A219086 -id:A219086 - cp's OEIS frontend

A123865 a(n) = n^4 - 1.

0, 15, 80, 255, 624, 1295, 2400, 4095, 6560, 9999, 14640, 20735, 28560, 38415, 50624, 65535, 83520, 104975, 130320, 159999, 194480, 234255, 279840, 331775, 390624, 456975, 531440, 614655, 707280, 809999, 923520, 1048575, 1185920, 1336335

Offset: 1

Reinhard Zumkeller, Oct 16 2006

a(n) mod 5 = 0 iff n mod 5 > 0: a(A008587(n)) = 4; a(A047201(n)) = 0; a(n) mod 5 = 4*(1-A079998(n)).

A129292(n) = number of divisors of a(n) that are not greater than n. - Reinhard Zumkeller, Apr 09 2007

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000583, A005563, A024002, A123866, A123867, A123868, A219086.

List([1..40], n-> n^4 -1); # G. C. Greubel, Aug 08 2019

[n^4 - 1: n in [1..40]]; // Vincenzo Librandi, May 01 2011

seq(n^4 -1, n=1..40); # G. C. Greubel, Aug 08 2019

Table[n^4-1,{n,40}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)

vector(40, n, n^4 -1) \\ G. C. Greubel, Aug 08 2019

[n^4 -1 for n in (1..40)] # G. C. Greubel, Aug 08 2019

G.f.: x^2*(15 + 5*x + 5*x^2 - x^3)/(1-x)^5. - Colin Barker, Jan 10 2012
-4*a(n+1) = -4*n*(n+2)*(n^2+2*n+2) = (n+n*i)*(n+2+n*i)*(n+(n+2)*i)*(n+2+(n+2)*i), where i is the imaginary unit. - Jon Perry, Feb 05 2014
From Vaclav Kotesovec, Feb 14 2015: (Start)
Sum_{n>=2} 1/a(n) = 7/8 - Pi*coth(Pi)/4 = A256919.
Sum_{n>=2} (-1)^n / a(n) = 1/8 - Pi/(4*sinh(Pi)). (End)
a(n) = A005563(A005563(n)). - Bruno Berselli, May 28 2015
E.g.f.: 1 + (-1 + x + 7*x^2 + 6*x^3 + x^4)*exp(x). - G. C. Greubel, Aug 08 2019
Product_{n>=2} (1 + 1/a(n)) = 4*Pi*csch(Pi). - Amiram Eldar, Jan 20 2021

A033455 Convolution of nonzero squares A000290 with themselves.

Original entry on oeis.org

1, 8, 34, 104, 259, 560, 1092, 1968, 3333, 5368, 8294, 12376, 17927, 25312, 34952, 47328, 62985, 82536, 106666, 136136, 171787, 214544, 265420, 325520, 396045, 478296, 573678, 683704, 809999, 954304, 1118480, 1304512, 1514513, 1750728, 2015538, 2311464

Offset: 1

N. J. A. Sloane

Total area of all square regions from an n X n grid. E.g., at n = 3, there are nine individual squares, four 2 X 2's and one 3 X 3, total area 9 + 16 + 9 = 34, hence a(3) = 34. - Jon Perry, Jul 29 2003
If X is an n-set and Y and Z disjoint 2-subsets of X then a(n) is equal to the number of 7-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Every fourth term is odd. However, there are no primes in the sequence. - Zak Seidov, Feb 28 2011
-120*a(n) is the real part of (n + n*i)*(n + 2 + n*i)*(n + (n + 2)i)*(n + 2+(n + 2)*i)*(n + 1 + (n + 1)*i), where i = sqrt(-1). - Jon Perry, Feb 05 2014
The previous formula rephrases the factorization of the 5th-order polynomial a(n) = (n+1)*((n+1)^4-1) = (n+1)*A123864(n+1) based on the factorization in A123865. - R. J. Mathar, Feb 08 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Abderrahim Arabi, Hacène Belbachir, Jean-Philippe Dubernard, Plateau Polycubes and Lateral Area, arXiv:1811.05707 [math.CO], 2018. See Column 1 Table 1 p. 8.
Milan Janjic, Two Enumerative Functions
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev. 19 (1977), no. 1, 90--99. MR0428678 (55 #1698). See Table 2. - _N. J. A. Sloane_, Mar 23 2014
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000290, A001249, A123865, A219086.

