A011765 -id:A011765 - cp's OEIS frontend

A002265 Nonnegative integers repeated 4 times.

0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 19

Offset: 0

N. J. A. Sloane

For n>=1 and i=sqrt(-1) let F(n) the n X n matrix of the Discrete Fourier Transform (DFT) whose element (j,k) equals exp(-2*Pi*i*(j-1)*(k-1)/n)/sqrt(n). The multiplicities of the four eigenvalues 1, i, -1, -i of F(n) are a(n+4), a(n-1), a(n+2), a(n+1), hence a(n+4) + a(n-1) + a(n+2) + a(n+1) = n for n>=1. E.g., the multiplicities of the eigenvalues 1, i, -1, -i of the DFT-matrix F(4) are a(8)=2, a(3)=0, a(6)=1, a(5)=1, summing up to 4. - Franz Vrabec, Jan 21 2005
Complement of A010873, since A010873(n)+4*a(n)=n. - Hieronymus Fischer, Jun 01 2007
For even values of n, a(n) gives the number of partitions of n into exactly two parts with both parts even. - Wesley Ivan Hurt, Feb 06 2013
a(n-4) counts number of partitions of (n) into parts 1 and 4. For example a(11) = 3 with partitions (44111), (41111111), (11111111111). - David Neil McGrath, Dec 04 2014
a(n-4) counts walks (closed) on the graph G(1-vertex; 1-loop, 4-loop) where order of loops is unimportant. - David Neil McGrath, Dec 04 2014
Number of partitions of n into 4 parts whose smallest 3 parts are equal. - Wesley Ivan Hurt, Jan 17 2021

V. Cizek, Discrete Fourier Transforms and their Applications, Adam Hilger, Bristol 1986, p. 61.

Todd Silvestri, Table of n, a(n) for n = 0..999
J. H. McClellan and T. W. Parks, Eigenvalue and Eigenvector Decomposition of the Discrete Fourier Transform, IEEE Trans. Audio and Electroacoust., Vol. AU-20, No. 1, March 1972, pp. 66-74.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A008615, A008621, A249356.
Zero followed by partial sums of A011765.
Partial sums: A130519. Other related sequences: A004526, A010872, A010873, A010874.
Third row of A180969.

[Floor(n/4): n in [0..80]]; // Vincenzo Librandi, Oct 28 2014

A002265:=n->floor(n/4); seq(A002265(n), n=0..100); # Wesley Ivan Hurt, Dec 10 2013

Table[Floor[n/4], {n, 0, 100}] (* Wesley Ivan Hurt, Dec 10 2013 *)
Table[{n,n,n,n},{n,0,20}]//Flatten (* Harvey P. Dale, Aug 08 2020 *)

