cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A011655 Period 3: repeat [0, 1, 1].

Original entry on oeis.org

0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

A binary m-sequence: expansion of reciprocal of x^2+x+1 (mod 2).
A Chebyshev transform of the Jacobsthal numbers A001045: if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))*A(x/(1+x^2)). - Paul Barry, Feb 16 2004
This is the r = 1 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.
This is the Fibonacci sequence (A000045) modulo 2. - Stephen Jordan (sjordan(AT)mit.edu), Sep 10 2007
For n > 0: a(n) = A084937(n-1) mod 2. - Reinhard Zumkeller, Dec 16 2007
This is also the Lucas numbers (A000032) mod 2. In general, this is the parity of any Lucas sequence associated with any pair (P,Q) when P and Q are odd; i.e., a(n) = U_n(P,Q) mod 2 = V_n(P,Q) mod 2. See Ribenboim. - Rick L. Shepherd, Feb 07 2009
Starting with offset 1: (1, 1, 0, 1, 1, 0, ...) = INVERTi transform of the tribonacci sequence A001590 starting (1, 2, 3, 6, 11, 20, 37, ...). - Gary W. Adamson, May 04 2009
From Reinhard Zumkeller, Nov 30 2009: (Start)
Characteristic function of numbers coprime to 3.
a(n) = 1 - A079978(n); a(A001651(n)) = 1; a(A008585(n)) = 0;
A000212(n) = Sum_{k=0..n} a(k)*(n-k). (End)
Sum_{k>0} a(k)/k/2^k = log(7)/3. - Jaume Oliver Lafont, Jun 01 2010
The sequence is the principal Dirichlet character of the reduced residue system mod 3 (the other is A102283). Associated Dirichlet L-functions are L(2,chi) = Sum_{n>=1} a(n)/n^2 = 4*Pi^2/27 = A214549, and L(3,chi) = Sum_{n>=1} a(n)/n^3 = 1.157536... = -(psi''(1/3) + psi''(2/3))/54 where psi'' is the tetragamma function. [Jolley eq 309 and arXiv:1008.2547, L(m = 3, r = 1, s)]. - R. J. Mathar, Jul 15 2010
a(n+1), n >= 0, is the sequence of the row sums of the Riordan triangle A158454. - Wolfdieter Lang, Dec 18 2010
Removing the first two elements and keeping the offset at 0, this is a periodic sequence (1, 0, 1, 1, 0, 1, ...). Its INVERTi transform is (1, -1, 2, -2, 2, -2, ...) with period (2,-2). - Gary W. Adamson, Jan 21 2011
Column k = 1 of triangle in A198295. - Philippe Deléham, Jan 31 2012
The set of natural numbers, A000027: (1, 2, 3, ...); is the INVERT transform of the signed periodic sequence (1, 1, 0, -1, -1, 0, 1, 1, 0, ...). - Gary W. Adamson, Apr 28 2013
Any integer sequence s(n) = |s(n-1) - s(n-2)| (equivalently, max(s(n-1), s(n-2)) - min(s(n-1), s(n-2))) for n > i + 1 with s(i) = j and s(i+1) = k, where j and k are not both 0, is or eventually becomes a multiple of this sequence, namely, the sequence repeat gcd(j, k), gcd(j, k), 0 (at some offset). In particular, if j and k are coprime, then s(n) is or eventually becomes this sequence (see, e.g., A110044). - Rick L. Shepherd, Jan 21 2014
For n >= 1, a(n) is also the characteristic function for rational g-adic integers (+n/3)A001651).%20See%20the%20definition%20in%20the%20Mahler%20reference,%20p.%207%20and%20also%20p.%2010.%20-%20_Wolfdieter%20Lang">g and also (-n/3)_g for all integers g >= 2 without a factor 3 (A001651). See the definition in the Mahler reference, p. 7 and also p. 10. - _Wolfdieter Lang, Jul 11 2014
Characteristic function for A007908(n+1) being divisible by 3. a(n) = bit flipped A007908(n+1) (mod 3) = bit flipped A079978(n). - Wolfdieter Lang, Jun 12 2017
Also Jacobi or Kronecker symbol (n/9) (or (n/9^e) for all e >= 1). - Jianing Song, Jul 09 2018
The binomial trans. is 0, 1, 3, 6, 11, 21, 42, 85, 171, 342,.. (see A024495). - R. J. Mathar, Feb 25 2023

Examples

			G.f. = x + x^2 + x^4 + x^5 + x^7 + x^8 + x^10 + x^11 + x^13 + x^14 + x^16 + x^17 + ...

References

  • S. W. Golomb, Shift-Register Sequences, Holden-Day, San Francisco, 1967.
  • H. D. Lueke, Korrelationssignale, Springer 1992, pp. 43-48.
  • F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier/North Holland, 1978, p. 408.
  • K. Mahler, p-adic numbers and their functions, 2nd ed., Cambridge University press, 1981.
  • Paulo Ribenboim, The Little Book of Big Primes. Springer-Verlag, NY, 1991, p. 46. [Rick L. Shepherd, Feb 07 2009]

Links

Crossrefs

Partial sums of A057078 give A011655(n+1).
Cf. A035191 (Mobius transform), A001590, A002487, A049347.
Cf. A000035, A011558, A097325, A109720, A145568, A166486, A168181, A168182, A168184, A168185. - Reinhard Zumkeller, Nov 30 2009
Cf. A000027, A000045, A004523 (partial sums), A057078 (first differences).
Cf. A007908, A079978 (bit flipped).
Cf. A011656 - A011751 for other binary m-sequences.
Cf. A002264.

