A006324 -id:A006324 - cp's OEIS frontend

A225010 T(n,k) = number of n X k 0..1 arrays with rows unimodal and columns nondecreasing.

2, 4, 3, 7, 9, 4, 11, 22, 16, 5, 16, 46, 50, 25, 6, 22, 86, 130, 95, 36, 7, 29, 148, 296, 295, 161, 49, 8, 37, 239, 610, 791, 581, 252, 64, 9, 46, 367, 1163, 1897, 1792, 1036, 372, 81, 10, 56, 541, 2083, 4166, 4900, 3612, 1716, 525, 100, 11, 67, 771, 3544, 8518, 12174, 11088, 6672, 2685, 715, 121, 12

Offset: 1

R. H. Hardin, Apr 23 2013

Table starts
..2...4...7...11....16.....22.....29......37......46.......56.......67
..3...9..22...46....86....148....239.....367.....541......771.....1068
..4..16..50..130...296....610...1163....2083....3544.....5776.....9076
..5..25..95..295...791...1897...4166....8518...16414....30086....52834
..6..36.161..581..1792...4900..12174...27966...60172...122464...237590
..7..49.252.1036..3612..11088..30738...78354..186142...416394...884236
..8..64.372.1716..6672..22716..69498..194634..505912..1233584..2845492
..9..81.525.2685.11517..43065.144111..439791.1241383..3276559..8157227
.10.100.715.4015.18832..76714.278707..920491.2803658..7963384.21280337
.11.121.946.5786.29458.129844.508937.1808521.5911763.17978389.51325352
From Charles A. Lane, Aug 22 2013: (Start)
The first column is also the coefficients of a in y''[x] - a*x^n*y[x] + b*en*y[x] = 0 where n = 0. The recursion yields coefficients of a, a*b*en, a*b^2*en^2 etc.
The second column is obtained when n=1, the third column when n=2. The final column is for n=10.
Example: Write a normal recursion for n=4. For convenience set x to 1. Running the recursion yields
1-(b en)/2+(b^2 en^2)/24+1/30 (a-(b^3 en^3)/24)+(-384 a b en+b^4 en^4)/40320+(2064 a b^2 en^2-b^5 en^5)/3628800+(120960 a^2-7104 a b^3 en^3+b^6 en^6)/479001600+(-4682880 a^2 b en+18984 a b^4 en^4-b^7 en^7)/87178291200+(54268416 a^2 b^2 en^2-43008 a b^5 en^5+b^8 en^8)/20922789888000.
The coefficient of a is 24, the coefficient of a b en is 384 and the coefficient of a b^2 en^2 is 2064. Dividing by 4! yields a sequence of 1,16,86... , the same as column 5 without the leading 1. There is a hint of unity among the oscillators. (End)

			Some solutions for n=3 k=4
..0..0..0..0....0..1..0..0....0..0..0..0....1..1..1..1....0..0..0..0
..0..0..0..0....0..1..1..0....0..0..0..0....1..1..1..1....1..1..0..0
..0..0..0..1....1..1..1..0....1..1..0..0....1..1..1..1....1..1..1..1

R. H. Hardin, Table of n, a(n) for n = 1..5304

Column 2 is A000290(n+1).
Column 3 is A002412(n+1).
Column 4 is A006324(n+1).
Row 1 is A000124.
Row 2 is A223718.
Row 3 is A223659.
Cf. A071920, A071921 (larger and reflected versions of table). - Alois P. Heinz, Sep 22 2013

T:= (n, k)-> add(binomial(k+2*j-1, 2*j), j=0..n):
seq(seq(T(n, 1+d-n), n=1..d), d=1..12);  # Alois P. Heinz, Sep 22 2013

T[n_, k_] := Sum[Binomial[k + 2*j - 1, 2*j], {j, 0, n}]; Table[T[n - k + 1, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Apr 07 2016, after Alois P. Heinz *)

Empirical: columns k=1..7 are polynomials of degree k.
Empirical: rows n=1..7 are polynomials of degree 2n.
T(n,k) = Sum_{j=0..n} C(k+2*j-1,2*j). - Alois P. Heinz, Sep 22 2013

A228461 Two-dimensional array read by antidiagonals: T(n,k) = number of arrays of maxima of three adjacent elements of some length n+2 0..k array.

