A002266 -id:A002266 - cp's OEIS frontend

A187243 Number of ways of making change for n cents using coins of 1, 5, and 10 cents.

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 9, 9, 9, 9, 9, 12, 12, 12, 12, 12, 16, 16, 16, 16, 16, 20, 20, 20, 20, 20, 25, 25, 25, 25, 25, 30, 30, 30, 30, 30, 36, 36, 36, 36, 36, 42, 42, 42, 42, 42, 49, 49, 49, 49, 49, 56, 56, 56, 56, 56, 64, 64, 64, 64, 64, 72, 72, 72, 72, 72

Offset: 0

T. D. Noe, Mar 07 2011

a(n) is the number of partitions of n into parts 1, 5, and 10. - Joerg Arndt, Feb 02 2017
From Gerhard Kirchner, Jan 25 2017: (Start)
There is a simple recurrence for solving such problems given coin values 1 = c(1) < c(2) < ... < c(k).
Let f(n, j), 1 < j <= k, be the number of ways of making change for n cents with coin values c(i), 1 <= i <= j. Then any number m of c(j)-coins with 0 <= m <= floor(n/c(j)) can be used, and the remaining amount of change to be made using coins of values smaller than c(j) will be n - m*c(j) cents. This leads directly to the recurrence formula with a(n) = f(n, k).
For k = 3 with c(1) = 1, c(2) = 5, c(3) = 10, the recurrence can be reduced to an explicit formula; see link "Derivation of formulas".
By the way, a(n) is also the number of ways of making change for n cents using coins of 2, 5, 10 cents and at most one 1-cent coin. That is because any coin combination is, as in the original problem, fixed by the numbers of 5-cent and 10-cent coins.
(End)

			From _Gerhard Kirchner_, Jan 25 2017: (Start)
Recurrence:
a(11)  = f(11, 3) = f(11 - 0, 2) + f(11 - 10, 2)
       = f(11 - 0, 1) + f(11 - 5, 1) + f(11 - 10, 1) + f(1, 2)
       = 1 + 1 + 1 + 1 = 4.
Explicitly: a(79) = (7 + 1)*(7 + 1 + 1) = 72.
(End)

Gerhard Kirchner, Table of n, a(n) for n = 0..1000
Gerhard Kirchner, Derivation of formulas
Index entries for sequences related to making change.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,-1,1).

Cf. A001299, A001300, A001306, A169718, A002266, A002620, A245573.

CoefficientList[ Series[ 1 / ((1 - x)(1 - x^5)(1 - x^10)), {x, 0, 75} ], x ]

Vec( 1/((1-x)*(1-x^5)*(1-x^10))+O(x^99)) \\ Charles R Greathouse IV, Aug 22 2011

a(n)=(n^2+16*n+97+10*(n\5+1)*(5*(n\5)+2-n))\100 \\ Tani Akinari, Sep 10 2015

a(n) = {my(q=n\10, s=(n%10)\5); (q+1)*(q+1+s); } \\ (Kirchner's explicit formula) Joerg Arndt, Feb 02 2017

G.f.: 1/((1-x)*(1-x^5)*(1-x^10)).
From Gerhard Kirchner, Jan 25 2017: (Start)
General recurrence: f(n, 1) = 1; j > 1: f(0, j) = 1 or f(n, j) = Sum_{m=0..floor(n/c(j))} f(n-m*c(j), j-1);
a(n) = f(n, k).
Note: f(n, j) = f(n, j-1) for n < c(j) => f(1, j) = 1.
Explicit formula:
a(n) = (q+1)*(q+1+s) with q = floor(n/10) and s = floor((n mod 10)/5). (End)
a(n) = A002620(A002266(n)+2) = floor((floor(n/5) + 2)^2 / 4). - Hoang Xuan Thanh, Jun 26 2025

A267541 Expansion of (2 + 4x + x^2 + x^3 + 2x^4 + x^5)/(1 - x - x^5 + x^6).