Programs

Formula

G.f.: (x + x^2) / (1 - x^3) = Sum_{k>0} (x^k - x^(3*k)).
G.f.: x / (1 - x / (1 + x / (1 + x / (1 - 2*x / (1 + x))))). - Michael Somos, Apr 02 2012
a(n) = a(n+3) = a(-n), a(3*n) = 0, a(3*n + 1) = a(3*n + 2) = 1 for all n in Z.
a(n) = (1/2)*( (-1)^(floor((2n + 4)/3)) + 1 ). - Mario Catalani (mario.catalani(AT)unito.it), Oct 22 2003
a(n) = Fibonacci(n) mod 2. - Paul Barry, Nov 12 2003
a(n) = (2/3)*(1 - cos(2*Pi*n/3)). - Ralf Stephan, Jan 06 2004
a(n) = 1 - a(n-1)*a(n-2), a(n) = n for n < 2. - Reinhard Zumkeller, Feb 28 2004
a(n) = 2*(1 - T(n, -1/2))/3 with Chebyshev's polynomials T(n, x) of the first kind; see A053120. - Wolfdieter Lang, Oct 18 2004
a(n) = n*Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k, k)*A001045(n-2k)/(n-k). - Paul Barry, Oct 31 2004
a(n) = A002487(n) mod 2. - Paul Barry, Jan 14 2005
From Bruce Corrigan (scentman(AT)myfamily.com), Aug 08 2005: (Start)
a(n) = n^2 mod 3.
a(n) = (1/3)*(2 - (r^n + r^(2*n))) where r = (-1 + sqrt(-3))/2. (End)
From Michael Somos, Sep 23 2005: (Start)
Euler transform of length 3 sequence [ 1, -1, 1].
Moebius transform is length 3 sequence [ 1, 0, -1].
Multiplicative with a(3^e) = 0^e, a(p^e) = 1 otherwise. (End)
From Hieronymus Fischer, Jun 27 2007: (Start)
a(n) = (4/3)*(|sin(Pi*(n-2)/3)| + |sin(Pi*(n-1)/3)|)*|sin(Pi*n/3)|.
a(n) = ((n+1) mod 3 + 1) mod 2 = (1 - (-1)^(n - 3*floor((n+1)/3)))/2. (End)
a(n) = 2 - a(n-1) - a(n-2) for n > 1. - Reinhard Zumkeller, Apr 13 2008
a(2*n+1) = a(n+1) XOR a(n), a(2*n) = a(n), a(1) = 1, a(0) = 0. - Reinhard Zumkeller, Dec 27 2008
Sum_{n>=1} a(n)/n^s = (1-1/3^s)*Riemann_zeta(s), s > 1. - R. J. Mathar, Jul 31 2010
a(n) = floor((4*n-5)/3) mod 2. - Gary Detlefs, May 15 2011
a(n) = (a(n-1) - a(n-2))^2 with a(0) = 0, a(1) = 1. - Francesco Daddi, Aug 02 2011
Convolution of A040000 with A049347. - R. J. Mathar, Jul 21 2012
G.f.: Sum_{k>0} x^A001651(k). - L. Edson Jeffery, Dec 05 2012
G.f.: x/(G(0) - x^2) where G(k) = 1 - x/(x + 1/(1 - x/G(k+1))); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 15 2013
For the general case: The characteristic function of numbers that are not multiples of m is a(n) = floor((n-1)/m) - floor(n/m) + 1, with m,n > 0. - Boris Putievskiy, May 08 2013
a(n) = sign(n mod 3). - Wesley Ivan Hurt, Jun 22 2013
a(n) = A000035(A000032(n)) = A000035(A000045(n)). - Omar E. Pol, Oct 28 2013
a(n) = (-n mod 3)^((n-1) mod 3). - Wesley Ivan Hurt, Apr 16 2015
a(n) = (2/3) * (1 - sin((Pi/6) * (4*n + 3))) for n >= 0. - Werner Schulte, Jul 20 2017
a(n) = a(n-1) XOR a(n-2) with a(0) = 0, a(1) = 1. - Chunqing Liu, Dec 18 2022
a(n) = floor((n+2)/3) - floor(n/3) = A002264(n+2) - A002264(n). - Aaron J Grech, Jul 30 2024
E.g.f.: 2*(exp(x) - exp(-x/2)*cos(sqrt(3)*x/2))/3. - Stefano Spezia, Mar 30 2025
Dirichlet g.f.: zeta(s)*(1-1/3^s). - R. J. Mathar, Aug 10 2025

Extensions

Better name from Omar E. Pol, Oct 28 2013

A007598 Squared Fibonacci numbers: a(n) = F(n)^2 where F = A000045.

Original entry on oeis.org

0, 1, 1, 4, 9, 25, 64, 169, 441, 1156, 3025, 7921, 20736, 54289, 142129, 372100, 974169, 2550409, 6677056, 17480761, 45765225, 119814916, 313679521, 821223649, 2149991424, 5628750625, 14736260449, 38580030724, 101003831721, 264431464441, 692290561600
Offset: 0

