot(sqrt(13/3)*Pi). - Amiram Eldar, Jan 26 2022

Edited by Ralf Stephan, Dec 16 2004

A005583 Coefficients of Chebyshev polynomials.

Original entry on oeis.org

2, 11, 36, 91, 196, 378, 672, 1122, 1782, 2717, 4004, 5733, 8008, 10948, 14688, 19380, 25194, 32319, 40964, 51359, 63756, 78430, 95680, 115830, 139230, 166257, 197316, 232841, 273296, 319176, 371008, 429352, 494802, 567987, 649572, 740259, 840788

Offset: 1

N. J. A. Sloane

If X is an n-set and Y a fixed 2-subset of X then a(n-5) is equal to the number of (n-5)-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007

a(n-1) = risefac(n,5)/5! - risefac(n,3)/3! is for n >= 1 also the number of independent components of a symmetric traceless tensor of rank 5 and dimension n. Here risefac is the rising factorial. Put a(0) = 0. - Wolfdieter Lang, Dec 10 2015

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), Table 22.7, p. 797.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..172
Milan Janjic, Two Enumerative Functions.
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972. [alternative scanned copy].
Richard K. Guy, Letter to N. J. A. Sloane, Feb 1988.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [broken link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
Index entries for sequences related to Chebyshev polynomials.

Cf. A000096, A000217, A000389, A051747, A127672.

Column 3 of A207606.

A005583:=-(-2+z)/(z-1)**6; # Simon Plouffe in his 1992 dissertation (this g.f. assumes offset 0)

conv(u,v)=local(w); w=vector(length(u),i,sum(j=1,i,u[j]*v[i+1-j])); w;
t(n)=n*(n+1)/2;
u=vector(10,i,t(i));
v=vector(10,i,t(i)-1);
conv(u,v)

a(n) = (1/120)*n*(n+9)*(n+3)*(n+2)*(n+1); \\ Joerg Arndt, Mar 05 2018

G.f.: x*(2-x)/(1-x)^6.
a(n) = binomial(n+4, n-1) + binomial(n+3, n-1) = (1/120)*n*(n+9)*(n+3)*(n+2)*(n+1).
a(n+1) = -A127672(10+n, n), n >= 0, with the coefficients of the Chebyshev C-polynomials A127672(n, k). - Wolfdieter Lang, Dec 10 2015
a(n) = Sum_{i=1..n} A000217(i)*A000096(n+1-i). - Bruno Berselli, Mar 05 2018
a(n) = binomial(n+3,5) + 2*binomial(n+3,4). - Yuchun Ji, May 23 2019
From Amiram Eldar, Feb 17 2023: (Start)
Sum_{n>=1} 1/a(n) = 40751/63504.
Sum_{n>=1} (-1)^(n+1)/a(n) = 1360*log(2)/63 - 922961/63504. (End)

More terms from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 07 1999

More terms from Zerinvary Lajos, Jul 21 2006

A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

Original entry on oeis.org

1, 12, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original name: The first summation of row 4 of Euler's triangle - a row that will recursively accumulate to the power of 4.

D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Dead link]
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (1).

For other sequences based upon MagicNKZ(n,k,z):
..... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7
---------------------------------------------------------------------------
z = 0 | A000007 | A019590 | .......MagicNKZ(n,k,0) = A008292(n,k+1) .......
z = 1 | A000012 | A040000 | A101101 | thisSeq | A101100 | ....... | .......
z = 2 | A000027 | A005408 | A008458 | A101103 | A101095 | ....... | .......
z = 3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | .......
z = 4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | .......
z = 5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181
z = 6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177
z = 7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523
z = 8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015
z = 9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541
Cf. A101095 for an expanded table and more about MagicNKZ.

MagicNKZ = Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 1, 1}, {k, 0, 34}]
Join[{1, 12, 23},LinearRecurrence[{1},{24},56]] (* Ray Chandler, Sep 23 2015 *)

a(k) = MagicNKZ(4,k,1) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n (cf. A101095). That is, a(k) = Sum_{j=0..k+1} (-1)^j*binomial(4, j)*(k-j+1)^4.
a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4. - Joerg Arndt, Nov 30 2014
G.f.: x*(1+11*x+11*x^2+x^3)/(1-x). - Colin Barker, Apr 16 2012

New name from Joerg Arndt, Nov 30 2014

Original Formula edited and Crossrefs table added by Danny Rorabaugh, Apr 22 2015

A254644 Fourth partial sums of fifth powers (A000584).