a(n)=n\4 \\ Charles R Greathouse IV, Dec 10 2013

def A002265(n): return n>>2 # Chai Wah Wu, Jul 27 2022

[floor(n/4) for n in range(0,84)] # Zerinvary Lajos, Dec 02 2009

a(n) = floor(n/4), n>=0;
G.f.: (x^4)/((1-x)*(1-x^4)).
a(n) = (2*n-(3-(-1)^n-2*(-1)^floor(n/2)))/8; also a(n) = (2*n-(3-(-1)^n-2*sin(Pi/4*(2*n+1+(-1)^n))))/8 = (n-A010873(n))/4. - Hieronymus Fischer, May 29 2007
a(n) = (1/4)*(n-(3-(-1)^n-2*(-1)^((2*n-1+(-1)^n)/4))/2). - Hieronymus Fischer, Jul 04 2007
a(n) = floor((n^4-1)/4*n^3) (n>=1); a(n) = floor((n^4-n^3)/(4*n^3-3*n^2)) (n>=1). - Mohammad K. Azarian, Nov 08 2007 and Aug 01 2009
For n>=4, a(n) = floor( log_4( 4^a(n-1) + 4^a(n-2) + 4^a(n-3) + 4^a(n-4) ) ). - Vladimir Shevelev, Jun 22 2010
a(n) = A180969(2,n). - Adriano Caroli, Nov 26 2010
a(n) = A173562(n)-A000290(n); a(n+2) = A035608(n)-A173562(n). - Reinhard Zumkeller, Feb 21 2010
a(n+1) = A140201(n) - A057353(n+1). - Reinhard Zumkeller, Feb 26 2011
a(n) = ceiling((n-3)/4), n >= 0. - Wesley Ivan Hurt, Jun 01 2013
a(n) = (2*n + (-1)^n + 2*sin(Pi*n/2) + 2*cos(Pi*n/2) - 3)/8. - Todd Silvestri, Oct 27 2014
E.g.f.: (x/4 - 3/8)*exp(x) + exp(-x)/8 + (sin(x)+cos(x))/4. - Robert Israel, Oct 30 2014
a(n) = a(n-1) + a(n-4) - a(n-5) with initial values a(3)=0, a(4)=1, a(5)=1, a(6)=1, a(7)=1. - David Neil McGrath, Dec 04 2014
a(n) = A004526(A004526(n)). - Bruno Berselli, Jul 01 2016
From Guenther Schrack, May 03 2019: (Start)
a(n) = (2*n - 3 + (-1)^n + 2*(-1)^(n*(n-1)/2))/8.
a(n) = a(n-4) + 1, a(k)=0 for k=0,1,2,3, for n > 3. (End)

A121262 The characteristic function of the multiples of four.

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Aug 23 2006, Aug 30 2007

Period 4: repeat [1, 0, 0, 0].
a(n) is also the number of partitions of n where each part is four (Since the empty partition has no parts, a(0) = 1). Hence a(n) is also the number of 2-regular graphs on n vertices such that each component has girth exactly four. - Jason Kimberley, Oct 01 2011
This sequence is the Euler transformation of A185014. - Jason Kimberley, Oct 01 2011
Number of permutations satisfying -k <= p(i) - i <= r and p(i)-i not in I, i = 1..n, with k = 1, r = 3, I = {0, 1, 2}. - Vladimir Baltic, Mar 07 2012

G. Balzarotti and P. P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 82.

Antti Karttunen, Table of n, a(n) for n = 0..65537
Vladimir Baltic, On the number of certain types of strongly restricted permutations, Applicable Analysis and Discrete Mathematics 4 (2010), 119-135
Steve Chow, 0,0,0,1,0,0,0,1 (deriving an explicit formula for the sequence) :YouTube Video, 2017.
Index entries for characteristic functions
Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

A011765 is another version of the same sequence.
Characteristic function of multiples of g: A000007 (g=0), A000012 (g=1), A059841 (g=2), A079978 (g=3), this sequence (g=4), A079998 (g=5), A079979 (g=6), A082784 (g=7). - Jason Kimberley, Oct 14 2011
Cf. A010873, A166486, A185014.

a121262 = (0 ^) . flip mod 4  -- Reinhard Zumkeller, Mar 04 2015
a121262_list = cycle [1,0,0,0]  -- Reinhard Zumkeller, Jan 06 2012

&cat [[1, 0, 0, 0]^^30]; // Wesley Ivan Hurt, Jul 07 2016

seq(op([1, 0, 0, 0]), n=0..50); # Wesley Ivan Hurt, Jul 07 2016

Table[Boole[IntegerQ[n/4]], {n,0,127}] (* Alonso del Arte, Jul 14 2013 *)