Views

Author

N. J. A. Sloane, Robert G. Wilson v

Keywords

Comments

a(n)*(-1)^(n+1) = (2*(1-T(n,-3/2))/5), n>=0, with Chebyshev's polynomials T(n,x) of the first kind, is the r=-1 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found. - Wolfdieter Lang, Oct 18 2004
From Giorgio Balzarotti, Mar 11 2009: (Start)
Determinant of power series with alternate signs of gamma matrix with determinant 1!.
a(n) = Determinant(A - A^2 + A^3 - A^4 + A^5 - ... - (-1)^n*A^n) where A is the submatrix A(1..2,1..2) of the matrix with factorial determinant.
A = [[1,1,1,1,1,1,...], [1,2,1,2,1,2,...], [1,2,3,1,2,3,...], [1,2,3,4,1,2,...], [1,2,3,4,5,1,...], [1,2,3,4,5,6,...], ...]; note: Determinant A(1..n,1..n) = (n-1)!.
a(n) is even with respect to signs of power of A.
See A158039...A158050 for sequence with matrix 2!, 3!, ... (End)
Equals the INVERT transform of (1, 3, 2, 2, 2, ...). Example: a(7) = 169 = (1, 1, 4, 9, 25, 64) dot (2, 2, 2, 2, 3, 1) = (2 + 2 + 8 + 18 + 75 + 64). - Gary W. Adamson, Apr 27 2009
This is a divisibility sequence.
a(n+1)*(-1)^n, n>=0, is the sequence of the alternating row sums of the Riordan triangle A158454. - Wolfdieter Lang, Dec 18 2010
a(n+1) is the number of tilings of a 2 X 2n rectangle with n tetrominoes of any shape, cf. A230031. - Alois P. Heinz, Nov 29 2013
This is the case P1 = 1, P2 = -6, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 31 2014
Differences between successive golden rectangle numbers A001654. - Jonathan Sondow, Nov 05 2015
a(n+1) is the number of 2 X n matrices that can be obtained from a 2 X n matrix by moving each element to an adjacent position, horizontally or vertically. This is because F(n+1) is the number of domino tilings of that matrix, therefore with a checkerboard coloring and two domino tilings we can move the black element of each domino of the first tiling to the white element of the same domino and similarly move the white element of each domino of the second tiling to the black element of the same domino. - Fabio Visonà, May 04 2022
In general, squaring the terms of a second-order linear recurrence with signature (c,d) will result in a third-order linear recurrence with signature (c^2+d,(c^2+d)*d,-d^3). - Gary Detlefs, Jan 05 2023

Examples

			G.f. = x + x^2 + 4*x^3 + 9*x^4 + 25*x^5 + 64*x^6 + 169*x^7 + 441*x^8 + ...

References

  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 8.
  • Ross Honsberger, Mathematical Gems III, M.A.A., 1985, p. 130.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Richard P. Stanley, Enumerative Combinatorics I, Example 4.7.14, p. 251.

Links

Crossrefs

Cf. A000032, A000045, A001254, A001622, A001654, A047946, A056570, A059260, A061646, A065885, A079291, A080097, A105393.
Bisection of A006498 and A074677. First differences of A001654.
Second row of array A103323.
Half of A175395.

Programs

  • GAP
    List([0..30], n -> Fibonacci(n)^2); # G. C. Greubel, Dec 10 2018
  • Haskell
    a007598 = (^ 2) . a000045  -- Reinhard Zumkeller, Sep 01 2013
  • Magma
    [Fibonacci(n)^2: n in [0..30]]; // Vincenzo Librandi, Apr 14 2011
  • Maple
    with(combinat): seq(fibonacci(n)^2, n=0..27); # Zerinvary Lajos, Sep 21 2007
  • Mathematica
    f[n_] := Fibonacci[n]^2; Array[f, 4!, 0] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2009 *)
LinearRecurrence[{2,2,-1},{0,1,1},41] (* Harvey P. Dale, May 18 2011 *)
  • PARI
    {a(n) = fibonacci(n)^2};
  • PARI
    concat(0, Vec(x*(1-x)/((1+x)*(1-3*x+x^2)) + O(x^30))) \\ Altug Alkan, Nov 06 2015
  • Python
    from sympy import fibonacci
def A007598(n): return fibonacci(n)**2 # Chai Wah Wu, Apr 14 2025
  • Sage
    [(fibonacci(n))^2 for n in range(0, 28)]# Zerinvary Lajos, May 15 2009
  • Sage
    [fibonacci(n)^2 for n in range(30)] # G. C. Greubel, Dec 10 2018

