A133872 -id:A133872 - cp's OEIS frontend

A042948 Numbers congruent to 0 or 1 (mod 4).

0, 1, 4, 5, 8, 9, 12, 13, 16, 17, 20, 21, 24, 25, 28, 29, 32, 33, 36, 37, 40, 41, 44, 45, 48, 49, 52, 53, 56, 57, 60, 61, 64, 65, 68, 69, 72, 73, 76, 77, 80, 81, 84, 85, 88, 89, 92, 93, 96, 97, 100, 101, 104, 105, 108

Offset: 0

N. J. A. Sloane

Maximum number of squares attacked by a bishop on an (n + 1) X (n + 1) chessboard. - Stewart Gordon, Mar 23 2001
Maximum vertex degree of the (n + 1) X (n + 1) bishop graph and black bishop graph. - Eric W. Weisstein, Jun 26 2017
Also number of squares attacked by a bishop on a toroidal chessboard. - Diego Torres (torresvillarroel(AT)hotmail.com), May 30 2001
Numbers n such that {1, 2, 3, ..., n-1, n} is a perfect Skolem set. - Emeric Deutsch, Nov 24 2006
The number of terms which lie on the principal diagonals of an n X n square spiral. - William A. Tedeschi, Mar 02 2008
Possible nonnegative discriminants of quadratic equation a*x^2 + b*x + c or discriminants of binary quadratic forms a*x^2 + b*x*y + c^y^2. - Artur Jasinski, Apr 28 2008
A133872(a(n)) = 1; complement of A042964. - Reinhard Zumkeller, Oct 03 2008
Partial sums are A035608. - Jaroslav Krizek, Dec 18 2009 [corrected by Werner Schulte, Dec 04 2023]
Nonnegative m for which floor(k*m/4) = k*floor(m/4), where k = 2 or 3. Example: 13 is in the sequence because floor(2*13/4) = 2*floor(13/4), and also floor(3*13/4) = 3*floor(13/4). - Bruno Berselli, Dec 09 2015
Also number of maximal cliques in the n X n white bishop graph. - Eric W. Weisstein, Dec 01 2017
The offset should have been 1. - Jianing Song, Oct 06 2018
Numbers k for which the binomial coefficient C(k,2) is even. - Tanya Khovanova, Oct 20 2018
Numbers m such that there exists a permutation (x(1), x(2), ..., x(m)) with all absolute differences |x(k) - k| distinct. - Jukka Kohonen, Oct 02 2021
Numbers m such that there exists a multiset of integers whose size is m, and sum and product are both -m. - Yifan Xie, Mar 25 2024

James Spahlinger, Table of n, a(n) for n = 0..10000
H. W. Gould, The inverse of a finite series and a third-order recurrent sequence, Fibonacci Quart., Vol. 44, No. 4 (2006), pp. 302-315. See p. 311.
M. J. Pelling and J. H. Steelman, E3269. Permutations with distinct displacements, (problem by Pelling and solution by Steelman), The American Mathematical Monthly, 96 (1989), 843-844.
T. Skolem, On certain distributions of integers in pairs with given differences, Math. Scand., Vol. 5 (1957), pp. 57-68.
Harry Tamvakis and O. P. Lossers, Amenable Numbers: 10454, The American Mathematical Monthly, Vol. 105, No. 4 (Apr., 1998), p. 368.
Leo Tavares, Illustration: Diamond Crosses
Eric Weisstein's World of Mathematics, Bishop Graph.
Eric Weisstein's World of Mathematics, Black Bishop Graph.
Eric Weisstein's World of Mathematics, Maximal Clique.
Eric Weisstein's World of Mathematics, Maximum Vertex Degree.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000290, A002620.

[n: n in [0..150]|n mod 4 in {0, 1}]; // Vincenzo Librandi, Dec 09 2015

a[0]:=0:a[1]:=1:for n from 2 to 100 do a[n]:=a[n-2]+4 od: seq(a[n], n=0..54); # Zerinvary Lajos, Mar 16 2008

Select[Range[0, 150], Or[Mod[#, 4] == 0, Mod[#, 4] == 1] &] (* Vincenzo Librandi, Dec 09 2015 *)
Table[(4 n - 5 - (-1)^n)/2, {n, 20}] (* Eric W. Weisstein, Dec 01 2017 *)
LinearRecurrence[{1, 1, -1}, {1, 4, 5}, {0, 20}] (* Eric W. Weisstein, Dec 01 2017 *)
CoefficientList[Series[x (1 + 3 x)/((-1 + x)^2 (1 + x)), {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)
{#, # + 1} & /@ (4 Range[0, 40]) // Flatten (* Harvey P. Dale, Jan 15 2024 *)

makelist(-1/2+1/2*(-1)^n+2*n, n, 0, 60); /* Martin Ettl, Nov 05 2012 */

PARI
```
a(n)=2*n-n%2;
```

concat(0, Vec(x*(1+3*x)/((1+x)*(1-x)^2) + O(x^100))) \\ Altug Alkan, Dec 09 2015

a(n) = A042963(n+1) - 1. [Corrected by Jianing Song, Oct 06 2018]
From Michael Somos, Jan 12 2000: (Start)
G.f.: x*(1 + 3*x)/((1 + x)*(1 - x)^2).
a(n) = a(n-1) + 2 + (-1)^n. (End)
a(n) = 4*n - a(n-1) - 3 with a(0) = 0. - Vincenzo Librandi, Nov 17 2010
a(n) = Sum_{k>=0} A030308(n,k)*A151821(k+1). - Philippe Deléham, Oct 17 2011
a(n) = floor((4/3)*floor(3*n/2)). - Clark Kimberling, Jul 04 2012
a(n) = n + 2*floor(n/2) = 2*n - (n mod 2). - Bruno Berselli, Apr 30 2016
E.g.f.: 2*exp(x)*x - sinh(x). - Stefano Spezia, Sep 09 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 + 3*log(2)/4. - Amiram Eldar, Dec 05 2021
a(n) = A000290(n) - 4*A002620(n-1). - Leo Tavares, Oct 04 2022

A133900 a(n) = period of the sequence {b(m), m>=0}, defined by b(m):=binomial(m+n,n) mod n.

