A227776 -id:A227776 - cp's OEIS frontend

A054554 a(n) = 4n^2 - 10n + 7.

1, 3, 13, 31, 57, 91, 133, 183, 241, 307, 381, 463, 553, 651, 757, 871, 993, 1123, 1261, 1407, 1561, 1723, 1893, 2071, 2257, 2451, 2653, 2863, 3081, 3307, 3541, 3783, 4033, 4291, 4557, 4831, 5113, 5403, 5701, 6007, 6321, 6643, 6973, 7311, 7657, 8011, 8373, 8743

Offset: 1

Enoch Haga, G. L. Honaker, Jr., Apr 10 2000

Move in 1-3 direction in a spiral organized like A068225 etc.
Equals binomial transform of [1, 2, 8, 0, 0, 0, ...]. - Gary W. Adamson, May 03 2008
Ulam's spiral (NE spoke). - Robert G. Wilson v, Oct 31 2011
Number of ternary strings of length 2*(n-1) that have one or no 0's, one or no 1's, and an even number of 2's. For n=2, the 3 strings of length 2 are 01, 10 and 22. For n=3, the 13 strings of length 4 are the 12 permutations of 0122 and 2222. - Enrique Navarrete, Jul 25 2025

Ivan Panchenko, Table of n, a(n) for n = 1..1000
James Grime and Brady Haran, Prime Spirals, Numberphile video (2013).
Scientific American, Cover of the March 1964 issue
Leo Tavares, Illustration: Hexagonal Dual Rays
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A054553, A068225, A054552, A054556, A054567, A054569, A033951.
Cf. A014105.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. A003215, A227776.

A054554:=n->4*n^2 -10*n + 7; seq(A054554(k),k=1..100); # Wesley Ivan Hurt, Nov 05 2013

f[n_] := 4n^2 -10n + 7; Array[f, 40] (* Vladimir Joseph Stephan Orlovsky, Sep 01 2008 *)
LinearRecurrence[{3,-3,1},{1,3,13},60] (* Harvey P. Dale, Jan 20 2025 *)

a(n)=4*n^2-10*n+7 \\ Charles R Greathouse IV, Nov 05 2013

a(n) = 8*n + a(n-1) - 14 with n > 1, a(1)=1. - Vincenzo Librandi, Aug 07 2010
G.f.: -x*(7*x^2+1)/(x-1)^3. - Colin Barker, Sep 21 2012
For n > 2, a(n) = A014105(n) + A014105(n-1). - Bruce J. Nicholson, May 07 2017
From Leo Tavares, Feb 21 2022: (Start)
a(n) = A003215(n-2) + 2*A000217(n-1). See Hexagonal Dual Rays illustration in links.
a(n) = A227776(n-1) - 4*A000217(n-1). (End)
a(k+1) = 4k^2 - 2k + 1 in the Numberphile video. - Frank Ellermann, Mar 11 2020
E.g.f.: exp(x)*(7 - 6*x + 4*x^2) - 7. - Stefano Spezia, Apr 24 2024

Edited by Frank Ellermann, Feb 24 2002

A287326 Triangle read by rows: T(n, k) = 6k(n-k) + 1; n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 13, 13, 1, 1, 19, 25, 19, 1, 1, 25, 37, 37, 25, 1, 1, 31, 49, 55, 49, 31, 1, 1, 37, 61, 73, 73, 61, 37, 1, 1, 43, 73, 91, 97, 91, 73, 43, 1, 1, 49, 85, 109, 121, 121, 109, 85, 49, 1, 1, 55, 97, 127, 145, 151, 145, 127, 97, 55, 1, 1, 61, 109, 145, 169, 181, 181, 169, 145, 109, 61, 1

Offset: 0

Kolosov Petro, Aug 31 2017

From Kolosov Petro, Apr 12 2020: (Start)
Let A(m, r) = A302971(m, r) / A304042(m, r).
Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.
Then T(n, k) = L(1, n, k), i.e T(n, k) is partial case of L(m, n, k) for m = 1.
T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