List([1..40], n-> ((n+1)^5 -(n+1))/30); # G. C. Greubel, Jul 05 2019

[(n^5-n)/30: n in [2..41]]; // Vincenzo Librandi, Mar 24 2014

Table[(n^5 - n)/30, {n,2,41}] (* Vladimir Joseph Stephan Orlovsky, Feb 28 2011 *)
CoefficientList[Series[(1+x)^2/(1-x)^6, {x,0,40}], x] (* Vincenzo Librandi, Mar 24 2014 *)

vector(40, n, ((n+1)^5 -n-1)/30) \\ G. C. Greubel, Jul 05 2019

[((n+1)^5 -n-1)/30 for n in (1..40)] # G. C. Greubel, Jul 05 2019

a(n-1) = n*(n^4 - 1)/30 = A061167(n)/30. - Henry Bottomley, Apr 18 2001
G.f.: x*(1+x)^2/(1-x)^6. - Philippe Deléham, Feb 21 2012
a(n) = Sum_{k=1..n+1} k^2*(n+1-k)^2. - Kolosov Petro, Feb 07 2019
E.g.f.: x*(30 +90*x +65*x^2 +15*x^3 +x^4)*exp(x)/30. - G. C. Greubel, Jul 05 2019

More terms from Vincenzo Librandi, Mar 24 2014

A011863 Nearest integer to (n/2)^4.

Original entry on oeis.org

0, 0, 1, 5, 16, 39, 81, 150, 256, 410, 625, 915, 1296, 1785, 2401, 3164, 4096, 5220, 6561, 8145, 10000, 12155, 14641, 17490, 20736, 24414, 28561, 33215, 38416, 44205, 50625, 57720, 65536, 74120, 83521, 93789, 104976, 117135, 130321, 144590

Offset: 0

R. K. Guy

First differences are in A019298.
The bisections are A000583 and A219086.
Number of ways to put n-1 copies of 1,2,3 into sets. [Zeilberger?]
s(n) is the number of 4-tuples (w,x,y,z) with all terms in {1,...,n} and |w-x| >= w + |y-z|; see A186707. - Clark Kimberling, May 24 2012

N. J. A. Sloane, Table of n, a(n) for n = 0..10000 (first 2000 terms from Vicenzo Librandi)
A. J. Guttmann, Indicators of solvability for lattice models, Discrete Math., 217 (2000), 167-189 (H_2 for square lattice of Section 6).
Doron Zeilberger, In How Many Ways Can You Reassemble Several Russian Dolls?, The Personal Journal of Shalosh B. Ekhad and Doron Zeilberger (2009); Local copy. [PDF file only, no active links]
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).

Cf. A000583, A002817, A019298, A106707, A212714, A219086.

[ (2*n^4-(1-(-1)^n))/32: n in [0..50] ];

Maple
```
seq(round((n/2)^4), n=0..40);
```

Round[(Range[40]/2)^4] (* or *) LinearRecurrence[{4,-5,0,5,-4,1},{0,1,5,16,39,81},40] (* Harvey P. Dale, Feb 07 2015 *)