Formula

G.f.: x*(1-x)/((1+x)*(1-3*x+x^2)).
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), n > 2. a(0)=0, a(1)=1, a(2)=1.
a(-n) = a(n) for all n in Z.
a(n) = A080097(n-2) + 1.
L.g.f.: 1/5*log((1+3*x+x^2)/(1-6*x+x^2)) = Sum_{n>=0} a(n)/n*x^n; special case of l.g.f. given in A079291. - Joerg Arndt, Apr 13 2011
a(0) = 0, a(1) = 1; a(n) = a(n-1) + Sum(a(n-i)) + k, 0 <= i < n where k = 1 when n is odd, or k = -1 when n is even. E.g., a(2) = 1 = 1 + (1 + 1 + 0) - 1, a(3) = 4 = 1 + (1 + 1 + 0) + 1, a(4) = 9 = 4 + (4 + 1 + 1 + 0) - 1, a(5) = 25 = 9 + (9 + 4 + 1 + 1 + 0) + 1. - Sadrul Habib Chowdhury (adil040(AT)yahoo.com), Mar 02 2004
a(n) = (2*Fibonacci(2*n+1) - Fibonacci(2*n) - 2*(-1)^n)/5. - Ralf Stephan, May 14 2004
a(n) = F(n-1)*F(n+1) - (-1)^n = A059929(n-1) - A033999(n).
Sum_{j=0..2*n} binomial(2*n,j)*a(j) = 5^(n-1)*A005248(n+1) for n >= 1 [P. Stanica]. Sum_{j=0..2*n+1} binomial(2*n+1,j)*a(j) = 5^n*A001519(n+1) [P. Stanica]. - R. J. Mathar, Oct 16 2006
a(n) = (A005248(n) - 2*(-1)^n)/5. - R. J. Mathar, Sep 12 2010
a(n) = (-1)^k*(Fibonacci(n+k)^2-Fibonacci(k)*Fibonacci(2*n+k)), for any k. - Gary Detlefs, Dec 13 2010
a(n) = 3*a(n-1) - a(n-2) + 2*(-1)^(n+1), n > 1. - Gary Detlefs, Dec 20 2010
a(n) = Fibonacci(2*n-2) + a(n-2). - Gary Detlefs, Dec 20 2010
a(n) = (Fibonacci(3*n) - 3*(-1)^n*Fibonacci(n))/(5*Fibonacci(n)), n > 0. - Gary Detlefs, Dec 20 2010
a(n) = (Fibonacci(n)*Fibonacci(n+4) - 3*Fibonacci(n)*Fibonacci(n+1))/2. - Gary Detlefs, Jan 17 2011
a(n) = (((3+sqrt(5))/2)^n + ((3-sqrt(5))/2)^n - 2*(-1)^n)/5; without leading zero we would have a(n) = ((3+sqrt(5))*((3+sqrt(5))/2)^n + (3-sqrt(5))*((3-sqrt(5))/2)^n + 4*(-1)^n)/10. - Tim Monahan, Jul 17 2011
E.g.f.: (exp((phi+1)*x) + exp((2-phi)*x) - 2*exp(-x))/5, with the golden section phi:=(1+sqrt(5))/2. From the Binet-de Moivre formula for F(n). - Wolfdieter Lang, Jan 13 2012
Starting with "1" = triangle A059260 * the Fibonacci sequence as a vector. - Gary W. Adamson, Mar 06 2012
a(0) = 0, a(1) = 1; a(n+1) = (a(n)^(1/2) + a(n-1)^(1/2))^2. - Thomas Ordowski, Jan 06 2013
a(n) + a(n-1) = A001519(n), n > 0. - R. J. Mathar, Mar 19 2014
From Peter Bala, Mar 31 2014: (Start)
a(n) = ( T(n,alpha) - T(n,beta) )/(alpha - beta), where alpha = 3/2 and beta = -1 and T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 3/2; 1, 1/2].
a(n) = U(n-1,i/2)*U(n-1,-i/2), where U(n,x) denotes the Chebyshev polynomial of the second kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials and 4th-order linear divisibility sequences. (End)
a(n) = (F(n+2)*F(n+3) - L(n)*L(n+1))/3 for F = A000045 and L = A000032. - J. M. Bergot, Jun 02 2014
0 = a(n)*(+a(n) - 2*a(n+1) - 2*a(n+2)) + a(n+1)*(+a(n+1) - 2*a(n+2)) + a(n+2)*(+a(n+2)) for all n in Z. - Michael Somos, Jun 03 2014
(F(n)*b(n+2))^2 + (F(n+1)*b(n-1))^2 = F(2*n+1)^3 = A001519(n+1)^3, with b(n) = a(n) + 2*(-1)^n and F(n) = A000045(n) (see Bruckman link). - Michel Marcus, Jan 24 2015
a(n) = 1/4*( a(n-2) - a(n-1) - a(n+1) + a(n+2) ). The same recurrence holds for A001254. - Peter Bala, Aug 18 2015
a(n) = F(n)*F(n+1) - F(n-1)*F(n). - Jonathan Sondow, Nov 05 2015
For n>2, a(n) = F(n-2)*(3*F(n-1) + F(n-3)) + F(2*n-5). Also, for n>2 a(n)=2*F(n-3)*F(n) + F(2*n-3) -(2)*(-1)^n. - J. M. Bergot, Nov 05 2015
a(n) = (F(n+2)^2 + L(n+1)^2) - 2*F(n+2)*L(n+1). - J. M. Bergot, Nov 08 2015
a(n) = F(n+3)^2 - 4*F(n+1)*F(n+2). - J. M. Bergot, Mar 17 2016
a(n) = (F(n-2)*F(n+2) + F(n-1)*F(n+1))/2. - J. M. Bergot, May 25 2017
4*a(n) = L(n+1)*L(n-1) - F(n+2)*F(n-2), where L = A000032. - Bruno Berselli, Sep 27 2017
a(n) = F(n+k)*F(n-k) + (-1)^(n+k)*a(k), for every integer k >= 0. - Federico Provvedi, Dec 10 2018
From Peter Bala, Nov 19 2019: (Start)
Sum_{n >= 3} 1/(a(n) - 1/a(n)) = 4/9.
Sum_{n >= 3} (-1)^n/(a(n) - 1/a(n)) = (10 - 3*sqrt(5))/18.
Conjecture: Sum_{n >= 1, n != 2*k+1} 1/(a(n) + (-1)^n*a(2*k+1)) = 1/a(4*k+2) for k = 0,1,2,.... (End)
Sum_{n>=1} 1/a(n) = A105393. - Amiram Eldar, Oct 22 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi (A001622) (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A001752 Expansion of 1/((1+x)*(1-x)^5).