Original entry on oeis.org

2, 3, 4, 4, 9, 7, 5, 16, 22, 11, 6, 25, 50, 46, 17, 7, 36, 95, 130, 91, 27, 8, 49, 161, 295, 310, 183, 44, 9, 64, 252, 581, 821, 736, 383, 72, 10, 81, 372, 1036, 1847, 2227, 1821, 819, 117, 11, 100, 525, 1716, 3703, 5615, 6254, 4673, 1749, 189, 12, 121, 715, 2685, 6812

Offset: 1

R. H. Hardin Aug 22 2013

There are two arrays (or lists, or vectors) involved, a length n+2 array with free elements from 0..k (thus (k+1)^(n+2) of them) and an array that is being enumerated of length n, each element of the latter being the maximum of three adjacent elements of the first array.

Many different first arrays can give the same second array.

			Table starts
...2....3.....4.....5......6......7.......8.......9......10.......11.......12
...4....9....16....25.....36.....49......64......81.....100......121......144
...7...22....50....95....161....252.....372.....525.....715......946.....1222
..11...46...130...295....581...1036....1716....2685....4015.....5786.....8086
..17...91...310...821...1847...3703....6812...11721...19117....29843....44914
..27..183...736..2227...5615..12453...25096...46941...82699...138699...223224
..44..383..1821..6254..17487..42386...92430..185727..349558...623513..1063283
..72..819..4673.18394..57303.151882..357510..768231.1535578..2893605..5191407
.117.1749.12107.55285.194064.567835.1453506.3357985.7152815.14263777.26930773
Some solutions for n=4 k=4
..3....4....4....3....3....4....3....4....3....0....3....3....4....2....0....2
..0....4....1....3....2....0....2....4....1....0....3....3....1....2....0....0
..4....1....1....0....1....0....4....0....0....0....2....1....4....2....2....3
..4....0....3....3....2....0....4....2....3....1....3....1....4....0....4....3

R. H. Hardin, Table of n, a(n) for n = 1..1700

Column 1 is A005252(n+3)
Column 2 is A217878
Column 3 is A217949.
A228464 is another column.
Row 1 is A000027(n+1)
Row 2 is A000290(n+1)
Row 3 is A002412(n+1)
Row 4 is A006324(n+1)
See A217883, A217954 for similar arrays.

Empirical for column k:
k=1: a(n) = 2*a(n-1) -a(n-2) +a(n-4)
k=2: a(n) = 3*a(n-1) -3*a(n-2) +a(n-3) +3*a(n-4) -a(n-5) +a(n-6) +a(n-7)
k=3: [order 10]
k=4: [order 13]
k=5: [order 16]
k=6: [order 19]
k=7: [order 22]
Empirical for row n:
n=1: a(n) = n + 1
n=2: a(n) = n^2 + 2*n + 1
n=3: a(n) = (2/3)*n^3 + (5/2)*n^2 + (17/6)*n + 1
n=4: a(n) = (1/3)*n^4 + 2*n^3 + (25/6)*n^2 + (7/2)*n + 1
n=5: a(n) = (2/15)*n^5 + (7/6)*n^4 + (25/6)*n^3 + (19/3)*n^2 + (21/5)*n + 1
n=6: [polynomial of degree 6]
n=7: [polynomial of degree 7]

Edited by N. J. A. Sloane, Sep 02 2013

A236770 a(n) = n(n + 1)(3n^2 + 3n - 2)/8.

