A002265 -id:A002265 - cp's OEIS frontend

A209007 T(n,k) = number of n-bead necklaces labeled with numbers -k..k not allowing reversal, with sum zero and first and second differences in -k..k.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 3, 1, 1, 2, 3, 5, 5, 2, 1, 2, 3, 10, 13, 10, 3, 1, 2, 7, 16, 31, 46, 23, 5, 1, 3, 7, 26, 71, 153, 157, 66, 7, 1, 3, 7, 38, 137, 409, 703, 608, 167, 10, 1, 3, 13, 55, 243, 923, 2313, 3393, 2245, 445, 13, 1, 3, 13, 75, 399, 1854, 6261, 13561, 16001

Offset: 1

R. H. Hardin, Mar 04 2012

Table starts
.1..1...1....1.....1.....1......1......1......1.......1.......1.......1......1
.1..1...1....2.....2.....2......2......3......3.......3.......3.......4......4
.1..1...3....3.....3.....7......7......7.....13......13......13......21.....21
.1..3...5...10....16....26.....38.....55.....75.....101.....131.....168....210
.1..5..13...31....71...137....243....399....619.....927....1329....1857...2525
.2.10..46..153...409...923...1854...3477...6034....9876...15590...23625..34577
.3.23.157..703..2313..6261..14701..31069..60429..109991..189627..312461.495497
.5.66.608.3393.13561.43228.116960.279721.607559.1221648.2305788.4128363

			Some solutions for n=6, k=6:
.-2...-2...-3...-1...-4...-2...-2...-2...-3...-2...-3...-2...-2...-2...-1...-3
.-2....2...-2...-1...-3...-2...-1....0...-3....0...-3...-1...-1...-1...-1...-3
..1....1....2....1....3...-1...-2....0....2...-2....3....1....2....1...-1....1
..2....0....3....0....3...-1....2...-1....2....0....4....2....2....1....1....0
..2....0....0....0....0....3....2....2....3....2....1...-1....0....0....0....3
.-1...-1....0....1....1....3....1....1...-1....2...-2....1...-1....1....2....2

R. H. Hardin, Table of n, a(n) for n = 1..184

Row 2 is A002265(n+4).

Row 3 is A002061(floor(n/3)+1).

Empirical for row n:
n=2: a(k) = a(k-1) + a(k-4) - a(k-5).
n=3: a(k) = a(k-1) + 2*a(k-3) - 2*a(k-4) - a(k-6) + a(k-7).
n=4: a(k) = 3*a(k-1) - 2*a(k-2) - 2*a(k-3) + 3*a(k-4) - a(k-5).

A010881 Simple periodic sequence: n mod 12.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Offset: 0

N. J. A. Sloane

The value of the rightmost digit in the base-12 representation of n. - Hieronymus Fischer, Jun 11 2007

			a(27) = 3 since 27 = 12*2+3.

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

Partial sums: A130490. Other related sequences A130481, A130482, A130483, A130484, A130485, A130486, A130487, A130488, A130489.

Mod[Range[0, 100], 12] (* Paolo Xausa, Feb 02 2024 *)

A010881(n)=n%12 \\ M. F. Hasler, Sep 25 2014

From Hieronymus Fischer, May 31 2007: (Start)
a(n) = n mod 12.
Complex representation: a(n) = (1/12)*(1-r^n)*Sum_{k=1..11} k*Product_{m=1..11, m<>k} (1-r^(n-m)) where r = exp(Pi/6*i) = (sqrt(3)+i)/2 and i = sqrt(-1).
Trigonometric representation: a(n) = (512/3)^2*(sin(n*Pi/12))^2*Sum_{k=1..11} k*Product_{m=1..11, m<>k} (sin((n-m)*Pi/12))^2.
G.f.: (Sum_{k=1..11} k*x^k)/(1-x^12).
G.f.: x*(11*x^12-12*x^11+1)/((1-x^12)*(1-x)^2). (End)
From Hieronymus Fischer, Jun 11 2007: (Start)
a(n) = (n mod 2)+2*(floor(n/2) mod 6) = A000035(n)+2*A010875(A004526(n)).
a(n) = (n mod 3)+3*(floor(n/3) mod 4) = A010872(n)+3*A010873(A002264(n)).
a(n) = (n mod 4)+4*(floor(n/4) mod 3) = A010873(n)+4*A010872(A002265(n)).
a(n) = (n mod 6)+6*(floor(n/6) mod 2) = A010875(n)+6*A000035(A152467(n)).
a(n) = (n mod 2)+2*(floor(n/2) mod 2)+4*(floor(n/4) mod 3) = A000035(n)+2*A000035(A004526(n))+4*A010872(A002265(n)). (End)
a(A001248(k) + 17) = 6 for k>2. - Reinhard Zumkeller, May 12 2010
a(n) = A034326(n+1)-1. - M. F. Hasler, Sep 25 2014