Original entry on oeis.org

2, 6, 7, 8, 10, 13, 17, 18, 19, 21, 24, 28, 29, 30, 32, 35, 39, 40, 41, 43, 46, 50, 51, 52, 54, 57, 61, 62, 63, 65, 68, 72, 73, 74, 76, 79, 83, 84, 85, 87, 90, 94, 95, 96, 98, 101, 105, 106, 107, 109, 112, 116, 117, 118, 120, 123, 127, 128, 129, 131, 134, 138, 139, 140

Offset: 0

Bruno Berselli, Jan 16 2016

Also, numbers that are congruent to {2, 6, 7, 8, 10} mod 11.
(m^k+1)/11 is a nonnegative integer when
. m is a member of this sequence and k is an odd multiple of 5 (A017329),
. m is a member of A017509 and k is odd but not multiple of 5 (A045572).
If k is even, (m^k+1)/11 is never an integer.
The product of two terms does not belong to the sequence.

			From the linear recurrence:
(-A267755) ..., -12, -9, -5, -4, -3, -1, 2, 6, 7, 8, 10, 13, ... (A267541)

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

Cf. A002266, A017329, A017509, A267755.

Cf. A088225: numbers congruent to {2,6,7,8} mod 11.

m:=70; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2+4*x+x^2+x^3+2*x^4+x^5)/(1-x-x^5+x^6)));

gf := (2+4*x+x^2+x^3+2*x^4+x^5)/((1-x)^2*(1+x+x^2+x^3+ x^4)): deg := 64: series(gf,x,deg): seq(coeff(%,x,n), n=0..deg-1); # Peter Luschny, Jan 19 2016

CoefficientList[Series[(2 + 4 x + x^2 + x^3 + 2 x^4 + x^5)/(1 - x - x^5 + x^6), {x, 0, 70}], x]
LinearRecurrence[{1, 0, 0, 0, 1, -1}, {2, 6, 7, 8, 10, 13}, 70]
Select[Range[150], MemberQ[{2, 6, 7, 8, 10}, Mod[#, 11]]&]

Vec((2+4*x+x^2+x^3+2*x^4+x^5)/(1-x-x^5+x^6)+O(x^70))

gf = (2+4*x+x^2+x^3+2*x^4+x^5)/((1-x)^2*(1+x+x^2+x^3+ x^4))
print(taylor(gf, x, 0, 63).list()) # Peter Luschny, Jan 19 2016

G.f.: (2 + 4*x + x^2 + x^3 + 2*x^4 + x^5)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(n-1) + a(n-5) - a(n-6).
a(-n) = -A267755(n-1).

A349841 Triangle T(n,k) built by placing all ones on the left edge, [1,0,0,0,0] repeated on the right edge, and filling the body using the Pascal recurrence T(n,k) = T(n-1,k) + T(n-1,k-1).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 1, 1, 5, 10, 10, 5, 2, 0, 1, 6, 15, 20, 15, 7, 2, 0, 1, 7, 21, 35, 35, 22, 9, 2, 0, 1, 8, 28, 56, 70, 57, 31, 11, 2, 0, 1, 9, 36, 84, 126, 127, 88, 42, 13, 2, 1

Offset: 0

Michael A. Allen, Dec 13 2021

This is the m=5 member in the sequence of triangles A007318, A059259, A118923, A349839, A349841 which have all ones on the left side, ones separated by m-1 zeros on the other side, and whose interiors obey Pascal's recurrence.
T(n,k) is the (n,n-k)-th entry of the (1/(1-x^5),x/(1-x)) Riordan array.
For n>0, T(n,n-1) = A002266(n+4).
For n>1, T(n,n-2) = A008732(n-2).
For n>2, T(n,n-3) = A122047(n-1).
Sums of rows give A349842.
Sums of antidiagonals give A349843.

			Triangle begins:
  1;
  1,   0;
  1,   1,   0;
  1,   2,   1,   0;
  1,   3,   3,   1,   0;
  1,   4,   6,   4,   1,   1;
  1,   5,  10,  10,   5,   2,   0;
  1,   6,  15,  20,  15,   7,   2,   0;
  1,   7,  21,  35,  35,  22,   9,   2,   0;
  1,   8,  28,  56,  70,  57,  31,  11,   2,   0;
  1,   9,  36,  84, 126, 127,  88,  42,  13,   2,   1;

Michael A. Allen and Kenneth Edwards, On Two Families of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.7.1.