Original entry on oeis.org

1, 4, 9, 16, 25, 72, 49, 64, 81, 400, 121, 864, 169, 784, 675, 256, 289, 2592, 361, 1600, 1323, 3872, 529, 3456, 625, 5408, 729, 3136, 841, 324000, 961, 1024, 9801, 18496, 6125, 31104, 1369, 23104, 13689, 32000, 1681, 254016, 1849, 15488, 30375, 33856

Offset: 1

Hieronymus Fischer, Oct 15 2007, Oct 20 2007

nonn

This is the analog of the sequence of Pisano periods (A001175) for binomial factors.
n^2 always divides a(n).
A prime p is a factor of a(n) if and only if it is a factor of n (i.e., a(n) and n have the same prime factors).

			a(3)=9 since binomial(m+3,3) mod 3, m>=0, is periodic with period length 3^2=9 (see A133883).
a(6)=72 since binomial(m+6,6) mod 6, m>=0, is periodic with period length 4*6^2=72 (see A133886).

Hieronymus Fischer, Table of n, a(n) for n = 1..111

Cf. A000040, A001175, A133872-A133880, A133882-A133890, A133910.
Cf. A133620-A133625, A133630, A038509, A133634-A133636.
Cf. A133905.

a(n)=n^2 if n is a prime or a power of a prime.

A133910 Period numbers of A133900 divided by n^2.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 4, 1, 6, 1, 4, 3, 1, 1, 8, 1, 4, 3, 8, 1, 6, 1, 8, 1, 4, 1, 360, 1, 1, 9, 16, 5, 24, 1, 16, 9, 20, 1, 144, 1, 8, 15, 16, 1, 18, 1, 16, 9, 8, 1, 16, 5, 28, 9, 16, 1, 360, 1, 16, 21, 1, 5, 288, 1, 16, 9, 1120, 1, 24, 1, 32, 9, 16, 7, 288, 1, 20, 1, 64, 1, 6048, 5, 32, 27, 8

Offset: 1

Hieronymus Fischer, Oct 20 2007

nonn

			a(6)=2, since A133900(6)/6^2=72/36=2.
a(18)=8, since A133900(18)/18^2=2592/324=8.

Cf. A000040, A100118, A046363, A133620, A133621, A133623, A133630, A133635.

Cf. A133872, A133880, A133890, A133900, A133911.

a(n)=A133900(n)/n^2.
a(n)=1, iff n is a prime or a power of a prime (including n=1).
If a prime p is a factor of a(n), then p is also a factor of n.

A042964 Numbers that are congruent to 2 or 3 mod 4.

Original entry on oeis.org

2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, 42, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 66, 67, 70, 71, 74, 75, 78, 79, 82, 83, 86, 87, 90, 91, 94, 95, 98, 99, 102, 103, 106, 107, 110, 111, 114, 115, 118, 119, 122, 123, 126, 127

Offset: 1

N. J. A. Sloane

Also numbers m such that binomial(m+2, m) mod 2 = 0. - Hieronymus Fischer, Oct 20 2007
Also numbers m such that floor(1+(m/2)) mod 2 = 0. - Hieronymus Fischer, Oct 20 2007
Partial sums of the sequence 2, 1, 3, 1, 3, 1, 3, 1, 3, 1, ... which has period 2. - Hieronymus Fischer, Oct 20 2007
In groups of four add and divide by two the odd and even numbers. - George E. Antoniou, Dec 12 2001
From Jeremy Gardiner, Jan 22 2006: (Start)
Comments on the "mystery calculator". There are 6 cards.
Card 0: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, ... (A005408 sequence).
Card 1: 2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, ... (this sequence).
Card 2: 4, 5, 6, 7, 12, 13, 14, 15, 20, 21, 22, 23, 28, 29, 30, 31, 36, 37, 38, 39, ... (A047566).
Card 3: 8, 9, 10, 11, 12, 13, 14, 15, 24, 25, 26, 27, 28, 29, 30, 31, 40, 41, 42, ... (A115419).
Card 4: 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 48, 49, 50, ... (A115420).
Card 5: 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, ... (A115421).
The trick: You secretly select a number between 1 and 63 from one of the cards. You indicate to me the cards on which that number appears; I tell you the number you selected!
The solution: I add together the first term from each of the indicated cards. The total equals the selected number. The numbers in each sequence all have a "1" in the same position in their binary expansion. Example: You indicate cards 1, 3 and 5. Your selected number is 2 + 8 + 32 = 42.
Numbers having a 1 in position 1 of their binary expansion. One of the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. (End)
Complement of A042948. - Reinhard Zumkeller, Oct 03 2008
Also the 2nd Witt transform of A040000 [Moree]. - R. J. Mathar, Nov 08 2008
In general, sequences of numbers congruent to {a,a+i} mod k will have a closed form of (k-2*i)*(2*n-1+(-1)^n)/4+i*n+a, from offset 0. - Gary Detlefs, Oct 29 2013
Union of A004767 and A016825; Fixed points of A098180. - Wesley Ivan Hurt, Jan 14 2014, Oct 13 2015

David Lovler, Table of n, a(n) for n = 1..10000
Maths Magic, Mystery Calculator.
Pieter Moree, The formal series Witt transform, Discr. Math. no. 295 vol. 1-3 (2005) 143-160.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000040, A133620, A133621, A133622, A133630, A133635.
Cf. A133872, A133882, A133890, A133900, A133910.
Card trick: A005408, A047566, A115419, A115420, A115421.
Cf. A004767, A016825 A040000, A042948, A047406, A098180.