A080239 Antidiagonal sums of triangle A035317.

Original entry on oeis.org

1, 1, 2, 3, 6, 9, 15, 24, 40, 64, 104, 168, 273, 441, 714, 1155, 1870, 3025, 4895, 7920, 12816, 20736, 33552, 54288, 87841, 142129, 229970, 372099, 602070, 974169, 1576239, 2550408, 4126648, 6677056, 10803704, 17480760, 28284465, 45765225, 74049690

Offset: 1

Paul Barry, Feb 11 2003

Convolution of Fibonacci sequence with sequence (1, 0, 0, 0, 1, 0, 0, 0, 1, ...).
There is an interesting relation between a(n) and the Fibonacci sequence f(n). Sqrt(a(4n-2)) = f(2n). By using this fact we can calculate the value of a(n) by the following (1),(2),(3),(4) and (5). (1) a(1) = 1. (2) If n = 2 (mod 4), then a(n) = f((n+2)/2)^2. (3) If n = 3 (mod 4), then a(n) = (f((n+5)/2)^2-2f((n+1)/2)^2-1)/3. (4) If n = 0 (mod 4), then a(n) = (f((n+4)/2)^2+f(n/2)^2-1)/3. (5) If n = 1 (mod 4), then a(n) = (2f((n+3)/2)^2-f((n-1)/2)^2+1)/3. - Hiroshi Matsui and Ryohei Miyadera, Aug 08 2006
Sequences of the form s(0)=a, s(1)=b, s(n) = s(n-1) + s(n-2) + k if n mod m = p, else s(n) = s(n-1) + s(n-2) will have a form fib(n-1)*a + fib(n)*b + P(x)*k. a(n) is the P(x) sequence for m=4...s(n) = fib(n)*a + fib(n-1)*b + a(n-4-p)*k. - Gary Detlefs, Dec 05 2010
A different formula for a(n) as a function of the Fibonacci numbers f(n) may be conjectured. The pattern is of the form a(n) = f(p)*f(p-q) - 1 if n mod 4 = 3, else f(p)*f(p-q) where p = 2,2,4,4,4,4,6,6,6,6,8,8,8,8... and q = 0,1,3,2,0,1,3,2,0,1,3,2... p(n) = 2 * A002265(n+4) = 2*(floor((n+3)/2) - floor((n+3)/4)) (see comment by Jonathan Vos Post at A002265). A general formula for sequences having period 4 with terms a,b,c,d is given in A121262 (the discrete Fourier transform, as for all periodic sequences) and is a function of t(n)= 1/4*(2*cos(n*Pi/2) + 1 + (-1)^n). r4(a,b,c,d,n) = a*t(n+3) + b*t(n+2) + c*t(n+1) + d*t(n). This same formula may be used to subtract the 1 at n mod 4 = 3. a(n) = f(p(n))*f(p(n) - r4(1,0,3,2,n)) - r4(0,0,1,0,n). - Gary Detlefs, Dec 09 2010
This sequence is the sequence B4,1 on p. 34 of "Pascal-like triangles and Fibonacci-like sequences" in the references. In this article the authors treat more general sequences that have this sequence as an example. - Hiroshi Matsui and Ryohei Miyadera, Apr 11 2014
It is easy to see that a(n) = a(n-4) + f(n), where f(n) is the Fibonacci sequence. By using this repeatedly we have for a natural number m
a(4m) =a(4) + f(4m) + f(4m-4) + ... + f(8),
a(4m+1) = a(1) + f(4m) + f(4m-4) + ... + f(5),
a(4m+2) = a(2) + f(4m) + f(4m-4) + ... + f(6) and
a(4m+3) = a(3) + f(4m) + f(4m-4) + ... + f(7).
- Wataru Takeshita and Ryohei Miyadera, Apr 11 2014
a(n-1) counts partially ordered partitions of (n-1) into (1,2,3,4) where the position (order) of 2's is unimportant. E.g., a(5)=6 (n-1)=4 These are (4),(31),(13),(22),(211,121,112=one),(1111). - David Neil McGrath, May 12 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
H. Matsui et al., Problem B-1019, Fibonacci Quarterly, Vol. 45, Number 2; 2007; p. 182.
H. Matsui and R. Miyadera et al., Pascal-like triangles and Fibonacci-like sequences, The Mathematical Gazette, Vol. 94, Number 529; March 2010; pp. 27-41.
Index entries for linear recurrences with constant coefficients, signature (1,1,0,1,-1,-1).