			Triangle begins:
  ----------------------------------------
  k=    0   1   2   3   4   5   6   7   8
  ----------------------------------------
  n=0:  1;
  n=1:  1,  1;
  n=2:  1,  7,  1;
  n=3:  1, 13, 13,  1;
  n=4:  1, 19, 25, 19,  1;
  n=5:  1, 25, 37, 37, 25,  1;
  n=6:  1, 31, 49, 55, 49, 31,  1;
  n=7:  1, 37, 61, 73, 73, 61, 37,  1;
  n=8:  1, 43, 73, 91, 97, 91, 73, 43,  1;

Georg Fischer, Table of n, a(n) for n = 0..495 [rows 0..10 and 12..30 from Kolosov Petro]
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, Polynomial identity involving binomial theorem and Faulhaber's formula, 2023.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.

Columns k=0..6 give A000012, A016921, A017533, A161705, A103214, A128470, A158065.
Column sums k=0..4 give A000027, A000567, A051866, A051872, A255185.
Row sums give A001093.
Various cases of L(m, n, k): This sequence (m=1), A300656(m=2), A300785(m=3). See comments for L(m, n, k).
Differences of cubes n^3 are T(A000124(n), 1).
Cf. A000124, A000578, A003154, A003215, A007318, A008458, A016969.
Cf. A038593, A055012, A068601, A077028, A094053, A166873, A227776.
Cf. A294317, A302971, A304042.

Flat(List([0..11],n->List([0..n],k->6*k*(n-k)+1))); # Muniru A Asiru, Oct 09 2018

/* As triangle */ [[6*k*(n-k) + 1: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 26 2018

T := (n, k) -> 6*k*(n-k) + 1:
seq(seq(T(n, k), k=0..n), n=0..11); # Muniru A Asiru, Oct 09 2018

T[n_, k_] := 6 k (n - k) + 1; Column[Table[T[n, k], {n, 0, 10}, {k, 0, n}], Center] (* Kolosov Petro, Jun 02 2019 *)

t(n, k) = 6*k*(n-k)+1
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
/* Print initial 9 rows of triangle as follows */
trianglerows(9) \\ Felix Fröhlich, Jan 09 2018

def A287326(n,k): return 6*k*(n-k) + 1
flatten([[A287326(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Sep 25 2024

T(n, k) = 6*k*(n-k) + 1.
G.f. of column k: n^k*(1+(6*k-1)*n)/(1-n)^2.
G.f.: (1 - x - x*y + 7*x^2*y)/((1 - x)^2*(1 - x*y)^2). - Stefano Spezia, Oct 09 2018 [Adapted by Stefano Spezia, Sep 25 2024]
From Kolosov Petro, Jun 05 2019: (Start)
T(n, k) = 1/2 * T(A294317(n, k), k) + 1/2.
T(n+1, k) = 2*T(n, k) - T(n-1, k), for n >= k.
T(n, k) = 6*A077028(n, k) - 5.
T(2n, n) = A227776(n).
T(2n+1, n) = A003154(n+1).
T(2n+3, n) = A166873(n+1).
Sum_{k=0..n-1} T(n, k) = Sum_{k=1..n} T(n, k) = A000578(n).
Sum_{k=1..n-1} T(n, k) = A068601(n).
(n+1)^3 - n^3 = T(A000124(n), 1). (End)
Sum_{k=0..n} (-1)^k*T(n, k) = (-1/2)*(1 + (-1)^n)*A016969(floor(n/2) - 1). - G. C. Greubel, Sep 25 2024

A080855 a(n) = (9n^2 - 3n + 2)/2.