Original entry on oeis.org

1, 36, 381, 2336, 10326, 36552, 110022, 292512, 704847, 1567852, 3263403, 6422208, 12046268, 21675408, 37608828, 63194304, 103199469, 164281524, 255573769, 389409504, 582206130, 855534680, 1237402530, 1763779680, 2480401755, 3444885756, 4729197591, 6422513536, 8634521016, 11499207456

Offset: 1

Luciano Ancora, Feb 05 2015

			Fifth differences:   1, 27,  93,  119,   120, (repeat 120) (A101100)
Fourth differences:  1, 28, 121,  240,   360,   480, ...   (A101095)
Third differences:   1, 29, 150,  390,   750,  1230, ...   (A101096)
Second differences:  1, 30, 180,  570,  1320,  2550, ...   (A101098)
First differences:   1, 31, 211,  781,  2101,  4651, ...   (A022521)
-------------------------------------------------------------------------
The fifth powers:    1, 32, 243, 1024,  3125,  7776, ...   (A000584)
-------------------------------------------------------------------------
First partial sums:  1, 33, 276, 1300,  4425, 12201, ...   (A000539)
Second partial sums: 1, 34, 310, 1610,  6035, 18236, ...   (A101092)
Third partial sums:  1, 35, 345, 1955,  7990, 26226, ...   (A101099)
Fourth partial sums: 1, 36, 381, 2336, 10326, 36552, ...   (this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials
Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers
Index entries for linear recurrences with constant coefficients, signature(10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A000539, A000584, A022521, A101092, A101095, A101096, A101098, A101099, A101100.

Cf. A101091 (fourth partial sums of fourth powers).

List([1..30], n-> Binomial(n+4,5)*(5*(n+2)^4 -35*(n+2)^2 +36)/126); # G. C. Greubel, Aug 28 2019

[Binomial(n+4,5)*(5*(n+2)^4 -35*(n+2)^2 +36)/126: n in [1..30]]; // G. C. Greubel, Aug 28 2019

seq(binomial(n+4,5)*(5*(n+2)^4 -35*(n+2)^2 +36)/126, n=1..30); # G. C. Greubel, Aug 28 2019

Table[n(1+n)(2+n)(3+n)(4+n)(-24 +20n +85n^2 +40n^3 +5n^4)/15120, {n, 30}] (* or *) Accumulate[Accumulate[Accumulate[Accumulate[Range[24]^5]]]] (* or *) CoefficientList[Series[(1 +26x +66x^2 +26x^3 +x^4)/(1-x)^10, {x, 0, 30}], x]
Nest[Accumulate,Range[30]^5,4] (* or *) LinearRecurrence[{10,-45,120, -210,252,-210,120,-45,10,-1}, {1,36,381,2336,10326,36552,110022,292512, 704847,1567852},30] (* Harvey P. Dale, May 08 2016 *)

vector(30, n, m=n+2; binomial(m+2,5)*(5*m^4 -35*m^2 +36)/126) \\ G. C. Greubel, Aug 28 2019

[binomial(n+4,5)*(5*(n+2)^4 -35*(n+2)^2 +36)/126 for n in (1..30)] # G. C. Greubel, Aug 28 2019

G.f.: x*(1 + 26*x + 66*x^2 + 26*x^3 + x^4)/(1 - x)^10.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(-24 + 20*n + 85*n^2 + 40*n^3 + 5*n^4)/15120.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + n^5.

Edited by Bruno Berselli, Feb 10 2015

A254682 Fifth partial sums of fifth powers (A000584).

Original entry on oeis.org

1, 37, 418, 2754, 13080, 49632, 159654, 452166, 1157013, 2724865, 5988268, 12410476, 24456744, 46132152, 83740980, 146935284, 250134753, 414416277, 669990046, 1059399550, 1641605680, 2497140360, 3734542890, 5498322570

Offset: 1

Luciano Ancora, Feb 12 2015

			Fifth differences:   1, 27,  93,  119,   120, (repeat 120) (A101100)
Fourth differences:  1, 28, 121,  240,   360,   480, ...   (A101095)
Third differences:   1, 29, 150,  390,   750,  1230, ...   (A101096)
Second differences:  1, 30, 180,  570,  1320,  2550, ...   (A101098)
First differences:   1, 31, 211,  781,  2101,  4651, ...   (A022521)
-------------------------------------------------------------------------
The fifth powers:    1, 32, 243, 1024,  3125,  7776, ...   (A000584)
-------------------------------------------------------------------------
First partial sums:  1, 33, 276, 1300,  4425, 12201, ...   (A000539)
Second partial sums: 1, 34, 310, 1610,  6035, 18236, ...   (A101092)
Third partial sums:  1, 35, 345, 1955,  7990, 26226, ...   (A101099)
Fourth partial sums: 1, 36, 381, 2336, 10326, 36552, ...   (A254644)
Fifth partial sums:  1, 37, 418, 2754, 13080, 49632, ...   (this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials.
Luciano Ancora, Pascal's triangle and recurrence relations for partial sums of m-th powers .
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Cf. A000539, A000584, A022521, A101092, A101095, A101096, A101098, A101099, A101100, A254644, A254681, A254683, A254684.