Cf. A000045, A004695, A026636, A026647, A035317.

List([1..40], n-> Sum([0..Int((n-1)/4)], k-> Fibonacci(n-4*k) )); # G. C. Greubel, Jul 13 2019

a080239 n = a080239_list !! (n-1)
a080239_list = 1 : 1 : zipWith (+)
   (tail a011765_list) (zipWith (+) a080239_list $ tail a080239_list)
-- Reinhard Zumkeller, Jan 06 2012

I:=[1,1,2,3,6,9]; [n le 6 select I[n] else Self(n-1)+Self(n-2)+Self(n-4)-Self(n-5)-Self(n-6): n in [1..50]]; // Vincenzo Librandi, Jun 07 2015

f:=proc(n) option remember; local t1; if n <= 2 then RETURN(1); fi: if n mod 4 = 1 then t1:=1 else t1:=0; fi: f(n-1)+f(n-2)+t1; end; [seq(f(n), n=1..100)]; # N. J. A. Sloane, May 25 2008
with(combinat): f:=n-> fibonacci(n): p:=n-> 2*(floor((n+3)/2)-floor((n+3)/4)): t:=n-> 1/4*(2*cos(n*Pi/2)+1+(-1)^n): r4:=(a,b,c,d,n)-> a*t(n+3)+b*t(n+2)+c*t(n+1)+d*t(n): seq(f(p(n))*f(p(n)-r4(1,0,3,2,n))-r4(0,0,1,0,n), n = 1..33); # Gary Detlefs, Dec 09 2010
with(combinat): a:=proc(n); add(fibonacci(n-4*k),k=0..floor((n-1)/4)) end: seq(a(n), n = 1..33); # Johannes W. Meijer, Apr 19 2012

(*f[n] is the Fibonacci sequence and a[n] is the sequence of A080239*) f[n_]:= f[n] =f[n-1] +f[n-2]; f[1]=1; f[2]=1; a[n_]:= Which[n==1, 1, Mod[n, 4]==2, f[(n+2)/2]^2, Mod[n, 4]==3, (f[(n+5)/2]^2 - 2f[(n + 1)/2]^2 -1)/3, Mod[n, 4]==0, (f[(n+4)/2]^2 + f[n/2]^2 -1)/3, Mod[n, 4] == 1, (2f[(n+3)/2]^2 -f[(n-1)/2]^2 +1)/3] (* Hiroshi Matsui and Ryohei Miyadera, Aug 08 2006 *)
a=0; b=0; lst={a,b}; Do[z=a+b+1; AppendTo[lst,z]; a=b; b=z; z=a+b; AppendTo[lst,z]; a=b; b=z; z=a+b; AppendTo[lst,z]; a=b; b=z; z=a+b; AppendTo[lst,z]; a=b; b=z,{n,4!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 16 2010 *)
(* Let f[n] be the Fibonacci sequence and a2[n] the sequence A080239 expressed by another formula discovered by Wataru Takeshita and Ryohei Miyadera *)
f=Fibonacci; a2[n_]:= Block[{m, s}, s = Mod[n, 4]; m = (n-s)/4;
Which[n==1, 1, n==2, 1, n==3, 2, s==0, 3 + Sum[f[4 i], {i, 2, m}], s == 1, 1 + Sum[f[4i+1], {i, 1, m}], s==2, 1 + Sum[f[4i+2], {i, 1, m}], s == 3, 2 + Sum[f[4i+3], {i, 1, m}]]]; Table[a2[n], {n, 1, 40}] (* Ryohei Miyadera, Apr 11 2014, minor update by Jean-François Alcover, Apr 29 2014 *)
LinearRecurrence[{1, 1, 0, 1, -1, -1}, {1, 1, 2, 3, 6, 9}, 41] (* Vincenzo Librandi, Jun 07 2015 *)

vector(40, n, f=fibonacci; sum(k=0,((n-1)\4), f(n-4*k))) \\ G. C. Greubel, Jul 13 2019

