A036487 -id:A036487 - cp's OEIS frontend

A036486 a(n) = ceiling((n^3)/2).

0, 1, 4, 14, 32, 63, 108, 172, 256, 365, 500, 666, 864, 1099, 1372, 1688, 2048, 2457, 2916, 3430, 4000, 4631, 5324, 6084, 6912, 7813, 8788, 9842, 10976, 12195, 13500, 14896, 16384, 17969, 19652, 21438, 23328, 25327, 27436, 29660, 32000, 34461, 37044

Offset: 0

N. J. A. Sloane, Dec 11 1999

a(n) is the number of compositions of even natural numbers into 3 parts < n. For example, a(2)=4 because compositions of even natural numbers into 3 parts < 2 are (0,0,0), (0,1,1), (1,0,1), and (1,1,0). a(3)=14 because compositions of even natural numbers into 3 parts <= 3 - 1 = 2 are (0,0,0), (0,1,1), (1,0,1), (1,1,0), (0,0,2), (0,2,0), (2,0,0), (1,1,2),(1,2,1),(2,1,1),(0,2,2),(2,0,2),(2,2,0) and (2,2,2). - Adi Dani, Jun 05 2011

Also the number of balls in a body-centered lattice cube with n layers. - K. G. Stier, Dec 26 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Cf. A036487.

[(2*n^3-(-1)^n+1)/4: n in [0..40]]; // Vincenzo Librandi, Jun 07 2011

[ seq(ceil((n^3)/2), n=0..100) ];
with (combinat):seq(count(Partition((n^3+1)), size=2), n=0..40); # Zerinvary Lajos, Mar 28 2008

Table[Ceiling[n^3/2], {n, 0, 40}] (* Wesley Ivan Hurt, May 21 2014 *)
LinearRecurrence[{3,-2,-2,3,-1},{0,1,4,14,32},50] (* Harvey P. Dale, Jan 14 2019 *)

a(n)=(2*n^3-(-1)^n+1)/4 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: x*(1+x+4*x^2) / ( (1+x)*(x-1)^4 ). - R. J. Mathar, Jun 06 2011
a(n) = (2*n^3 - (-1)^n + 1)/4. - Bruno Berselli, Jun 07 2011
a(n) = n^3 - A036487(n), where n^3 is the number of compositions of natural numbers into 3 parts < n. - R. J. Mathar, Jun 07 2011
a(n) = (n^3 + (n mod 2))/2. - Wesley Ivan Hurt, May 21 2014
E.g.f.: (x*(1 + 3*x + x^2)*cosh(x) + (1 + x + 3*x^2 + x^3)*sinh(x))/2. - Stefano Spezia, Sep 09 2022

A131476 a(n) = floor(n^3/3).

Original entry on oeis.org

0, 0, 2, 9, 21, 41, 72, 114, 170, 243, 333, 443, 576, 732, 914, 1125, 1365, 1637, 1944, 2286, 2666, 3087, 3549, 4055, 4608, 5208, 5858, 6561, 7317, 8129, 9000, 9930, 10922, 11979, 13101, 14291, 15552, 16884, 18290, 19773, 21333, 22973

Offset: 0

Mohammad K. Azarian, Jul 27 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,2,-3,3,-1).

Cf. A000982, A008619, A004526, A007590, A036487.

[Floor(n^3/3): n in [0..50]]; // Vincenzo Librandi, May 08 2011

Table[Quotient[n^3,3],{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, May 07 2011 *)

a(n)=n^3\3 \\ Charles R Greathouse IV, May 08 2011

From R. J. Mathar, Dec 19 2008: (Start)
G.f.: x^2*(2 + 3*x + x^3)/((1 - x)^4*(1 + x + x^2)).
a(n) = (A057078(n) - A024001(n))/3. (End)
a(n) = (3*n^3 + 3*cos(2*Pi*n/3) + sqrt(3)*sin(2*Pi*n/3) - 3)/9. - Vladimir Reshetnikov, Oct 09 2016
a(n) = (n - 1)*n*(n + 1)/3 + floor(n/3). - Bruno Berselli, Jun 08 2017

A131479 a(n) = floor(n^4/4).