a(n)=!(n%4) \\ Charles R Greathouse IV, Oct 25 2012

a(n) = (1/4)*(2*cos(n*Pi/2) + 1 + (-1)^n).
Additive with a(p^e) = 1 if p = 2 and e > 1, 0 otherwise.
Sequence shifted right by 2 is additive with a(p^e) = 1 if p = 2 and e = 1, 0 otherwise.
a(n) = 1 - (C(n + 1, n + (-1)^(n+1)) mod 2).
a(n) = 0^(n mod 4). - Reinhard Zumkeller, Sep 30 2008
a(n) = !(n%4). - Jaume Oliver Lafont, Mar 01 2009
a(n) = (1/4)*(1 + I^n + (-1)^n + (-I)^n). - Paolo P. Lava, May 04 2010
a(n) = ((n-1)^k mod 4 - (n-1)^(k-1) mod 4)/2, k > 2. - Gary Detlefs, Feb 21 2011
a(n) = floor(1/2*cos(n*Pi/2) + 1/2). - Gary Detlefs, May 16 2011
G.f.: 1/(1 - x^4); a(n) = (1 + (-1)^n)*(1 + i^((n-1)*n))/4, where i = sqrt(-1). - Bruno Berselli, Sep 28 2011
a(n) = floor(((n+3) mod 4)/3). - Gary Detlefs, Dec 29 2011
a(n) = floor(n/4) - floor((n-1)/4). - Tani Akinari, Oct 25 2012
a(n) = ceiling( (1/2)*cos(Pi*n/2) ). - Wesley Ivan Hurt, May 31 2013
a(n) = ((1+(-1)^(n/2))*(1+(-1)^n))/4. - Bogart B. Strauss, Jul 14 2013
a(n) = C(n-1,3) mod 2. - Wesley Ivan Hurt, Oct 07 2014
a(n) = (((n+1) mod 4) mod 3) mod 2. - Ctibor O. Zizka, Dec 11 2014
a(n) = (sin(Pi*(n+1)/2)^2)/2 + sin(Pi*(n+1)/2)/2. - Mikael Aaltonen, Jan 02 2015
E.g.f.: (cos(x) + cosh(x))/2. - Vaclav Kotesovec, Feb 15 2015
a(n) = a(n-4) for n>3. - Wesley Ivan Hurt, Jul 07 2016
a(n) = (1-sqrt(2)*cos(n*Pi/2-3*Pi/4))/2 * cos(n*Pi/2). - (found by Steve Chow) Iain Fox, Nov 16 2017
a(n) = 1-A166486(n). - Antti Karttunen, Jul 29 2018
a(n) = (1-(-1)^A142150(n+1))/2. - Adriano Caroli, Sep 28 2019

More terms from Antti Karttunen, Jul 29 2018

A000749 a(n) = 4a(n-1) - 6a(n-2) + 4*a(n-3), n > 3, with a(0)=a(1)=a(2)=0, a(3)=1.

Original entry on oeis.org

0, 0, 0, 1, 4, 10, 20, 36, 64, 120, 240, 496, 1024, 2080, 4160, 8256, 16384, 32640, 65280, 130816, 262144, 524800, 1049600, 2098176, 4194304, 8386560, 16773120, 33550336, 67108864, 134225920, 268451840, 536887296, 1073741824, 2147450880

Offset: 0

N. J. A. Sloane

Number of strings over Z_2 of length n with trace 1 and subtrace 1.
Same as number of strings over GF(2) of length n with trace 1 and subtrace 1.
Also expansion of bracket function.
a(n) is also the number of induced subgraphs with odd number of edges in the complete graph K(n-1). - Alessandro Cosentino (cosenal(AT)gmail.com), Feb 02 2009
From Gary W. Adamson, Mar 13 2009: (Start)
M^n * [1,0,0,0] = [A038503(n), a(n), A038505(n), A038504(n)];
where M = the 4 X 4 matrix [1,1,0,0; 0,1,1,0; 0,0,1,1; 1,0,0,1].
Sum of the 4 terms = 2^n.
Example; M^6 * [1,0,0,0] = [16, 20, 16, 12] sum = 64 = 2^6. (End)
Binomial transform of the period 4 repeat: [0,0,0,1], which is the same as A011765 with offset 0. - Wesley Ivan Hurt, Dec 30 2015
{A038503, A038504, A038505, A000749} is the difference analog of the hyperbolic functions of order 4, {h_1(x), h_2(x), h_3(x), h_4(x)}. For a definition see the reference "Higher Transcendental Functions" and the Shevelev link. - Vladimir Shevelev, Jun 14 2017
This is the p-INVERT of (1,1,1,1,1,...) for p(S) = 1 - S^4; see A291000.  - Clark Kimberling, Aug 24 2017

			a(4;1,1)=4 since the four binary strings of trace 1, subtrace 1 and length 4 are { 0111, 1011, 1101, 1110 }.

