A005900 -id:A005900 - cp's OEIS frontend

A004467 a(n) = n(11n^2 - 5)/6.

0, 1, 13, 47, 114, 225, 391, 623, 932, 1329, 1825, 2431, 3158, 4017, 5019, 6175, 7496, 8993, 10677, 12559, 14650, 16961, 19503, 22287, 25324, 28625, 32201, 36063, 40222, 44689, 49475, 54591, 60048

Offset: 0

Albert D. Rich (Albert_Rich(AT)msn.com)

3-dimensional analog of centered polygonal numbers, that is: centered hendecagonal pyramidal numbers (see Deza paper in References).

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

[n*(11*n^2-5)/6: n in [0..50]]; // Vincenzo Librandi, May 15 2011

Table[n(11n^2-5)/6,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
LinearRecurrence[{4,-6,4,-1},{0,1,13,47},80] (* Harvey P. Dale, Sep 22 2013 *)

a(n)=n*(11*n^2-5)/6 \\ Charles R Greathouse IV, Sep 28 2011

G.f.: x*(1+9*x+x^2)/(1-x)^4. - Colin Barker, Jan 08 2012
a(0)=0, a(1)=1, a(2)=13, a(3)=47; for n>3, a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Sep 22 2013
E.g.f.: (x/6)*(6 + 33*x + 11*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A062025 a(n) = n(13n^2 - 7)/6.

Original entry on oeis.org

0, 1, 15, 55, 134, 265, 461, 735, 1100, 1569, 2155, 2871, 3730, 4745, 5929, 7295, 8856, 10625, 12615, 14839, 17310, 20041, 23045, 26335, 29924, 33825, 38051, 42615, 47530, 52809, 58465, 64511, 70960, 77825, 85119, 92855, 101046, 109705, 118845, 128479, 138620, 149281

Offset: 0

N. J. A. Sloane, Aug 02 2001

Harry J. Smith, Table of n, a(n) for n = 0..1000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Table[n(13n^2-7)/6,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)

a(n) = n*(13*n^2 - 7)/6 \\ Harry J. Smith, Jul 29 2009

From G. C. Greubel, Sep 01 2017: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: (x + 11*x^2 + x^3)/(1 - x)^4.
E.g.f.: (x/6)*(6 + 39*x + 13*x^2)*exp(x). (End)

A063523 a(n) = n(8n^2 - 5)/3.

Original entry on oeis.org

0, 1, 18, 67, 164, 325, 566, 903, 1352, 1929, 2650, 3531, 4588, 5837, 7294, 8975, 10896, 13073, 15522, 18259, 21300, 24661, 28358, 32407, 36824, 41625, 46826, 52443, 58492, 64989, 71950, 79391, 87328, 95777, 104754, 114275, 124356, 135013, 146262, 158119, 170600

Offset: 0

N. J. A. Sloane, Aug 02 2001

Also as a(n)=(1/6)*(16*n^3-10*n), n>0: structured octagonal anti-diamond numbers (vertex structure 17) (Cf. A100187 = alternate vertex; A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004

Harry J. Smith, Table of n, a(n) for n = 0..1000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Table[n(8n^2-5)/3,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
LinearRecurrence[{4,-6,4,-1},{0,1,18,67},81] (* or *) CoefficientList[ Series[ (x+14 x^2+x^3)/(x-1)^4,{x,0,80}],x] (* Harvey P. Dale, Jul 11 2011 *)

a(n) = n*(8*n^2 - 5)/3 \\ Harry J. Smith, Aug 25 2009

a(0)=0, a(1)=1, a(2)=18, a(3)=67, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)- a(n-4). - Harvey P. Dale, Jul 11 2011
G.f.: (x+14*x^2+x^3)/(x-1)^4. - Harvey P. Dale, Jul 11 2011
E.g.f.: (x/3)*(3 + 24*x + 8*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A069038 Expansion of g.f. x*(1+x)^4/(1-x)^6.