[sum(fibonacci(n-4*k) for k in (0..floor((n-1)/4))) for n in (1..40)] # G. C. Greubel, Jul 13 2019

G.f.: x/((1-x^4)(1 - x - x^2)) = x/(1 - x - x^2 - x^4 + x^5 + x^6).
a(n) = a(n-1) + a(n-2) + a(n-4) - a(n-5) - a(n-6).
a(n) = Sum_{j=0..floor(n/2)} Sum_{k=0..floor((n-j)/2)} binomial(n-j-2k, j-2k) for n>=0.
Another recurrence is given in the Maple code.
If n mod 4 = 1 then a(n) = a(n-1) + a(n-2) + 1, else a(n)= a(n-1) + a(n-2). - Gary Detlefs, Dec 05 2010
a(4n) = A058038(n) = Fibonacci(2n+2)*Fibonacci(2n).
a(4n+1) = A081016(n) = Fibonacci(2n+2)*Fibonacci(2n+1).
a(4n+2) = A049682(n+1) = Fibonacci(2n+2)^2.
a(4n+3) = A081018(n+1) = Fibonacci(2n+2)*Fibonacci(2n+3).
a(n) = 8*a(n-4) - 8*a(n-8) + a(n-12), n>12. - Gary Detlefs, Dec 10 2010
a(n+1) = a(n) + a(n-1) + A011765(n+1). - Reinhard Zumkeller, Jan 06 2012
a(n) = Sum_{k=0..floor((n-1)/4)} Fibonacci(n-4*k). - Johannes W. Meijer, Apr 19 2012

A180969 Array read by antidiagonals: a(k,n) = natural numbers each repeated 2^k times.

Original entry on oeis.org

0, 1, 0, 2, 0, 0, 3, 1, 0, 0, 4, 1, 0, 0, 0, 5, 2, 0, 0, 0, 0, 6, 2, 1, 0, 0, 0, 0, 7, 3, 1, 0, 0, 0, 0, 0, 8, 3, 1, 0, 0, 0, 0, 0, 0, 9, 4, 1, 0, 0, 0, 0, 0, 0, 0, 10, 4, 2, 0, 0, 0, 0, 0, 0, 0, 0, 11, 5, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 12, 5, 2, 1, 0, 0, 0, 0

Offset: 0

Adriano Caroli, Nov 17 2010

Generalization of P. Barry's (2003) formula in A004526.

			Sequence gives the antidiagonals of the infinite square array with rows indexed by k and columns indexed by n:
0  1  2  3  4  5  6  7  8  9 10 11 12 13...
0  0  1  1  2  2  3  3  4  4   5   5   6...
0  0  0  0  1  1  1  1  2  2   2   2   3...
0  0  0  0  0  0  0  0  1  1   1   1   1...
0  0  0  0  0  0  0  0  0  0   0   0   0...
...........................................

Danny Rorabaugh, Table of n, a(n) for n = 0..10000

Cf. A001477, A004526, A002265, A132292.

function v=A180969(k,n,q)
% n=vector of natural numbers 0,1,...,n
% v=vector in which each n is repeated k times
% q=q-th term of v from where to start
if k==0;v=n+q;return;end
v=A180969(k-1,n,q);
% calculate repetition only if v terms are not all zeros
if any(v); v=v/2+((-1).^v-1)/4;end
% Adriano Caroli, Nov 28 2010

Table[Floor[#/2^k] &[n - k], {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 30 2019 *)

matrix(10, 20, k, n, k--; n--; floor(n/2^k)) \\ Michel Marcus, Sep 09 2019

a(k,n) = (n/2^k) + Sum_{j=1..k} ((-1)^a(j-1,n) - 1)/2^(k-j+2).

a(k,n) = floor(n/2^k). - Adriano Caroli, Sep 30 2019

A029581 Numbers in which all digits are composite.