a(n)=round((n/2)^4) \\ Charles R Greathouse IV, Jun 23 2011

G.f.: x^2*(1 + x + x^2)/((1 - x)^5*(1+x)).
a(n) = +4*a(n-1) -5*a(n-2) +5*a(n-4) -4*a(n-5) +a(n-6). - R. J. Mathar, Dec 07 2010
a(n)+a(n+1) = A002817(n). - R. J. Mathar, Dec 19 2008
a(n) = n^4/16 - 1/32 + (-1)^n/32 - R. J. Mathar, Dec 07 2010, adapted to added a(0) by Hugo Pfoertner, Dec 29 2019
a(n) = (2*A000583(n) + (-1)^n - 1)/32. - Bruno Berselli, Dec 07 2010, adapted to added a(0) by Hugo Pfoertner, Dec 29 2019
n*(n^2+n+2)*a(n+1) = 4*(n^2+2*n+2)*a(n)+(n+2)*(n^2+3*n+4)*a(n-1). Holonomic Ansatz with smallest order of recurrence. - Thotsaporn Thanatipanonda, Dec 12 2010
a(n) = floor(n^4/8)/2. - Gary Detlefs, Feb 19 2011, adapted to added a(0) by Hugo Pfoertner, Dec 29 2019
a(n) = A212714(n)/2, n >= 0. - Wolfdieter Lang, Oct 03 2016, adapted to added a(0) by Hugo Pfoertner, Dec 29 2019
E.g.f.: (1/32)*exp(-x)*(1 + exp(2*x)*(-1 + 2*x + 14*x^2 + 12*x^3 + 2*x^4)). - Stefano Spezia, Dec 29 2019
Sum_{n>=2} 1/a(n) = 6 + Pi^4/90 - 2*Pi*tanh(Pi/2). - Amiram Eldar, Aug 13 2022

Missing a(0) added by N. J. A. Sloane, Dec 29 2019. As a result some of the comments and formulas will need to be adjusted.

A219085 a(n) = floor((n + 1/2)^3).

Original entry on oeis.org

0, 3, 15, 42, 91, 166, 274, 421, 614, 857, 1157, 1520, 1953, 2460, 3048, 3723, 4492, 5359, 6331, 7414, 8615, 9938, 11390, 12977, 14706, 16581, 18609, 20796, 23149, 25672, 28372, 31255, 34328, 37595, 41063, 44738, 48627, 52734, 57066

Offset: 0

Clark Kimberling, Dec 20 2012

a(n) is the number k such that {k^p} < 1/2 < {(k+1)^p}, where p = 1/3 and { } = fractional part.  In general, suppose that f is a continuous strictly increasing downward concave function, with f(1)>=0 and f(k)+1/2 not an integer.  Let J(k) denote the inequality {f(k)} < 1/2 < {f(k+1)}, where {}= fractional part; equivalently, [{f(k)} + 1/2] = 0 and [{f(k+1)} + 1/2] = 1, where [ ] = floor.  Thus J(k) holds if the integer nearest f(k+1) exceeds the integer nearest f(k), so that k can be regarded as a "jump point for f".  The solutions of J(k) are the numbers [g(n)+1/2)] for n >= 0, where g = (inverse of f).
Conjecture:  if d is a positive integer and f(x) = x^(1/d), then the solutions of J(k) form a linearly recurrent sequence.
This conjecture was proved by David Moews; see Problem 21 in "Unsolved Problems and Rewards".  - Clark Kimberling, Feb 06 2013
Guide to related sequences:
f(x) ....... jump sequence ... linear recurrence order
x^(1/2) .... A002378 ......... 3
x^(1/3) .... A219085 ......... 7
x^(2/3) .... A203302 ......... (not linearly recurrent)
x^(1/4) .... A219086 ......... 5
x^(3/4) .... A219087 ......... (not linearly recurrent)
x^(1/5) .... A219088 ......... 21
x^(1/6) .... A219089 ......... 21
x^(1/7) .... A219090 ..........71
x^(1/8) .... A219091 ......... 23
log(x) ..... A219092 ......... (not linearly recurrent)
log_2(x) ... A084188 ......... (not linearly recurrent)

			Let p=1/3.  Then
3^p=1.44... and 4^p=1.58..., so 3 is a jump point.
15^p=2.46... and 16^p=2.51..., so 15 is a jump point.

Clark Kimberling, Table of n, a(n) for n = 0..10000
Clark Kimberling, Unsolved Problems and Rewards, Problem 21.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1,1,-3,3,-1).

Cf. A002378, A203302.