Original entry on oeis.org

0, 1, 10, 51, 180, 501, 1182, 2471, 4712, 8361, 14002, 22363, 34332, 50973, 73542, 103503, 142544, 192593, 255834, 334723, 432004, 550725, 694254, 866295, 1070904, 1312505, 1595906, 1926315, 2309356, 2751085, 3258006, 3837087, 4495776, 5242017, 6084266

Offset: 0

Vladeta Jovovic, Apr 03 2002

Hyun Kwang Kim asserts that every nonnegative integer can be represented by the sum of no more than 14 of these numbers. - Jonathan Vos Post, Nov 16 2004
If Y_i (i=1,2,3,4) are 2-blocks of a (n+4)-set X then a(n-4) is the number of 9-subsets of X intersecting each Y_i (i=1,2,3,4). - Milan Janjic, Oct 28 2007
Starting with 1 = binomial transform of [1, 9, 32, 56, 48, 16, 0, 0, 0, ...], where (1, 9, 32, 56, 48, 16) = row 5 of the Chebyshev triangle A081277. Also = row 5 of the array in A142978. - Gary W. Adamson, Jul 19 2008
Starting with the term 1 this is the self-convolution of A001844(n). - Anton Zakharov, Sep 02 2016

H. S. M. Coxeter, Regular Polytopes, New York: Dover Publications, 1973.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Milan Janjic, Two Enumerative Functions
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000332, A014820, A005900.
Cf. A142978, A081277.
Cf. A001844.

[n*(2*n^4 + 10*n^2 + 3)/15: n in [0..40]]; // Vincenzo Librandi, May 22 2011

al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n), r=0..d-1); end; [seq(be(5,n),n=0..100)];

CoefficientList[Series[x (1 + x)^4/(1 - x)^6, {x, 0, 32}], x] (* Michael De Vlieger, Sep 02 2016 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{0,1,10,51,180,501},40] (* Harvey P. Dale, Jun 19 2021 *)

concat(0, Vec(x*(1+x)^4/(1-x)^6 + O(x^99))) \\ Altug Alkan, Sep 02 2016

{a(n) = n * (2*n^4 + 10*n^2 + 3) / 15}; /* Michael Somos, Jun 17 2018 */

Recurrence: 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).
a(n) = n*(2*n^4 + 10*n^2 + 3)/15. - Jonathan Vos Post, Nov 16 2004
a(n) = C(n+4,5) + 4*C(n+3,5) + 6*C(n+2,5) + 4*C(n+1,5) + C(n,5).
Sum_{n>=1} 1/((1/15)*n*(2*n^4 + 10*n^2 + 3)*n!) = hypergeom([1, 1, 1+i*sqrt(10-2*sqrt(19))*(1/2), 1-i*sqrt(10-2*sqrt(19))*(1/2), 1+i*sqrt(10+2*sqrt(19))*(1/2), 1-i*sqrt(10+2*sqrt(19))*(1/2)], [2, 2, 2+i*sqrt(10-2*sqrt(19))*(1/2), 2-i*sqrt(10-2*sqrt(19))*(1/2), 2+i*sqrt(10+2*sqrt(19))*(1/2), 2-i*sqrt(10+2*sqrt(19))*(1/2)], 1) = 1.05351734968093116819345664995829700099916... - Stephen Crowley, Jul 14 2009
a(n) = a(n-1) + A014820(n-1) + A014820(n-2). - Bruce J. Nicholson, Apr 18 2018
a(n) = 10*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018
Euler transform of length 2 sequence [10, -4]. - Michael Somos, Jun 19 2018
Sum_{k >= 1} (-1)^k/(a(k)*a(k+1)) = 10*log(2) - 41/6 = 1/(10 + 2/(10 + 6/(10 + ... + n*(n-1)/(10 + ...)))). See A142983. Cf. A005900 and A014820. - Peter Bala, Mar 08 2024
E.g.f.: exp(x)*x*(15 + 60*x + 60*x^2 + 20*x^3 + 2*x^4)/15. - Stefano Spezia, Mar 10 2024

A059722 a(n) = n(2n^2 - 2*n + 1).