Higher Transcendental Functions, Bateman Manuscript Project, Vol. 3, ed. A. Erdelyi, 1983 (chapter XVIII).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
H. W. Gould, Binomial coefficients, the bracket function and compositions with relatively prime summands, Fib. Quart. 2(4) (1964), 241-260.
Maran van Heesch, The multiplicative complexity of symmetric functions over a field with characteristic p, Thesis, 2014.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
F. Ruskey, Strings over Z_2 with given trace and subtrace
F. Ruskey, Strings over GF(2) with given trace and subtrace
Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4).

Cf. A000748, A000750, A001659, A006090, A007877, A009545, A011765.
Cf. A021913, A038503, A038504, A038505, A133209, A133212, A291000.
Sequences of the form 1/((1-x)^m - x^m): A000079 (m=1,2), A024495 (m=3), this sequence (m=4), A049016 (m=5), A192080 (m=6), A049017 (m=7), A290995 (m=8), A306939 (m=9).

a000749 n = a000749_list !! n
a000749_list = 0 : 0 : 0 : 1 : zipWith3 (\u v w -> 4 * u - 6 * v + 4 * w)
   (drop 3 a000749_list) (drop 2 a000749_list) (drop 1 a000749_list)
-- Reinhard Zumkeller, Jul 15 2013

I:=[0,0,0,1]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Dec 31 2015

A000749 := proc(n) local k; add(binomial(n,4*k+3),k=0..floor(n/4)); end;
A000749:=-1/((2*z-1)*(2*z**2-2*z+1)); # Simon Plouffe in his 1992 dissertation
a:= n-> if n=0 then 0 else (Matrix(3, (i,j)-> if (i=j-1) then 1 elif j=1 then [4,-6,4][i] else 0 fi)^(n-1))[1,3] fi: seq(a(n), n=0..33); # Alois P. Heinz, Aug 26 2008
# Alternatively:
s := sqrt(2): h := n -> [0,-s,-2,-s,0,s,2,s][1+(n mod 8)]:
a := n -> `if`(n=0,0,(2^n+2^(n/2)*h(n))/4):
seq(a(n),n=0..33); # Peter Luschny, Jun 14 2017

Join[{0},LinearRecurrence[{4,-6,4},{0,0,1},40]] (* Harvey P. Dale, Mar 31 2012 *)
CoefficientList[Series[x^3/(1 -4x +6x^2 -4x^3), {x,0,80}], x] (* Vincenzo Librandi, Dec 31 2015 *)

PARI
```
a(n)=sum(k=0,n\4,binomial(n,4*k+3))
```

@CachedFunction
def a(n): # a = A000749
    if (n<4): return (n//3)
    else: return 4*a(n-1) -6*a(n-2) +4*a(n-3)
[a(n) for n in range(41)] # G. C. Greubel, Apr 11 2023

G.f.: x^3/((1-x)^4 - x^4).
a(n) = Sum_{k=0..n} binomial(n, 4*k+3).
a(n) = a(n-1) + A038505(n-2) = 2*a(n-1) + A009545(n-2) for n>=2.
Without the two initial zeros, binomial transform of A007877. - Henry Bottomley, Jun 04 2001
From Paul Barry, Aug 30 2004: (Start)
a(n) = (2^n - 2^(n/2+1)*sin(Pi*n/4) - 0^n)/4.
a(n+1) is the binomial transform of A021913. (End)
a(n; t, s) = a(n-1; t, s) + a(n-1; t+1, s+t+1) where t is the trace and s is the subtrace.
Without the initial three zeros, = binomial transform of [1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 3, ...]. - Gary W. Adamson, Jun 19 2008
From Vladimir Shevelev, Jun 14 2017: (Start)
1) For n>=1, a(n) = (1/4)*(2^n + i*(1+i)^n - i*(1-i)^n), where i=sqrt(-1);
2) a(n+m) = a(n)*H_1(m) + H_3(n)*H_2(m) + H_2(n)*H_3(m) + H_1(n)*a(m),
where H_1 = A038503, H_2 = A038504, H_3 = A038505. (End)
a(n) = (2^n - 2*A009545(n) - [n=0])/4. - G. C. Greubel, Apr 11 2023

Additional comments from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 22 2002
New definition from Paul Curtz, Oct 29 2007
Edited by N. J. A. Sloane, Jun 13 2008

A130519 a(n) = Sum_{k=0..n} floor(k/4). (Partial sums of A002265.)