Original entry on oeis.org

1, 4, 11, 24, 46, 80, 130, 200, 295, 420, 581, 784, 1036, 1344, 1716, 2160, 2685, 3300, 4015, 4840, 5786, 6864, 8086, 9464, 11011, 12740, 14665, 16800, 19160, 21760, 24616, 27744, 31161, 34884, 38931, 43320, 48070, 53200, 58730, 64680, 71071, 77924, 85261
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Define a unit column of a binary matrix to be a column with only one 1. a(n) = number of 3 X n binary matrices with 1 unit column up to row and column permutations (if offset is 1). - Vladeta Jovovic, Sep 09 2000
Generally, number of 3 X n binary matrices with k=0,1,2,... unit columns, up to row and column permutations, is the coefficient of x^k in 1/6*(Z(S_n; 5 + 3*x,5 + 3*x^2, ...) + 3*Z(S_n; 3 + x,5 + 3*x^2,3 + x^3,5 + 3*x^4, ...) + 2*Z(S_n; 2,2,5 + 3*x^3,2,2,5 + 3*x^6, ...)), where Z(S_n; x_1,x_2,...,x_n) is the cycle index of symmetric group S_n of degree n.
First differences of a(n) give number of minimal 3-covers of an unlabeled n-set that cover 4 points of that set uniquely (if offset is 4).
Transform of tetrahedral numbers, binomial(n+3,3), under the Riordan array (1/(1-x^2), x). - Paul Barry, Apr 16 2005
Equals triangle A152205 as an infinite lower triangular matrix * [1, 2, 3, ...]. - Gary W. Adamson, Feb 14 2010
With a leading zero, number of all possible octahedra of any size, formed when intersecting a regular tetrahedron by planes parallel to its sides and dividing its edges into n equal parts. - V.J. Pohjola, Sep 13 2012
With 2 leading zeros and offset 1, the sequence becomes 0,0,1,4,11,... for n=1,2,3,... Call this b(n). Consider the partitions of n into two parts (p,q) with p <= q. Then b(n) is the total volume of the family of rectangular prisms with dimensions p, |q - p| and |q - p|. - Wesley Ivan Hurt, Apr 14 2018
Conjecture: For n > 2, a(n-3) is the absolute value of the coefficient of the term [x^(n-2)] in the characteristic polynomial of an n X n square matrix M(n) defined as the n-th principal submatrix of the array A010751 whose general element is given by M[i,j] = floor((j - i + 1)/2). - Stefano Spezia, Jan 12 2022
Consider the following drawing of the complete graph on n vertices K_n: Vertices 1, 2, ..., n are on a straight line. Any pair of nonconsecutive vertices (i, j) with i < j is connected by a semicircle that goes above the line if i is odd, and below if i is even. With four leading zeros and offset 1, a(n) gives the number of edge crossings of the aforementioned drawing of K_n. - Carlo Francisco E. Adajar, Mar 17 2022

Examples

			There are 4 binary 3 X 2 matrices with 1 unit column up to row and column permutations:
  [0 0] [0 0] [0 1] [0 1]
  [0 0] [0 1] [0 1] [0 1]
  [0 1] [1 1] [1 0] [1 1].
For n=5, the numbers of the octahedra, starting from the smallest size, are Te(5)=35, Te(3)=10, and Te(1)=1, the sum being 46. Te denotes the tetrahedral number A000292. - _V.J. Pohjola_, Sep 13 2012

References

  • T. A. Saaty, The Minimum Number of Intersections in Complete Graphs, Proc. Natl. Acad. Sci. USA., 52 (1964), 688-690.

Links

Crossrefs

Cf. A000217, A000292, A000332, A000929, A002620, A004526, A010751, A056885, A057524, A108561, A152205, A216172, A216173, A216175.
Cf. A001753 (partial sums), A002623 (first differences), A158454 (signed column k=2), A169792 (binomial transform).

Programs

  • Magma
    [Floor(((n+3)^2-1)*((n+3)^2-3)/48): n in [0..40]]; // Vincenzo Librandi, Aug 15 2011
  • Maple
    A001752:=n->(3*(-1)^n+93+168*n+100*n^2+24*n^3+2*n^4)/96:
seq(A001752(n), n=0..50); # Wesley Ivan Hurt, Apr 01 2015
  • Mathematica
    a = {1, 4}; Do[AppendTo[a, a[[n - 2]] + (n*(n + 1)*(n + 2))/6], {n, 3, 10}]; a
(* Number of octahedra *) nnn = 100; Teo[n_] := (n - 1) n (n + 1)/6
Table[Sum[Teo[n - nn], {nn, 0, n - 1, 2}], {n, 1, nnn}]
(* V.J. Pohjola, Sep 13 2012 *)
LinearRecurrence[{4,-5,0,5,-4,1},{1,4,11,24,46,80},50] (* Harvey P. Dale, Feb 07 2019 *)
  • PARI
    a(n)=if(n<0,0,((n+3)^2-1)*((n+3)^2-3)/48-if(n%2,1/16))
  • PARI
    a(n)=(n^4+12*n^3+50*n^2+84*n+46)\/48 \\ Charles R Greathouse IV, Sep 13 2012

Formula

a(n) = floor(((n+3)^2 - 1)*((n+3)^2 - 3)/48).
G.f.: 1/((1+x)*(1-x)^5).
a(n) - 2*a(n-1) + a(n-2) = A002620(n+2).
a(n) + a(n-1) = A000332(n+4).
a(n) - a(n-2) = A000292(n+1).
a(n) = Sum_{k=0..n} (-1)^(n-k)*C(k+4, 4). - Paul Barry, Jul 01 2003
a(n) = (3*(-1)^n + 93 + 168*n + 100*n^2 + 24*n^3 + 2*n^4)/96. - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004
From Paul Barry, Apr 16 2005: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(n-2k+3, 3).
a(n) = Sum_{k=0..n} binomial(k+3, 3)*(1-(-1)^(n+k-1))/2. (End)
a(n) = A108561(n+5,n) for n > 0. - Reinhard Zumkeller, Jun 10 2005
From Wesley Ivan Hurt, Apr 01 2015: (Start)
a(n) = 4*a(n-1) - 5*a(n-2) + 5*(n-4) - 4*a(n-5) + a(n-6).
a(n) = Sum_{i=0..n+3} (n+3-i) * floor(i^2/2)/2. (End)
Boas-Buck recurrence: a(n) = (1/n)*Sum_{p=0..n-1} (5 + (-1)^(n-p))*a(p), n >= 1, a(0) = 1. See the Boas-Buck comment in A158454 (here for the unsigned column k = 2 with offset 0). - Wolfdieter Lang, Aug 10 2017
Convolution of A000217 and A004526. - R. J. Mathar, Mar 29 2018
E.g.f.: ((48 + 147*x + 93*x^2 + 18*x^3 + x^4)*cosh(x) + (45 + 147*x + 93*x^2 + 18*x^3 + x^4)*sinh(x))/48. - Stefano Spezia, Jan 12 2022

Extensions

Formulae corrected by Bruno Berselli, Sep 13 2012

A115141 Convolution of A115140 with itself.