Original entry on oeis.org

0, 1, 12, 51, 145, 330, 651, 1162, 1926, 3015, 4510, 6501, 9087, 12376, 16485, 21540, 27676, 35037, 43776, 54055, 66045, 79926, 95887, 114126, 134850, 158275, 184626, 214137, 247051, 283620, 324105, 368776, 417912, 471801, 530740, 595035, 665001, 740962

Offset: 0

Bruno Berselli, Jan 31 2014

After 0, first trisection of A011779 and right border of A177708.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Partial sums of A004188.
Cf. A000217, A000332, A011779, A177708.
Cf. similar sequences on the polygonal numbers: A002817(n) = A000217(A000217(n)); A000537(n) = A000290(A000217(n)); A037270(n) = A000217(A000290(n)); A062392(n) = A000384(A000217(n)).
Cf. sequences of the form A000217(m)+k*A000332(m+2): A062392 (k=12); A264854 (k=11); A264853 (k=10); this sequence (k=9); A006324 (k=8); A006323 (k=7); A000537 (k=6); A006322 (k=5); A006325 (k=4), A002817 (k=3), A006007 (k=2), A006522 (k=1).
Cf. A232713, A260810.

[n*(n+1)*(3*n^2+3*n-2)/8: n in [0..40]];

Table[n (n + 1) (3 n^2 + 3 n - 2)/8, {n, 0, 40}]
LinearRecurrence[{5,-10,10,-5,1},{0,1,12,51,145},40] (* Harvey P. Dale, Aug 22 2016 *)

for(n=0, 40, print1(n*(n+1)*(3*n^2+3*n-2)/8", "));

G.f.: x*(1 + 7*x + x^2)/(1 - x)^5.
a(n) = a(-n-1) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A000326(A000217(n)).
a(n) = A000217(n) + 9*A000332(n+2).
Sum_{n>=1} 1/a(n) = 2 + 4*sqrt(3/11)*Pi*tan(sqrt(11/3)*Pi/2) = 1.11700627139319... . - Vaclav Kotesovec, Apr 27 2016

A112742 a(n) = n^2*(n^2 - 1)/3.

Original entry on oeis.org

0, 0, 4, 24, 80, 200, 420, 784, 1344, 2160, 3300, 4840, 6864, 9464, 12740, 16800, 21760, 27744, 34884, 43320, 53200, 64680, 77924, 93104, 110400, 130000, 152100, 176904, 204624, 235480, 269700, 307520, 349184, 394944, 445060, 499800, 559440

Offset: 0

Matthew T. Cornick (maruth(AT)gmail.com), Sep 16 2005

Second derivative of the n-th Chebyshev polynomial (of the first kind) evaluated at x=1.
The second derivative at x=-1 is just (-1)^n * a(n).
The difference between two consecutive terms generates the sequence a(n+1) - a(n) = A002492(n).
Consider the partitions of 2n into two parts (p,q) where p <= q. Then a(n) is the total volume of the family of rectangular prisms with dimensions p, |q-p| and |q-p|. - Wesley Ivan Hurt, Apr 15 2018

			a(4)=80 because
C_4(x) = 1 - 8x^2 + 8x^4,
C'_4(x) = -16x + 32x^3,
C''_4(x) = -16 + 96x^2,
C''_4(1) = -16 + 96 = 80.

Eric Weisstein's World of Mathematics, Chebyshev polynomials of the first kind
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000914, A002415, A002492.

Table[D[ChebyshevT[n, x], {x, 2}], {n, 0, 100}] /. x -> 1

a(n)=n^2*(n^2-1)/3 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = (n-1)*n^2*(n+1)/3 = 4*A002415(n).
a(n) = 2*( A000914(n-1) + C(n+1,4) ). - David Scambler, Nov 27 2006
From Colin Barker, Jan 26 2012: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: 4*x^2*(1+x)/(1-x)^5. (End)
E.g.f.: exp(x)*x^2*(6 + 6*x + x^2)/3. - Stefano Spezia, Dec 11 2021
a(n) = A053126(n+2) - A006324(n-1). - Yasser Arath Chavez Reyes, Feb 22 2024

A002299 Binomial coefficients C(2*n+5,5).