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 3, 4, 6, 8, 10, 12, 15, 18, 21, 24, 28, 32, 36, 40, 45, 50, 55, 60, 66, 72, 78, 84, 91, 98, 105, 112, 120, 128, 136, 144, 153, 162, 171, 180, 190, 200, 210, 220, 231, 242, 253, 264, 276, 288, 300, 312, 325, 338, 351, 364, 378, 392, 406, 420, 435, 450

Offset: 0

Hieronymus Fischer, Jun 01 2007

Complementary to A130482 with respect to triangular numbers, in that A130482(n) + 4*a(n) = n(n+1)/2 = A000217(n).
Disregarding the first three 0's the resulting sequence a'(n) is the sum of the positive integers <= n that have the same residue modulo 4 as n. This is the additive counterpart of the quadruple factorial numbers. - Peter Luschny, Jul 06 2011
From Heinrich Ludwig, Dec 23 2017: (Start)
Column sums of (shift of rows = 4):
  1 2 3 4 5 6  7  8  9 10 11 12 13 14 ...
          1 2  3  4  5  6  7  8  9 10 ...
                     1  2  3  4  5  6 ...
                                 1  2 ...
  .......................................
  ---------------------------------------
  1 2 3 4 6 8 10 12 15 18 21 24 28 32 ...
  shift of rows = 1 see A000217
  shift of rows = 2 see A002620
  shift of rows = 3 see A001840
  shift of rows = 5 see A130520
(End)
Conjecture: a(n+2) is the maximum effective weight of a numerical semigroup S of genus n (see Nathan Pflueger). - Stefano Spezia, Jan 04 2019

			G.f. = x^4 + 2*x^5 + 3*x^6 + 4*x^7 + 6*x^8 + 8*x^9 + 10*x^10 + 12*x^11 + ...
[ n] a(n)
---------
[ 4] 1
[ 5] 2
[ 6] 3
[ 7] 4
[ 8] 1 + 5
[ 9] 2 + 6
[10] 3 + 7
[11] 4 + 8

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Bakir Farhi, On the Representation of the Natural Numbers as the Sum of Three Terms of the Sequence floor(n^2/a), Journal of Integer Sequences, Vol. 16 (2013), Article 13.6.4.
Darren Glass and Joshua Wagner, Arithmetical Structures on Paths With a Doubled Edge, arXiv:1903.01398 [math.CO], 2019.
Nathan Pflueger, On non-primitive Weierstrass points, Alg. Number Th. 12 (2018) 1923-1947 and arXiv:1608.0566 [math.AG], 2016.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Cf. A000217, A001840, A002264, A002265, A002266, A002620, A004526, A010872, A010873, A010874, A130481, A130483, A130520.

Cf. A000290, A007590, A000212, A118015, A056827, A056834, A056838, A056865.

a:=List([0..65],n->Sum([0..n],k->Int(k/4)));; Print(a); # Muniru A Asiru, Jan 04 2019

[Round(n*(n-2)/8): n in [0..70]]; // Vincenzo Librandi, Jun 25 2011

quadsum := n -> add(k, k = select(k -> k mod 4 = n mod 4, [$1 .. n])):
A130519 := n ->`if`(n<3,0,quadsum(n-3)); seq(A130519(n),n=0..58); # Peter Luschny, Jul 06 2011

a[ n_] := Quotient[ (n - 1)^2, 8]; (* Michael Somos, Oct 14 2011 *)

