A087509 -id:A087509 - cp's OEIS frontend

A008133 a(n) = floor(n/3)*floor((n+1)/3).

0, 0, 0, 1, 1, 2, 4, 4, 6, 9, 9, 12, 16, 16, 20, 25, 25, 30, 36, 36, 42, 49, 49, 56, 64, 64, 72, 81, 81, 90, 100, 100, 110, 121, 121, 132, 144, 144, 156, 169, 169, 182, 196, 196, 210, 225, 225, 240, 256, 256, 272, 289, 289, 306, 324, 324, 342, 361, 361, 380, 400, 400, 420, 441, 441, 462, 484, 484, 506

Offset: 0

N. J. A. Sloane

Oblong numbers and squares are subsequences: a(A016789(n)) = A002378(n); a(A008585(n)) = a(A016777(n)) = A000290(n). - Reinhard Zumkeller, Oct 09 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Vladimir Baltić, Applications of the finite state automata for counting restricted permutations and variations, Yugoslav Journal of Operations Research, 22 (2012), Number 2, 183-198; alternative link. - From _N. J. A. Sloane_, Jan 02 2013
Index entries for linear recurrences with constant coefficients, signature (1,0,2,-2,0,-1,1).

Cf. A000290, A008217, A008585, A016777, A016789, A087509.

a008133 n = a008133_list !! n
a008133_list = zipWith (*) (tail ts) ts where ts = map (`div` 3) [0..]
-- Reinhard Zumkeller, Oct 09 2011

[Floor(n/3)*Floor((n+1)/3): n in [0..60]]; // Vincenzo Librandi, Aug 20 2011

Table[Floor[n/3]Floor[(n+1)/3],{n,0,100}] (* or *) LinearRecurrence[{1,0,2,-2,0,-1,1},{0,0,0,1,1,2,4},100] (* Harvey P. Dale, Sep 21 2024 *)

a(n) = floor(n/3)*floor((n+1)/3); /* Joerg Arndt, Mar 31 2013 */

From Paul Barry, Sep 14 2003: (Start)
Partial sums of A087509.
a(n+1) = Sum_{j=0..n} Sum_{k=0..j} [mod(j*k, 3)=2], where [] is the Iverson bracket. (End)
Empirical g.f.: -x^3*(x^2+1) / ((x-1)^3*(x^2+x+1)^2). - Colin Barker, Mar 31 2013
From Amiram Eldar, May 10 2025: (Start)
Sum_{n>=3} 1/a(n) = Pi^2/3 + 1.
Sum_{n>=3} (-1)^(n+1)/a(n) = 2*log(2)-1. (End)

A087508 Number of k such that mod(k*n,3) = 1 for 0 <= k <= n.

Original entry on oeis.org

0, 1, 1, 0, 2, 2, 0, 3, 3, 0, 4, 4, 0, 5, 5, 0, 6, 6, 0, 7, 7, 0, 8, 8, 0, 9, 9, 0, 10, 10, 0, 11, 11, 0, 12, 12, 0, 13, 13, 0, 14, 14, 0, 15, 15, 0, 16, 16, 0, 17, 17, 0, 18, 18, 0, 19, 19, 0, 20, 20, 0, 21, 21, 0, 22, 22, 0, 23, 23, 0, 24, 24, 0, 25, 25, 0, 26, 26, 0, 27, 27, 0, 28, 28, 0

Offset: 0

Paul Barry, Sep 11 2003

			a(4) = 2 because k=1 and k=4 satisfy the equation.

G. C. Greubel, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A000027, A087507, A087509.

I:=[0,1,1,0,2,2]; [n le 6 select I[n] else 2*Self(n-3) - Self(n-6): n in [1..100]]; // Vincenzo Librandi, Sep 22 2015

LinearRecurrence[{0,0,2,0,0,-1}, {0,1,1,0,2,2}, 100] (* Vincenzo Librandi, Sep 22 2015 *)
Table[PadRight[{0},3,n],{n,30}]//Flatten (* Harvey P. Dale, Jan 27 2021 *)

concat(0,Vec((1+x)/(1-x^3)^2 +O(x^99))) \\ Charles R Greathouse IV, Oct 24 2014

a(n) = sum(k=0, n, Mod(k*n, 3)==1); \\ Michel Marcus, Sep 27 2017

@CachedFunction
def A087508(n):
    if (n<6): return (0,1,1,0,2,2)[n]
    else: return 2*A087508(n-3) - A087508(n-6)
[A087508(n) for n in (0..100)] # G. C. Greubel, Sep 02 2022

a(n) = A000027(n) - A087509(n) - A087507(n).
a(n) = (2/3)*(floor(n/3)+1)*(1-cos(2*Pi*n/3)).
G.f.: x*(1 + x)/(1 - x^3)^2. - Arkadiusz Wesolowski, May 28 2013
a(n) = sin(n*Pi/3)*((4n+6)*sin(n*Pi/3)-sqrt(3)*cos(n*Pi))/9. - Wesley Ivan Hurt, Sep 24 2017