[2*n+((-1)^(n-1)-1)/2 : n in [1..100]]; // Wesley Ivan Hurt, Oct 13 2015

[n: n in [1..150] | n mod 4 in [2, 3]]; // Vincenzo Librandi, Oct 13 2015

A042964:=n->2*n+((-1)^(n-1)-1)/2; seq(A042964(n), n=1..100); # Wesley Ivan Hurt, Jan 07 2014

Flatten[Table[4n + {2, 3}, {n, 0, 31}]] (* Alonso del Arte, Feb 07 2013 *)
Select[Range[200],MemberQ[{2,3},Mod[#,4]]&] (* or *) LinearRecurrence[ {1,1,-1},{2,3,6},90] (* Harvey P. Dale, Nov 28 2018 *)

PARI
```
a(n)=2*n+2-n%2
```

Vec((2+x+x^2)/((1-x)*(1-x^2)) + O(x^100)) \\ Altug Alkan, Oct 13 2015

a(n) = A047406(n)/2.
From Michael Somos, Jan 12 2000: (Start)
G.f.: x*(2+x+x^2)/((1-x)*(1-x^2)).
a(n) = a(n-1) + 2 + (-1)^n. (End)
a(n) = 2n if n is odd, otherwise n = 2n - 1. - Amarnath Murthy, Oct 16 2003
a(n) = (3 + (-1)^(n-1))/2 + 2*(n-1) = 2n + 2 - (n mod 2). - Hieronymus Fischer, Oct 20 2007
A133872(a(n)) = 0. - Reinhard Zumkeller, Oct 03 2008
a(n) = 4*n - a(n-1) - 3 (with a(1) = 2). - Vincenzo Librandi, Nov 17 2010
a(n) = 2*n + ((-1)^(n-1) - 1)/2. - Gary Detlefs, Oct 29 2013
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 - log(2)/4. - Amiram Eldar, Dec 05 2021
E.g.f.: 1 + ((4*x - 1)*exp(x) - exp(-x))/2. - David Lovler, Aug 08 2022

Edited by N. J. A. Sloane, Jun 30 2008 at the suggestion of R. J. Mathar

Corrected by Jaroslav Krizek, Dec 18 2009

A021913 Period 4: repeat [0, 0, 1, 1].

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

Decimal expansion of 1/909.
Lexicographically earliest de Bruijn sequence for n = 2 and k = 2.
Except for first term, binary expansion of the decimal number 1/10 = 0.000110011001100110011... in base 2. - Benoit Cloitre, May 18 2002
Content of #2 binary placeholder when n is converted from decimal to binary. a(n) = n*(n-1)/2 mod 2. Example: a(7) = 1 since 7 in binary is 1 -1- 1 and (7*6/2) mod 2 = 1. - Anne M. Donovan (anned3005(AT)aol.com), Sep 15 2003
Expansion in any base b of 1/((b-1)*(b^2+1)) = 1/(b^3-b^2+b-1). E.g., 1/5 in base 2, 1/20 in base 3, 1/51 in base 4, etc. - Franklin T. Adams-Watters, Nov 07 2006
Except for first term, parity of the triangular numbers A000217. - Omar E. Pol, Jan 17 2012
Except for first term, more generally: 1) Parity of the k-polygonal numbers, if k is odd (Cf. A139600, A139601). 2) Parity of the generalized k-gonal numbers, for even k >= 6. - Omar E. Pol, Feb 05 2012
Except for first term, parity of Recamán's sequence A005132. - Omar E. Pol, Apr 13 2012
Inverse binomial transform of A000749(n+1). - Wesley Ivan Hurt, Dec 30 2015
Least significant bit of tribonacci numbers (A000073). - Andres Cicuttin, Apr 04 2016

			G.f. = x^2 + x^3 + x^6 + x^7 + x^10 + x^11 + x^14 + x^15 + x^18 + x^19 + ...;
1/909 = 0.001100110011001 ...

Guenther Schrack, Table of n, a(n) for n = 0..1000
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (1,-1,1).

Cf. A000073, A000217, A000749, A005132, A007877, A056594, A062158, A133872, A139600, A139601.

&cat [[0, 0, 1, 1]^^30]; // Vincenzo Librandi, Dec 31 2015

A021913:=n->floor((n mod 4)/2); seq(A021913(n), n=0..100); # Wesley Ivan Hurt, Feb 28 2014

Table[Floor[Mod[n,4]/2], {n,0,100}] (* Wesley Ivan Hurt, Feb 28 2014 *)
a[ n_] := Mod[ Quotient[ n, 2], 2]; (* Michael Somos, Feb 28 2014 *)
LinearRecurrence[{1,-1,1},{0,0,1}, 100] (* Ray Chandler, Aug 25 2015 *)
CoefficientList[Series[x^2(1+x)/(1-x^4), {x, 0, 100}], x] (* Vincenzo Librandi, Dec 31 2015 *)
(1-(-1)^Binomial[Range[0, 100], 2])/2 (* G. C. Greubel, Apr 03 2019 *)

{a(n) = n \ 2 % 2}; /* Michael Somos, Feb 28 2014 */

x='x+O('x^99); concat([0, 0], Vec(x^2/(1-x+x^2-x^3))) \\ Altug Alkan, Apr 04 2016