makelist(floor((n-1)^2/8), n, 0, 70); /* Stefano Spezia, Jan 04 2019 */

{a(n) = (n - 1)^2 \ 8}; /* Michael Somos, Oct 14 2011 */

def A130519(n): return (n-1)**2>>3  # Chai Wah Wu, Jul 30 2022

G.f.: x^4/((1-x^4)*(1-x)^2) = x^4/((1+x)*(1+x^2)*(1-x)^3).
a(n) = +2*a(n-1) -1*a(n-2) +1*a(n-4) -2*a(n-5) +1*a(n-6).
a(n) = floor(n/4)*(n - 1 - 2*floor(n/4)) = A002265(n)*(n - 1 - 2*A002265(n)).
a(n) = (1/2)*A002265(n)*(n - 2 + A010873(n)).
a(n) = floor((n-1)^2/8). - Mitch Harris, Sep 08 2008
a(n) = round(n*(n-2)/8) = round((n^2-2*n-1)/8) = ceiling((n+1)*(n-3)/8). - Mircea Merca, Nov 28 2010
a(n) = A001972(n-4), n>3. - Franklin T. Adams-Watters, Jul 10 2009
a(n) = a(n-4)+n-3, n>3. - Mircea Merca, Nov 28 2010
Euler transform of length 4 sequence [ 2, 0, 0, 1]. - Michael Somos, Oct 14 2011
a(n) = a(2-n) for all n in Z. - Michael Somos, Oct 14 2011
a(n) = A214734(n, 1, 4). - Renzo Benedetti, Aug 27 2012
a(4n) = A000384(n), a(4n+1) = A001105(n), a(4n+2) = A014105(n), a(4n+3) = A046092(n). - Philippe Deléham, Mar 26 2013
a(n) = Sum_{i=1..ceiling(n/2)-1} (i mod 2) * (n - 2*i - 1). - Wesley Ivan Hurt, Jan 23 2014
a(n) = ( 2*n^2-4*n-1+(-1)^n+2*((-1)^((2*n-1+(-1)^n)/4)-(-1)^((6*n-1+(-1)^n)/4)) )/16 = ( 2*n*(n-2) - (1-(-1)^n)*(1-2*i^(n*(n-1))) )/16, where i=sqrt(-1). - Luce ETIENNE, Aug 29 2014
E.g.f.: (1/8)*((- 1 + x)*x*cosh(x) + 2*sin(x) + (- 1 - x + x^2)*sinh(x)). - Stefano Spezia, Jan 15 2019
a(n) = (A002620(n-1) - A011765(n+1)) / 2, for n > 0. - Yuchun Ji, Feb 05 2021
Sum_{n>=4} 1/a(n) = Pi^2/12 + 5/2. - Amiram Eldar, Aug 13 2022

Partially edited by R. J. Mathar, Jul 11 2009

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

A170819 a(n) = product of distinct primes of the form 4k-1 that divide n.

Original entry on oeis.org

1, 1, 3, 1, 1, 3, 7, 1, 3, 1, 11, 3, 1, 7, 3, 1, 1, 3, 19, 1, 21, 11, 23, 3, 1, 1, 3, 7, 1, 3, 31, 1, 33, 1, 7, 3, 1, 19, 3, 1, 1, 21, 43, 11, 3, 23, 47, 3, 7, 1, 3, 1, 1, 3, 11, 7, 57, 1, 59, 3, 1, 31, 21, 1, 1, 33, 67, 1, 69, 7, 71, 3, 1, 1, 3, 19, 77, 3, 79, 1, 3, 1, 83, 21, 1

Offset: 1

N. J. A. Sloane, Dec 23 2009

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A007947, A011765, A065339, A097706, A170817, A170818, A363340.

A170819 := proc(n) a := 1 ; for p in numtheory[factorset](n) do if p mod 4 = 3 then a := a*p ; end if; end do: a ; end proc:
seq(A170819(n),n=1..20) ; # R. J. Mathar, Jun 07 2011

Array[Times @@ Select[FactorInteger[#][[All, 1]], Mod[#, 4] == 3 &] &, 85] (* Michael De Vlieger, Feb 19 2019 *)

for(n=1,99, t=select(x->x%4==3, factor(n)[,1]); print1(prod(i=1,#t,t[i])","))

Multiplicative with a(p^e) = p^A011765(p+1), e > 0. - R. J. Mathar, Jun 07 2011

a(n) = A007947(A097706(n)) = A097706(A007947(n)). - Peter Munn, Jul 06 2023

Extended with PARI program by M. F. Hasler, Dec 23 2009

A292576 Permutation of the natural numbers partitioned into quadruples [4k-1, 4k-3, 4k-2, 4k], k > 0.