Original entry on oeis.org

1, 4, 16, 37, 67, 106, 154, 211, 277, 352, 436, 529, 631, 742, 862, 991, 1129, 1276, 1432, 1597, 1771, 1954, 2146, 2347, 2557, 2776, 3004, 3241, 3487, 3742, 4006, 4279, 4561, 4852, 5152, 5461, 5779, 6106, 6442, 6787, 7141, 7504, 7876, 8257, 8647, 9046

Offset: 0

Paul Barry, Feb 23 2003

The old definition of this sequence was "Generalized polygonal numbers".
Row T(3,n) of A080853.
Equals binomial transform of [1, 3, 9, 0, 0, 0, ...] - Gary W. Adamson, Apr 30 2008
a(n) is also the least weight of self-conjugate partitions having n different parts such that each part is congruent to 2 modulo 3. The first such self-conjugate partitions, corresponding to a(n)=1,2,3,4, are 2+2, 5+5+2+2+2, 8+8+5+5+5+2+2+2, 11+11+8+8+8+5+5+5+2+2+2. - Augustine O. Munagi, Dec 18 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=3, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 3, a(n-1) = -coeff(charpoly(A,x), x^(n-2)). - Milan Janjic, Jan 27 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8.
A. O. Munagi, Pairing conjugate partitions by residue classes, Discrete Math., 308 (2008), 2492-2501.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Leo Tavares, Illustration: Hexagonal Tri-Rays
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A027468, A038764.
Cf. A283394 (see Crossrefs section).
Cf. A003215, A000217, A227776.

List([0..50],n->(9*n^2-3*n+2)/2); # Muniru A Asiru, Nov 02 2018

[(9*n^2 - 3*n +2)/2: n in [0..50]]; // G. C. Greubel, Nov 02 2018

seq((9*n^2-3*n+2)/2,n=0..50); # Muniru A Asiru, Nov 02 2018

s = 1; lst = {s}; Do[s += n + 2; AppendTo[lst, s], {n, 1, 500, 9}]; lst (* Zerinvary Lajos, Jul 11 2009 *)
Table[(9n^2-3n+2)/2,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1}, {1,4,16}, 50] (* Harvey P. Dale, Jul 24 2013 *)

a(n)=binomial(3*n,2)+1 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (1 + x + 7*x^2)/(1 - x)^3.
a(n) = 9*n + a(n-1) - 6 with n > 0, a(0)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = n*A005448(n+1) - (n-1)*A005448(n), with A005448(0)=1. - Bruno Berselli, Jan 15 2013
a(0)=1, a(1)=4, a(2)=16; for n > 2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jul 24 2013
a(n) = A152947(3*n+1). - Franck Maminirina Ramaharo, Jan 10 2018
E.g.f.: (2 + 6*x + 9*x^2)*exp(x)/2. - G. C. Greubel, Nov 02 2018
From Leo Tavares, Feb 20 2022: (Start)
a(n) = A003215(n-1) + 3*A000217(n). See Hexagonal Tri-Rays illustration in links.
a(n) = A227776(n) - 3*A000217(n). (End)

Definition replaced with the closed form by Bruno Berselli, Jan 15 2013

A357196 Number of regions in a hexagon when n internal hexagons are drawn between the 6n points that divide each side into n+1 equal parts.

Original entry on oeis.org

1, 7, 25, 55, 97, 151, 217, 295, 385, 475, 601, 715, 865, 1015, 1159, 1351, 1537, 1735, 1945, 2131, 2401, 2647, 2905, 3115, 3457, 3751, 4057, 4357, 4705, 5005, 5401, 5767, 6133, 6535, 6925, 7303, 7777, 8215, 8653, 9025, 9601, 10051, 10585, 11071, 11587, 12151, 12697, 13171, 13825, 14395, 14989

Offset: 0

Scott R. Shannon, Sep 17 2022

nonn

Unlike similar dissections of the triangle and square, see A356984 and A357058, there is no obvious pattern for n values that yield hexagons with non-simple intersections; these n values begin 9, 11, 14, 19, 23, 27, 29, 32, 34, 35, 38, 39, 41, 43, ... .

Scott R. Shannon, Image for n = 1.
Scott R. Shannon, Image for n = 2.
Scott R. Shannon, Image for n = 5.
Scott R. Shannon, Image for n = 9. This is the first term that forms hexagons with non-simple intersections.
Scott R. Shannon, Image for n = 50.
Scott R. Shannon, Image for n = 150.

Cf. A357197 (vertices), A357198 (edges), A331931, A356984 (triangle), A357058 (square).

Cf. A227776 (6*n^2 + 1).

a(n) = A357198(n) - A357197(n) + 1 by Euler's formula.