Original entry on oeis.org

1, 21, 126, 462, 1287, 3003, 6188, 11628, 20349, 33649, 53130, 80730, 118755, 169911, 237336, 324632, 435897, 575757, 749398, 962598, 1221759, 1533939, 1906884, 2349060, 2869685, 3478761, 4187106, 5006386, 5949147, 7028847, 8259888, 9657648, 11238513

Offset: 0

N. J. A. Sloane, Eric Lane

Number of standard tableaux of shape (2n+1,1^5). - Emeric Deutsch, May 30 2004

G. C. Greubel, Table of n, a(n) for n = 0..5000 (terms 0..200 from Vincenzo Librandi)
Milan Janjic, Two Enumerative Functions.
J. M. Landsberg and L. Manivel, The sextonions and E7 1/2, Adv. Math., Vol. 201, No. 1 (2006), pp. 143-179. [Th. 7.2(i), case a=1]
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000389, A053127.

[Binomial(2*n+5,5): n in [0..30]]; // Vincenzo Librandi, Oct 07 2011

Table[Binomial[2*n + 5, 5], {n, 0, 50}] (* G. C. Greubel, Nov 23 2017 *)

a(n)=n*(8*n^4+60*n^3+170*n^2+225*n+137)/30+1 \\ Charles R Greathouse IV, Apr 18 2012

a(n) = A000389(2*n+5).
G.f.: (1+15*x+15*x^2+x^3)/(1-x)^6 = (1+x)*(x^2+14*x+1)/(1-x)^6.
E.g.f.: (30 + 600*x + 1275*x^2 + 730*x^3 + 140*x^4 + 8*x^5)*exp(x)/30. - G. C. Greubel, Nov 23 2017
Sum_{n>=0} (-1)^n/a(n) = 5*(10/3 - Pi). - Matthieu Pluntz, Oct 08 2019
Sum_{n>=0} 1/a(n) = 40*log(2) - 80/3. - Amiram Eldar, Jan 03 2022
From Peter Bala, Sep 03 2023: (Start)
a(n) = Sum_{0 <= i <= j <= n} (j+1)*(2*i+1)^2.
a(n) = (n+2)*(2*n+5)/(n*(2*n-1))*a(n-1) with a(0) := 1. (End)
a(n) = 2*A225007(n) - A006324(n+1). - Yasser Arath Chavez Reyes, Feb 27 2024

A162011 A sequence related to the recurrence relations of the right hand columns of the EG1 triangle A162005.

Original entry on oeis.org

1, -1, 1, -11, 19, -9, 1, -46, 663, -3748, 7711, -6606, 2025, 1, -130, 6501, -163160, 2236466, -17123340, 71497186, -154127320, 174334221, -98986050, 22325625, 1, -295, 36729, -2549775, 109746165, -3080128275, 57713313405, -727045264875

Offset: 1

Johannes W. Meijer, Jun 27 2009

The recurrence relation RR(n) = 0 of the n-th right hand column can be found with RR(n) = expand( product((1-(2*k-1)^2*z)^(n-k+1),k=1..n),z) = 0 and replacing z^p by a(n-p).

The polynomials in the numerators of the generating functions GF(z) of the coefficients that precede the a(n), a(n-1), a(n-2) and a(n-3) sequences, see A000012, A006324, A162012 and A162013, are symmetrical. This phenomenon leads to the sequence [1, 1, 6, 1, 19, 492, 1218, 492, 19 , 9, 3631, 115138, 718465, 1282314, 718465, 115138, 3631, 9].

			The recurrence relations for the first few right hand columns:
n = 1: a(n) = 1*a(n-1)
n = 2: a(n) = 11*a(n-1)-19*a(n-2)+9*a(n-3)
n = 3: a(n) = 46*a(n-1)-663*a(n-2)+3748*a(n-3)-7711*a(n-4)+6606*a(n-5)-2025*a(n-6)
n = 4: a(n) = 130*a(n-1)-6501*a(n-2)+163160*a(n-3)-2236466*a(n-4)+17123340*a(n-5)-71497186*a(n-6)+154127320*a(n-7)-174334221*a(n-8)+98986050*a(n-9)-22325625*a(n-10)

A000012, A004004 (2x), A162008, A162009 and A162010 are the first five right hand columns of EG1 triangle A162005.
A000124 (the Lazy Caterer's sequence) gives the number of terms of the RR(n).
A006324, A162012 and A162013 equal the absolute values of the coefficients that precede the a(n-1), a(n-2) and a(n-3) factors of the RR(n).

nmax:=5; for n from 1 to nmax do RR(n) := expand(product((1-(2*k-1)^2*z)^(n-k+1), k=1..n), z) od: T:=1: for n from 1 to nmax do for m from 0 to(n)*(n+1)/2 do a(T):= coeff(RR(n), z, m): T:=T+1 od: od: seq(a(k), k=1..T-1);

RR(n) = expand( product((1-(2*k-1)^2*z)^(n-k+1),k=1..n),z) with n = 1, 2, 3, .. . The coefficients of these polynomials lead to the sequence given above.

