A080859 -id:A080859 - cp's OEIS frontend

A015237 a(n) = (2n - 1)n^2.

0, 1, 12, 45, 112, 225, 396, 637, 960, 1377, 1900, 2541, 3312, 4225, 5292, 6525, 7936, 9537, 11340, 13357, 15600, 18081, 20812, 23805, 27072, 30625, 34476, 38637, 43120, 47937, 53100, 58621, 64512

Offset: 0

N. J. A. Sloane

Structured hexagonal prism numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of divisors of 60^(n-1) for n>0. - J. Lowell, Aug 30 2008
The sum of the 2*n+1 numbers between n*(n+1) and (n+1)*(n+2) gives a(n+1). - J. M. Bergot, Apr 17 2013
Partial sums of A080859. - J. M. Bergot, Jul 03 2013
a(n) = number of 2 X 2 matrices having all elements in {0..n} with determinant = permanent. - Indranil Ghosh, Dec 26 2016
Number of additions and multiplications needed to multiply two n X n matrices naively. - Charles R Greathouse IV, Jan 19 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
M. Janjic and B. Petkovic, A counting function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.
M. Janjic, B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A100177 (structured prisms); A100145 (more on structured numbers).
Cf. A000578, A045991, A000384, A080859 (first diffs), A103220 (partial sums).
Cf. similar sequences, with the formula (k*n-k+2)*n^2/2, listed in A262000.
Cf. A036970, A272378.

List([0..40],n->(2*n-1)*n^2); # Muniru A Asiru, Feb 05 2019

[(2*n-1)*n^2: n in [0..40]]; // Vincenzo Librandi, Jul 19 2011

[(2*n-1)*n^2$n=0..40]; # Muniru A Asiru, Feb 05 2019

RecurrenceTable[{a[0]==0, a[1]==1, a[2]==12, a[n]== 3*a[n-1] - 3*a[n-2] + a[n-3] + 12}, a, {n, 30}] (* G. C. Greubel, Jul 31 2015 *)
Table[(2 n - 1) n^2, {n, 0, 40}] (* Bruno Berselli, Sep 08 2015 *)

a(n)=(2*n-1)*n^2 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = A000578(n) + A045991(n). - Zerinvary Lajos, Jun 11 2008
a(n) = A199771(2*n-1) for n > 0. - Reinhard Zumkeller, Nov 23 2011
G.f.: x*(1+8*x+3*x^2)/(1-x)^4. - Colin Barker, Jun 08 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 12, a(0)=1, a(1)=1, a(2)=12. - G. C. Greubel, Jul 31 2015
E.g.f.: x*(2*x^2 + 5*x + 1)*exp(x). - G. C. Greubel, Jul 31 2015
a(n) = Sum_{i=0..n-1} n*(4*i+1) for n>0. - Bruno Berselli, Sep 08 2015
Sum_{n>=1} 1/a(n) = 4*log(2) - Pi^2/6. - Vaclav Kotesovec, Oct 04 2016
a(n) = Sum_{i=n^2-n+1..n^2+n-1} i. - Wesley Ivan Hurt, Dec 27 2016
From Peter Bala, Jan 30 2019: (Start)
Let a(n,x) = Product_{k = 0..n} (x - k)/(x + k). Then for positive integer x we have (2*x - 1)*x^2 = Sum_{n >= 0} ((n+1)^5 + n^5)*a(n,x) and (2*x - 1)*x = Sum_{n >= 0} ((n+1)^4 - n^4)*a(n,x). Both identities are also valid for complex x in the half-plane Re(x) > 2. See the Bala link in A036970. Cf. A272378. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi - Pi^2/12 - 2*log(2). - Amiram Eldar, Jul 12 2020

A186424 Odd terms in A186423.