Conjecture: a(n) = 6*n^2 + 1 for hexagons that only contain simple intersections when cut by n internal hexagons.

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

A158547 a(n) = 24*n^2 + 1.

Original entry on oeis.org

1, 25, 97, 217, 385, 601, 865, 1177, 1537, 1945, 2401, 2905, 3457, 4057, 4705, 5401, 6145, 6937, 7777, 8665, 9601, 10585, 11617, 12697, 13825, 15001, 16225, 17497, 18817, 20185, 21601, 23065, 24577, 26137, 27745, 29401, 31105, 32857, 34657, 36505, 38401, 40345

Offset: 0

Vincenzo Librandi, Mar 21 2009

The identity (24*n^2 + 1)^2 - (144*n^2 + 12)*(2*n)^2 = 1 can be written as a(n)^2 - A158546(n)*A005843(n)^2 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Leo Tavares, Illustration: Stellar Stars.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A008594, A158546, A227776.

I:=[1, 25, 97]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 14 2012

LinearRecurrence[{3, -3, 1}, {1, 25, 97}, 50] (* Vincenzo Librandi, Feb 14 2012 *)

for(n=0, 40, print1(24*n^2+1", ")); \\ Vincenzo Librandi, Feb 14 2012

G.f.: (1 + 22*x + 25*x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 02 2023: (Start)
Sum_{n>=0} 1/a(n) = 1/2 + coth(Pi/(2*sqrt(6)))*Pi/(4*sqrt(6)).
Sum_{n>=0} (-1)^n/a(n) = 1/2 + cosech(Pi/(2*sqrt(6)))*Pi/(4*sqrt(6)). (End)
From Elmo R. Oliveira, Jan 16 2025: (Start)
E.g.f.: exp(x)*(1 + 24*x + 24*x^2).
a(n) = A227776(2*n). (End)

Comment rewritten, a(0) added by R. J. Mathar, Oct 16 2009

A366551 Number of distinct characteristic polynomials for 3 X 3 matrices with entries from {0, 1, ..., n}.

Original entry on oeis.org

1, 32, 513, 4407, 21393, 86620, 242057, 673623, 1467642, 3107487, 5836467, 11108595, 18102935, 31327359, 48505904, 74802671, 110297111, 166721570, 230270840

Offset: 0

Robert P. P. McKone, Oct 13 2023

Robert P. P. McKone, The distinct characteristic polynomials for a(0)-a(6).

Cf. A366448 (2 X 2 matrices), A367978 (4 X 4 matrices).
Cf. A366158 (determinants), A227776 (2nd order coefficients), A016777 (traces).
Cf. A272659.

mat[n_Integer?Positive] := mat[n] = Array[m, {n, n}]; flatMat[n_Integer?Positive] := flatMat[n] = Flatten[mat[n]]; charPolyMat[n_Integer?Positive] := charPolyMat[n] = FullSimplify[CoefficientList[Expand[CharacteristicPolynomial[mat[n], x]], x]]; a[d_Integer?Positive, 0] = 1; a[d_Integer?Positive, n_Integer?Positive] := a[d, n] = Length[DeleteDuplicates[Flatten[Table[Evaluate[charPolyMat[d]], ##] & @@ Table[{flatMat[d][[i]], 0, n}, {i, 1, d^2}], d^2 - 1]]]; Table[a[3, n], {n, 0, 7}]

import itertools
def a(n):
    ans, W = set(), itertools.product(range(n+1), repeat=9)
    for w in W: ans.add(Matrix(ZZ, 3, 3, w).charpoly())
    return len(ans)  # Robin Visser, May 08 2025

a(n) <= A366158(n) * A227776(n) * A016777(n).

a(12)-a(18) from Robin Visser, May 08 2025

A158479 a(n) = 36*n^2 + 6.