Original entry on oeis.org

0, 1, 10, 39, 100, 205, 366, 595, 904, 1305, 1810, 2431, 3180, 4069, 5110, 6315, 7696, 9265, 11034, 13015, 15220, 17661, 20350, 23299, 26520, 30025, 33826, 37935, 42364, 47125, 52230, 57691, 63520, 69729, 76330, 83335, 90756, 98605, 106894, 115635, 124840

Offset: 0

Henry Bottomley, Feb 07 2001

Mean of the first four nonnegative powers of 2n+1, i.e., ((2n+1)^0 + (2n+1)^1 + (2n+1)^2 + (2n+1)^3)/4. E.g., a(2) = (1 + 3 + 9 + 27)/4 = 10.
Equatorial structured meta-diamond numbers, the n-th number from an equatorial structured n-gonal diamond number sequence. There are no 1- or 2-gonal diamonds, so 1 and (n+2) are used as the first and second terms since all the sequences begin as such. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Starting with offset 1 = row sums of triangle A143803. - Gary W. Adamson, Sep 01 2008
Form an array from the antidiagonals containing the terms in A002061 to give antidiagonals 1; 3,3; 7,4,7; 13,8,8,13; 21,14,9,14,21; and so on. The difference between the sum of the terms in n+1 X n+1 matrices and those in n X n matrices is a(n) for n>0. - J. M. Bergot, Jul 08 2013
Sum of the numbers from (n-1)^2 to n^2. - Wesley Ivan Hurt, Sep 08 2014

Harry J. Smith, Table of n, a(n) for n = 0..1000
Bruno Berselli, Table of sequences with the formula m*(m+1)^2 - (k+2)*m^2 (table includes this sequence for k = 2-m, m >= 0).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000330, A005900, A081436, A100178, A100179, A059722: "equatorial" structured diamonds; A000447: "polar" structured meta-diamond; A006484 for other structured meta numbers; and A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004

Cf. A000217, A000290, A001844, A002061, A002414, A053698, A059721, A059723, A136392, A143803.

[n*(2*n^2 - 2*n + 1) : n in [0..50]]; // Wesley Ivan Hurt, Sep 08 2014

A059722:=n->n*(2*n^2 - 2*n + 1): seq(A059722(n), n=0..50); # Wesley Ivan Hurt, Sep 08 2014

Table[n (2 n^2 - 2 n + 1), {n, 0, 50}] (* Wesley Ivan Hurt, Sep 08 2014 *)

a(n) = { n*(2*n^2 - 2*n + 1) } \\ Harry J. Smith, Jun 28 2009

a(n) = A053698(2*n-1)/4.
a(n) = Sum_{j=1..n} ((n+j-1)^2-j^2+1). - Zerinvary Lajos, Sep 13 2006
From R. J. Mathar, Sep 02 2008: (Start)
G.f.: x*(1 + x)*(1 + 5*x)/(1 - x)^4.
a(n) = A002414(n-1) + A002414(n).
a(n+1) - a(n) = A136392(n+1). (End)
a(n) = (A000290(n) + A000290(n+1)) * (A000217(n+1) - A000217(n)). - J. M. Bergot, Nov 02 2012
a(n) = n * A001844(n-1). - Doug Bell, Aug 18 2015
a(n) = A000217(n^2) - A000217(n^2-2*n). - Bruno Berselli, Jun 26 2018
E.g.f.: exp(x)*x*(1 + 4*x + 2*x^2). - Stefano Spezia, Jun 20 2021

Edited with new definition by N. J. A. Sloane, Aug 29 2008

A066357 Number of ordered (i.e., planar) trees on 2n edges with every subtree at the root having an even number of edges.