Original entry on oeis.org

4, 6, 8, 9, 44, 46, 48, 49, 64, 66, 68, 69, 84, 86, 88, 89, 94, 96, 98, 99, 444, 446, 448, 449, 464, 466, 468, 469, 484, 486, 488, 489, 494, 496, 498, 499, 644, 646, 648, 649, 664, 666, 668, 669, 684, 686, 688, 689, 694, 696, 698, 699, 844, 846, 848

Offset: 1

N. J. A. Sloane

If n is represented as a zerofree base-4 number (see A084544) according to n=d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n) = Sum_{j=0..m} c(d(j))*10^j, where c(k)=4,6,8,9 for k=1..4. - Hieronymus Fischer, May 30 2012

			From _Hieronymus Fischer_, May 30 2012: (Start)
a(1000) = 88649.
a(10^4) = 6468989
a(10^5) = 449466489. (End)

Hieronymus Fischer, Table of n, a(n) for n = 1..10000
Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
Index entries for 10-automatic sequences.

Cf. A002808, A001744, A046034, A084544, A084984, A017042, A001743, A014261, A014263, A202267, A202268.

[n: n in [1..1000] | Set(Intseq(n)) subset [4, 6, 8, 9]]; // Vincenzo Librandi, Dec 17 2018

Table[FromDigits/@Tuples[{4, 6, 8, 9}, n], {n, 3}] // Flatten (* Vincenzo Librandi, Dec 17 2018 *)

From Hieronymus Fischer, May 30 and Jun 25 2012: (Start)
a(n) = Sum_{j=0..m-1} (2*b(j) mod 8 + 4 + floor(b(j)/4) - floor((b(j)+1)/4))*10^j, where m = floor(log_4(3*n+1)), b(j) = floor((3*n+1-4^m)/(3*4^j)).
Also: a(n) = Sum_{j=0..m-1} (A010877(2*b(j)) + 4 + A002265(b(j)) - A002265(b(j)+1))*10^j.
Special values:
a(1*(4^n-1)/3) = 4*(10^n-1)/9.
a(2*(4^n-1)/3) = 2*(10^n-1)/3.
a(3*(4^n-1)/3) = 8*(10^n-1)/9.
a(4*(4^n-1)/3) = 10^n-1.
a(n) < 4*(10^log_4(3*n+1)-1)/9, equality holds for n=(4^k-1)/3, k > 0.
a(n) < 4*A084544(n), equality holds iff all digits of A084544(n) are 1.
a(n) > 2*A084544(n).
Lower and upper limits:
lim inf a(n)/10^log_4(n) = 1/10*10^log_4(3) = 0.62127870, for n --> inf.
lim sup a(n)/10^log_4(n) = 4/9*10^log_4(3) = 2.756123868970, for n --> inf.
where 10^log_4(n) = n^1.66096404744...
G.f.: g(x) = (x^(1/3)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(4/3)*(1-z(j))*(4 + 6z(j) + 8*z(j)^2 + 9*z(j)^3)/(1-z(j)^4), where z(j) = x^4^j.
Also: g(x) = (1/(1-x))*(4*h_(4,0)(x) + 2*h_(4,1)(x) + 2*h_(4,2)(x) + h_(4,3)(x) - 9*h_(4,4)(x)), where h_(4,k)(x) = Sum_{j>=0} 10^j*x^((4^(j+1)-1)/3)*(x^(k*4^j)/(1-x^4^(j+1)). (End)
Sum_{n>=1} 1/a(n) = 1.039691381254753739202528087006945643166147087095114911673083135126969046250... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024

Offset corrected by Arkadiusz Wesolowski, Oct 03 2011

A010883 Simple periodic sequence: repeat 1,2,3,4.