Original entry on oeis.org

0, 0, 4, 20, 64, 156, 324, 600, 1024, 1640, 2500, 3660, 5184, 7140, 9604, 12656, 16384, 20880, 26244, 32580, 40000, 48620, 58564, 69960, 82944, 97656, 114244, 132860, 153664, 176820, 202500, 230880, 262144, 296480, 334084, 375156

Offset: 0

Mohammad K. Azarian, Jul 27 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).

Cf. A000982, A004526, A007590, A008619, A011863, A036487.

Cf. A131478.

[Floor(n^4/4): n in [0..60]]; // Vincenzo Librandi, Jun 16 2011

seq(op([4*k^4, 2*(k^3*(k+1)+k*(k+1)^3)]),k=0..100); # Robert Israel, Jun 01 2015

Table[Floor[n^4/4], {n, 0, 20}] (* Enrique Pérez Herrero, May 31 2015 *)

vector(50, n, n--; n^4\4) \\ Michel Marcus, Jun 02 2015

def A131479(n): return n**4>>2 # Chai Wah Wu, Jan 31 2023

From R. J. Mathar, Dec 19 2008: (Start)
G.f.: 4*x^2*(1+x+x^2)/((1+x)*(1-x)^5).
a(n) = 4*A011863(n-1). (End)
a(n) = floor(n^2/2)*ceiling(n^2/2) = A007590(n) * A000982(n). - Enrique Pérez Herrero, May 31 2015
Sum_{n>=2} 1/a(n) = Sum_{n>=1} 1/(4n^4) + Sum_{n>=1} 1/(2n*(n+1)*(2n^2+2n+1)) = Zeta(4)/4 + (3-Pi*tanh(Pi/2))/2. - Enrique Pérez Herrero, May 31 2015
a(2*k) = 4*k^4; a(2*k+1) = 2*(k^3*(k+1) + k*(k+1)^3). - Robert Israel, Jun 01 2015
E.g.f.: (x*(x^3 + 6*x^2 + 7*x + 1)*cosh(x) + (x^4 + 6*x^3 + 7*x^2 + x - 1)*sinh(x))/4. - Stefano Spezia, Feb 18 2023

A131478 a(n) = ceiling(n^4/4).

Original entry on oeis.org

0, 1, 4, 21, 64, 157, 324, 601, 1024, 1641, 2500, 3661, 5184, 7141, 9604, 12657, 16384, 20881, 26244, 32581, 40000, 48621, 58564, 69961, 82944, 97657, 114244, 132861, 153664, 176821, 202500, 230881, 262144, 296481, 334084, 375157, 419904, 468541, 521284

Offset: 0

Mohammad K. Azarian, Jul 27 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).

Cf. A000982, A004526, A007590, A008619, A036487, A058919.

Cf. A131479.

[Ceiling(n^4/4) : n in [0..50]]; // Vincenzo Librandi, Oct 01 2011

Ceiling[Range[0,40]^4/4] (* Harvey P. Dale, May 17 2019 *)
CoefficientList[Series[(x(x^3 + 6x^2 + 7x + 1)Cosh[x]+ (x^4 + 6x^3 + 7x^2 + x + 3)Sinh[x])/4,{x,0,35}],x]Table[n!,{n,0,35}] (* Stefano Spezia, Feb 19 2023 *)

vector(50, n, n--;ceil(n^4/4)) \\ Michel Marcus, Jun 16 2015

def A131478(n): return n**4+3>>2 # Chai Wah Wu, Jan 30 2023

From R. J. Mathar, Dec 19 2008: (Start)
G.f.: x*(1 + 10*x^2 + x^4)/((1 - x)^5*(1 + x)).
a(n) + a(n+1) = A058919(n+1). (End)
a(n) = floor(n^4/4 + 3/4). - Bruno Berselli, Dec 21 2017
E.g.f.: (x*(x^3 + 6*x^2 + 7*x + 1)*cosh(x) + (x^4 + 6*x^3 + 7*x^2 + x + 3)*sinh(x))/4. - Stefano Spezia, Feb 18 2023