Original entry on oeis.org

1, 1, 6, 53, 554, 6362, 77580, 986253, 12927170, 173452334, 2370742868, 32892031042, 462030186916, 6557906929108, 93909078262808, 1355087936016957, 19684187540818866, 287612514032460070, 4224238030616082948, 62329883931236020470, 923519220367120779820

Offset: 0

Louis Shapiro, Feb 01 2002

Row sums of A078990. First column of A079513.
a(n) is the number of walks from (0,0) to (2n,2n) using steps (0,1) and (1,0) which never stray below the line y=x and which avoid the points (m,m) m odd. - Paul Boddington, Mar 14 2003
Series reversion of Sum_{n>0} -a(n)(-x)^n is g.f. of A005900.
a(n) is the number of linear extensions of the one-level grid poset G[(0^n), (1^(n-1)), (1^(n-1))]. The definition of a one-level grid poset can be found in the Pan links. - Ran Pan, Jul 05 2016
These numbers have the same parity as the Catalan numbers C(n), that is, a(n) is even except when n has the form 2^m - 1. This follows immediately from the formula a(n) = C(2*n+1) + 2*C(2*n) - 2^(2*n + 1)*C(n) given below by Callan. We conjecture that a(n) and C(n) have the same 2-adic valuation (checked up to n = 100). - Peter Bala, Aug 02 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..200
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
Nantel Bergeron, Cesar Ceballos, Vincent Pilaud, Hopf dreams, arXiv:1807.03044 [math.CO], 2018. See p. 17.
A. de Mier and M. Noy, A solution to the tennis ball problem, arXiv:math/0311242 [math.CO], 2003.
J.-G. Luque and J.-Y. Thibon, Noncommutative Symmetric Functions Associated with a Code, Lazard Elimination and Witt Vectors, arXiv:math/0607254 [math.CO], 2006; Discrete Math. Theor. Comput. Sci. 9 (2007), no. 2, 59-72.
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344 (p. 333).
Ran Pan, Problem 1, Project P.
Ran Pan, Algorithmic Solution to Problem 1 (and linear extensions of general one-level grid-like posets), Project P.
A. Regev, Enumerating triangulations by parallel diagonals, arXiv:1208.3915 [math.CO], 2012, J. Int. Seq. 15 (2012) #12.8.5

Cf. A078990, A079513, A001448, A005900, A079489, A274644, A274763.

[1] cat [(&+[Binomial(4*n,k)*Binomial(3*n-k-2,n-k-1)/n: k in [0..n]]): n in [1..30]]; // G. C. Greubel, Jan 15 2019

gf := (1-sqrt(1-4*z)-sqrt(1+4*z)+sqrt(1-16*z^2))/(z*(sqrt(1-4*z)-sqrt(1+4*z))):s := series(gf, z, 80): for i from 0 to 50 by 2 do printf(`%d,`,coeff(s,z,i)) od: # James Sellers, Feb 11 2002
a := n -> `if`(n=0,1,binomial(3*n-2,n-1)*hypergeom([1-n,-4*n],[2-3*n], -1)/n): seq(simplify(a(n)),n=0..20); # Peter Luschny, Oct 15 2015

CoefficientList[Series[2/(1 + 4 Sqrt[x]/(Sqrt[1 + 4 Sqrt[x]] - Sqrt[1 - 4 Sqrt[x]])), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 21 2014 *)

a(n)=local(A); if(n<1,n==0,A=sqrt(1+4*x+O(x^(2*n+2))); A-=subst(A,x,-x); polcoeff(((2*A-8*x)/A^2)^2,2*n))

vector (100, n, n--; if(n<1, 1, sum(k=0, n, binomial(4*n,k)*binomial(3*n-k-2,n-k-1)/n))) \\ Altug Alkan, Oct 07 2015

[1] + [sum(binomial(4*n,k)*binomial(3*n-k-2,n-k-1)/n for k in (0..n)) for n in (1..30)] # G. C. Greubel, Jan 15 2019