Other members of sequence of triangles: A007318, A059259, A118923, A349839.
Columns: A000012, A001477, A000217, A000292, A000332, A323228.
Diagonals: A079998, A002266, A008732, A122047.

Flatten[Table[CoefficientList[Series[(1 - x*y)/((1 - (x*y)^5)(1 - x - x*y)), {x, 0, 20}, {y, 0, 10}], {x, y}][[n+1,k+1]],{n,0,10},{k,0,n}]]

G.f.: (1-x*y)/((1-(x*y)^5)(1-x-x*y)) in the sense that T(n,k) is the coefficient of x^n*y^k in the series expansion of the g.f.
T(n,0) = 1.
T(n,n) = delta(n mod 5,0).
T(n,1) = n-1 for n>0.
T(n,2) = (n-1)*(n-2)/2 for n>1.
T(n,3) = (n-1)*(n-2)*(n-3)/6 for n>2.
T(n,4) = (n-1)*(n-2)*(n-3)*(n-4)/24 for n>3.
T(n,5) = C(n-1,5) + 1 for n>4.
T(n,6) = C(n-1,6) + n - 6 for n>5.
For 0 <= k < n, T(n,k) = (n-k)*Sum_{j=0..floor(k/5)} binomial(n-5*j,n-k)/(n-5*j).
The g.f. of the n-th subdiagonal is 1/((1-x^5)(1-x)^n).

A008648 Molien series of 3 X 3 upper triangular matrices over GF( 5 ).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 11, 11, 11, 11, 11, 13, 13, 13, 13, 13, 15, 15, 15, 15, 15, 18, 18, 18, 18, 18, 21, 21, 21, 21, 21, 24, 24, 24, 24, 24

Offset: 0

N. J. A. Sloane

a(n) is the number of partitions of n into parts 1, 5, and 25. - Joerg Arndt, Sep 07 2019

D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 105.

G. C. Greubel, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 221
Index entries for Molien series
Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 0, -1, 1).

Cf. A002266.

R:=PowerSeriesRing(Integers(), 70); Coefficients(R!( 1/((1-x)*(1-x^5)*(1-x^25)) )); // G. C. Greubel, Sep 06 2019

seq(coeff(series(1/((1-x)*(1-x^5)*(1-x^25)), x, n+1), x, n), n = 0 .. 70); # modified by G. C. Greubel, Sep 06 2019

CoefficientList[Series[1/((1-x)*(1-x^5)*(1-x^25)), {x,0,70}], x] (* G. C. Greubel, Sep 06 2019 *)

my(x='x+O('x^70)); Vec(1/((1-x)*(1-x^5)*(1-x^25))) \\ G. C. Greubel, Sep 06 2019

def A008648_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(1/((1-x)*(1-x^5)*(1-x^25))).list()
A008648_list(70) # G. C. Greubel, Sep 06 2019

G.f.: 1/((1-x)*(1-x^5)*(1-x^25)).

A010885 Period 6: repeat [1, 2, 3, 4, 5, 6].

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3

Offset: 0

N. J. A. Sloane

Partial sums are given by A130484(n)+n+1. - Hieronymus Fischer, Jun 08 2007

41152/333333 = 0.123456123456123456... [Eric Desbiaux, Nov 03 2008]

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Cf. A010872, A010873, A010874, A010875, A010876, A004526, A002264, A002265, A002266.

Cf. A177158 (decimal expansion of (103+2*sqrt(4171))/162). [From Klaus Brockhaus, May 03 2010]

&cat[[1..6]: n in [0..20]]; // Wesley Ivan Hurt, Jun 17 2016

A010885:=n->(21-3*cos(n*Pi)-4*sqrt(3)*cos((1-4*n)*Pi/6)-12*sin((1+2*n)*Pi/6))/6: seq(A010885(n), n=0..100); # Wesley Ivan Hurt, Jun 17 2016