Original entry on oeis.org

1, 3, 11, 17, 33, 43, 67, 81, 113, 131, 171, 193, 241, 267, 323, 353, 417, 451, 523, 561, 641, 683, 771, 817, 913, 963, 1067, 1121, 1233, 1291, 1411, 1473, 1601, 1667, 1803, 1873, 2017, 2091, 2243, 2321, 2481, 2563, 2731, 2817, 2993, 3083, 3267, 3361, 3553, 3651

Offset: 0

Reinhard Zumkeller, Feb 21 2011

Sum of odd square and half of even square. - Vladimir Joseph Stephan Orlovsky, May 20 2011

Numbers m such that 6*m-2 is a square. - Bruno Berselli, Apr 29 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A186421, A186423, A203568, A091999, A080859, A126587, A053253.

a186424 n = a186424_list !! n
a186424_list = filter odd a186423_list

Table[If[OddQ[n],n^2+((n+1)^2)/2,(n^2)/2+(n+1)^2],{n,0,100}] (* Vladimir Joseph Stephan Orlovsky, May 20 2011 *)

def A186424(n): return (n*(3*n + 2) + 1 if n&1 else n*(3*n + 4) + 2)>>1 # Chai Wah Wu, Jan 31 2023

From R. J. Mathar, Feb 28 2011: (Start)
G.f.: ( -1-2*x-6*x^2-2*x^3-x^4 ) / ( (1+x)^2*(x-1)^3 ).
a(n) = 3*(1+2*n+2*n^2)/4 + (-1)^n*(1+2*n)/4. (End)
a(n+2) = a(n) + A091999(n+2).
Union of A080859 and A126587: a(2*n) = A080859(n) and a(2*n+1) = A126587(n+1).
From Peter Bala, Feb 13 2021: (Start)
Appears to be the sequence of exponents in the following series expansion:
Sum_{n >= 0} (-1)^n * x^n/Product_{k = 1..n} 1 - x^(2*k-1) = 1 - x - x^3 + x^11 + x^17 - x^33 - x^43 + + - - .... Cf. A053253.
More generally, for nonnegative integer N, we appear to have the identity
Product_{j = 1..N} 1/(1 + x^(2*j-1))*( P(N,x) + Sum_{n >= 1} (-1)^n * x^((2*N+1)*n-N)/Product_{k = 1..n} 1 - x^(2*k-1) ) = 1 - x - x^3 + x^11 + x^17 - x^33 - x^43 + + - - ..., where P(N,x) is a polynomial in x of degree N^2 - 1, with the first few values given empirically by
P(0,x) = 0, P(1,x) = 1, P(2,x) = 1 - x^2 + x^3, P(3,x) = 1 - x^2 + x^5 - x^7 + x^8 and P(4,x) =  1 - x^2 - x^4 + x^5 + x^8 - x^9 + x^12 - x^14 + x^15. Cf. A203568. (End)
E.g.f.: ((2 + 5*x + 3*x^2)*cosh(x) + (1 + 7*x + 3*x^2)*sinh(x))/2. - Stefano Spezia, May 08 2021
Sum_{n>=0} 1/a(n) = sqrt(2)*Pi*sinh(sqrt(2)*Pi/3)/(1+2*cosh(sqrt(2)*Pi/3)). - Amiram Eldar, May 11 2025

A080853 Square array of generalized polygonal numbers, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 4, 1, 1, 4, 9, 7, 1, 1, 5, 16, 19, 11, 1, 1, 6, 25, 37, 33, 16, 1, 1, 7, 36, 61, 67, 51, 22, 1, 1, 8, 49, 91, 113, 106, 73, 29, 1, 1, 9, 64, 127, 171, 181, 154, 99, 37, 1, 1, 10, 81, 169, 241, 276, 265, 211, 129, 46, 1, 1, 11, 100, 217, 323, 391, 406, 365, 277

Offset: 0

Paul Barry, Feb 23 2003

			Rows begin with n>=0, k>=0
1 1 1 1 1 ...
1 2 4 7 11 ...
1 3 9 19 33 ...
1 4 16 37 67 ...
1 5 25 61 113 ...

Rows include A000124, A058331, A080855, A080856, A080857, A080919.

Columns include A000290, A003215, A003215, A080859, A080860, A080861.