Original entry on oeis.org

42, 150, 330, 582, 906, 1302, 1770, 2310, 2922, 3606, 4362, 5190, 6090, 7062, 8106, 9222, 10410, 11670, 13002, 14406, 15882, 17430, 19050, 20742, 22506, 24342, 26250, 28230, 30282, 32406, 34602, 36870, 39210, 41622, 44106, 46662, 49290, 51990, 54762, 57606, 60522

Offset: 1

Vincenzo Librandi, Mar 20 2009

The identity (12*n^2 + 1)^2 - (36*n^2 + 6)*(2*n)^2 = 1 can be written as A158480(n)^2 - a(n)*A005843(n)^2 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A158480, A227776.

I:=[42, 150, 330]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]];

LinearRecurrence[{3,-3,1},{42,150,330},40]

a(n)=36*n^2+6 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f: 6*x*(7+4*x+x^2)/(1-x)^3.
From Amiram Eldar, Mar 05 2023: (Start)
Sum_{n>=1} 1/a(n) = (coth(Pi/sqrt(6))*Pi/sqrt(6) - 1)/12.
Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/sqrt(6))*Pi/sqrt(6))/12. (End)
From Elmo R. Oliveira, Jan 15 2025: (Start)
E.g.f.: 6*(exp(x)*(6*x^2 + 6*x + 1) - 1).
a(n) = 6*A227776(n). (End)

A243813 Table read by antidiagonals: T(n,k) is the curvature (truncated to integer) of a circle in a variation of nested Pappus chains (see Comments for details).

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 1, 1, 1, 2, 9, 1, 1, 1, 1, 3, 13, 1, 1, 1, 1, 2, 5, 19, 1, 1, 1, 1, 1, 3, 7, 25, 1, 1, 1, 1, 1, 2, 4, 9, 33, 1, 1, 1, 1, 1, 1, 2, 5, 11, 41, 1, 1, 1, 1, 1, 1, 2, 3, 6, 14, 51, 1, 1, 1, 1, 1, 1, 1, 2, 4, 7, 17, 61, 1, 1, 1, 1, 1, 1, 1, 2, 3, 5, 9, 21

Offset: 0

Kival Ngaokrajang, Jun 11 2014

Refer to the construction rule used in A243618. For this case, the curvature is defined by (-1/k, 1/(k-1), 1), the circle radius will diverge to infinity (zero curvature). The integral curvatures appearing as periodic, i.e., 2, 6, 6, 10, 30, 42, 28, 12, ..., = A083482(k-1). The integral curvatures seem to align as some sequence, e.g., 3, 7, 13, 21, 31, 43, ..., = A002061(k) and 9, 25, 49, ..., = A016754(k-1). See illustration.

			Table begins:
  n/k  2   3   4   5   6   7  ...
   0   1   1   1   1   1   1  ...
   1   1   1   1   1   1   1  ...
   2   3   1   1   1   1   1  ...
   3   5   2   1   1   1   1  ...
   4   9   3   2   1   1   1  ...
   5  13   5   3   2   1   1  ...
   6  19   7   4   2   2   1  ...
   7  25   9   5   3   2   2  ...
   8  33  11   6   4   3   2  ...
   9  41  14   7   5   3   2  ...
  10  51  17   9   6   4   3  ...
  11  61  21  11   7   5   3  ...
  12  73  25  13   8   5   4  ...
  ...

Kival Ngaokrajang, Illustration for k = 2..7

Cf. A243618, A083482, A002061, A016754.
Cf. Column 1 = A080827(n), column 2 = A056827(n) + 1.
Cf. Integral curvature in column 1..6: [A058331, A227776, A056107, A212656, A158558, A158604].

A372118 Square array A(n, k) = ((k+2)^(n+2) - 2 * (k+1)^(n+2) + k^(n+2))/2 for k, n >= 0 read by ascending antidiagonals.