From Paul Barry, Aug 30 2004: (Start)
G.f.: x^2*(1 + x)/(1 - x^4).
a(n) = 1/2 - cos(Pi*n/2)/2 - sin(Pi*n/2)/2.
a(n) = a(n-1) - a(n-2) + a(n-3) for n > 2. (End)
a(n+2) = Sum_{k=0..n} b(k), with b(k) = A056594(k) (partial sums of S(n,x) Chebyshev polynomials at x=0).
a(n) = -a(n-2) + 1, for n >= 2 with a(0) = a(1) = 0.
G.f.: x^2/((1 - x)*(1 + x^2)) = x^2/(1 - x + x^2 - x^3).
From Jaume Oliver Lafont, Dec 05 2008: (Start)
a(n) = 1/2 - sin((2n+1)*Pi/4)/sqrt(2).
a(n) = 1/2 - cos((2n-1)*Pi/4)/sqrt(2). (End)
a(n) = floor((n mod 4)/2). - Reinhard Zumkeller, Apr 15 2011
Euler transform of length 4 sequence [1, -1, 0, 1]. - Michael Somos, Feb 28 2014
a(1-n) = a(n) for all n in Z. - Michael Somos, Feb 28 2014
From Wesley Ivan Hurt, Jul 22 2016: (Start)
a(n) = a(n-4) for n > 3.
a(n) = A133872(n+2).
a(n) + a(n+1) = A007877(n). (End)
E.g.f.: (exp(x) - sin(x) - cos(x))/2. - Ilya Gutkovskiy, Jul 11 2016
a(n) = (1 - (-1)^(n*(n-1)/2))/2. - Guenther Schrack, Feb 28 2019

Chebyshev comment from Wolfdieter Lang, Sep 10 2004

A011848 a(n) = floor(binomial(n, 2)/2).

Original entry on oeis.org

0, 0, 0, 1, 3, 5, 7, 10, 14, 18, 22, 27, 33, 39, 45, 52, 60, 68, 76, 85, 95, 105, 115, 126, 138, 150, 162, 175, 189, 203, 217, 232, 248, 264, 280, 297, 315, 333, 351, 370, 390, 410, 430, 451, 473, 495, 517, 540, 564, 588, 612, 637, 663, 689, 715, 742, 770, 798

Offset: 0

N. J. A. Sloane, Dec 11 1996

Column sums of an array of the odd numbers repeatedly shifted 4 places to the right:
1 3 5 7  9 11 13 15 17...
         1  3  5  7  9...
                     1...
.........................
-------------------------
1 3 5 7 10 14 18 22 27...
Floor of the area under the polygon connecting the lattice points (n, floor(n/2)) from 0..n. - Wesley Ivan Hurt, Jun 09 2014
Beginning with a(4)=3, the sequence might be called the "off-axis" Ulam-Spiral numbers because they are the numbers in ascending order on the horizontal and vertical spokes (heading outward) starting with the first turning points on the spiral (i.e., 3, 5, 7 and 10). That is, starting with: 3 (upward); 5 (leftward); 7 (downward) and 10 (rightward). These are A033991 (starting at a(1)), A007742 (starting at a(1)), A033954 (starting at a(1)) and A001107 (starting at a(2)), respectively. These quadri-sections are summarized in the formulas of Sep 26 2015. - Bob Selcoe, Oct 05 2015
Conjecture: For n = 2, a(n) is the greatest k such that A123663(k) < A000217(n - 2). - Peter Kagey, Nov 18 2016
a(n) is also the matching number of the n-triangular graph, (n-1)-triangular honeycomb queen graph, (n-1)-triangular honeycomb bishop graphs, and (for n > 7) (n-1)-triangular honeycomb obtuse knight graphs. - Eric W. Weisstein, Jun 02 2017 and Apr 03 2018
After 0, 0, 0, add 1, then add 2 three times, then add 3, then add 4 three times, then add 5, etc.; i.e., first differences are A004524 = (0, 0, 0, 1, 2, 2, 2, 3, 4, 4, 4, 5, ...). - M. F. Hasler, May 09 2018
Let s(0) = s(1) = 1, s(-1) = s(2) = x, and s(n+2)*s(n-2) = s(n+1)*s(n-1) + s(n)^2 for all n in Z. Then s(n) = p(n) / x^e(n) is a Laurent polynomial in x with p(n) a polynomial with nonnegative integer coefficients of degree a(n) for all n in Z. If x = 1, then s(n) = p(n) = A006720(n+1). - Michael Somos, Mar 22 2023

			G.f. = x^3 + 3*x^4 + 5*x^5 + 7*x^6 + 10*x^7 + 14*x^8 + 18*x^9 + 22*x^10 + ...
p(0) = p(1) = 1, p(2) = 1 + x, p(3) = 1 + x + x^3, p(4) = 1 + 2*x + 2*x^2 + x^3 + x^5. - _Michael Somos_, Mar 22 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Kyu-Hwan Lee and Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
Eric Weisstein's World of Mathematics, Matching Number.
Eric Weisstein's World of Mathematics, Triangular Graph.
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

A column of triangle A011857.
First differences are in A004524.
Cf. A007318, A033991, A007742, A033954, A001107, A006720, A035608 (bisection), A156859 (bisection).