Flatten[Table[Range[6],{n,15}]] (* Harvey P. Dale, Aug 01 2011 *)
PadRight[{},90,Range[6]] (* Harvey P. Dale, Sep 23 2019 *)

a(n) = 1 + (n mod 6). - Paolo P. Lava, Nov 21 2006
a(n) = A010875(n)+1. G.f.: g(x)=(Sum_{0<=k<6} (k+1)*x^k)/(1-x^6). Also g(x)=(6*x^7-7*x^6+1)/((1-x^6)*(1-x)^2). - Hieronymus Fischer, Jun 08 2007
From Wesley Ivan Hurt, Jun 17 2016: (Start)
G.f.: (1+2*x+3*x^2+4*x^3+5*x^4+6*x^5)/(1-x^6).
a(n) = (21-3*cos(n*Pi)-4*sqrt(3)*cos((1-4*n)*Pi/6)-12*sin((1+2*n)*Pi/6))/6.
a(n) = a(n-6) for n>5. (End)

A010889 Simple periodic sequence: repeat 1,2,3,4,5,6,7,8,9,10.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1

Offset: 0

N. J. A. Sloane

Partial sums are given by A130488(n)+n+1. - Hieronymus Fischer, Jun 08 2007

Continued fraction expansion of (232405+sqrt(71216963807))/348378. [From Klaus Brockhaus, May 15 2010]

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,1).

Cf. A010872, A010873, A010874, A010875, A010876, A010877, A010878, A004526, A002264, A002265, A002266.

Cf. A177933 (decimal expansion of (232405+sqrt(71216963807))/348378). [From Klaus Brockhaus, May 15 2010]

PadRight[{},120,Range[10]] (* Harvey P. Dale, Feb 22 2015 *)

def a(n): return n % 10 + 1 # Paul Muljadi, Aug 06 2024

a(n) = 1 + (n mod 10) - Paolo P. Lava, Nov 21 2006
From Hieronymus Fischer, Jun 08 2007: (Start)
a(n) = A010879(n)+1.
G.f.: (Sum_{k=0..9} (k+1)*x^k)/(1-x^10).
G.f.: (10x^11-11x^10+1)/((1-x^10)(1-x)^2). (End)

More terms from Klaus Brockhaus, May 15 2010

A194222 a(n) = floor(Sum_{k=1..n} frac(k/5)), where frac() = fractional part.

Original entry on oeis.org

0, 0, 1, 2, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 6, 6, 7, 8, 8, 8, 8, 9, 10, 10, 10, 10, 11, 12, 12, 12, 12, 13, 14, 14, 14, 14, 15, 16, 16, 16, 16, 17, 18, 18, 18, 18, 19, 20, 20, 20, 20, 21, 22, 22, 22, 22, 23, 24, 24, 24, 24, 25, 26, 26, 26, 26, 27, 28, 28, 28, 28, 29, 30

Offset: 1

Clark Kimberling, Aug 19 2011

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Cf. A118015.

seq(floor((n+1)/5)+floor((n+2)/5), n=1..80); # Ridouane Oudra, Dec 14 2021

r = 1/5;
a[n_] := Floor[Sum[FractionalPart[k*r], {k, 1, n}]]
Table[a[n], {n, 1, 90}]    (* A194222 *)
s[n_] := Sum[a[k], {k, 1, n}]
Table[s[n], {n, 1, 100}]   (* A118015 *)
LinearRecurrence[{1,0,0,0,1,-1},{0,0,1,2,2,2},80] (* Harvey P. Dale, Jun 06 2024 *)

From Chai Wah Wu, Jun 10 2020: (Start)
a(n) = a(n-1) + a(n-5) - a(n-6) for n > 6.
G.f.: x^3*(x + 1)/((x-1)^2*(1+x+x^2+x^3+x^4)). (End)
a(n) = floor((n+1)/5) + floor((n+2)/5). - Ridouane Oudra, Dec 14 2021
a(n) = A002266(n+1)+A002266(n+2). - R. J. Mathar, Nov 21 2023

A246720 Number A(n,k) of partitions of n into parts of the k-th list of distinct parts in the order given by A246688; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 2, 0, 1, 0, 1, 1, 0, 2, 1, 1, 0, 1, 0, 1, 1, 3, 0, 1, 0, 1, 1, 0, 2, 0, 3, 1, 1, 0, 1, 0, 1, 0, 2, 0, 4, 0, 1, 0, 1, 0, 1, 1, 1, 2, 1, 4, 1, 1, 0, 1, 1, 0, 1, 2, 0, 3, 0, 5, 0, 1, 0, 1, 1, 2, 0, 1, 2, 0, 3, 0, 5, 1, 1, 0