A192396 Square array T(n, k) = floor(((k+1)^n - (1+(-1)^k)/2)/2) read by antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 1, 0, 0, 4, 4, 2, 0, 0, 8, 13, 8, 2, 0, 0, 16, 40, 32, 12, 3, 0, 0, 32, 121, 128, 62, 18, 3, 0, 0, 64, 364, 512, 312, 108, 24, 4, 0, 0, 128, 1093, 2048, 1562, 648, 171, 32, 4, 0, 0, 256, 3280, 8192, 7812, 3888, 1200, 256, 40, 5, 0

Offset: 0

Adi Dani, Jun 29 2011

T(n,k) is the number of compositions of odd natural numbers into n parts <=k.

			T(2,4)=12: there are 12 compositions of odd natural numbers into 2 parts <=4
  1: (0,1), (1,0);
  3: (1,2), (2,1), (0,3), (3,0);
  5: (1,4), (4,1), (2,3), (3,2);
  7: (3,4), (4,3).
The table starts
    0,  0,   0,   0,    0,    0, ... A000004;
    0,  1,   1,   2,    2,    3, ... A004526;
    0,  2,   4,   8,   12,   18, ... A007590;
    0,  4,  13,  32,   62,  108, ... A036487;
    0,  8,  40, 128,  312,  648, ... A191903;
    0, 16, 121, 512, 1562, 3888, ... A191902;
    .        .      .       .    ...
with columns: A000004, A000079, A003462, A004171, A128531, A081341, ... .
Antidiagonal triangle begins:
  0;
  0,  0;
  0,  1,   0;
  0,  2,   1,   0;
  0,  4,   4,   2,   0;
  0,  8,  13,   8,   2,   0;
  0, 16,  40,  32,  12,   3,  0;
  0, 32, 121, 128,  62,  18,  3,  0;
  0, 64, 364, 512, 312, 108, 24,  4,  0;

G. C. Greubel, Antidiagonals n = 0..50, flattened
Adi Dani, Restricted compositions of natural numbers

Rows: A000004, A004526, A007590, A036487, A191902, A191902.

Columns: A000004, A000079, A003462, A004171, A081341, A128531.

A192396:= func< n,k | Floor(((k+1)^n - (1+(-1)^k)/2)/2) >;
[A192396(n-k,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 11 2023

A192396 := proc(n,k) (k+1)^n-(1+(-1)^k)/2 ; floor(%/2) ; end proc:
seq(seq( A192396(d-k,k),k=0..d),d=0..10) ; # R. J. Mathar, Jun 30 2011

T[n_, k_]:= Floor[((k+1)^n - (1+(-1)^k)/2)/2];
Table[T[n-k,k], {n,0,12}, {k,0,n}]//Flatten

def A192396(n,k): return ((k+1)^n - ((k+1)%2))//2
flatten([[A192396(n-k,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 11 2023

A131477 a(n) = ceiling(n^3/3).

Original entry on oeis.org

0, 1, 3, 9, 22, 42, 72, 115, 171, 243, 334, 444, 576, 733, 915, 1125, 1366, 1638, 1944, 2287, 2667, 3087, 3550, 4056, 4608, 5209, 5859, 6561, 7318, 8130, 9000, 9931, 10923, 11979, 13102, 14292, 15552, 16885, 18291, 19773, 21334, 22974

Offset: 0

Mohammad K. Azarian, Jul 27 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3, -3, 2, -3, 3, -1).

Cf. A000982, A008619, A004526, A007590, A036487.

[Ceiling(n^3/3) : n in [0..50]]; // Vincenzo Librandi, Oct 01 2011

Ceiling[Range[0,50]^3/3] (* Harvey P. Dale, Oct 19 2013 *)

From R. J. Mathar, Dec 19 2008: (Start)
G.f.: x*(1 + 3*x^2 + 2*x^3)/((1 - x)^4*(1 + x + x^2)).
a(n) = (A001093(n) - A049347(n))/3. (End)
a(n) = floor(n^3/3 + 2/3). - Bruno Berselli, Dec 21 2017

A173704 Partial sums of floor(n^3/2).