List([0..60],n->Int(Binomial(n,2)/2)); # Muniru A Asiru, Apr 05 2018

a011848 n = if n < 2 then 0 else flip div 2 $ a007318 n 2
-- Reinhard Zumkeller, Mar 04 2015

[ Floor(n*(n-1)/4) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014

seq(floor(binomial(n,2)/2), n=0..57); # Zerinvary Lajos, Jan 12 2009

Table[Floor[n (n - 1)/4], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jun 28 2011 *)
CoefficientList[Series[x^3/((1 + x^2) (1 - x)^3), {x, 0, 70}], x] (* Vincenzo Librandi, Jun 21 2013 *)
LinearRecurrence[{3, -4, 4, -4, 1}, {0, 0, 1, 3, 5}, {0, 20}] (* Eric W. Weisstein, Jun 02 2017 *)
Table[Floor[Binomial[n, 2]/2], {n, 0, 20}] (* Eric W. Weisstein, Jun 02 2017 *)
Table[1/4 (-1 + (-1 + n) n + Cos[n Pi/2] + Sin[n Pi/2]), {n, 0, 20}] (* Eric W. Weisstein, Jun 02 2017 *)
Floor[Binomial[Range[0, 20], 2]/2] (* Eric W. Weisstein, Apr 03 2018 *)

PARI
```
a(n) = binomial(n, 2)\2;
```

vector(100, n, n--; floor(n*(n-1)/4)) \\ Altug Alkan, Sep 30 2015

def a(n): return n*(n-1)//4 # Christoph B. Kassir, Oct 07 2022

[floor(binomial(n,2)/2) for n in range(0,58)] # Zerinvary Lajos, Dec 01 2009

G.f.: x^3*(1-x^2)/((1-x)^3*(1-x^4)).
G.f.: x^3/((1+x^2)*(1-x)^3). - Jon Perry, Mar 31 2004
a(n) = +3*a(n-1) -4*a(n-2) +4*a(n-3) -3*a(n-4) +a(n-5). - R. J. Mathar, Apr 15 2010
a(n) = floor((n/(1+e^(1/n)))^2). - Richard R. Forberg, Jun 19 2013
a(n) = floor(n*(n-1)/4). - T. D. Noe, Jun 20 2013
a(n) = (1/4) * ( n^2 - n - 1 + (-1)^floor(n/2) ). - Ralf Stephan, Aug 11 2013
a(n) = A054925(n) - A133872(n+2). - Wesley Ivan Hurt, Jun 09 2014
a(4*n) = A033991(n). a(4*n+1) = A007742(n). a(4*n+2) = A033954(n). a(4*n+3) = A001107(n+1). - Bob Selcoe, Sep 26 2015
E.g.f.: (sin(x) + cos(x) + (x^2 - 1)*exp(x))/4. - Ilya Gutkovskiy, Nov 18 2016
A054925(n) = a(-n). A035608(n) = a(2*n+1). Wesley Ivan Hurt, Jun 09 2014
A156859(n) = a(2*n+2). - Michael Somos, Nov 18 2016
Euler transform of length 4 sequence [ 3, -1, 0, 1]. - Michael Somos, Nov 18 2016
From Amiram Eldar, Mar 18 2022: (Start)
Sum_{n>=3} 1/a(n) = 40/9 - 2*Pi/3.
Sum_{n>=3} (-1)^(n+1)/a(n) = 32/9 - 4*log(2). (End)
0 = a(n+2)*(a(n)*(a(n) -6*a(n+1) +4*a(n+2)) +a(n+1)*(8*a(n+1) -10*a(n+2)) + 3*a(n+2)^2) +a(n+3)*(a(n)*(+a(n) -2*a(n+1)) +a(n+2)*(2*a(n+1) -a(n+2))) for all n in Z. - Michael Somos, Mar 22 2023
2*a(n) + 2*a(n-2) = (n-1)*(n-2). - R. J. Mathar, Feb 12 2024

A052551 Expansion of 1/((1 - x)(1 - 2x^2)).

Original entry on oeis.org

1, 1, 3, 3, 7, 7, 15, 15, 31, 31, 63, 63, 127, 127, 255, 255, 511, 511, 1023, 1023, 2047, 2047, 4095, 4095, 8191, 8191, 16383, 16383, 32767, 32767, 65535, 65535, 131071, 131071, 262143, 262143, 524287, 524287, 1048575, 1048575, 2097151, 2097151