Table[n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (- 2 + 5 n + n^2) (9 + 10 n + 2 n^2)/60480, {n,24}] (* or *)
CoefficientList[Series[(- 1 - 26 x - 66 x^2 - 26 x^3 - x^4)/(- 1 + x)^11, {x,0,23}], x]
Nest[Accumulate,Range[30]^5,5] (* or *) LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,37,418,2754,13080,49632,159654,452166,1157013,2724865,5988268},30] (* Harvey P. Dale, Jan 30 2019 *)

a(n)=n*(1+n)*(2+n)*(3+n)*(4+n)*(5+n)*(-2+5*n+n^2)*(9+10*n+2*n^2)/60480 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (- x - 26*x^2 - 66*x^3 - 26*x^4 - x^5)/(- 1 + x)^11.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(- 2 + 5*n + n^2)*(9 + 10*n + 2*n^2)/60480.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + n^5.
Sum_{n>=1} 1/a(n) = 475867/180 - (2560/13)*sqrt(7)*Pi*tan(sqrt(7)*Pi/2) + (210/13)*sqrt(3/11)*Pi*tan(sqrt(33)*Pi/2). - Amiram Eldar, Jan 27 2022

A005584 Coefficients of Chebyshev polynomials.

Original entry on oeis.org

2, 13, 49, 140, 336, 714, 1386, 2508, 4290, 7007, 11011, 16744, 24752, 35700, 50388, 69768, 94962, 127281, 168245, 219604, 283360, 361790, 457470, 573300, 712530, 878787, 1076103, 1308944, 1582240, 1901416, 2272424, 2701776, 3196578, 3764565, 4414137, 5154396

Offset: 1

N. J. A. Sloane

If X is an n-set and Y a fixed 2-subset of X then a(n-6) is equal to the number of (n-6)-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007

a(n-1) = risefac(n+1,6)/6! - risefac(n+1,4)/4! is for n >=1 also the number of independent components of a symmetric traceless tensor of rank 6 and dimension n. Here risefac is the rising factorial. Put a(0) = 0. - Wolfdieter Lang, Dec 10 2015

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 797.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Milan Janjic, Two Enumerative Functions.
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Richard K. Guy, Letter to N. J. A. Sloane, Feb 1988.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A127672, A005581, A005582, A005583.

List([1..40], n-> (n+11)*Binomial(n+4,5)/6); # G. C. Greubel, Aug 27 2019

[n*(n+11)*(n+4)*(n+3)*(n+2)*(n+1)/720: n in [1..40]]; // Vincenzo Librandi, Jun 15 2011

[seq(binomial(n,6)+2*binomial(n,5), n=5..35)]; # Zerinvary Lajos, Jul 26 2006
A005584:=(-2+z)/(z-1)**7; # Simon Plouffe in his 1992 dissertation

Table[2*Binomial[n+4, 5] + Binomial[n+4, 6], {n,40}] (* Vladimir Joseph Stephan Orlovsky, Jun 14 2011, modified by G. C. Greubel, Aug 27 2019 *)
Table[(n+11)*Pochhammer[n, 5]/6!, {n,40}] (* G. C. Greubel, Aug 27 2019 *)

a(n)=n*(n+11)*(n+4)*(n+3)*(n+2)*(n+1)/720 \\ Charles R Greathouse IV, Jun 14 2011

[(n+11)*rising_factorial(n,5)/factorial(6) for n in (1..40)] # G. C. Greubel, Aug 27 2019

G.f.: x*(2-x) / (1-x)^7.
a(n) = binomial(n+5, n-1) + binomial(n+4, n-1) = 1/720*n*(n+11)*(n+4)*(n+3)*(n+2)*(n+1).
a(n) = binomial(n,6) + 2*binomial(n,5), n >= 5. - Zerinvary Lajos, Jul 26 2006
a(n+1) = A127672(12+n, n), n >= 0, where A127672 gives the coefficients of Chebyshev's C polynomials. See the Abramowitz-Stegun reference. - Wolfdieter Lang, Dec 10 2015
From G. C. Greubel, Aug 27 2019: (Start)
a(n) = (n+11)*Pochhammer(n, 5)/6!.
E.g.f.: x*(1440 +3240*x +1920*x^2 +420*x^3 +36*x^4 +x^5)*exp(x)/6!. (End)
From Amiram Eldar, Feb 17 2023: (Start)
Sum_{n>=1} 1/a(n) = 1303391/2134440.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4160*log(2)/77 - 78994697/2134440. (End)

More terms from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 07 1999

A101095 Fourth difference of fifth powers (A000584).