Original entry on oeis.org

1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1

Offset: 0

N. J. A. Sloane

Partial sums are given by A130482(n) + n + 1. - Hieronymus Fischer, Jun 08 2007

1234/9999 = 0.123412341234... - Eric Desbiaux, Nov 03 2008

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

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

Cf. A177037 (decimal expansion of (9+2*sqrt(39))/15). - Klaus Brockhaus, May 01 2010

PadRight[{},120,{1,2,3,4}] (* Harvey P. Dale, Aug 02 2016 *)

a(n)=(n-1)%4+1 \\ Charles R Greathouse IV, Jun 11 2015

def A010883(n): return 1 + (n & 3) # Chai Wah Wu, May 25 2022

a(n) = 1 + (n mod 4). - Paolo P. Lava, Nov 21 2006
From Hieronymus Fischer, Jun 08 2007: (Start)
a(n) = A010873(n) + 1.
Also a(n) = (1/2)*(5 - (-1)^n - 2*(-1)^((2*n - 1 + (-1)^n)/4)).
G.f.: g(x) = (4*x^3 + 3*x^2 + 2*x + 1)/(1 - x^4) = (4*x^5 - 5*x^4 + 1)/((1 - x^4)*(1-x)^2). (End)
a(n) = 5/2 - cos(Pi*n/2) - sin(Pi*n/2) - (-1)^n/2. - R. J. Mathar, Oct 08 2011

A129756 Repetitions of odd numbers four times.

Original entry on oeis.org

1, 1, 1, 1, 3, 3, 3, 3, 5, 5, 5, 5, 7, 7, 7, 7, 9, 9, 9, 9, 11, 11, 11, 11, 13, 13, 13, 13, 15, 15, 15, 15, 17, 17, 17, 17, 19, 19, 19, 19, 21, 21, 21, 21, 23, 23, 23, 23, 25, 25, 25, 25, 27, 27, 27, 27, 29, 29, 29, 29, 31, 31, 31, 31, 33, 33, 33, 33, 35, 35, 35, 35, 37, 37, 37, 37

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, May 15 2007

Conjecture: number of roots of P(x) = x^n - x^(n-1) - x^(n-2) - ... - x - 1 in the right half-plane. - Michel Lagneau, Apr 09 2013

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

Cf. A002265, A083219.

[1+2*Floor(n/4): n in [0..100]]; // Bruno Berselli, Jul 26 2014

I:=[1,1,1,1,3,3,3,3,5]; [n le 9 select I[n] else Self(n-1)+Self(n-4)-Self(n-5): n in [1..100]]; // Vincenzo Librandi, Jul 25 2014

Table[1 + 2 Floor[n/4], {n, 0, 100}] (* Bruno Berselli, Jul 26 2014 *)
CoefficientList[Series[(1 + x^4)/(-1 + x)^2/(1 + x)/(x^2 + 1), {x, 0, 100}], x] (* Vincenzo Librandi, Jul 26 2014 *)

def A129756(n): return (n>>1)|1 # Chai Wah Wu, Jan 31 2023

a(n) = (Sum_{k=0..n} (k+1)*cos((n-k)*Pi/2)) + (1/4)*(2*cos(n*Pi/2) + 1 + (-1)^n) - 1, with n >= 0.
a(n) = 1 + 2*floor(n/4) = 1 + 2*A002265(n). - R. J. Mathar, Jun 10 2007
G.f.: (1+x^4)/((-1+x)^2*(1+x)*(x^2+1)). - R. J. Mathar, Nov 18 2007
a(n) = -1 + Sum_{k=0..n} ((1/12)*(-5*(k mod 4) + ((k+1) mod 4) + ((k+2) mod 4) + 7*((k+3) mod 4))). - Paolo P. Lava, Aug 21 2009
a(n) = n - A083219(n). - Michel Lagneau, Apr 09 2013
a(n) = (2*n + 1 + 2*cos(n*Pi/2) + cos(n*Pi) + 2*sin(n*Pi/2))/4. - Wesley Ivan Hurt, Oct 02 2017
From Stefano Spezia, May 26 2021: (Start)
E.g.f.: (cos(x) + cosh(x) + sin(x) + x*(cosh(x) + sinh(x)))/2.
a(n) = a(n-1) + a(n-4) - a(n-5) for n > 4. (End)

A132292 Integers repeated 8 times: a(n) = floor((n-1)/8).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10

Offset: 1

Mohammad K. Azarian, Nov 06 2007

Also floor((n^8-1)/(8*n^7)).

Cf. A004526, A002264, A002265, A002266, A054895.