A087507 #{0<=k<=n: k*n is divisible by 3}.

Original entry on oeis.org

1, 1, 1, 4, 2, 2, 7, 3, 3, 10, 4, 4, 13, 5, 5, 16, 6, 6, 19, 7, 7, 22, 8, 8, 25, 9, 9, 28, 10, 10, 31, 11, 11, 34, 12, 12, 37, 13, 13, 40, 14, 14, 43, 15, 15, 46, 16, 16, 49, 17, 17, 52, 18, 18, 55, 19, 19, 58, 20, 20, 61, 21, 21, 64, 22, 22, 67, 23, 23, 70, 24, 24, 73, 25, 25, 76

Offset: 0

Paul Barry, Sep 11 2003

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

Cf. A016777 (trisection).

Vec((2*x^3+x^2+x+1)/((x-1)^2*(x^2+x+1)^2) + O(x^100)) \\ Colin Barker, May 02 2015

a(n) = sum{k=0..n, if (mod(kn, 3)=0, 1, 0) }.
a(n) = floor(n/3)(5/3+4/3cos(2Pi*/3))+1.
a(n)+A087508(n)+A087509(n) = n.
a(n) = 2*a(n-3)-a(n-6) for n>5. - Colin Barker, May 02 2015
G.f.: (2*x^3+x^2+x+1) / ((x-1)^2*(x^2+x+1)^2). - Colin Barker, May 02 2015

A077814 a(n) = #{0<=k<=n: mod(k*n,4)=2}.

Original entry on oeis.org

0, 0, 1, 1, 0, 1, 3, 2, 0, 2, 5, 3, 0, 3, 7, 4, 0, 4, 9, 5, 0, 5, 11, 6, 0, 6, 13, 7, 0, 7, 15, 8, 0, 8, 17, 9, 0, 9, 19, 10, 0, 10, 21, 11, 0, 11, 23, 12, 0, 12, 25, 13, 0, 13, 27, 14, 0, 14, 29, 15, 0, 15, 31, 16, 0, 16, 33, 17, 0, 17, 35, 18, 0, 18, 37, 19, 0, 19, 39, 20, 0, 20, 41, 21, 0

Offset: 0

John W. Layman, Dec 03 2002

Coefficients in the unique expansion of e/4 = Sum_{n>=1} a(n)/n!, where a(n) satisfies 0<=a(n)

			a(6) = #{1, 3, 5} = 3.

Index entries for linear recurrences with constant coefficients, signature (2,-3,4,-3,2,-1).

Cf. A009947, A087509, A087620.

a = Table[0, {i, 1, 50}]; x = Exp[1]/4; For[n = 2, n <= 51, n++, { an = 0; While [(x >= (1/n!)) && (an < (n - 1)), {an++, x = x - (1/n!)} ]}; a[[n - 1]] = an;]; a
LinearRecurrence[{2,-3,4,-3,2,-1},{0,0,1,1,0,1},90] (* Harvey P. Dale, Apr 07 2025 *)

a(n)=0 if n=4k, a(n)=k if n=4k+1, a(n)=2k+1 if n=4k+2 and a(n)=k+1 if n=4k+3.
a(n) = floor(n!*e/4) - n*floor((n-1)!*e/4). - Benoit Cloitre, Dec 07 2002
a(n) = Sum_{k=0..n} if(mod(k*n, 4)=2, 1, 0). - Paul Barry, Sep 10 2003
O.g.f.: x^2*(1-x+x^2)/((x-1)^2*(1+x^2)^2). - R. J. Mathar, Jun 13 2008
From Wesley Ivan Hurt, May 30 2015: (Start)
a(n) = 2*a(n-1)-3*a(n-2)+4*a(n-3)-3*a(n-4)+2*a(n-5)-a(n-6), n>6.
a(n) = (-1)^((1-2*n-(-1)^n)/4)*((-1)^n-2*n*(-1)^((2*n+3+(-1)^n)/4)+n*(-1)^((1+(-1)^n)/2)+n*(-1)^((2*n+1+(-1)^n)/2)-1)/8. (End)

More terms from Benoit Cloitre, Dec 07 2002

A136493 Triangle of coefficients of characteristic polynomials of symmetrical pentadiagonal matrices of the type (1,-1,1,-1,1).