A162012 The sequence of the absolute values of the a(n-2) coefficients of A162011.

Original entry on oeis.org

0, 19, 663, 6501, 36729, 149842, 491274, 1375206, 3413982, 7710813, 16133689, 31690659, 59028879, 105082068, 179893252, 297641916, 477906924, 747198807, 1140797259, 1704931921, 2499346773, 3600290694, 5103978990, 7130572930

Offset: 1

Johannes W. Meijer, Jun 27 2009

Equals the absolute values of the coefficients that precede the a(n-2) factors of the recurrence relations RR(n) of A162011.

Cf. A006324 [a(n-1)] and A162013 [a(n-3)].

nmax:=26; for n from 1 to nmax do RR(n) := expand(product((1-(2*k-1)^2*z)^(n-k+1),k=1..n),z) od: T:=1: for n from 1 to nmax do a(T):=coeff(RR(n),z,2): T:=T+1 od: seq(a(k),k=1..T-1);

a(n) = (20*n^8+80*n^7+4*n^6-268*n^5-155*n^4+230*n^3+131*n^2-42*n)/360
Recurrence relation sum((-1)^k*binomial(9,k)*a(n-k), k= 0 .. 9) = 0
GF(z) = z*(19+492*z+1218*z^2+492*z^3+19*z^4)/(1-z)^9

A162013 The sequence of the absolute values of the a(n-3) coefficients of A162011.

Original entry on oeis.org

0, 9, 3748, 163160, 2549775, 22768402, 141820764, 685234196, 2738273230, 9438613635, 28894483904, 80240970524, 205377597269, 490460693060, 1103418293480, 2356809738456, 4809498575164, 9426116131517, 17820475867500

Offset: 1

Johannes W. Meijer, Jun 27 2009

Equals the absolute values of the coefficients that precede the a(n-3) factors of the recurrence relations RR(n) of A162011.

Cf. A006324 [a(n-1)] and A162012 [a(n-2)].

nmax:=21; for n from 1 to nmax do RR(n) := expand(product((1-(2*k-1)^2*z)^(n-k+1),k=1..n),z) od: T:=1: for n from 1 to nmax do a(T):=coeff(-RR(n),z,3): T:=T+1 od: seq(a(k),k=1..T-1);

a(n) = (280*n^12+1680*n^11-252*n^10-16660*n^9-13758*n^8+63408*n^7+68705*n^6-104265*n^5-111657*n^4+66997*n^3+56682*n^2-11160*n)/45360
Recurrence relation sum((-1)^k*binomial(13,k)*a(n-k), k= 0..13) = 0
GF(z) = z*(9+3631*z+115138*z^2+718465*z^3+1282314*z^4+718465*z^5+115138*z^6+ 3631*z^7+ 9*z^8)/(1-z)^13

A264854 a(n) = n(n + 1)(11n^2 + 11n - 10)/24.

Original entry on oeis.org

0, 1, 14, 61, 175, 400, 791, 1414, 2346, 3675, 5500, 7931, 11089, 15106, 20125, 26300, 33796, 42789, 53466, 66025, 80675, 97636, 117139, 139426, 164750, 193375, 225576, 261639, 301861, 346550, 396025, 450616, 510664, 576521, 648550, 727125, 812631, 905464, 1006031

Offset: 0

Ilya Gutkovskiy, Nov 26 2015

Partial sums of centered 11-gonal (or hendecagonal) pyramidal numbers.

OEIS Wiki, Figurate numbers
Eric Weisstein's World of Mathematics, Pyramidal Number
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A004467.