Original entry on oeis.org

3, 1, 2, 4, 7, 5, 6, 8, 11, 9, 10, 12, 15, 13, 14, 16, 19, 17, 18, 20, 23, 21, 22, 24, 27, 25, 26, 28, 31, 29, 30, 32, 35, 33, 34, 36, 39, 37, 38, 40, 43, 41, 42, 44, 47, 45, 46, 48, 51, 49, 50, 52, 55, 53, 54, 56, 59, 57, 58, 60, 63, 61, 62

Offset: 1

Guenther Schrack, Sep 19 2017

nonn

Partition the natural number sequence into quadruples starting with (1,2,3,4); swap the second and third elements, then swap the first and the second element; repeat for all quadruples.

Guenther Schrack, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).
Index entries for sequences that are permutations of the natural numbers

Inverse: A056699(n+1) - 1 for n > 0.
Sequence of fixed points: A008586(n) for n > 0.
Subsequences:
   elements with odd index: A042964(A103889(n)) for n > 0.
   elements with even index: A042948(n) for n > 0.
   odd elements: A166519(n) for n>0.
   indices of odd elements: A042963(n) for n > 0.
   even elements: A005843(n) for n>0.
   indices of even elements: A014601(n) for n > 0.
Sum of pairs of elements:
   a(n+2) + a(n) = A163980(n+1) = A168277(n+2) for n > 0.
Difference between pairs of elements:
   a(n+2) - a(n) = (-1)^A011765(n+3)*A091084(n+1) for n > 0.
Compound relations:
   a(n) = A284307(n+1) - 1 for n > 0.
   a(n+2) - 2*a(n+1) + a(n) = (-1)^A011765(n)*A132400(n+1) for n > 0.
Compositions:
   a(n) = A116966(A080412(n)) for n > 0.
   a(A284307(n)) = A256008(n) for n > 0.
   a(A042963(n)) = A166519(n-1) for n > 0.
   A256008(a(n)) = A056699(n) for n > 0.

a = [3 1 2 4]; % Generate b-file
max = 10000;
for n := 5:max
   a(n) = a(n-4) + 4;
end;

for(n=1, 10000, print1(n + ((-1)^(n*(n-1)/2)*(2 - (-1)^n) - (-1)^n)/2, ", "))

a(1)=3, a(2)=1, a(3)=2, a(4)=4, a(n) = a(n-4) + 4 for n > 4.
O.g.f.: (2*x^3 + x^2 - 2*x + 3)/(x^5 - x^4 - x + 1).
a(n) = n + ((-1)^(n*(n-1)/2)*(2-(-1)^n) - (-1)^n)/2.
a(n) = n + (cos(n*Pi/2) - cos(n*Pi) + 3*sin(n*Pi/2))/2.
a(n) = n + n mod 2 + (ceiling(n/2)) mod 2 - 2*(floor(n/2) mod 2).
Linear recurrence: a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
First Differences, periodic: (-2, 1, 2, 3), repeat; also (-1)^A130569(n)*A068073(n+2) for n > 0.

A348264 a(n) is the number of iterations that n requires to reach a fixed point under the map x -> A348173(x).

Original entry on oeis.org

0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0

Offset: 1

Amiram Eldar, Oct 09 2021

nonn

a(n) first differs from A011765(n+2) at n = 84.
The fixed points are terms of A348004, so a(n) = 0 if and only if n is a term of A348004.
Conjecture: essentially partial sums of A219977 (verified for n <= 5000).

			a(1) = 0 since 1 is in A348004.
a(2) = 1 since there is one iteration of the map x -> A348173(x) starting from 2: 2 -> 1.
a(84) = 2 since there are 2 iterations of the map x -> A348173(x) starting from 84: 84 -> 78 -> 39.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A047994, A011765, A348004, A348173.

f[p_, e_] := p^e - 1; uphi[1] = 1; uphi[n_] := Times @@ f @@@ FactorInteger[n]; s[n_] := Plus @@ DeleteDuplicates[uphi /@ Select[Divisors[n], CoprimeQ[#, n/#] &]]; a[n_] := -2 + Length@ FixedPointList[s, n]; Array[a, 100]

A058034 Number of numbers whose cube root rounds to n.