[(n - 1) div 8 : n in [1..90]]; // Vincenzo Librandi, Oct 05 2018

A132292:=n->floor((n-1)/8); seq(A132292(n), n=1..100); # Wesley Ivan Hurt, Feb 27 2014

Table[Floor[(n-1)/8], {n, 100}] (* Wesley Ivan Hurt, Feb 27 2014 *)
Table[PadRight[{},8,n],{n,0,10}]//Flatten (* Harvey P. Dale, Apr 13 2020 *)

a(n)=(n-1)\8 \\ Charles R Greathouse IV, Jun 18 2013

def A132292(n): return n-1>>3 # Chai Wah Wu, Jul 27 2022

Also, a(n) = floor((n^8-n^7)/(8n^7-7n^6)). - Mohammad K. Azarian, Nov 18 2007
a(n) = A180969(3,n).
a(n) = (r - 8 + 4*sin(r*Pi/8))/16 where r = 2*n - 1 - 2*cos(n*Pi/2) - cos(n*Pi) + 2*sin(n*Pi/2). - Wesley Ivan Hurt, Oct 04 2018

Offset corrected by Mohammad K. Azarian, Nov 20 2008

New name from Wesley Ivan Hurt, Jun 17 2013

A130909 Simple periodic sequence (n mod 16).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Offset: 0

Hieronymus Fischer, Jun 11 2007

The value of the rightmost digit in the base-16 representation of n. Also, the equivalent value of the two rightmost digits in the base-4 representation of n. Also, the equivalent value of the four rightmost digits in the base-2 representation of n.

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

Cf. partial sums A130910. Other related sequences A010872, A010873, A130481, A130482, A130483, A130486.

See A010877 for a general formula in terms of powers of -1 (for period 2^k).

a(n)=n%16 \\ Charles R Greathouse IV, Jul 13 2016

def A130909(n): return n&15 # Chai Wah Wu, Jan 18 2023

a(n) = n mod 16 = n-16*floor(n/16).
G.f.: g(x) = (Sum_{k=1..15} k*x^k)/(1-x^16).
G.f.: g(x) = x(15x^16-16x^15+1)/((1-x^16)(1-x)^2).
a(n) = A000035(n) + 2*A010877(A004526(n)).
a(n) = A010873(n) + 4*A010873(A002265(n)).
a(n) = A010877(n) + 8*A000035(floor(n/8)).
a(n) = (1/2)*(15 - ( - 1)^n - 2*( - 1)^(b/4) - 4*( - 1)^((b - 2 + 2*( - 1)^(b/4))/8) - 8*( - 1)^((b - 6 + ( - 1)^n + 2*( - 1)^(b/4) + 4*( - 1)^((b - 2 + 2*( - 1)^(b/4))/8))/16)) where b = 2n - 1 + ( - 1)^n.
a(n) = n mod 2+2*(floor(n/2)mod 2)+4*(floor(n/4)mod 2)+8*(floor(n/8)mod 2).
a(n) = (1/2)*(15-(-1)^n-2*(-1)^floor(n/2)-4*(-1)^floor(n/4)-8*(-1)^floor(n/= 8)).
Complex representation: a(n) = (1/16)*(1-r^n)*sum{1<=k<16, k*product{1<=m<16,m<>k, (1-r^(n-m))}} where r=exp(Pi/8*i)=(sqrt(2+sqrt(2))+i*sqrt(2-sqrt(2)))/2 and i=sqrt(-1).
Trigonometric representation: a(n) = 2^22*(sin(n*Pi/16))^2*sum{1<=k<16, k*product{1<=m<16,m<>k, (sin((n-m)*Pi/16))^2}}.
a(n) = (1/2)*(15-(-1)^p(0,n)-2*(-1)^p(1,n)-4*(-1)^p(2,n)-8*(-1)^p(3,n)) where p(k,n) is defined recursively by p(0,n)=n, p(k,n)=1/4*(2*p(k-1,n)-1+(-1)^p(k-1,n)).

A025767 Expansion of 1/((1-x)(1-x^3)(1-x^4)).