Mathematica
```
Table[Floor[(n + 1/2)^3], {n, 0, 100}]
```

a(n)=n^3 + (6*n^2 + 3*n)\4 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = floor((n + 1/2)^3).
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3) +a(n-4) -3*a(n-5) +3*a(n-6) -a(n-7).
G.f.: (3*x +6*x^2 +6*x^3 +7*x^4 +x^5 +x^6)/(u*v), where u = (1 - x)^4, v = 1 + x + x^2 + x^3.
a(n) = (n + 1/2)^3 + (2*i^(n*(n-1))+(-1)^n-4)/8, where i=sqrt(-1). - Bruno Berselli, Dec 21 2012

A175110 a(n) = ((2*n+1)^4+1)/2.

Original entry on oeis.org

1, 41, 313, 1201, 3281, 7321, 14281, 25313, 41761, 65161, 97241, 139921, 195313, 265721, 353641, 461761, 592961, 750313, 937081, 1156721, 1412881, 1709401, 2050313, 2439841, 2882401, 3382601, 3945241, 4575313, 5278001, 6058681

Offset: 0

R. J. Mathar, Feb 13 2010

Binomial transform of 1,40,232,384,192,0,0,.. (0 continued). Convolution of the finite sequence 1,36,118,36,1 with A000332, dropping zeros.
Hypotenuse of Pythagorean triangles with smallest side a square: A016754(n)^2 + (a(n)-1)^2 = a(n)^2. - Martin Renner, Nov 12 2011
a(n) is also the first integer in a sum of (2*n + 1)^4 consecutive integers that equal (2*n + 1)^8. See A016756 and A016760. - Patrick J. McNab, Dec 26 2016

Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 54.

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

Cf. A000332, A016756, A016760. Partial sums of A117216.

I:=[1, 41, 313, 1201, 3281]; [n le 5 select I[n] else 5*Self(n-1) - 10*Self(n-2) + 10*Self(n-3) - 5*Self(n-4) + Self(n-5): n in [1..40]]; // Vincenzo Librandi, Dec 19 2012

A175110:=n->((2*n+1)^4+1)/2: seq(A175110(n), n=0..50); # Wesley Ivan Hurt, Apr 13 2017

CoefficientList[Series[(1 + 36*x + 118*x^2 + 36*x^3 + x^4)/(1-x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 19 2012 *)
Table[((2 n + 1)^4 + 1)/2, {n, 0, 29}] (* Michael De Vlieger, Dec 26 2016 *)
LinearRecurrence[{5,-10,10,-5,1},{1,41,313,1201,3281},40] (* Harvey P. Dale, Jan 01 2022 *)

a(n)=((2*n+1)^4+1)/2 \\ Charles R Greathouse IV, Oct 16 2015

a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5).
G.f.: (1+36*x+118*x^2+36*x^3+x^4)/ (1-x)^5.
a(n)-a(n-1) = A117216(n).
a(n) = 8*A001844(n) * A000217(n) + 1 = 8*A219086(n) + 1. - Bruce J. Nicholson, Apr 13 2017

A317297 a(n) = (n - 1)(4n^2 - 8*n + 5).

Original entry on oeis.org

0, 5, 34, 111, 260, 505, 870, 1379, 2056, 2925, 4010, 5335, 6924, 8801, 10990, 13515, 16400, 19669, 23346, 27455, 32020, 37065, 42614, 48691, 55320, 62525, 70330, 78759, 87836, 97585, 108030, 119195, 131104, 143781, 157250, 171535, 186660, 202649, 219526, 237315, 256040, 275725, 296394, 318071

Offset: 1

Omar E. Pol, Sep 01 2018

Conjecture: For n > 1, a(n) is the maximum eigenvalue of a 2*(n-1) X 2*(n-1) square matrix M defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even (see A317614). - Stefano Spezia, Dec 27 2018

Connections can be made to A022144 and A010014. Namely, a formula for A022144 is (2*n+1)^2 - (2*n-1)^2. A formula for A010014 is (2*n+1)^3 - (2*n-1)^3. The general form can be represented by (2*n+1)^d - (2*n-1)^d, where d designates the number of dimensions. When d is 4, a(n) = ((2*(n-1)+1)^4 - (2*(n-1)-1)^4)/16, namely the general form shifted by 1 and divided by 16 is a(n). - Yigit Oktar, Aug 16 2024