For n>0, a(n) = Sum_{r=1..n} C(2*r-1)*a(n-r). Here C(2*r-1) is a Catalan number (A000108). - Paul Boddington, Mar 14 2003
G.f.: 2/(1+4*sqrt(x)/(sqrt(1+4*sqrt(x))-sqrt(1-4*sqrt(x)))).
D-finite with recurrence a(n)*(2*n-1)*(n+1)n = a(n-1)*(32*n^2 - 64*n + 39)*2*n - a(n-2)*(2*n-3)*(4*n-5)*(4*n-7)*16, n>1.
a(0) = 1,a(n) = (1/n)*Sum_{k=0..n} C(4*n,k)*C(3*n-k-2,n-k-1), n>1. - Paul Barry, Apr 09 2007
a(n) = ((2^(4*n))/Gamma(1/2)) * ((6*(2*n+1)*Gamma(2*n+1/2)/Gamma(2*n+3))-2*Gamma(n+1/2)/Gamma(n+2)). - David Dickson (dcmd(AT)unimelb.edu.au), Nov 10 2009
Convolution of A079489 with itself: (1, 6, 53, 554, ...) = (1, 3, 22, 211, ...)*(1, 3, 22, 211, ...).
Proof. Working with Dyck paths, we must show that Dyck paths of size (semilength) 2n, all of whose components (constituent primitive Dyck paths) have even size, are equinumerous with ordered pairs of nonempty Dyck paths of total size 2n in each of which the first component is of odd size and all other components (if any) are of even size. Given a Dyck path P of the former class, use the first return decomposition to write P (uniquely) as the concatenation of U A_1 A_2 ... A_j O E D Q where U denotes upstep, D denotes downstep, A_1,...,A_j are all primitive Dyck paths of even size with j>=0, O is a primitive Dyck path of odd size, E is a Dyck path of even size, and Q is a Dyck path in which all components are of even size. Then P -> (O A_1 A_2 ... A_j, U E D Q) is the desired bijection. QED - David Callan, Apr 11 2012
a(n) = C(2*n+1) + 2*C(2*n) - 2^(2*n+1)*C(n), where C(n) is the Catalan number A000108. This formula can be obtained by manipulating generating functions. The equivalence of this formula and the Barry (Apr 09 2007) sum can be established by the WZ method with a second-order operator. A combinatorial interpretation of the Barry sum would be nice. - David Callan, Apr 10 2012
a(n) ~ (3-2*sqrt(2)) * 2^(4*n) / (n^(3/2) * sqrt(2*Pi)). - Vaclav Kotesovec, Mar 21 2014
exp( Sum_{n >= 1} binomial(4*n,2*n)*x^n/n ) = 1 + 6*x + 53*x^2 + 554*x^3 + ... is an o.g.f. for this sequence omitting the initial term. See A001448. - Peter Bala, Oct 02 2015
a(n) = binomial(3*n-2,n-1)*hypergeom([1-n,-4*n],[2-3*n],-1)/n for n>=1. - Peter Luschny, Oct 15 2015
a(n) = 3*(2*n+1) /(2*n+2) /(4*n+1) *binomial(4*n+2,2*n+1) -4^n /(2*n+1) *binomial(2*n+2,n+1) [Merlini et al F_n formula] - R. J. Mathar, Oct 01 2021

A074279 n appears n^2 times.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 1

Jon Perry, Sep 21 2002

Since the last occurrence of n comes one before the first occurrence of n+1 and the former is at Sum_{i=0..n} i^2 = A000330(n), we have a(A000330(n)) = a(n*(n+1)*(2n+1)/6) = n and a(1+A000330(n)) = a(1+(n*(n+1)*(2n+1)/6)) = n+1. So the current sequence is, loosely speaking, the inverse function of the square pyramidal sequence A000330. A000330 has many alternative formulas, thus yielding many alternative formulas for the current sequence. - Jonathan Vos Post, Mar 18 2006

Partial sums of A253903. - Jeremy Gardiner, Jan 14 2018

			This can be viewed also as an irregular table consisting of successively larger square matrices:
  1;
  2, 2;
  2, 2;
  3, 3, 3;
  3, 3, 3;
  3, 3, 3;
  4, 4, 4, 4;
  4, 4, 4, 4;
  4, 4, 4, 4;
  4, 4, 4, 4;
  etc.
When this is used with any similarly organized sequence, a(n) is the index of the matrix in whose range n is. A121997(n) (= A237451(n)+1) and A238013(n) (= A237452(n)+1) would then yield the index of the column and row within that matrix.