Original entry on oeis.org

1, -2, -1, -2, -5, -14, -42, -132, -429, -1430, -4862, -16796, -58786, -208012, -742900, -2674440, -9694845, -35357670, -129644790, -477638700, -1767263190, -6564120420, -24466267020, -91482563640, -343059613650, -1289904147324, -4861946401452, -18367353072152
Offset: 0

Views

Author

Wolfdieter Lang, Jan 13 2006

Keywords

Comments

This is the so-called A-sequence for the Riordan triangles A053122, A110162, A129818, A158454 and signed A158909. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. Wolfdieter Lang, Dec 20 2010. [Revised, Nov 13 2012, Nov 22 2012 and Oct 22 2019]
a(n)*(-1)^n is the A-sequence for the Riordan triangle A111125. - Wolfdieter Lang, Jun 26 2011

Examples

			G.f. = 1 - 2*x - x^2 - 2*x^3 - 5*x^4 - 14*x^5 - 42*x^6 - 132*x^7 - 429*x^8 + ...

Links

Crossrefs

Cf. A000108, A115140, A115142 - A115149.

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (1-2*x+Sqrt(1-4*x))/2 )); // G. C. Greubel, Feb 12 2019
  • Mathematica
    a[n_] := -First[ ListConvolve[ cc = Array[ CatalanNumber, n-1, 0], cc]]; a[0] = 1; a[1] = -2; Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Oct 21 2011 *)
CoefficientList[Series[(1-2*x+Sqrt[1-4*x])/2, {x, 0, 30}], x] (* G. C. Greubel, Feb 12 2019 *)
  • PARI
    {a(n) = if( n<1, n==0, -(n==1) -binomial( 2*n-2, n-1) / n)} /* Michael Somos, Mar 28 2012 */
  • Sage
    ((1-2*x+sqrt(1-4*x))/2).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 12 2019

Formula

O.g.f.: 1/c(x)^2 = (1-x) - x*c(x) with the o.g.f. c(x) = (1-sqrt(1-4*x) )/(2*x) of A000108 (Catalan numbers).
a(0)=1, a(1)=-2, a(n) = -C(n-1), n>=2, with C(n):=A000108(n) (Catalan). The start [1, -2] is row n=2 of signed A034807 (signed Lucas polynomials). See A115149 and A034807 for comments.
The convolution inverse is A000108(x)^2. - Michael Somos, Mar 28 2012
REVERT transform is A069271. - Michael Somos, Mar 28 2012
EULER transform of -A060165. - Michael Somos, Mar 28 2012
D-finite with recurrence: n*a(n) +2*(-2*n+3)*a(n-1)=0. - R. J. Mathar, Feb 21 2020

A001769 Expansion of 1/((1+x)*(1-x)^7).

Original entry on oeis.org

1, 6, 22, 62, 148, 314, 610, 1106, 1897, 3108, 4900, 7476, 11088, 16044, 22716, 31548, 43065, 57882, 76714, 100386, 129844, 166166, 210574, 264446, 329329, 406952, 499240, 608328, 736576, 886584, 1061208, 1263576, 1497105, 1765518, 2072862, 2423526, 2822260, 3274194, 3784858
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A002620, A002623, A001752, A001753 (first differences), A158454 (signed column k=3), A001779 (partial sums), A169794 (binomial transf.).

Programs

  • Magma
    [(4*n^6+96*n^5+910*n^4+4320*n^3+10696*n^2+12864*n+5715)/5760+(-1)^n/128: n in [0..40]]; // Vincenzo Librandi, Aug 15 2011
  • Mathematica
    CoefficientList[Series[1/((1+x)(1-x)^7),{x,0,30}],x] (* or *) LinearRecurrence[ {6,-14,14,0,-14,14,-6,1},{1,6,22,62,148,314,610,1106},40] (* Harvey P. Dale, May 24 2015 *)
  • PARI
    a(n)=(4*n^6+96*n^5+910*n^4+4320*n^3+10696*n^2+12864*n)\/5760+1 \\ Charles R Greathouse IV, Apr 17 2012

Formula

From Paul Barry, Jul 01 2003: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k)*C(k+6, 6).
a(n) = (4*n^6 +96*n^5 +910*n^4 +4320*n^3 +10696*n^2 +12864*n+5715)/5760+(-1)^n/128. (End)
Boas-Buck recurrence: a(n) = (1/n)*Sum_{p=0..n-1} (7 + (-1)^(n-p))*a(p), n >= 1, a(0) = 1. See the Boas-Buck comment in A046521 (here for the unsigned column k = 3 with offset 0).
a(n)+a(n+1) = A000579(n+7). - R. J. Mathar, Jan 06 2021

A123521 Triangle read by rows: T(n,k)=number of tilings of a 2 X n grid with k pieces of 1 X 2 tiles (in horizontal position) and 2n-2k pieces of 1 X 1 tiles (0<=k<=n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 4, 4, 1, 6, 11, 6, 1, 1, 8, 22, 24, 9, 1, 10, 37, 62, 46, 12, 1, 1, 12, 56, 128, 148, 80, 16, 1, 14, 79, 230, 367, 314, 130, 20, 1, 1, 16, 106, 376, 771, 920, 610, 200, 25, 1, 18, 137, 574, 1444, 2232, 2083, 1106, 295, 30, 1, 1, 20, 172, 832, 2486, 4744, 5776, 4352, 1897, 420, 36
Offset: 0