Offset: 0

Alois P. Heinz, Sep 02 2014

The first lists of distinct parts in the order given by A246688 are: 0:[], 1:[1], 2:[2], 3:[1,2], 4:[3], 5:[1,3], 6:[4], 7:[1,4], 8:[2,3], 9:[5], 10:[1,2,3], 11:[1,5], 12:[2,4], 13:[6], 14:[1,2,4], 15:[1,6], 16:[2,5], 17:[3,4], 18:[7], 19:[1,2,5], 20:[1,3,4], ... .

			Square array A(n,k) begins:
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1,  1, 1, 1, 1,  1, ...
  0, 1, 0, 1, 0, 1, 0, 1, 0, 0,  1, 1, 0, 0,  1, ...
  0, 1, 1, 2, 0, 1, 0, 1, 1, 0,  2, 1, 1, 0,  2, ...
  0, 1, 0, 2, 1, 2, 0, 1, 1, 0,  3, 1, 0, 0,  2, ...
  0, 1, 1, 3, 0, 2, 1, 2, 1, 0,  4, 1, 2, 0,  4, ...
  0, 1, 0, 3, 0, 2, 0, 2, 1, 1,  5, 2, 0, 0,  4, ...
  0, 1, 1, 4, 1, 3, 0, 2, 2, 0,  7, 2, 2, 1,  6, ...
  0, 1, 0, 4, 0, 3, 0, 2, 1, 0,  8, 2, 0, 0,  6, ...
  0, 1, 1, 5, 0, 3, 1, 3, 2, 0, 10, 2, 3, 0,  9, ...
  0, 1, 0, 5, 1, 4, 0, 3, 2, 0, 12, 2, 0, 0,  9, ...
  0, 1, 1, 6, 0, 4, 0, 3, 2, 1, 14, 3, 3, 0, 12, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0-11, 13-20, 23 give: A000007, A000012, A059841, A008619, A079978, A008620, A121262, A008621, A103221, A079998, A001399, A002266(n+5), A079979, A008642, A097992, A008616, A008679, A082784, A000115, A025767, A008676.
Main diagonal gives A246721.
Cf. A246688, A246690 (the same for compositions).

b:= proc(n, i) b(n, i):= `if`(n=0, [[]], `if`(i>n, [],
      [map(x->[i, x[]], b(n-i, i+1))[], b(n, i+1)[]]))
    end:
f:= proc() local i, l; i, l:=0, [];
      proc(n) while n>=nops(l)
        do l:=[l[], b(i, 1)[]]; i:=i+1 od; l[n+1]
      end
    end():
g:= proc(n, l) option remember; `if`(n=0, 1, `if`(l=[], 0,
      add(g(n-l[-1]*j, subsop(-1=NULL, l)), j=0..n/l[-1])))
    end:
A:= (n, k)-> g(n, f(k)):
seq(seq(A(n, d-n), n=0..d), d=0..16);

b[n_, i_] := b[n, i] = If[n == 0, {{}}, If[i > n, {}, Join[Prepend[#, i]& /@ b[n - i, i + 1], b[n, i + 1]]]];
f = Module[{i = 0, l = {}}, Function[n, While[ n >= Length[l], l = Join[l, b[i, 1]]; i++ ]; l[[n + 1]]]];
g[n_, l_] := g[n, l] = If[n == 0, 1, If[l == {}, 0, Sum[g[n - l[[-1]] j, ReplacePart[l, -1 -> Nothing]], {j, 0, n/l[[-1]]}]]];
A[n_, k_] := g[n, f[k]];
Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 16}] // Flatten (* Jean-François Alcover, Dec 07 2020, after Alois P. Heinz *)

A008496 a(n) = floor(n/5)floor((n+1)/5)floor((n+2)/5).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 2, 4, 8, 8, 8, 12, 18, 27, 27, 27, 36, 48, 64, 64, 64, 80, 100, 125, 125, 125, 150, 180, 216, 216, 216, 252, 294, 343, 343, 343, 392, 448, 512, 512, 512, 576, 648, 729, 729, 729, 810, 900, 1000