Original entry on oeis.org

1, -1, 1, 1, -2, 0, -1, 3, 0, 0, 1, -4, 1, 2, 0, -1, 5, -3, -5, 1, 1, 1, -6, 6, 8, -5, -2, 1, -1, 7, -10, -10, 14, 4, -4, 0, 1, -8, 15, 10, -29, -4, 12, 0, 0, -1, 9, -21, -7, 50, -4, -30, 4, 4, 0, 1, -10, 28, 0, -76, 28, 61, -20, -15, 2, 1

Offset: 0

Roger L. Bagula, Mar 21 2008

From Georg Fischer, Mar 29 2021: (Start)
The pentadiagonal matrices have 1 in the main diagonal, -1 in the first lower and upper diagonal, 1 in the second lower and upper diagonal, and 0 otherwise.
The linear recurrences that yield A124805, A124806, A124807 and similar can be derived from the rows of this triangle (the first element of a row must be removed and multiplied onto the remaining elements).
This observation extends to other sequences. For example the linear recurrence signature (5,-6,2,4,0) of A124698 "Number of base 5 circular n-digit numbers with adjacent digits differing by 1 or less" can be derived from the coefficients of the characteristic polynomial of a tridiagonal (type -1,1,-1) 5 X 5 matrix.
(End)

			Triangle begins:
   1;
  -1,   1;
   1,  -2,   0;
  -1,   3,   0,   0;
   1,  -4,   1,   2,   0;
  -1,   5,  -3,  -5,   1,  1;
   1,  -6,   6,   8,  -5, -2,   1;
  -1,   7, -10, -10,  14,  4,  -4,   0;
   1,  -8,  15,  10, -29, -4,  12,   0,   0;
  -1,   9, -21,  -7,  50, -4, -30,   4,   4,  0;
   1, -10,  28,   0, -76, 28,  61, -20, -15,  2,  1;

Anthony Ralston and Philip Rabinowitz, A First Course in Numerical Analysis, 1978, ISBN 0070511586, see p. 256.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A011658, A087509, A124805 ff., A124696 ff., A124999 ff., A161680, A181983.

T[n_, m_]:= Piecewise[{{-1, 1+m==n || m==1+n}, {1, 2+m==n || m==n || m==2+n}}];
MO[d_]:= Table[T[n, m], {n,d}, {m,d}];
CL[n_]:= CoefficientList[CharacteristicPolynomial[MO[n], x], x];
Join[{{1}}, Table[Reverse[CL[n]], {n,10}]]//Flatten
(* For the signature of A124698 added by Georg Fischer, Mar 29 2021 : *)
Reverse[CoefficientList[CharacteristicPolynomial[{{1,-1,0,0,0}, {-1, 1,-1,0,0}, {0,-1,1,-1,0}, {0,0,-1,1,-1}, {0,0,0,-1,1}}, x], x]]

Sum_{k=1..n} T(n, k) = (-1)^(n mod 3) * A087509(n+1) + [n=1].
From G. C. Greubel, Aug 01 2023: (Start)
T(n, n) = A011658(n+2).
T(n, 1) = (-1)^(n-1).
T(n, 2) = A181983(n-1).
T(n, 3) = (-1)^(n-3)*A161680(n-3). (End)

Edited by Georg Fischer, Mar 29 2021

Showing 1-5 of 5 results.

cp's OEIS Frontend

A008133 a(n) = floor(n/3)*floor((n+1)/3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

PARI

Formula

A087508 Number of k such that mod(k*n,3) = 1 for 0 <= k <= n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

SageMath

Formula

A087507 #{0<=k<=n: k*n is divisible by 3}.

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

Formula

A077814 a(n) = #{0<=k<=n: mod(k*n,4)=2}.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A136493 Triangle of coefficients of characteristic polynomials of symmetrical pentadiagonal matrices of the type (1,-1,1,-1,1).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

Extensions