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Equals row sums of triangle A137865. - Gary W. Adamson, Feb 18 2008
Also, the decimal representation of the diagonal from the corner to the origin of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 566", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Jul 05 2017
Number of nonempty subsets of {1,2,...,n+1} that contain only odd numbers. a(0) = a(1) = 1: {1}; a(6) = a(7) = 15: {1}, {3}, {5}, {7}, {1,3}, {1,5}, {1,7}, {3,5}, {3,7}, {5,7}, {1,3,5}, {1,3,7}, {1,5,7}, {3,5,7}, {1,3,5,7}. - Enrique Navarrete, Mar 16 2018
Number of nonempty subsets of {1,2,...,n+2} that contain only even numbers. a(0) = a(1) = 1: {2}; a(4) = a(5) = 7: {2}, {4}, {6}, {2,4}, {2,6}, {4,6}, {2,4,6}. - Enrique Navarrete, Mar 26 2018
Doubling of A000225(n+1), n >= 0 entries. First differences give A077957. - Wolfdieter Lang, Apr 08 2018
a(n-2) is the number of achiral rows or cycles of length n partitioned into two sets or the number of color patterns using exactly 2 colors. An achiral row or cycle is equivalent to its reverse. Two color patterns are equivalent if the colors are permuted. For n = 4, the a(n-2) = 3 row patterns are AABB, ABAB, and ABBA; the cycle patterns are AAAB, AABB, and ABAB. For n = 5, the a(n-2) = 3 patterns for both rows and cycles are AABAA, ABABA, and ABBBA. For n = 6, the a(n-2) = 7 patterns for rows are AAABBB, AABABB, AABBAA, ABAABA, ABABAB, ABBAAB, and ABBBBA; the cycle patterns are AAAAAB, AAAABB, AAABAB, AAABBB, AABAAB, AABABB, and ABABAB. - Robert A. Russell, Oct 15 2018
For integers m > 1, the expansion of 1/((1 - x)*(1 - m*x^2)) generates a(n) = (sqrt(m)^(n + 1)*((-1)^n*(sqrt(m) - 1) + sqrt(m) + 1) - 2)/(2*(m - 1)). It appears, for integer values of n >= 0 and m > 1, that it could be simplified in the integral domain a(n) = (m^(1 + floor(n/2)) - 1)/(m - 1). - Federico Provvedi, Nov 23 2018
From Werner Schulte, Mar 04 2019: (Start)
More generally: For some fixed integers q and r > 0 the expansion of A(q,r; x) = 1/((1-x)*(1-q*x^r)) generates coefficients a(q,r; n) = (q^(1+floor(n/r))-1)/(q-1) for n >= 0; the special case q = 1 leads to a(1,r; n) = 1 + floor(n/r).
The a(q,r; n) satisfy for n > r a linear recurrence equation with constant coefficients. The signature vector is given by the sum of two vectors v and w where v has terms 1 followed by r zeros, i.e., (1,0,0,...,0), and w has r-1 leading zeros followed by q and -q, i.e., (0,0,...,0,q,-q).
Let a_i(q,r; n) be the convolution inverse of a(q,r; n). The terms are given by the sum a_i(q,r; n) = b(n) + c(n) for n >= 0 where b(n) has terms 1 and -1 followed by infinitely zeros, i.e., (1,-1,0,0,0,...), and c(n) has r leading zeros followed by -q, q and infinitely zeros, i.e., (0,0,...,0,-q,q,0,0,0,...).
Here is an example for q = 3 and r = 5: The expansion of A(3,5; x) = 1/((1-x)*(1-3*x^5)) = Sum_{n>=0} a(3,5; n)*x^n generates the sequence of coefficients (a(3,5; n)) = (1,1,1,1,1,4,4,4,4,4,13,13,13,13,13,40,...) where r = 5 controls the repetition and q = 3 the different values.
The a(3,5; n) satisfy for n > 5 the linear recurrence equation with constant coefficients and signature (1,0,0,0,0,0) + (0,0,0,0,3,-3) = (1,0,0,0,3,-3).
The convolution inverse a_i(3,5; n) has terms (1,-1,0,0,0,0,0,0,0,...) + (0,0,0,0,0,-3,3,0,0,...) = (1,-1,0,0,0,-3,3,0,0,...).
For further examples and informations see A014983 (q,r = -3,1), A077925 (q,r = -2,1), A000035 (q,r = -1,1), A000012 (q,r = 0,1), A000027 (q,r = 1,1), A000225 (q,r = 2,1), A003462 (q,r = 3,1), A077953 (q,r = -2,2), A133872 (q,r = -1,2), A004526 (q,r = 1,2), A052551 (this sequence with q,r = 2,2), A077886 (q,r = -2,3), A088911 (q,r = -1,3), A002264 (q,r = 1,3) and A077885 (q,r = 2,3). The offsets might be different.
(End)
a(n) is the number of palindromes of length n over the alphabet {1,2} containing the letter 1. More generally, the number of palindromes of length n over the alphabet {1,2,...,k} containing the letter 1 is given by k^ceiling(n/2)-(k-1)^ceiling(n/2). - Sela Fried, Dec 10 2024

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 488
Donatella Merlini and Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.3.
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
S. Wolfram, A New Kind of Science
Wolfram Research, Wolfram Atlas of Simple Programs
Index entries for sequences related to cellular automata
Index to 2D 5-Neighbor Cellular Automata
Index to Elementary Cellular Automata
Index entries for linear recurrences with constant coefficients, signature (1,2,-2).

Cf. A000225, A077957, A136865, A289404, A289405, A032085, A050686.
Column 2 (offset by two) of A304972.
Cf. A000225 (oriented), A056326 (unoriented), and A122746(n-2) (chiral) for rows.
Cf. A056295 (oriented), A056357 (unoriented), and A059053 (chiral) for cycles.

Flat(List([1..21],n->[2^n-1,2^n-1])); # Muniru A Asiru, Oct 16 2018

[2^Floor(n/2)-1: n in [2..50]]; // Vincenzo Librandi, Aug 16 2011

spec := [S,{S=Prod(Sequence(Prod(Z,Union(Z,Z))),Sequence(Z))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);

Table[StirlingS2[Floor[n/2] + 2, 2], {n, 0, 50}] (* Robert A. Russell, Dec 20 2017 *)
Drop[LinearRecurrence[{1, 2, -2}, {0, 1, 1}, 50], 1] (* Robert A. Russell, Oct 14 2018 *)
CoefficientList[Series[1/((1-x)*(1-2*x^2)), {x, 0, 50}], x] (* Stefano Spezia, Oct 16 2018 *)
2^(1+Floor[(Range[0,50])/2])-1 (* Federico Provvedi, Nov 22 2018 *)
((-1)^#(Sqrt[2]-1)+Sqrt[2]+1)2^((#-1)/2)-1&@Range[0, 50] (* Federico Provvedi, Nov 23 2018 *)