A080853 := proc(n,k)
    binomial(k,0)+n*binomial(k,1)+n^2*binomial(k,2) ;
end proc:
seq( seq(A080853(d-k,k),k=0..d),d=0..12) ; # R. J. Mathar, Oct 01 2021

T(n, k)=C(k, 0)+C(k, 1)n+C(k, 2)n^2=(n^2*k^2-(n^2-2n)*k+2)/2 =(k(k-1)*n^2+2k*n+2)/2
Row n has g.f. (1+(n-2)x+(n^2-n+1)x^2)/(1-x)^3.
Column k has g.f. (C(k-1, 0)+(C(k+1, 2)-2)*x+C(k-1, 2)*x^2)/(1-x)^3.
Diagonals are given by (n^4+(2k-1)*n^3+((k-1)^2+1)*n^2+(1-(k-1)^2)*n+2)/2.
Antidiagonal sums are 1, 2, 4, 9, 22, 53, 119,... = (d+1)*(2*d^4-7*d^3+27*d^2-22*d+120)/120 = sum_{k=0..d} T(d-k,k), first differences in  A116701, d>=0. - R. J. Mathar, Oct 01 2021

A220083 a(n) = (15n^2 + 9n + 2)/2.

Original entry on oeis.org

1, 13, 40, 82, 139, 211, 298, 400, 517, 649, 796, 958, 1135, 1327, 1534, 1756, 1993, 2245, 2512, 2794, 3091, 3403, 3730, 4072, 4429, 4801, 5188, 5590, 6007, 6439, 6886, 7348, 7825, 8317, 8824, 9346, 9883, 10435, 11002, 11584, 12181, 12793, 13420, 14062

Offset: 0

Bruno Berselli, Dec 10 2012

Sequence related to the heptagonal numbers (A000566) by a(n) = n*A000566(n)-(n-1)*A000566(n-1).
Other similar sequences:
A005408(m) = (m+1)*A001477(m+1)-m*A001477(m), A001477 = nonn. integers;
A000326(m) = m*A000217(n)-(m-1)*A000217(m-1), A000217 = triangular numbers;
A003215(m) = (m+1)*A000290(m+1)-n*A000290(m), A000290 = square numbers;
A081267(m) = (m+1)*A000326(m+1)-n*A000326(m), A000326 = pentagonal numbers;
A080859(m) = (m+1)*A000384(m+1)-n*A000384(m), A000384 = hexagonal numbers;
A214675(m) = m*A000567(m)-(m-1)*A000567(m-1), A000567 = octagonal numbers.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000326, A003215, A005408, A006597, A080859, A081267, A214675.

/* By first comment: */  A000566:=func; [n*A000566(n)-(n-1)*A000566(n-1): n in [1..45]];

[(15*n^2 + 9*n + 2)/2: n in [0..50]]; // Vincenzo Librandi, Aug 18 2013

A220083:=n->(15*n^2 + 9*n + 2)/2; seq(A220083(n), n=0..100); # Wesley Ivan Hurt, Nov 14 2013

Table[(15 n^2 + 9 n + 2)/2, {n, 0, 45}]
CoefficientList[Series[(1 + 10 x + 4 x^2) / (1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)

A220083(n):=(15*n^2 + 9*n + 2)/2$ makelist(A220083(n),n,0,20); /* Martin Ettl, Dec 11 2012 */

a(n)=(15*n^2+9*n+2)/2 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (1+10*x+4*x^2)/(1-x)^3.
Sum( a(i), i=0..n ) = A006597(n+1).
a(n) + a(-n) = A010005(n) for n>0.

A201279 a(n) = 6n^2 + 10n + 5.

Original entry on oeis.org

5, 21, 49, 89, 141, 205, 281, 369, 469, 581, 705, 841, 989, 1149, 1321, 1505, 1701, 1909, 2129, 2361, 2605, 2861, 3129, 3409, 3701, 4005, 4321, 4649, 4989, 5341, 5705, 6081, 6469, 6869, 7281, 7705, 8141, 8589, 9049, 9521, 10005, 10501, 11009, 11529, 12061

Offset: 0

Eleonora Echeverri-Toro, Nov 29 2011

Numbers n where 6n-5 is a square of a number type 6n-1.
Also sequence found by reading the line from 5, in the direction 5, 21,..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Jul 18 2012
The spiral mentioned above naturally appears on a "graphene" like lattice (planar net 6^3). The opposite diagonal is A080859. - Yuriy Sibirmovsky, Oct 04 2016
First differences of A048395. - Leo Tavares, Nov 24 2021 [Corrected by Omar E. Pol, Dec 26 2021]

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Yuriy Sibirmovsky, Diagonals of a 'graphene' number spiral.
Leo Tavares, Illustration: Diamond Frame Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A136392, A080859.

Cf. A003154, A022144, A016754, A046092, A048395.