Original entry on oeis.org

1, 3, 12, 27, 49, 75, 108, 147, 193, 243, 300, 363, 433, 507, 588, 675, 769, 867, 972, 1083, 1201, 1323, 1452, 1587, 1729, 1875, 2028, 2187, 2353, 2523, 2700, 2883, 3073, 3267, 3468, 3675, 3889, 4107, 4332, 4563, 4801, 5043, 5292, 5547, 5809, 6075, 6348

Offset: 0

Henry Bottomley, Nov 22 2000

			a(2)=12 since the cube roots of 4, 5, 6, ..., 15 all lie between 1.5 and 2.5.

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

Cf. A003215 for number whose floor (or ceiling) of the cube root is n, A004277 for number whose square root rounds to n.

[n mod 4 eq 0 select 3*n^2+1 else 3*n^2: n in [0..80]]; // Vincenzo Librandi, Dec 25 2015

seq(1 + floor((n+1/2)^3) - ceil((n-1/2)^3), n = 0 .. 100);

Table[SeriesCoefficient[-(3 x^5 + 6 x^4 + 6 x^3 + 7 x^2 + x + 1)/((x - 1)^3 (x + 1) (x^2 + 1)), {x, 0, n}], {n, 0, 46}] (* Michael De Vlieger, Dec 24 2015 *)
LinearRecurrence[{2, -1, 0, 1, -2, 1}, {1, 3, 12, 27, 49, 75}, 50] (* Vincenzo Librandi, Dec 25 2015 *)

Vec(-(3*x^5+6*x^4+6*x^3+7*x^2+x+1)/((x-1)^3*(x+1)*(x^2+1)) + O(x^100)) \\ Colin Barker, Jul 04 2014

a(n) = 3n^2+1 if n == 0 (mod 4), 3n^2 otherwise.
a(n) = A033428(n)+A011765(n) = A034131(n-1)-A034131(n-2).
a(n) = (1+(-1)^n+(-i)^n+i^n+12*n^2)/4 where i=sqrt(-1). - Colin Barker, Jul 04 2014
G.f.: -(3*x^5+6*x^4+6*x^3+7*x^2+x+1) / ((x-1)^3*(x+1)*(x^2+1)). - Colin Barker, Jul 04 2014

A058936 Decomposition of Stirling's S(n,2) based on associated numeric partitions.

Original entry on oeis.org

0, 1, 3, 8, 3, 30, 20, 144, 90, 40, 840, 504, 420, 5760, 3360, 2688, 1260, 45360, 25920, 20160, 18144, 403200, 226800, 172800, 151200, 72576, 3991680, 2217600, 1663200, 1425600, 1330560, 43545600, 23950080, 17740800, 14968800, 13685760, 6652800, 518918400

Offset: 1

Alford Arnold, Jan 11 2001

These values also appear in a wider context when counting elements of finite groups by cycle structure. For example, the alternating group on four symbols has 12 elements; eight associated with the partition 3+1, three associated with 2+2 and the identity associated with 1+1+1+1. The cross-referenced sequences are all associated with similar numeric partitions and "M2" weights.

			Triangle begins:
  0;
  1;
  3;
  8, 3;
  30, 20;
  144, 90, 40;
  840, 504, 420;
  ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 831.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Cf. A000012, A000035, A000027, A004526, A022003, A008619, A000217, A007997, A001399, A011765 A008620, A027656, A002620, A000292, A008627.

From Sean A. Irvine, Sep 05 2022: (Start)
T(1,1) = 0.
T(n,k) = n! / (k * (n-k)) for 1 <= k < n/2.
T(2n,n) = (2*n)! / (2*n^2).
(End)

More terms from Sean A. Irvine, Sep 05 2022

cp's OEIS Frontend

A002265 Nonnegative integers repeated 4 times.

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A121262 The characteristic function of the multiples of four.

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A000749 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), n > 3, with a(0)=a(1)=a(2)=0, a(3)=1.

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A130519 a(n) = Sum_{k=0..n} floor(k/4). (Partial sums of A002265.)

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A080239 Antidiagonal sums of triangle A035317.

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A000749 a(n) = 4a(n-1) - 6a(n-2) + 4*a(n-3), n > 3, with a(0)=a(1)=a(2)=0, a(3)=1.