Original entry on oeis.org

0, 0, 4, 17, 49, 111, 219, 390, 646, 1010, 1510, 2175, 3039, 4137, 5509, 7196, 9244, 11700, 14616, 18045, 22045, 26675, 31999, 38082, 44994, 52806, 61594, 71435, 82411, 94605, 108105, 123000, 139384, 157352, 177004, 198441, 221769, 247095, 274531, 304190, 336190, 370650, 407694, 447447, 490039, 535601, 584269, 636180, 691476, 750300, 812800

Offset: 0

Mircea Merca, Nov 25 2010

nonn

Partial sums of A036487.

			a(4) = floor(1/2) + floor(8/2) + floor(27/2) + floor(64/2) = 49.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).

Cf. A036487.

[Round((n^4+2*n^3+n^2-2*n)/8): n in [0..40]]; // Vincenzo Librandi, Jun 22 2011

A173704 := proc(n) (n^4+2*n^3+n^2-2*n-1+(-1)^n)/8 ; end proc:

Table[Sum[Floor[k^3/2], {k, 0, n}], {n,0,50}] (* G. C. Greubel, Nov 23 2016 *)
Accumulate[Floor[Range[0,50]^3/2]] (* Harvey P. Dale, Jun 22 2025 *)

a(n) = Sum_{k=0..n} floor(k^3/2).
a(n) = round((n^4+2*n^3+n^2-2*n)/8).
a(n) = round((n^4+2*n^3+n^2-2*n-1)/8).
a(n) = floor((n^4+2*n^3+n^2-2*n)/8).
a(n) = ceiling((n-1)*(n+1)*(n^2+2*n+2)/8).
a(n) = a(n-2)+(n-1)*(2*n^2-n+2)/2, n>1.
From R. J. Mathar, Nov 26 2010: (Start)
a(n) = 4*a(n-1) - 5*a(n-2) + 5*a(n-4) - 4*a(n-5) + a(n-6).
G.f.: -x^2*(4+x+x^2) / ( (1+x)*(x-1)^5 ).
a(n) = (n^4 + 2*n^3 + n^2 - 2*n - 1 + (-1)^n)/8. (End)

Maple program replaced by R. J. Mathar, Nov 26 2010

A184632 Floor(1/{(8+n^4)^(1/4)}), where {}=fractional part.

Original entry on oeis.org

1, 4, 13, 32, 62, 108, 171, 256, 364, 500, 665, 864, 1098, 1372, 1687, 2048, 2456, 2916, 3429, 4000, 4630, 5324, 6083, 6912, 7812, 8788, 9841, 10976, 12194, 13500, 14895, 16384, 17968, 19652, 21437, 23328, 25326, 27436, 29659, 32000, 34460, 37044, 39753, 42592, 45562, 48668, 51911, 55296, 58824, 62500, 66325, 70304, 74438, 78732, 83187, 87808, 92596, 97556, 102689, 108000

Offset: 1

Clark Kimberling, Jan 18 2011

nonn

Index entries for linear recurrences with constant coefficients, signature (3, -2, -2, 3, -1).

Cf. A184536, A036487.

p[n_]:=FractionalPart[(n^4+8)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1},LinearRecurrence[{3,-2,-2,3,-1},{4,13,32,62,108},59]] (* Ray Chandler, Aug 02 2015 *)

a(n)=floor(1/{(8+n^4)^(1/4)}), where {}=fractional part.

It appears that a(n)=3a(n-1)-2a(n-2)-2a(n-3)+3a(n-4)-a(n-5) for n>=7.

cp's OEIS Frontend

A036486 a(n) = ceiling((n^3)/2).

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A131476 a(n) = floor(n^3/3).

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A131479 a(n) = floor(n^4/4).

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A131478 a(n) = ceiling(n^4/4).

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A192396 Square array T(n, k) = floor(((k+1)^n - (1+(-1)^k)/2)/2) read by antidiagonals.

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A131477 a(n) = ceiling(n^3/3).

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A173704 Partial sums of floor(n^3/2).

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