Cf. similar sequences provided by the partial sums of centered k-gonal pyramidal numbers: A006522 (k=1), A006007 (k=2), A002817 (k=3), A006325 (k=4), A006322 (k=5), A000537 (k=6), A006323 (k=7), A006324 (k=8), A236770 (k=9), A264853 (k=10), this sequence (k=11), A062392 (k=12), A264888 (k=13).

[n*(n+1)*(11*n^2+11*n-10)/24: n in [0..50]]; // Vincenzo Librandi, Nov 27 2015

Table[n (n + 1) (11 n^2 + 11 n - 10)/24, {n, 0, 50}]

a(n)=n*(n+1)*(11*n^2+11*n-10)/24 \\ Charles R Greathouse IV, Jul 26 2016

G.f.: x*(1 + 9*x + x^2)/(1 - x)^5.
a(n) = Sum_{k = 0..n} A004467(k).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Nov 27 2015

A270695 Alternating sum of centered octagonal pyramidal numbers.

Original entry on oeis.org

0, -1, 9, -26, 58, -107, 179, -276, 404, -565, 765, -1006, 1294, -1631, 2023, -2472, 2984, -3561, 4209, -4930, 5730, -6611, 7579, -8636, 9788, -11037, 12389, -13846, 15414, -17095, 18895, -20816, 22864, -25041, 27353, -29802, 32394, -35131, 38019

Offset: 0

Ilya Gutkovskiy, Mar 21 2016

G. C. Greubel, Table of n, a(n) for n = 0..1000
OEIS Wiki, Centered pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (-3,-2,2,3,1).

Cf. A000447 (centered octagonal pyramidal numbers).

Cf. A000330, A006324 (partial sums of centered octagonal pyramidal numbers), A019298, A232599.

[((-1)^n*(4*n^2 - 1)*(2*n + 3) + 3)/12 : n in [0..40]]; // Wesley Ivan Hurt, Mar 21 2016

A270695:= n-> ((-1)^n*(4*n^2 -1)*(2*n+3) +3)/12: seq(A270695(n), n=0..40); # G. C. Greubel, Apr 02 2021

LinearRecurrence[{-3, -2, 2, 3, 1}, {0, -1, 9, -26, 58}, 39]
Table[((-1)^n (4 n^2 - 1) (2 n + 3) + 3)/12, {n, 0, 38}]

x='x+O('x^100); concat(0, Vec(-x*(1-6*x+x^2)/((1-x)*(1+x)^4))) \\ Altug Alkan, Mar 21 2016

[((-1)^n*(4*n^2 -1)*(2*n+3) +3)/12 for n in (0..40)] # G. C. Greubel, Apr 02 2021

G.f.: -x*(1 - 6*x + x^2)/((1 - x)*(1 + x)^4).
E.g.f.: (1/12)*(3*exp(x) - (3 + 18*x - 36*x^2 + 8*x^3)*exp(-x)).
a(n) = -3*a(n-1) - 2*a(n-2) + 2*a(n-3) + 3*a(n-4) + a(n-5).
a(n) = ((-1)^n*(4*n^2 - 1)*(2*n + 3) + 3)/12.

a(6)=179 inserted by Georg Fischer, Apr 03 2019

cp's OEIS Frontend

A225010 T(n,k) = number of n X k 0..1 arrays with rows unimodal and columns nondecreasing.

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A228461 Two-dimensional array read by antidiagonals: T(n,k) = number of arrays of maxima of three adjacent elements of some length n+2 0..k array.

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A236770 a(n) = n*(n + 1)*(3*n^2 + 3*n - 2)/8.

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A112742 a(n) = n^2*(n^2 - 1)/3.

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A002299 Binomial coefficients C(2*n+5,5).

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A162011 A sequence related to the recurrence relations of the right hand columns of the EG1 triangle A162005.

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A162012 The sequence of the absolute values of the a(n-2) coefficients of A162011.

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A162013 The sequence of the absolute values of the a(n-3) coefficients of A162011.

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A236770 a(n) = n(n + 1)(3n^2 + 3n - 2)/8.

A264854 a(n) = n(n + 1)(11n^2 + 11n - 10)/24.