A066023 -id:A066023 - cp's OEIS frontend

A098547 a(n) = n^3 + n^2 + 1.

1, 3, 13, 37, 81, 151, 253, 393, 577, 811, 1101, 1453, 1873, 2367, 2941, 3601, 4353, 5203, 6157, 7221, 8401, 9703, 11133, 12697, 14401, 16251, 18253, 20413, 22737, 25231, 27901, 30753, 33793, 37027, 40461, 44101, 47953, 52023, 56317, 60841, 65601, 70603, 75853

Offset: 0

Douglas Winston (douglas.winston(AT)srupc.com), Oct 26 2004

G. C. Greubel, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000578, A001093, A011379, A027444, A033431, A033562, A034262, A053698, A061317, A066023, A071568.

Cf. A000217, A081423.

[(n^3+n^2+1): n in [1..60]]; // Vincenzo Librandi, Apr 06 2011

with(combinat): seq(fibonacci(3,n)+n^3, n=0..40); # Zerinvary Lajos, May 25 2008

Table[n^3+n^2+1,{n,0,100}] (* Vladimir Joseph Stephan Orlovsky, Mar 09 2011 *)

Vec(-(x^3-7*x^2+x-1)/(x-1)^4 + O(x^100)) \\ Colin Barker, Aug 29 2014

From Colin Barker, Aug 29 2014: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: (1 - x + 7*x^2 - x^3)/(1-x)^4. (End)
a(n) = A081423(n) + A000217(n-1). - Bruce J. Nicholson, Jan 06 2019
E.g.f.: exp(x)*(1 + 2*x + 4*x^2 + x^3). - Elmo R. Oliveira, Apr 20 2025

A099721 a(n) = n^2(2n+1).

Original entry on oeis.org

0, 3, 20, 63, 144, 275, 468, 735, 1088, 1539, 2100, 2783, 3600, 4563, 5684, 6975, 8448, 10115, 11988, 14079, 16400, 18963, 21780, 24863, 28224, 31875, 35828, 40095, 44688, 49619, 54900, 60543, 66560, 72963, 79764, 86975, 94608, 102675, 111188, 120159, 129600

Offset: 0

Douglas Winston (douglas.winston(AT)srupc.com), Nov 07 2004

For a right triangle with sides of lengths 8*n^3 + 12*n^2 + 8*n + 2, 4*n^4 + 8*n^3 + 4*n^2, and 4*n^4 + 8*n^3 + 12*n^2 + 8*n + 2, dividing the area by the perimeter gives a(n). - J. M. Bergot, Jul 30 2013
This sequence is the difference between the centered icosahedral (or cuboctahedral) numbers (A005902(n)) and the centered octagonal pyramidal numbers (A000447(n+1)). - Peter M. Chema, Jan 09 2016
a(n) is the sum of the integers in the closed interval (n-1)*n to n*(n+1). - J. M. Bergot, Apr 19 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Row n=3 of A229079.
Cf. A000578, A001093, A011379, A015237, A027444, A033431, A033562, A034262, A053698, A061317, A066023, A071568, A098547.
Cf. A000290, A000447, A002378, A005408, A005902, A024196.

[n^2*(2*n+1): n in [0..50]]; // Vincenzo Librandi, May 01 2011

A099721 := proc(n) n^2*(2*n+1) ; end proc:
seq(A099721(n),n=0..10) ;

a[n_]:=2*n^3+n^2; (* Vladimir Joseph Stephan Orlovsky, Dec 21 2008 *)
LinearRecurrence[{4,-6,4,-1},{0,3,20,63},40] (* Harvey P. Dale, Aug 19 2022 *)

a(n) = ceil(sum(i=n^2-(n-1), n^2+(n-1), if(!issquare(4*i+1), (2*i+1+sqrt(4*i+1))/2, 0))); \\ Michel Marcus, Nov 14 2014, after Richard R. Forberg

G.f.: x*(3 + 8*x + x^2)/(x-1)^4.
a(n) = A024196(n) - A024196(n-1). - Philippe Deléham, May 07 2012
a(n) = ceiling(Sum_{i=n^2-(n-1)..n^2+(n-1)} s(i)), for n > 0 and integer i, where s(i) are the real solutions to x = i + sqrt(x), and the summation range excludes the integer solutions which occur where i is an oblong number (A002378). The fractional portion of the summation converges to 2/3 for large n. If s(i) is replaced with i, then the summation equals n^2*(2*n-1) = A015237. - Richard R. Forberg, Oct 15 2014
a(n) = A005902(n) - A000447(n+1). - Peter M. Chema, Jan 09 2016
From Amiram Eldar, May 17 2022: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/6 + 4*log(2) - 4.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/12 - Pi - 2*log(2) + 4. (End)
From Elmo R. Oliveira, Aug 08 2025: (Start)
E.g.f.: x*(1 + 2*x)*(3 + x)*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = A000290(n)*A005408(n). (End)

A184628 Floor(1/frac((4+n^4)^(1/4))), where frac(x) is the fractional part of x.

Original entry on oeis.org

2, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507, 85184, 91125, 97336, 103823, 110592, 117649, 125000, 132651, 140608, 148877, 157464, 166375

Offset: 1

Clark Kimberling, Jan 18 2011

Is a(n) = A066023(n) for n>=2? R. J. Mathar, Jan 28 2011

Ray Chandler, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

Cf. A000578, A066023, A184536.

p[n_]:=FractionalPart[(n^4+4)^(1/4)];
  q[n_]:=Floor[1/p[n]]; Table[q[n],{n,1,80}]
  FindLinearRecurrence[Table[q[n],{n,1,1000}]]
Join[{2}, LinearRecurrence[{4, -6, 4, -1}, {8, 27, 64, 125}, 54]] (* Ray Chandler, Aug 01 2015 *)

a(n) = floor(1/{(4+n^4)^(1/4)}), where {}=fractional part.
It appears that a(n)=4a(n-1)-6a(n-2)+4a(n-3)-a(n-4) for n>=6 and that a(n)=n^3 for n>=2.
Empirical g.f.: x*(x^4-4*x^3+7*x^2+2) / (x-1)^4. - Colin Barker, Sep 06 2014

cp's OEIS Frontend

A098547 a(n) = n^3 + n^2 + 1.

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A099721 a(n) = n^2(2n+1).

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A184628 Floor(1/frac((4+n^4)^(1/4))), where frac(x) is the fractional part of x.

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cp's OEIS Frontend

A098547 a(n) = n^3 + n^2 + 1.

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Author

Keywords

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Programs

Magma

Maple

Mathematica

PARI

Formula

A099721 a(n) = n^2*(2*n+1).

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Author

Keywords

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Crossrefs

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Magma

Maple

Mathematica

PARI

Formula

A184628 Floor(1/frac((4+n^4)^(1/4))), where frac(x) is the fractional part of x.

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Mathematica

Formula

A099721 a(n) = n^2(2n+1).