Original entry on oeis.org

1, 3, 1, 7, 6, 1, 15, 25, 9, 1, 31, 90, 55, 12, 1, 63, 301, 285, 97, 15, 1, 127, 966, 1351, 660, 151, 18, 1, 255, 3025, 6069, 4081, 1275, 217, 21, 1, 511, 9330, 26335, 23772, 9751, 2190, 295, 24, 1, 1023, 28501, 111645, 133057, 70035, 19981, 3465, 385, 27, 1

Offset: 0

Werner Schulte, Apr 19 2024

Depending on some fixed integer m >= 0 we define a family of square arrays A(m; n, k) = (Sum_{i=0..m} (-1)^i * binomial(m, i) * (k + m - i)^(n+m)) / m! for k, n >= 0. Special cases are: A004248 (m=0), A343237 (m=1) and this array (m=2). The A(m; n, k) satisfy: A(m; n, k) = (k+m) * A(m; n-1, k) + A(m-1; n, k) with initial values A(0; n, k) = k^n and A(m; 0, k) = 1.
Further properties are conjectures:
  (1) O.g.f. of column k is Prod_{i=k..k+m} 1 / (1 - i * t);
  (2) E.g.f. of row n is exp(x) * (Sum_{k=0..n} binomial(k+m, m) * A048993(n+m, k+m) * x^k);
  (3) The LU decompositions of these arrays are given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L, where L is defined L(m; n, k) = A048993(n+m, k+m) * (k+m)! / m!, i.e., A(m; n, k) = Sum_{i=0..k} L(m; n, i) * binomial(k, i).
The three conjectures are true, see links. - Sela Fried, Jul 07 2024

			Square array A(n, k) starts:
n\k :    0     1       2       3        4         5         6         7
=======================================================================
  0 :    1     1       1       1        1         1         1         1
  1 :    3     6       9      12       15        18        21        24
  2 :    7    25      55      97      151       217       295       385
  3 :   15    90     285     660     1275      2190      3465      5160
  4 :   31   301    1351    4081     9751     19981     36751     62401
  5 :   63   966    6069   23772    70035    170898    365001    706104
  6 :  127  3025   26335  133057   481951   1398097   3463615   7628545
  7 :  255  9330  111645  724260  3216795  11075670  31794105  79669320
  etc.

Paolo Xausa, Table of n, a(n) for n = 0..11324 (first 150 antidiagonals, flattened).
Sela Fried, On an integer sequence related to Euler's formula for the Stirling numbers of the second kind, 2024.

Rows: A000012 (n=0), A008585 (n=1), A227776 (n=2).
Columns: A000225 (k=0), A000392 (k=1), A016269 (k=2), A016753 (k=3), A016103 (k=4), A019757 (k=5), A020570 (k=6), A020782 (k=7).
Main diagonal: A281596(n+2).
Cf. A007318, A048993, A004248, A343237.

A372118[n_, k_] := ((k+2)^(n+2) - 2*(k+1)^(n+2) + k^(n+2))/2;
Table[A372118[n-k, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Jul 10 2024 *)

A(n, k) = ((k+2)^(n+2) - 2 * (k+1)^(n+2) + k^(n+2))/2

A(n, k) = (k+2) * A(n-1, k) + (k+1)^(n+1) - k^(n+1) for n > 0.
Conjectures:
  (1) O.g.f. of column k is Prod_{i=k..k+2} 1 / (1 - i * t);
  (2) E.g.f. of row n is exp(x) * (Sum_{k=0..n} binomial(k+2, 2) * A048993(n+2, k+2) * x^k);
  (3) The LU decomposition of this array is given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L, where L is defined L(n, k) = A048993(n+2, k+2) * (k+2)! / 2!, i.e., A(n, k) = Sum_{i=0..k} L(n, i) * binomial(k, i).
The three conjectures are true. See comments. - Sela Fried, Jul 09 2024

cp's OEIS Frontend

A054554 a(n) = 4*n^2 - 10*n + 7.

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A287326 Triangle read by rows: T(n, k) = 6*k*(n-k) + 1; n >= 0, 0 <= k <= n.

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A080855 a(n) = (9*n^2 - 3*n + 2)/2.

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A357196 Number of regions in a hexagon when n internal hexagons are drawn between the 6n points that divide each side into n+1 equal parts.

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

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A158547 a(n) = 24*n^2 + 1.

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A054554 a(n) = 4n^2 - 10n + 7.

A287326 Triangle read by rows: T(n, k) = 6k(n-k) + 1; n >= 0, 0 <= k <= n.

A080855 a(n) = (9n^2 - 3n + 2)/2.

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.