Antti Karttunen, Table of n, a(n) for n = 1..9455 (Table of squares from 1 X 1 to 30 X 30, flattened)
Y.-F. S. Petermann, J.-L. Remy and I. Vardi, Discrete derivatives of sequences, Adv. in Appl. Math. 27 (2001), 562-84.

Cf. A000217, A000330, A002024, A003881, A006331, A050446, A050447, A000537, A006003, A005900, A064866, A237451, A237452.

Table[n, {n, 0, 6}, {n^2}] // Flatten (* Arkadiusz Wesolowski, Jan 13 2013 *)

A074279_vec(N=9)=concat(vector(N,i,vector(i^2,j,i))) \\ Note: This creates a vector; use A074279_vec()[n] to get the n-th term. - M. F. Hasler, Feb 17 2014

a(n) = my(k=sqrtnint(3*n,3)); k + (6*n > k*(k+1)*(2*k+1)); \\ Kevin Ryde, Sep 03 2025

from sympy import integer_nthroot
def A074279(n): return (m:=integer_nthroot(3*n,3)[0])+(6*n>m*(m+1)*((m<<1)+1)) # Chai Wah Wu, Nov 04 2024

For 1 <= n <= 650, a(n) = floor((3n)^(1/3)+1/2). - Mikael Aaltonen, Jan 05 2015
a(n) = 1 + floor( t(n) + 1 / ( 12 * t(n) ) - 1/2 ), where t(n) = (sqrt(3888*(n-1)^2-1) / (8*3^(3/2)) + 3 * (n-1)/2 ) ^(1/3). - Mikael Aaltonen, Mar 01 2015
a(n) = floor(t + 1/(12*t) + 1/2), where t = (3*n - 1)^(1/3). - Ridouane Oudra, Oct 30 2023
a(n) = m+1 if n > m(m+1)(2m+1)/6 and a(n) = m otherwise where m = floor((3n)^(1/3)). - Chai Wah Wu, Nov 04 2024
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/4 (A003881). - Amiram Eldar, Jun 30 2025

Offset corrected from 0 to 1 by Antti Karttunen, Feb 08 2014

A069039 Expansion of g.f. x*(1+x)^5/(1-x)^7.

Original entry on oeis.org

0, 1, 12, 73, 304, 985, 2668, 6321, 13504, 26577, 48940, 85305, 142000, 227305, 351820, 528865, 774912, 1110049, 1558476, 2149033, 2915760, 3898489, 5143468, 6704017, 8641216, 11024625, 13933036, 17455257, 21690928, 26751369, 32760460, 39855553, 48188416, 57926209

Offset: 0

Vladeta Jovovic, Apr 03 2002

Figurate numbers based on the 6-dimensional regular convex polytope called the 6-dimensional cross-polytope, or 6-dimensional hyperoctahedron, which is represented by the Schlaefli symbol {3, 3, 3, 3, 4}. It is the dual of the 6-dimensional hypercube. Kim asserts that every nonnegative integer can be represented by the sum of no more than 19 of these 6-crosspolytope numbers. - Jonathan Vos Post, Nov 16 2004

Starting with 1 = binomial transform of [1, 11, 50, 120, 160, 112, 32, 0, 0, 0, ...] where (1, 11, 50, 120, 160, 112, 32) = row 6 of the Chebyshev triangle A081277. Also = row 6 of the array in A142978. - Gary W. Adamson, Jul 19 2008

H. S. M. Coxeter, Regular Polytopes, New York: Dover, 1973.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 240.
Jonathan Vos Post, "4-Dimensional Jonathan numbers: polytope numbers and Centered polytope numbers of Higher Than 3 Dimensions", Draft 1.5 of 9 a.m., 12 March 2004, circulated by e-mail.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Milan Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75.
Jonathan Vos Post, Table of polytope numbers, Sorted, Through 1,000,000.
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Similar sequence: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A099193 (m=7), A099195 (m=8), A099196 (m=9), A099197 (m=10).
Cf. A000332.
Cf. A081277, A142978.