[6*n^2 + 10*n + 5: n in [0..60]]; // Vincenzo Librandi, Dec 01 2011

LinearRecurrence[{3,-3,1},{5,21,49},50] (* Vincenzo Librandi, Dec 01 2011 *)
Table[6 n^2 + 10 n + 5, {n, 0, 44}] (* or *)
CoefficientList[Series[(1 + x) (5 + x)/(1 - x)^3, {x, 0, 44}], x] (* Michael De Vlieger, Oct 04 2016 *)

a(n)=6*n^2+10*n+5 \\ Charles R Greathouse IV, Nov 29 2011

G.f.: (1+x)*(5+x)/(1-x)^3. - Colin Barker, Jan 09 2012
a(n) = 1 + A033579(n+1). - Omar E. Pol, Jul 18 2012
a(n) = (n+1)*A001844(n+1)-n*A001844(n). [Bruno Berselli, Jan 15 2013]
From Leo Tavares, Nov 24 2021: (Start)
a(n) = A003154(n+2) - A022144(n+1). See Diamond Frame Stars illustration.
a(n) = A016754(n) + A046092(n+1). (End)

A202804 a(n) = n(6n+4).

Original entry on oeis.org

0, 10, 32, 66, 112, 170, 240, 322, 416, 522, 640, 770, 912, 1066, 1232, 1410, 1600, 1802, 2016, 2242, 2480, 2730, 2992, 3266, 3552, 3850, 4160, 4482, 4816, 5162, 5520, 5890, 6272, 6666, 7072, 7490, 7920, 8362, 8816, 9282, 9760, 10250, 10752, 11266, 11792, 12330

Offset: 0

Jeremy Gardiner, Dec 24 2011

Sequence found by reading the line from 0, in the direction 0, 10, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A033579 in the same spiral. - Omar E. Pol, Jul 17 2012

Partial sums give A163815. - Leo Tavares, Feb 25 2022

Jeremy Gardiner, Table of n, a(n) for n = 0..3000
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Leo Tavares, Illustration: Star Halves.
Pavlos Vavolas, Cellular automaton model of cardiac arrhythmias, University of Sheffield Department of Computer Science (2005), (see page 43). [broken link, abstract]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A003154, A005408, A033580, A033581, A049453, A163815, A195319.

Cf. A001318, A033579, A045944, A080859.

A202804:=n->n*(6*n+4): seq(A202804(n), n=0..100); # Wesley Ivan Hurt, Apr 09 2017

Table[n(6n+4),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,10,32},50] (* Harvey P. Dale, Dec 28 2015 *)

x='x + O('x^50); concat([0], Vec(-2*x*(5 + x)/(x - 1)^3)) \\ Indranil Ghosh, Apr 10 2017

a(n) = 2*n(3*n+2) = 6*n^2 + 4*n = 2*A045944(n).
a(n) = A080859(n) - 1. - Omar E. Pol, Jul 18 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Dec 28 2015
G.f.: 2*x*(5 + x)/(1 - x)^3. - Indranil Ghosh, Apr 10 2017
a(n) = A003154(n+1) - A005408(n). - Leo Tavares, Feb 25 2022
From Amiram Eldar, Mar 01 2022: (Start)
Sum_{n>=1} 1/a(n) = (Pi/sqrt(3) - 3*log(3) + 3)/8.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(4*sqrt(3)) - 3/8. (End)
E.g.f.: 2*exp(x)*x*(5 + 3*x). - Elmo R. Oliveira, Dec 12 2024

A272378 a(n) = n(6n^2 - 8*n + 3).

Original entry on oeis.org

0, 1, 22, 99, 268, 565, 1026, 1687, 2584, 3753, 5230, 7051, 9252, 11869, 14938, 18495, 22576, 27217, 32454, 38323, 44860, 52101, 60082, 68839, 78408, 88825, 100126, 112347, 125524, 139693, 154890, 171151, 188512, 207009, 226678, 247555, 269676, 293077

Offset: 0

Vincenzo Librandi, Apr 29 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014 (page 16).
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1)

Cf. A000384, A015237, A272379, A272380.