Views

Author

Emeric Deutsch, Oct 16 2006

Keywords

Comments

Also the triangle of the coefficients of the squares of the Fibonacci polynomials. Row n has 1+2*floor(n/2) terms. Sum of terms in row n = (Fibonacci(n+1))^2 (A007598).
From Michael A. Allen, Jun 24 2020: (Start)
T(n,k) is the number of tilings of an n-board (a board with dimensions n X 1) using k (1/2, 1/2)-fence tiles and 2*(n-k) half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal). A (1/2, 1/2)-fence is a tile composed of two 1/2 X 1 pieces separated by a gap of width 1/2.
T(n,k) is the (n, (n-k))-th entry of the (1/(1-x^2), x/(1-x)^2) Riordan array.
(-1)^(n+k)*T(n,k) is the (n, (n-k))-th entry of the (1/(1-x^2), x/(1+x)^2) Riordan array (A158454). (End)

Examples

			T(3,1)=4 because the 1 X 2 tile can be placed in any of the four corners of the 2 X 3 grid.
The irregular triangle begins as:
  1;
  1;
  1,  2,   1;
  1,  4,   4;
  1,  6,  11,   6,    1;
  1,  8,  22,  24,    9;
  1, 10,  37,  62,   46,   12,    1;
  1, 12,  56, 128,  148,   80,   16;
  1, 14,  79, 230,  367,  314,  130,   20,    1;
  1, 16, 106, 376,  771,  920,  610,  200,   25;
  1, 18, 137, 574, 1444, 2232, 2083, 1106,  295,  30,  1;
  1, 20, 172, 832, 2486, 4744, 5776, 4352, 1897, 420, 36;

References

  • Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.

Links

Crossrefs

Cf. A007598, A158454.
Other triangles related to tiling using fences: A059259, A157897, A335964.

Programs

  • Magma
    function A123521(n,k)
  if k eq 0 then return 1;
  elif k eq 1 then return 2*(n-1);
  else return A123521(n-2,k-2) + Binomial(2*n-k-1, 2*n-2*k-1);
  end if; return A123521;
end function;
[A123521(n,k): k in [0..2*Floor(n/2)], n in [0..14]]; // G. C. Greubel, Sep 01 2022
  • Maple
    G:=(1-t*z)/(1+t*z)/(1-z-2*t*z+t^2*z^2): Gser:=simplify(series(G,z=0,14)): for n from 0 to 11 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 11 do seq(coeff(P[n],t,k),k=0..2*floor(n/2)) od; # yields sequence in triangular form
  • Mathematica
    Block[{T}, T[0, 0]= T[1, 0]= 1; T[n_, k_]:= Which[k==0, 1, k==1, 2(n-1), True, T[n -2, k-2] + Binomial[2n-k-1, 2n-2k-1]]; Table[T[n, k], {n, 0, 14}, {k, 0, 2 Floor[n/2]}]] // Flatten (* Michael De Vlieger, Jun 24 2020 *)
  • SageMath
    @CachedFunction
def T(n,k): # T = A123521
    if (k==0): return 1
    elif (k==1): return 2*(n-1)
    else: return T(n-2, k-2) + binomial(2*n-k-1, 2*n-2*k-1)
flatten([[T(n,k) for k in (0..2*(n//2))] for n in (0..12)]) # G. C. Greubel, Sep 01 2022

Formula

G.f.: G = (1-t*z)/((1+t*z)*(1-z-2*t*z+t^2*z^2)). G = 1/(1-g), where g = z+t^2*z^2+2*t*z^2/(1-t*z) is the g.f. of the indecomposable tilings, i.e., of those that cannot be split vertically into smaller tilings. The row generating polynomials are P(n) = (Fibonacci(n))^2. They satisfy the recurrence relation P(n) = (1+t)*(P(n-1) + t*P(n-2)) - t^3*P(n-3).
T(n,k) = T(n-2,k-2) + binomial(2*n-k-1, 2*n-2*k-1). - Michael A. Allen, Jun 24 2020

A299989 Triangle read by rows: T(n,0) = 0 for n >= 0; T(n,2*k+1) = A152842(2*n,2*(n-k)) and T(n,2*k) = A152842(2*n,2*(n-k)+1) for n >= k > 0.

Original entry on oeis.org

0, 1, 0, 3, 4, 1, 0, 9, 24, 22, 8, 1, 0, 27, 108, 171, 136, 57, 12, 1, 0, 81, 432, 972, 1200, 886, 400, 108, 16, 1, 0, 243, 1620, 4725, 7920, 8430, 5944, 2810, 880, 175, 20, 1, 0, 729, 5832, 20898, 44280, 61695, 59472, 40636, 19824, 6855, 1640, 258, 24, 1
Offset: 0

Views

Author

Franck Maminirina Ramaharo, Feb 26 2018

Keywords

Comments

T(n,k) is the number of state diagrams having k components of n connected summed trefoil knots.
Row sums gives A001018.

Examples

			The triangle T(n, k) begins:
n\k 0     1      2      3       4       5       6      7        8       9
0:  0     1
1:  0     3      4      1
2:  0     9     24     22       8       1
3:  0    27    108    171     136      57      12       1
4:  0    81    432    972    1200     886     400     108      16       1

References

  • V. I. Arnold, Topological Invariants of Plane Curves and Caustics, American Math. Soc., 1994.

Links

Crossrefs

Row 2: row 5 of A158454.
Row 3: row 2 of A220665.
Row 4: row 5 of A219234.