Offset: 0

N. J. A. Sloane

nonn

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

Cf. A008382, A008497. - R. J. Mathar, Apr 16 2010

List([0..55], n-> Int(n/5)*Int((n+1)/5)*Int((n+2)/5) ); # G. C. Greubel, Nov 08 2019

[&*[Floor((n+j)/5): j in [0..2]]: n in [0..55]]; // G. C. Greubel, Nov 08 2019

seq( mul(floor((n+j)/5), j=0..2), n=0..55); # G. C. Greubel, Nov 08 2019

LinearRecurrence[{1,0,0,0,3,-3,0,0,0,-3,3,0,0,0,1,-1}, {0,0,0,0,0,1,1,1, 2,4,8,8,8,12,18,27},60] (* or *) Table[Times@@Thread[Floor[(n +{0,1,2} )/5]],{n,0,60}] (* Harvey P. Dale, Apr 09 2018 *)
Product[Floor[(Range[55] +j-1)/5], {j,0,2}] (* G. C. Greubel, Nov 08 2019 *)

vector(56, n, prod(j=0,2, (n+j-1)\5) ) \\ G. C. Greubel, Nov 08 2019

[product(floor((n+j)/5) for j in (0..2)) for n in (0..55)] # G. C. Greubel, Nov 08 2019

From R. J. Mathar, Apr 16 2010: (Start)
a(n) = A002266(n)*A008497(n+1).
a(n) = a(n-1) +3*a(n-5) -3*a(n-6) -3*a(n-10) +3*a(n-11) +a(n-15) -a(n-16).
G.f.: x^5*(1+x+x^2)*(x^6-x^5+2*x^3-x+1)/((x^4+x^3+x^2+x+1)^3 *(x-1)^4). (End)

A008497 a(n) = floor(n/5)*floor((n+1)/5).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 4, 4, 4, 4, 6, 9, 9, 9, 9, 12, 16, 16, 16, 16, 20, 25, 25, 25, 25, 30, 36, 36, 36, 36, 42, 49, 49, 49, 49, 56, 64, 64, 64, 64, 72, 81, 81, 81, 81, 90, 100, 100, 100, 100, 110, 121, 121, 121, 121

Offset: 0

N. J. A. Sloane

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

Cf. A002266.

List([0..55], n-> Int(n/5)*Int((n+1)/5) ); # G. C. Greubel, Nov 08 2019

[&*[Floor((n+j)/5): j in [0..1]]: n in [0..55]]; // G. C. Greubel, Nov 08 2019

seq( mul(floor((n+j)/5), j=0..1), n=0..55); # G. C. Greubel, Nov 08 2019

Times@@@Partition[Floor[Range[0,60]/5],2,1] (* or *) LinearRecurrence[ {1,0,0,0,2,-2,0,0,0,-1,1},{0,0,0,0,0,1,1,1,1,2,4},60] (* Harvey P. Dale, Feb 01 2015 *)
Product[Floor[(Range[55] +j-1)/5], {j,0,1}] (* G. C. Greubel, Nov 08 2019 *)

a(n) = (n\5)*((n+1)\5); \\ Michel Marcus, Jan 06 2017

vector(56, n, prod(j=0,1, (n+j-1)\5) ) \\ G. C. Greubel, Nov 08 2019

[product(floor((n+j)/5) for j in (0..1)) for n in (0..55)] # G. C. Greubel, Nov 08 2019

From R. J. Mathar, Apr 16 2010: (Start)
a(n) = A002266(n)*A002266(n+1).
a(n)= a(n-1) + 2*a(n-5) - 2*a(n-6) - a(n-10) + a(n-11).
G.f.: x^5*(1+x^4)/ ((x^4+x^3+x^2+x+1)^2 * (1-x)^3). (End)
From Amiram Eldar, May 10 2025: (Start)
Sum_{n>=5} 1/a(n) = 2*Pi^2/3 + 1.
Sum_{n>=5} (-1)^(n+1)/a(n) = 2*log(2)-1. (End)

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A008496 a(n) = floor(n/5)floor((n+1)/5)floor((n+2)/5).