x='x+O('x^50); Vec(1/((1-x)*(1-2*x^2))) \\ Altug Alkan, Mar 19 2018

[2^(floor(n/2)) -1 for n in (2..50)] # G. C. Greubel, Mar 04 2019

G.f.: 1/((1 - x)*(1 - 2*x^2)).
Recurrence: a(1) = 1, a(0) = 1, -2*a(n) - 1 + a(n+2) = 0.
a(n) = -1 + Sum((1/2)*(1 + 2*alpha)*alpha^(-1 - n)) where the sum is over alpha = the two roots of -1 + 2*x^2.
a(n) = A016116(n+2) - 1. - R. J. Mathar, Jun 15 2009
a(n) = A060546(n+1) - 1. - Filip Zaludek, Dec 10 2016
From Robert A. Russell, Oct 15 2018: (Start)
a(n-2) = S2(floor(n/2)+1,2), where S2 is the Stirling subset number A008277.
a(n-2) = 2*A056326(n) - A000225(n) = A000225(n) - 2*A122746(n-2) = A056326(n) - A122746(n-2).
a(n-2) = 2*A056357(n) - A056295(n) = A056295(n) - 2*A059053(n) = A056357(n) - A059053(n). (End)
From Federico Provvedi, Nov 22 2018: (Start)
a(n) = 2^( 1 + floor(n/2) ) - 1.
a(n) = ( (-1)^n*(sqrt(2)-1) + sqrt(2) + 1 ) * 2^( (n - 1)/2 ) - 1.  (End)
E.g.f.: 2*cosh(sqrt(2)*x) + sqrt(2)*sinh(sqrt(2)*x) - cosh(x) - sinh(x). - Franck Maminirina Ramaharo, Nov 23 2018

More terms from James Sellers, Jun 06 2000

A088911 Period 6: repeat [1, 1, 1, 0, 0, 0].

Original entry on oeis.org

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

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Oct 22 2003

For periodic sequences having a period of 2*k and composed of k ones followed by k zeros we have a(n) = floor(((n+k) mod 2*k)/k). Sequences of this form are A000035(n+1) (k=1), A133872(n) (k=2), this sequence (k=3), A131078(n) (k=4), and A112713(n-1) (k=5). - Gary Detlefs, May 17 2011

Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (1,0,-1,1).

Cf. A000035, A010701, A079979, A133872, A131078, A112713.

&cat [[1, 1, 1, 0, 0, 0]^^30]; // Wesley Ivan Hurt, Jul 05 2016

seq(op([1, 1, 1, 0, 0, 0]), n=0..40); # Wesley Ivan Hurt, Jul 05 2016

CoefficientList[Series[(1 + x + x^2)/(1 - x^6), {x, 0, 50}], x]
Flatten[Table[{1,1,1,0,0,0},{20}]] (* Harvey P. Dale, Jul 17 2011 *)

a(n)=n%6<3 \\ Jaume Oliver Lafont, Mar 17 2009

def A088911(n): return int(n % 6 < 3) # Chai Wah Wu, May 25 2022

G.f.: (1+x+x^2)/(1-x^6) = 1/((1-x)*(1+x)*(1-x+x^2)).
a(n) = a(n-6) for n>=6, a(0)=a(1)=a(2)=1, a(3)=a(4)=a(5)=0.
a(n) = ((-1)^floor((5*n + 2)/3) + 1)/2 = ( (-1)^floor(n/3) + 1 )/2. [Simplified by Bruno Berselli, Jul 09 2013]
a(n) = Sum_{k=0..floor(n/2)} U(n-2k, 1/2). - Paul Barry, Nov 15 2003
From Paul Barry, Mar 14 2004: (Start)
Partial sums of expansion of 1/(1+x^3), see A131531.
a(n) = 2*sin(Pi*n/3 + Pi/6)/3 + cos(Pi*n)/6 + 1/2. (End)
a(n) = floor(((n+3) mod 6)/3).
a(n) = floor((5*n-1)/3) mod 2. - Gary Detlefs, May 17 2011
a(n) = 1/2 + cos(Pi*n/3)/3 + sin(Pi*n/3)/sqrt(3) + (-1)^n/6. - R. J. Mathar, Oct 08 2011
a(n) = floor(((n+2)^2)/3) mod 2. - Wesley Ivan Hurt, Jun 29 2013
a(n) = A079979(n) + A079979(n-1) + A079979(n-2). - R. J. Mathar, Jul 10 2015
a(n) = a(n-1) - a(n-3) + a(n-4) for n > 3. - Wesley Ivan Hurt, Jul 05 2016
a(n) = 2*floor(n/6) - floor(n/3) + 1. - Ridouane Oudra, Dec 14 2021
E.g.f.: (2*cosh(x) + exp(x/2)*(cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2)) + sinh(x))/3. - Stefano Spezia, Aug 04 2025

A212336 Expansion of 1/(1 - 23x + 23x^2 - x^3).

Original entry on oeis.org

1, 23, 506, 11110, 243915, 5355021, 117566548, 2581109036, 56666832245, 1244089200355, 27313295575566, 599648413462098, 13164951800590591, 289029291199530905, 6345479454589089320, 139311518709760434136, 3058507932160140461673

Offset: 0

Bruno Berselli, Jun 08 2012

Partial sums of A077421.

Bruno Berselli, Table of n, a(n) for n = 0..200
Robert K. Moniot, A Family of Integer Sequences, 2021.
Index entries for linear recurrences with constant coefficients, signature (23,-23,1).

Sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3): A334673 (k=24), A212336 (k=23), A212335 (k=22), A097833 (k=21), A097832 (k=20), A049664 (k=19), A097831-A097829 (k=18,17,16), A076139 (k=15), A097828-A097826 (k=14,13,12), A097784 (k=11), A092420 (k=10), A076765 (k=9), A092521 (k=8), A053142 (k=7), A089817(k=6), A061278 (k=5), A027941 (k=4), A000217 (k=3), A021823 (k=2), A133872 (k=1), A079978 (k=0).

m:=17; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/(1-23*x+23*x^2-x^3)));

I:=[1,23,506]; [n le 3 select I[n] else 23*Self(n-1)-23*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Aug 18 2013

a:= n-> (<<0|1|0>, <0|0|1>, <1|-23|23>>^n. <<1, 23, 506>>)[1, 1]:
seq(a(n), n=0..20);  # Alois P. Heinz, Jun 15 2012

CoefficientList[Series[1/(1 - 23 x + 23 x^2 - x^3), {x, 0, 16}], x]
LinearRecurrence[{23, -23, 1}, {1, 23, 506}, 20] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(coeff(taylor(1/(1-23*x+23*x^2-x^3), x, 0, n), x, n), n, 0, 16);