Programs

  • Mathematica
    row[n_] := CoefficientList[x*(x^2 + 4*x + 3)^n, x]; Array[row, 7, 0] // Flatten (* Jean-François Alcover, Mar 16 2018 *)
  • Maxima
    g(x, y) := taylor(x/(1 - y*(x^2 + 4*x + 3)), y, 0, 10)$
a : makelist(ratcoef(g(x, y), y, n), n, 0, 10)$
T : []$
for i:1 thru 11 do
  T : append(T, makelist(ratcoef(a[i], x, n), n, 0, 2*i - 1))$
T;
  • PARI
    T(n, k) = polcoeff(x*(x^2 + 4*x + 3)^n, k);
tabf(nn) = for (n=0, nn, for (k=0, 2*n+1, print1(T(n, k), ", ")); print); \\ Michel Marcus, Mar 03 2018

Formula

T(n,k) = coefficients of x*(x^2 + 4*x + 3)^n.
T(n,k) = T(n-1,k-2) + 4*T(n-1,k-1) + 3*T(n-1,k), with T(n,0) = 0, T(n,1) = 3^n and T(n,2) = 4*n*3^(n-1).
T(n,n+k+1) = A152842(2*n,n+k) and T(n,n-k) = A152842(2*n,n+k+1), for n >= k >= 0.
T(n,1) = A000244(n).
T(n,2) = A120908(n).
T(n,n+1) = A069835(n).
T(n,2*n-1) = A139272(n).
T(n,2*n) = A008586(n).
T(n,2*n-2) = A140138(4*n) = A185872(2n,2) for n >= 1.
G.f.: x/(1 - y*(x^2 + 4*x + 3)).

Extensions

Typo in row 6 corrected by Jean-François Alcover, Mar 16 2018

A001780 Expansion of 1/((1+x)(1-x)^9).

Original entry on oeis.org

1, 8, 37, 128, 367, 920, 2083, 4352, 8518, 15792, 27966, 47616, 78354, 125136, 194634, 295680, 439791, 641784, 920491, 1299584, 1808521, 2483624, 3369301, 4519424, 5998876, 7885280, 10270924, 13264896
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A001769, A158454 (signed column k=4), A001779 (first differences), A169796 (binomial trans.).

Programs

Formula

a(n) = 431*n/168 + (-1)^n/512 + 391*n^3/288 + 26011*n^2/10080 + 797*n^4/1920 + 11*n^5/144 + n^6/120 + n^7/2016 + n^8/80640 + 511/512. - R. J. Mathar, Mar 15 2011
Boas-Buck recurrence: a(n) = (1/n)*Sum_{p=0..n-1} (9 + (-1)^(n-p))*a(p), n >= 1, a(0) = 1. See the Boas-Buck comment in A046521 (here for the unsigned column k = 4 with offset 0). - Wolfdieter Lang, Aug 10 2017
a(n)+a(n+1) = A000581(n+9) . - R. J. Mathar, Jan 06 2021

A001786 Expansion of 1/((1+x)*(1-x)^11).

Original entry on oeis.org

1, 10, 56, 230, 771, 2232, 5776, 13672, 30086, 62292, 122464, 230252, 416394, 727672, 1233584, 2035176, 3276559, 5159726, 7963384, 12066626, 17978389, 26373776, 38138464, 54422576, 76705564, 106873832, 147313024, 201017112, 271716644, 364028752, 483631776, 637467632, 833975341
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A001780, A158454 (signed column k=5).
Eleventh column of A112465.

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 50);
Coefficients(R!( 1/((1+x)*(1-x)^11) )); // G. C. Greubel, Apr 20 2025
  • Mathematica
    CoefficientList[Series[1/((1+x)(1-x)^11),{x,0,50}],x] (* Vincenzo Librandi, Feb 24 2012 *)
LinearRecurrence[{10,-44,110,-165,132,0,-132,165,-110,44,-10,1},{1,10,56,230,771,2232, 5776,13672,30086,62292,122464,230252},30] (* Harvey P. Dale, Oct 22 2015 *)
  • SageMath
    def A001786_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 1/((1+x)*(1-x)^11) ).list()
print(A001786_list(50)) # G. C. Greubel, Apr 20 2025

Formula

Boas-Buck recurrence: a(n) = (1/n)*Sum_{p=0..n-1} (11 + (-1)^(n-p))*a(p), n >= 1, a(0) = 1. See the Boas-Buck comment in A046521 (here for the unsigned column k = 5 with offset 0). - Wolfdieter Lang, Aug 10 2017

A128495 Coefficient table for sums of squares of Chebyshev's S-Polynomials.

Original entry on oeis.org

1, 1, 1, 2, -1, 1, 2, 3, -3, 1, 3, -3, 8, -5, 1, 3, 6, -16, 17, -7, 1, 4, -6, 30, -45, 30, -9, 1, 4, 10, -50, 103, -98, 47, -11, 1, 5, -10, 80, -211, 269, -183, 68, -13, 1, 5, 15, -120, 399, -651, 588, -308, 93, -15, 1, 6, -15, 175, -707, 1432, -1644, 1136, -481, 122, -17, 1, 6, 21, -245, 1190, -2920, 4132, -3608
Offset: 0

Views

Author

Wolfdieter Lang Apr 04 2007

Keywords

Comments

See A049310 for the coefficient table of Chebyshev's S(n,x)=U(n,x/2) polynomials.
The triangle for the coefficients of x^2 in S(n,x)^2 is A158454. - Wolfdieter Lang, Oct 18 2012

Examples

			[1]; [1,1]; [2,-1,1]; [2,3,-3,1]; [3,-3,8,-5,1]; [3,6,-16,17,-7,1]; ...
Row polynomial S(2;4,x)=3-3*x^2+8*x^4-5*x^6+x^8 = sum(S(k,x)^2,k=0..4).
(4+2-T(4+1,x/2)*U(4+1,x/2))/(2*(1-(x/2)^2))= S(2;4,x)

Links

Crossrefs

Row sums (signed) look like: A004523. Row sums (unsigned): A128496.
Cf. A128494 =S(1; n, m).

Formula

S(2;n,x):=sum(S(k,x)^2,k=0..n)=sum(a(n,m)*x^(2*m),m=0..n), n>=0.
a(n,m)=[x^m](n+2-T(n+1,x/2)*U(n+1,x/2))/(2*(1-(x/2)^2)).
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