Original entry on oeis.org

1, 28, 121, 240, 360, 480, 600, 720, 840, 960, 1080, 1200, 1320, 1440, 1560, 1680, 1800, 1920, 2040, 2160, 2280, 2400, 2520, 2640, 2760, 2880, 3000, 3120, 3240, 3360, 3480, 3600, 3720, 3840, 3960, 4080, 4200, 4320, 4440, 4560, 4680, 4800, 4920, 5040, 5160, 5280

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original Name: Shells (nexus numbers) of shells of shells of shells of the power of 5.

The (Worpitzky/Euler/Pascal Cube) "MagicNKZ" algorithm is: MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n, with k>=0, n>=1, z>=0. MagicNKZ is used to generate the n-th accumulation sequence of the z-th row of the Euler Triangle (A008292). For example, MagicNKZ(3,k,0) is the 3rd row of the Euler Triangle (followed by zeros) and MagicNKZ(10,k,1) is the partial sums of the 10th row of the Euler Triangle. This sequence is MagicNKZ(5,k-1,2).

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Archive Machine link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Fourth differences of A000584, third differences of A022521, second differences of A101098, and first differences of A101096.
For other sequences based upon MagicNKZ(n,k,z):
...... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7  |  n = 8
--------------------------------------------------------------------------------------
z =  0 | A000007 | A019590 | .......  MagicNKZ(n,k,0) = T(n,k+1) from A008292  .......
z =  1 | A000012 | A040000 | A101101 | A101104 | A101100 | ....... | ....... | .......
z =  2 | A000027 | A005408 | A008458 | A101103 | thisSeq | ....... | ....... | .......
z =  3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | ....... | .......
z =  4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | ....... | .......
z =  5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181 | .......
z =  6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177 | A255182
z =  7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523 | A255178
z =  8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015 | A022524
z =  9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541 | A001016
z = 10 | A000582 | A054333 | A254469 | A254681 | A254644 | A254640 | A250212 | A000542
z = 11 | A001287 | A054334 | A254869 | A254470 | A254682 | A254645 | A254641 | A253636
z = 12 | A001288 | A057788 | ....... | A254870 | A254471 | A254683 | A254646 | A254642
z = 13 | A010965 | ....... | ....... | ....... | A254871 | A254472 | A254684 | A254647
z = 14 | A010966 | ....... | ....... | ....... | ....... | A254872 | ....... | .......
--------------------------------------------------------------------------------------
Cf. A047969.

I:=[1,28,121,240,360]; [n le 5 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, May 07 2015

MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 5, 5}, {z, 2, 2}, {k, 0, 34}]
CoefficientList[Series[(1 + 26 x + 66 x^2 + 26 x^3 + x^4)/(1 - x)^2, {x, 0, 50}], x] (* Vincenzo Librandi, May 07 2015 *)
Join[{1,28,121,240},Differences[Range[50]^5,4]] (* or *) LinearRecurrence[{2,-1},{1,28,121,240,360},50] (* Harvey P. Dale, Jun 11 2016 *)

a(n)=if(n>3, 120*n-240, 33*n^2-72*n+40) \\ Charles R Greathouse IV, Oct 11 2015

[1,28,121]+[120*(k-2) for k in range(4,36)] # Danny Rorabaugh, Apr 23 2015

a(k+1) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n; n = 5, z = 2.
For k>3, a(k) = Sum_{j=0..4} (-1)^j*binomial(4, j)*(k - j)^5 = 120*(k - 2).
a(n) = 2*a(n-1) - a(n-2), n>5. G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (1-x)^2. - Colin Barker, Mar 01 2012

MagicNKZ material edited, Crossrefs table added, SeriesAtLevelR material removed by Danny Rorabaugh, Apr 23 2015

Name changed and keyword 'uned' removed by Danny Rorabaugh, May 06 2015

cp's OEIS Frontend

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.

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A101102 Fifth partial sums of cubes (A000578).

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A101099 Third partial sums of fifth powers (A000584).

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A101091 Fourth partial sums of fourth powers (A000583).

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A005583 Coefficients of Chebyshev polynomials.

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A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

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A001844 Centered square numbers: a(n) = 2n(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.