al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n), r=0..d-1); end; [seq(be(6,n),n=0..100)];

a[n_] := n^2*(2*n^4 + 20*n^2 + 23)/45; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jan 29 2014 *)
CoefficientList[Series[x (1+x)^5/(1-x)^7,{x,0,40}],x] (* or *) LinearRecurrence[ {7,-21,35,-35,21,-7,1},{0,1,12,73,304,985,2668},40] (* Harvey P. Dale, Aug 05 2018 *)

x='x+O('x^100); concat(0, Vec(x*(1+x)^5/(1-x)^7)) \\ Altug Alkan, Dec 14 2015

Recurrence: 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).
a(n) = (n^2)*(2*n^4 + 20*n^2 + 23 )/45. - Jonathan Vos Post, Nov 16 2004
From Stephen Crowley, Jul 14 2009: (Start)
Sum_{n >= 1} 1/a(n) = -5*(Sum(_alpha*(77*_alpha^2+655)*Psi(1-_alpha), _alpha = RootOf(2*_Z^4+20*_Z^2+23)))*(1/3174)+15*Pi^2*(1/46)=1.10203455013915915542552577192042916250524...
Sum_{n>=1} 1/(a(n)*n!) = hypergeom([1, 1, 1, 1-a, 1+b, 1-b, 1+a], [2, 2, 2, 2+b, 2-b, 2+a, 2-a], 1) = 1.04409584723862654376639417281585634150689... where a = (i/2)*sqrt(20+6*sqrt(6)), b = (i/2)*sqrt(20-6*sqrt(6)), and i = sqrt(-1). (End)
a(n) = 12*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018
E.g.f.: exp(x)*x*(45 + 225*x + 300*x^2 + 150*x^3 + 30*x^4 + 2*x^5)/45. - Stefano Spezia, Mar 10 2024

A099193 a(n) = n(4n^6 + 70n^4 + 196n^2 + 45)/315.

Original entry on oeis.org

0, 1, 14, 99, 476, 1765, 5418, 14407, 34232, 74313, 149830, 284075, 511380, 880685, 1459810, 2340495, 3644272, 5529233, 8197758, 11905267, 16970060, 23784309, 32826266, 44673751, 60018984, 79684825, 104642486, 136030779, 175176964, 223619261, 283131090, 355747103

Offset: 0

Jonathan Vos Post, Nov 16 2004

Kim asserts that every nonnegative integer can be represented by the sum of no more than 21 of these numbers.

Starting with 1 = binomial transform of [1, 13, 72, 220, 400, 432, 256, 0, 0, 0, ...], where (1, 13, 72, 220, 400, 432, 256) = row 7 of the Chebyshev triangle A081277. Also = row 7 of the array in A142978. - Gary W. Adamson, Jul 19 2008

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75.
Index entries for linear recurrences with constant coefficients, signature (8, -28, 56, -70, 56, -28, 8, -1).

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099195 (m=8), A099196 (m=9), A099197 (m=10).
Cf. A000332.
Cf. A142978, A081277.

Table[SeriesCoefficient[x (1 + x)^6/(1 - x)^8, {x, 0, n}], {n, 0, 31}] (* Michael De Vlieger, Dec 14 2015 *)

concat(0, Vec(x*(1+x)^6/(1-x)^8 + O(x^40))) \\ Michel Marcus, Dec 14 2015

a(n) = n*(4*n^6 + 70*n^4 + 196*n^2 + 45)/315.
G.f.: x*(1+x)^6/(1-x)^8. - R. J. Mathar, Jul 18 2009
a(n) = 14*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018

More terms from Michel Marcus, Dec 14 2015

A035597 Number of points of L1 norm 3 in cubic lattice Z^n.