PARI
```
Vec(1/(1-23*x+23*x^2-x^3)+O(x^17))
```

[(1/20)*(-1 +21*chebyshev_U(n, 11) -chebyshev_U(n-1, 11)) for n in (0..30)] # G. C. Greubel, Feb 07 2022

G.f.: 1/((1-x)*(1 - 22*x + x^2)).
a(n) = (((6+sqrt(30))^(2*n+3) + (6-sqrt(30))^(2*n+3))/6^(n+1) - 12)/240.
a(n) = a(-n-3) = 23*a(n-1) - 23*a(n-2) + a(n-3).
a(n)*a(n+2) = a(n+1)*(a(n+1)-1).
a(n+1) - 11*a(n) = A133285(n+2).
11*a(n+1) - a(n) = (1/5)*A157096(n+2).
a(n) = (1/20)*(-1 + 21*ChebyshevU(n, 11) - ChebyshevU(n-1, 11)). - G. C. Greubel, Feb 07 2022

A282011 Number T(n,k) of k-element subsets of [n] having an even sum; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 2, 4, 6, 3, 0, 1, 3, 6, 10, 9, 3, 0, 1, 3, 9, 19, 19, 9, 3, 1, 1, 4, 12, 28, 38, 28, 12, 4, 1, 1, 4, 16, 44, 66, 60, 40, 20, 5, 0, 1, 5, 20, 60, 110, 126, 100, 60, 25, 5, 0, 1, 5, 25, 85, 170, 226, 226, 170, 85, 25, 5, 1, 1, 6, 30, 110, 255, 396, 452, 396, 255, 110, 30, 6, 1

Offset: 0

Alois P. Heinz, Feb 04 2017

Row n is symmetric if and only if n mod 4 in {0,3} (or if T(n,n) = 1).

			T(5,0) = 1: {}.
T(5,1) = 2: {2}, {4}.
T(5,2) = 4: {1,3}, {1,5}, {2,4}, {3,5}.
T(5,3) = 6: {1,2,3}, {1,2,5}, {1,3,4}, {1,4,5}, {2,3,5}, {3,4,5}.
T(5,4) = 3: {1,2,3,4}, {1,2,4,5}, {2,3,4,5}.
T(5,5) = 0.
T(7,7) = 1: {1,2,3,4,5,6,7}.
Triangle T(n,k) begins:
  1;
  1, 0;
  1, 1,  0;
  1, 1,  1,   1;
  1, 2,  2,   2,   1;
  1, 2,  4,   6,   3,   0;
  1, 3,  6,  10,   9,   3,   0;
  1, 3,  9,  19,  19,   9,   3,   1;
  1, 4, 12,  28,  38,  28,  12,   4,   1;
  1, 4, 16,  44,  66,  60,  40,  20,   5,   0;
  1, 5, 20,  60, 110, 126, 100,  60,  25,   5,  0;
  1, 5, 25,  85, 170, 226, 226, 170,  85,  25,  5, 1;
  1, 6, 30, 110, 255, 396, 452, 396, 255, 110, 30, 6, 1;

Alois P. Heinz, Rows n = 0..200, flattened
Johann Cigler, Some remarks on Rogers-Szegö polynomials and Losanitsch's triangle, arXiv:1711.03340 [math.CO], 2017.
Johann Cigler, Some Pascal-like triangles, 2018.

Columns k=0..10 give (offsets may differ): A000012, A004526, A002620, A005993, A005994, A032092, A032093, A018211, A018212, A282077, A282078.
Row sums give A011782.
Main diagonal gives A133872(n+1).
Lower diagonals T(n+j,n) for j=1..10 give: A004525(n+1), A282079, A228705, A282080, A282081, A282082, A282083, A282084, A282085, A282086.
T(2n,n) gives A119358.
Cf. A007318, A057711, A087447, A159916.

b:= proc(n, s) option remember; expand(
      `if`(n=0, s, b(n-1, s)+x*b(n-1, irem(s+n, 2))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n, 1)):
seq(T(n), n=0..16);

Flatten[Table[Sum[Binomial[Ceiling[n/2],2j]Binomial[Floor[n/2],k-2j],{j,0,Floor[(n+1)/4]}],{n,0,10},{k,0,n}]] (* Indranil Ghosh, Feb 26 2017 *)

a(n,k)=sum(j=0,floor((n+1)/4),binomial(ceil(n/2),2*j)*binomial(floor(n/2),k-2*j));
tabl(nn)={for(n=0,nn,for(k=0,n,print1(a(n,k),", "););print(););} \\ Indranil Ghosh, Feb 26 2017

T(n,k) = Sum_{j=0..floor((n+1)/4)} C(ceiling(n/2),2*j) * C(floor(n/2),k-2*j).
T(n,k) = A007318(n,k) - A159916(n,k).
Sum_{k=0..n} k * T(n,k) = A057711(n-1) for n>0.
Sum_{k=0..n} (k+1) * T(n,k) = A087447(n) + [n=2].

cp's OEIS Frontend

A042948 Numbers congruent to 0 or 1 (mod 4).

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A133900 a(n) = period of the sequence {b(m), m>=0}, defined by b(m):=binomial(m+n,n) mod n.

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A133910 Period numbers of A133900 divided by n^2.

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A042964 Numbers that are congruent to 2 or 3 mod 4.

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A021913 Period 4: repeat [0, 0, 1, 1].

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A011848 a(n) = floor(binomial(n, 2)/2).

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A052551 Expansion of 1/((1 - x)(1 - 2x^2)).

A212336 Expansion of 1/(1 - 23x + 23x^2 - x^3).