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

First bisection of A006003.
Nonzero terms give the row sums of A007607.
Conjecture: 0 together with a bisection of A246697.
Cf. A219086 (partial sums).
Cf. A008595, A033430, A135453, A317614.
Cf. A010014, A022144 (see comments)

Table[(n - 1) (4 n^2 - 8 n + 5), {n, 1, 50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {0, 5, 34, 111}, 50] (* or *) CoefficientList[Series[x (5 + 14 x + 5 x^2)/(1 - x)^4, {x, 0, 50}], x] (* Stefano Spezia, Sep 01 2018 *)

PARI
```
a(n) = (n - 1)*(4*n^2 - 8*n + 5)
```

concat(0, Vec(x^2*(5 + 14*x + 5*x^2)/(1 - x)^4 + O(x^50))) \\ Colin Barker, Sep 01 2018

a(n) = 4*n^3 - 12*n^2 + 13*n - 5 = A033430(n) - A135453(n) + A008595(n) - 5.
G.f.: x^2*(5 + 14*x + 5*x^2)/(1 - x)^4. - Colin Barker, Sep 01 2018
a(n) = 4*a(n - 1) - 6*a(n - 2) + 4*a(n - 3) - a(n - 4) for n > 4. - Stefano Spezia, Sep 01 2018
E.g.f.: exp(x)*(5*x + 12*x^2 + 4*x^3). - Stefano Spezia, Jan 15 2019
a(n) = ((2*(n-1)+1)^4 - (2*(n-1)-1)^4)/16. - Yigit Oktar, Aug 16 2024

A359498 a(n) = ((2*n+1)^8 - 1)/32.

Original entry on oeis.org

0, 205, 12207, 180150, 1345210, 6698715, 25491585, 80090332, 217992420, 530736345, 1181964355, 2447218290, 4768371582, 8825923015, 15632700405, 26652844920, 43950269320, 70371105957, 109764982935, 167250289390, 249528913410, 365256258675, 525472668457, 744102708180

Offset: 0

Jianing Song, Jan 03 2023

a(n) and A000217(n) have the same parity.

Jianing Song, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. {((2*n+1)^2^k - 1)/2^(k+2)}: A000217 (k=1), A219086 (k=2), this sequence (k=3), A359499 (k=4).

((2*Range[0, 25] + 1)^8 - 1)/32 (* Paolo Xausa, Jan 23 2025 *)

PARI
```
a(n) = ((2*n+1)^8 - 1)/32
```

def A359498(n): return ((n<<1)+1)**8-1>>5 # Chai Wah Wu, Jan 15 2023

a(n) = A000217(n) * A219086(n) * A175110(n) = A219086(n) * A175110(n).

A359499 a(n) = ((2*n+1)^16 - 1)/64.

Original entry on oeis.org

0, 672605, 2384185791, 519264540150, 28953440450810, 717964529118315, 10397134518487185, 102631380558013916, 760331123057294820, 4506897086994080745, 22352635785031020755, 95822037745015603890, 363797880709171295166, 1246350673076132966615, 3910101151255427324805

Offset: 0

Jianing Song, Jan 03 2023

a(n) and A000217(n) have the same parity.

Jianing Song, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (17,-136,680,-2380,6188,-12376,19448,-24310,24310,-19448,12376,-6188,2380,-680,136,-17,1).

Cf. {((2*n+1)^2^k - 1)/2^(k+2)}: A000217 (k=1), A219086 (k=2), A359498 (k=3), this sequence (k=4).

Table[((2*n + 1)^16 - 1)/64, {n, 0, 15}] (* Paolo Xausa, Oct 04 2024 *)

PARI
```
a(n) = ((2*n+1)^16 - 1)/64
```

def A359499(n): return ((n<<1)+1)**16-1>>6 # Chai Wah Wu, Jan 15 2023

a(n) = A000217(n) * A219086(n) * A175110(n) * A359844(n) = A219086(n) * A175110(n) * A359844(n) = A359498(n) * A359499(n).