Magma
```
[n*(6*n^2 - 8*n + 3): n in [0..50]];
```

Table[n (6 n^2 - 8 n + 3), {n, 0, 50}]
LinearRecurrence[{4,-6,4,-1},{0,1,22,99},40] (* Harvey P. Dale, Dec 29 2017 *)

vector(100, n, n--; n*(6*n^2 - 8*n + 3)) \\ Altug Alkan, Apr 29 2016

for n in range(0,10**3):print(n*(6*n**2-8*n+3),end=", ") # Soumil Mandal, Apr 30 2016

G.f.: x*(1 + 18*x + 17*x^2)/(1 - x)^4.
E.g.f.: x*(1 + 10*x + 6*x^2)*exp(x).
a(n) = n*A080859(n+1).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), for n>3.
See page 7 in Brent's paper:
a(n) = n^2*A000384(n) - n*(n-1)*A000384(n-1).
A272379(n) = n^2*a(n) - n*(n-1)*a(n-1).
From Peter Bala, Jan 30 2019: (Start)
Let a(n,x) = Product_{k = 0..n} (x - k)/(x + k). Then for positive integer x we have x^2*(6*x^2 - 8*x + 3) = Sum_{n >= 0} ((n+1)^7 + n^7)*a(n,x) and x*(6*x^2 - 8*x + 3) = Sum_{n >= 0} ((n+1)^6 - n^6)*a(n,x). Both identities are also valid for complex x in the half-plane Re(x) > 7/2. See the Bala link in A036970. Cf. A272379. (End)

A319384 a(n) = a(n-1) + 2a(n-2) - 2a(n-3) - a(n-4) + a(n-5), a(0)=1, a(1)=5, a(2)=9, a(3)=21, a(4)=29.

Original entry on oeis.org

1, 5, 9, 21, 29, 49, 61, 89, 105, 141, 161, 205, 229, 281, 309, 369, 401, 469, 505, 581, 621, 705, 749, 841, 889, 989, 1041, 1149, 1205, 1321, 1381, 1505, 1569, 1701, 1769, 1909, 1981, 2129, 2205, 2361, 2441, 2605, 2689, 2861, 2949, 3129, 3221, 3409, 3505, 3701, 3801, 4005, 4109, 4321, 4429, 4649, 4761, 4989, 5105, 5341, 5461

Offset: 0

Paul Curtz, Sep 18 2018

The two bisections A136392(n+1)=1,9,29,61, ... and A201279(n)=5,21,49, ... are in the hexagonal spiral based on 2*n+1:
.
                    67--65--63--61
                    /             \
                  69  33--31--29  59
                  /   /         \   \
                71  35  11---9  27  57
                /   /   /     \   \   \
              73  37  13   1   7  25  55
                  /   /   /   /   /   /
                39  15   3---5  23  53
                  \   \         /   /
                  41  17--19--21  51
                    \             /
                    43--45--47--49
.
A201279(n) - A136892(n) = 20*n.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A001844, A032766, A104585.
In the spiral: A003154(n+1), A080859, A126587, A136392, A201279, A227776.
Partial sums of A382154.

[(6*n^2 + 6*n + 5 - (2*n + 1)*(-1)^n)/4 : n in [0..50]]; // Wesley Ivan Hurt, Jan 19 2021

Table[(6 n^2 + 6 n + 5 - (2 n + 1)*(-1)^n)/4, {n, 0, 80}] (* Wesley Ivan Hurt, Jan 07 2021 *)

Vec((1 + x^2)*(1 + 4*x + x^2) / ((1 - x)^3*(1 + x)^2) + O(x^50)) \\ Colin Barker, Jun 05 2019

def A319384(n): return (n*(3*n+4)+3 if n&1 else n*(3*n+2)+2)>>1 # Chai Wah Wu, Mar 25 2025

a(2*n) = A136392(n+1), a(2*n+1) = A201279(n).
a(-n) = a(n).
a(2*n) + a(2*n+1) = 6*A001844(n).
a(n) = (6*n^2 + 6*n + 5 - (2*n + 1)*(-1)^n)/4. - Wesley Ivan Hurt, Oct 04 2018
G.f.: (1 + x^2)*(1 + 4*x + x^2) / ((1 - x)^3*(1 + x)^2). - Colin Barker, Jun 05 2019
a(n) = A104585(n) + A032766(n+1). - Alex W. Nowak, Jan 08 2021