Original entry on oeis.org

0, 2, 12, 38, 88, 170, 292, 462, 688, 978, 1340, 1782, 2312, 2938, 3668, 4510, 5472, 6562, 7788, 9158, 10680, 12362, 14212, 16238, 18448, 20850, 23452, 26262, 29288, 32538, 36020, 39742, 43712, 47938, 52428, 57190, 62232, 67562

Offset: 0

N. J. A. Sloane

Sums of the first n terms > 0 of A001105 in palindromic arrangement. a(n) = Sum_{i=1 .. n} A001105(i) + Sum_{i=1 .. n-1} A001105(i), e.g. a(3) = 38 = 2 + 8 + 18 + 8 + 2; a(4) = 88 = 2 + 8 + 18 + 32 + 18 + 8 + 2. - Klaus Purath, Jun 19 2020

Apart from multiples of 3, all divisors of n are also divisors of a(n), i.e. if n is not divisible by 3, a(n) is divisible by n. All divisors d of a(n) for d !== 0 (mod) 3 are also divisors of a(abs(n-d)) and a(n+d). For all n congruent to 0,2,7 (mod 9) a(n) is divisible by 3. If n is divisible by 3^k, a(n) is divisible by 3^(k-1). - Klaus Purath, Jul 24 2020

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
J. H. Conway and N. J. A. Sloane, Low-dimensional lattices. VII. Coordination sequences, Proc. Roy. Soc. Lond. A 458 (1996) 2369-2389.
Milzn Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.2.
Milan Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - _N. J. A. Sloane_, Feb 13 2013
Milan Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
R. J. Mathar, Bivariate generating functions for non-attacking Wazirs on rectangular boards, viXra:2404.0122 (2024) page 14
Joan Serra-Sagrista, Enumeration of lattice points in l_1 norm, Inf. Proc. Lett. 76 (1-2) (2000) 39-44.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Partial sums of A069894.
Cf. A001105, A005900, A084570, A097869.
Cf. A001844, A001845, A005899.
Column 3 of A035607, A266213, A343599.
Row 3 of A113413, A119800, A122542.

[(4*n^3 + 2*n)/3: n in [0..40]]; // Vincenzo Librandi, Sep 19 2011

f := proc(n,m) local i; sum( 2^i*binomial(n,i)*binomial(m-1,i-1),i=1..min(n,m)); end; # n=dimension, m=norm

Table[(4n^3+2n)/3,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,2,12,38},41] (* Harvey P. Dale, Sep 18 2011 *)

a(n) = (4*n^3 + 2*n)/3.
a(n) = 2*A005900(n). - R. J. Mathar, Dec 05 2009
a(0)=0, a(1)=2, a(2)=12, a(3)=38, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). G.f.: (2*x*(x+1)^2)/(x-1)^4. - Harvey P. Dale, Sep 18 2011
a(n) = -a(-n), a(n+1) = A097869(4n+3) = A084570(2n+1). - Bruno Berselli, Sep 20 2011
a(n) = 2*n*Hypergeometric2F1(1-n,1-k,2,2), where k=3. Also, a(n) = A001845(n) - A001844(n). - Shel Kaphan, Feb 26 2023
a(n) = A005899(n)*n/3. - Shel Kaphan, Feb 26 2023
a(n) = A006331(n)+A006331(n-1). - R. J. Mathar, Aug 12 2025

cp's OEIS Frontend

A004467 a(n) = n*(11*n^2 - 5)/6.

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A062025 a(n) = n*(13*n^2 - 7)/6.

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A063523 a(n) = n*(8*n^2 - 5)/3.

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A069038 Expansion of g.f. x*(1+x)^4/(1-x)^6.

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A059722 a(n) = n*(2*n^2 - 2*n + 1).

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A066357 Number of ordered (i.e., planar) trees on 2n edges with every subtree at the root having an even number of edges.

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A074279 n appears n^2 times.

A004467 a(n) = n(11n^2 - 5)/6.

A062025 a(n) = n(13n^2 - 7)/6.

A063523 a(n) = n(8n^2 - 5)/3.

A059722 a(n) = n(2n^2 - 2*n + 1).

A099193 a(n) = n(4n^6 + 70n^4 + 196n^2 + 45)/315.