A219087 a(n) = floor((n + 1/2)^(4/3)).

Original entry on oeis.org

0, 1, 3, 5, 7, 9, 12, 14, 17, 20, 22, 25, 29, 32, 35, 38, 42, 45, 48, 52, 56, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99, 103, 107, 112, 116, 121, 125, 130, 134, 139, 143, 148, 152, 157, 162, 167, 172, 176, 181, 186, 191, 196, 201, 206, 211, 216, 221, 227

Offset: 0

Clark Kimberling, Jan 01 2013

a(n) is the number k such that {k^p} < 1/2 < {(k+1)^p}, where p = 3/4 and { } = fractional part. Equivalently, the jump sequence of f(x) = x^(3/4), in the sense that these are the nonnegative integers k for which round(k^p) < round((k+1)^p). For details and a guide to related sequences, see A219085.

Clark Kimberling, Table of n, a(n) for n = 0..10000

Cf. A219085, A219086.

Table[Floor[(n + 1/2)^(4/3)], {n, 0, 100}]

a(n) = floor((n + 1/2)^(4/3)).

A305290 Numbers k such that 4*k + 1 is a perfect cube, sorted by absolute values.

Original entry on oeis.org

0, -7, 31, -86, 182, -333, 549, -844, 1228, -1715, 2315, -3042, 3906, -4921, 6097, -7448, 8984, -10719, 12663, -14830, 17230, -19877, 22781, -25956, 29412, -33163, 37219, -41594, 46298, -51345, 56745, -62512, 68656, -75191, 82127, -89478, 97254, -105469, 114133, -123260, 132860

Offset: 1

Bruno Berselli, May 29 2018

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (-3,-2,2,3,1).

Cf. A016755.
Cf. A000290: k such that 4*k is a square.
Cf. A002378: k such that 4*k+1 is a square.
Cf. A033431: k such that 4*k is a nonnegative cube.
Cf. A305291: k such that 4*k+3 is a cube.
Cf. A141046: k such that 4*k is a fourth power.
Cf. 4*A219086: k such that 4*k+1 is a fourth power.

seq(coeff(series(x^2*(-7+10*x-7*x^2)/((1-x)*(1+x)^4), x,50),x,n),n=1..45); # Muniru A Asiru, May 31 2018

LinearRecurrence[{-3, -2, 2, 3, 1}, {0, -7, 31, -86, 182}, 45] (* Jean-François Alcover, Jun 04 2018 *)

concat(0, Vec(-x^2*(7 - 10*x + 7*x^2) / ((1 - x)*(1 + x)^4) + O(x^40))) \\ Colin Barker, Jun 04 2018

G.f.: x^2*(-7 + 10*x - 7*x^2)/((1 - x)*(1 + x)^4).
a(n) = -3*a(n-1) - 2*a(n-2) + 2*a(n-3) + 3*a(n-4) + a(n-5).
a(n) = (-1 - A016755(n-1)*(-1)^n)/4.
a(n) + a(-n) = (-1)^n*2^((1-(-1)^n)/2).
(n - 2)*(4*n^2 - 16*n + 19)*a(n) + (12*n^2 - 36*n + 31)*a(n-1) - (n - 1)*(4*n^2 - 8*n + 7)*a(n-2) = 0.
From Colin Barker, May 30 2018: (Start)
a(n) = n*(4*n^2 + 6*n + 3)/2 for n even.
a(n) = -(n + 1)*(4*n^2 + 2*n + 1)/2 for n odd.
(End)

cp's OEIS Frontend

A123865 a(n) = n^4 - 1.

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A033455 Convolution of nonzero squares A000290 with themselves.

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A011863 Nearest integer to (n/2)^4.

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Magma

Maple

Mathematica

PARI

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A219085 a(n) = floor((n + 1/2)^3).

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PARI

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A175110 a(n) = ((2*n+1)^4+1)/2.

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Magma

Maple

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PARI

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A317297 a(n) = (n - 1)*(4*n^2 - 8*n + 5).

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A317297 a(n) = (n - 1)(4n^2 - 8*n + 5).