More terms from N. J. A. Sloane, Mar 23 2025

A340970 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = Sum_{j=0..n} k^j * binomial(n,j) * binomial(2*j,j).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 11, 1, 1, 7, 33, 45, 1, 1, 9, 67, 245, 195, 1, 1, 11, 113, 721, 1921, 873, 1, 1, 13, 171, 1593, 8179, 15525, 3989, 1, 1, 15, 241, 2981, 23649, 95557, 127905, 18483, 1, 1, 17, 323, 5005, 54691, 361449, 1137709, 1067925, 86515, 1

Offset: 0

Seiichi Manyama, Feb 01 2021

			Square array begins:
  1,   1,     1,     1,      1,       1, ...
  1,   3,     5,     7,      9,      11, ...
  1,  11,    33,    67,    113,     171, ...
  1,  45,   245,   721,   1593,    2981, ...
  1, 195,  1921,  8179,  23649,   54691, ...
  1, 873, 15525, 95557, 361449, 1032801, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..3 give A000012, A026375, A084771, A340973.
Rows n=0..2 give A000012, A005408, A080859.
Main diagonal gives A340971.
Cf. A340968.

T[n_, k_] := Sum[If[j == k == 0, 1, k^j] * Binomial[n, j] * Binomial[2*j, j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Feb 01 2021 *)

T(n, k) = sum(j=0, n, k^j*binomial(n, j)*binomial(2*j, j));

T(n, k) = polcoef((1+(2*k+1)*x+(k*x)^2)^n, n);

G.f. of column k: 1/sqrt((1 - x) * (1 - (4*k+1)*x)).
T(n,k) = [x^n] (1+(2*k+1)*x+(k*x)^2)^n.
n * T(n,k) = (2*k+1) * (2*n-1) * T(n-1,k) - (4*k+1) * (n-1) * T(n-2,k) for n > 1.
E.g.f. of column k: exp((2*k+1)*x) * BesselI(0,2*k*x). - Ilya Gutkovskiy, Feb 01 2021
From Seiichi Manyama, Aug 19 2025: (Start)
T(n,k) = (1/4)^n * Sum_{j=0..n} (4*k+1)^j * binomial(2*j,j) * binomial(2*(n-j),n-j).
T(n,k) = Sum_{j=0..n} (-k)^j * (4*k+1)^(n-j) * binomial(n,j) * binomial(2*j,j). (End)

A354594 a(n) = n^2 + 2*floor(n/2)^2.

Original entry on oeis.org

0, 1, 6, 11, 24, 33, 54, 67, 96, 113, 150, 171, 216, 241, 294, 323, 384, 417, 486, 523, 600, 641, 726, 771, 864, 913, 1014, 1067, 1176, 1233, 1350, 1411, 1536, 1601, 1734, 1803, 1944, 2017, 2166, 2243, 2400, 2481, 2646, 2731, 2904

Offset: 0

David Lovler, Jun 01 2022

The first bisection is A033581, the second bisection is A080859. - Bernard Schott, Jun 07 2022

David Lovler, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000290, A008794, A033581, A080859, A247375.

Cf. A213037, A322744 (diagonal), A354595, A354596.

a[n_] := n^2 + 2 Floor[n/2]^2
Table[a[n], {n, 0, 90}]    (* A354594 *)
LinearRecurrence[{1, 2, -2, -1, 1}, {0, 1, 6, 11, 24}, 60]

PARI
```
a(n) = n^2 + 2*(n\2)^2;
```

a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5), n >= 5.
a(n) = A000290(n) + 2*A008794(n).
G.f.: x*(1 + 5*x + 3*x^2 + 3*x^3)/((1 - x)^3*(1 + x)^2).
E.g.f.: (x*(1 + 3*x)*cosh(x) + (1 + 3*x + 3*x^2)*sinh(x))/2. - Stefano Spezia, Jun 07 2022

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A015237 a(n) = (2n - 1)n^2.

A220083 a(n) = (15n^2 + 9n + 2)/2.

A202804 a(n) = n(6n+4).

A272378 a(n) = n(6n^2 - 8*n + 3).

A319384 a(n) = a(n-1) + 2a(n-2) - 2a(n-3) - a(n-4) + a(n-5), a(0)=1, a(1)=5, a(2)=9, a(3)=21, a(4)=29.