Original entry on oeis.org

1, 1, 1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 15, 17, 18, 20, 22, 24, 26, 28, 30, 33, 35, 37, 40, 43, 45, 48, 51, 54, 57, 60, 63, 67, 70, 73, 77, 81, 84, 88, 92, 96, 100, 104, 108, 113, 117, 121, 126, 131, 135, 140, 145, 150, 155, 160, 165, 171, 176, 181, 187, 193, 198

Offset: 0

N. J. A. Sloane

Apply the Riordan array (1/(1-x^4),x) to floor((n+3)/3). - Paul Barry, Jan 20 2006
Number of partitions of n into parts 1, 3, and 4. - David Neil McGrath, Aug 30 2014
Also, a(n-4) is equal to the number of partitions mu of n of length 3 such that mu_1-mu_2 is even and mu_2-mu_3 is odd or vice versa (see below example). - John M. Campbell, Jan 29 2016
With four 0's prepended and offset 0, a(n) is the number of partitions of n into four parts whose 2nd and 3rd largest parts are equal. - Wesley Ivan Hurt, Jan 05 2021

			The a(4)=3 partitions of 4 into parts 1, 3, and 4 are (4), (3,1), and (1,1,1,1). - _David Neil McGrath_, Aug 30 2014
From _John M. Campbell_, Jan 29 2016: (Start)
Letting n=12, there are a(n-4)=a(8)=6 partitions mu of n=12 of length 3 such that mu_1-mu_2 is even and mu_2-mu_3 is odd or vice versa:
(10,1,1) |- n
(8,3,1) |- n
(7,3,2) |- n
(6,5,1) |- n
(6,3,3) |- n
(5,5,2) |- n
(End)

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

A008621(n) = A002265(n+4) = a(n) - a(n-3).

[Floor(n^2/24 + n/3 + 1): n in [0..70]]; // Vincenzo Librandi, Aug 31 2014

A056594 := proc(n) op(1+(n mod 4),[1,0,-1,0]) ; end proc:
A061347 := proc(n) op(1+(n mod 3),[-2,1,1]) ; end proc:
A025767 := proc(n) n^2/24+n/3+83/144+(-1)^n/16 +A061347(n+1)/9 +A056594(n)/4 ; end proc: # R. J. Mathar, Mar 31 2011

Table[Floor[n^2/24 + n/3 + 1], {n, 0, 60}] (* Vincenzo Librandi, Aug 31 2014 *)

PARI
```
a(n)=if(n<0,0,(n^2+8*n)\24+1)
```

{a(n) = round( ((n + 4)^2 - 1) / 24 )}; /* Michael Somos, Nov 09 2007 */

Vec(1/((1-x)*(1-x^3)*(1-x^4)) + O(x^80)) \\ Michel Marcus, Jan 29 2016

G.f.: 1/((1-x)*(1-x^3)*(1-x^4)).
a(n) = floor(n^2/24+n/3+1).
a(n) = Sum_{k=0..floor(n/4)} floor((n-4*k+3)/3). - Paul Barry, Jan 20 2006
Euler transform of length 4 sequence [1, 0, 1, 1]. - Michael Somos, Nov 09 2007
a(n) = a(-8 - n) for all n in Z. - Michael Somos, Nov 09 2007
a(n) = n^2/24 + n/3 + 83/144 + (-1)^n/16 + A061347(n+1)/9 + A056594(n)/4. - R. J. Mathar, Mar 31 2011
a(n) = a(n-1)+a(n-3)-a(n-5)-a(n-7)+a(n-8). - David Neil McGrath, Aug 30 2014
a(n) = Sum_{k=1..floor((n+4)/4)} Sum_{j=k..floor((n+4-k)/3)} Sum_{i=j..floor((n+4-j-k)/2)} [j = i], where [ ] is the Iverson bracket. - Wesley Ivan Hurt, Jan 17 2021
a(n)-a(n-1) = A008679(n). - R. J. Mathar, Jun 23 2021
a(n)-a(n-4) = A008620(n). - R. J. Mathar, Jun 23 2021

cp's OEIS Frontend

A209007 T(n,k) = number of n-bead necklaces labeled with numbers -k..k not allowing reversal, with sum zero and first and second differences in -k..k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A010881 Simple periodic sequence: n mod 12.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A080239 Antidiagonal sums of triangle A035317.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

Sage

Formula

A180969 Array read by antidiagonals: a(k,n) = natural numbers each repeated 2^k times.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

MATLAB

Mathematica

PARI

Formula

A029581 Numbers in which all digits are composite.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Formula

Extensions

A010883 Simple periodic sequence: repeat 1,2,3,4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A129756 Repetitions of odd numbers four times.

Views

Author

Keywords

A025767 Expansion of 1/((1-x)(1-x^3)(1-x^4)).