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.

Previous Showing 11-20 of 26 results. Next

A001351 Associated Mersenne numbers.

Original entry on oeis.org

0, 1, 3, 1, 3, 11, 9, 8, 27, 37, 33, 67, 117, 131, 192, 341, 459, 613, 999, 1483, 2013, 3032, 4623, 6533, 9477, 14311, 20829, 30007, 44544, 65657, 95139, 139625, 206091, 300763, 439521, 646888, 948051, 1385429, 2033193, 2983787, 4366197, 6397723, 9387072
Offset: 0

Views

Author

Keywords

Comments

From Peter Bala, Sep 15 2019: (Start)
This is a linear divisibility sequence of order 6 (Haselgrove, p. 21). It is a particular case of a family of divisibility sequences studied by Roettger et al. The o.g.f. has the form x*d/dx(f(x)/(x^3*f(1/x))) where f(x) = x^3 - x^2 - 1.
More generally, if f(x) = 1 + P*x + Q*x^2 + x^3 or f(x) = -1 + P*x + Q*x^2 + x^3, where P and Q are integers, then the rational function x*d/dx(f(x)/(x^3*f(1/x))) is the generating function for a linear divisibility sequence of order 6. Cf. A001945. There are corresponding results when f(x) is a monic quartic polynomial with constant term 1. (End)

References

  • 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).

Crossrefs

Programs

  • Magma
    I:=[0,1,3,1,3,11]; [n le 6 select I[n] else Self(n-1) - Self(n-2) + 3*Self(n-3) - Self(n-4) + Self(n-5) - Self(n-6): n in [1..50]]; // Vincenzo Librandi, Sep 23 2015
  • Maple
    A001351:=z*(z^2-z+1)*(z^2+3*z+1)/(z^3+z-1)/(z^3-z^2-1); # conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    LinearRecurrence[{1, -1, 3, -1, 1, -1}, {0, 1, 3, 1, 3, 11}, 50] (* Vincenzo Librandi, Sep 23 2015 *)

Formula

a(n) = a(n-1) - a(n-2) + 3*a(n-3) - a(n-4) + a(n-5) - a(n-6) for n >= 6. - Sean A. Irvine, Sep 23 2015
a(n) = (alpha^n - 1)*(beta^n - 1)*(gamma^n - 1) where alpha, beta and gamma are the zeros of x^3 - x^2 - 1. - Peter Bala, Sep 15 2019

Extensions

More terms from Vincenzo Librandi, Sep 23 2015

A129744 a(n) = -(u^n-1)*(v^n-1) with u = 1+sqrt(2), v = 1-sqrt(2).

Original entry on oeis.org

2, 4, 14, 32, 82, 196, 478, 1152, 2786, 6724, 16238, 39200, 94642, 228484, 551614, 1331712, 3215042, 7761796, 18738638, 45239072, 109216786, 263672644, 636562078, 1536796800, 3710155682, 8957108164, 21624372014, 52205852192
Offset: 1

Views

Author

N. J. A. Sloane, May 13 2007

Keywords

Crossrefs

Programs

  • Maple
    u:=1+sqrt(2): v:=1-sqrt(2): a:=n->expand(-(u^n-1)*(v^n-1)): seq(a(n),n=1..33); # Emeric Deutsch, May 13 2007
  • Mathematica
    Table[Simplify[ -((1 + Sqrt[2])^n - 1)*((1 - Sqrt[2])^n - 1)], {n, 1, 30}] (* Stefan Steinerberger, May 15 2007 *)
  • PARI
    w = quadgen(8); vector(30, n, -((1+w)^n-1)*((1-w)^n-1)) \\ Michel Marcus, Mar 21 2015
    
  • PARI
    Vec(2*x*(1+x^2)/((x^2+2*x-1)*(-1+x)*(1+x))+O(x^99)) \\ Charles R Greathouse IV, Nov 13 2015

Formula

a(2n) = A002203(2n)-2. a(2n+1) = A002203(2n+1). - R. J. Mathar, corrected Dec 05 2007.
G.f.: 2*x*(1+x^2)/((x^2+2*x-1)*(-1+x)*(1+x)).
From Peter Bala, Mar 19 2015: (Start)
a(n) = -det(I - M^n) where I is the 2X2 identity matrix and M = [2, 1; 1, 0]. Cf. A001350.
a(n) = 2*A113224(n-1).
This is divisibility sequence, that is, if n | m then a(n) | a(m).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*Sum_{n >= 1} Pell(n) *x^n. (End)
a(n) = 2*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) for n > 4. - Seiichi Manyama, Jun 07 2018

Extensions

More terms from Emeric Deutsch and Stefan Steinerberger, May 13 2007

A152152 a(n) = Product_{k=1..n} (1 + 4*sin(2*Pi*k/n)^2).

Original entry on oeis.org

0, 1, 1, 16, 25, 121, 256, 841, 2025, 5776, 14641, 39601, 102400, 271441, 707281, 1860496, 4862025, 12752041, 33362176, 87403801, 228765625, 599074576, 1568239201, 4106118241, 10749542400, 28143753121, 73680216481, 192900153616
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Nov 26 2008

Keywords

Crossrefs

Programs

  • Magma
    [(1-Lucas(n)+(-1)^n)^2: n in [0..30]]; // G. C. Greubel, Mar 13 2019
    
  • Mathematica
    Table[(1 + Fibonacci[n] - 2*Fibonacci[n+1] + (-1)^n)^2, {n, 0, 30}]
  • PARI
    {a(n) = (1-fibonacci(n-1)-fibonacci(n+1)+(-1)^n)^2}; \\ G. C. Greubel, Mar 13 2019
    
  • Sage
    [(1-lucas_number2(n,1,-1)+(-1)^n)^2 for n in (0..30)] # G. C. Greubel, Mar 13 2019

Formula

a(n) = Product_{k=1..n} (1 + 4*sin(2*Pi*k/n)^2).
a(n) = (1 + Fibonacci(n) - 2*Fibonacci(n + 1) + (-1)^n)^2.
G.f.: -x*(x^6 -2*x^5 +10*x^4 -14*x^3 +10*x^2 -2*x +1)/((x -1)*(x +1)*(x^2 -3*x +1)*(x^2 -x -1)*(x^2 +x -1)). - Colin Barker, Apr 13 2014
a(n) = A001350(n)^2. - Colin Barker, Apr 13 2014
a(n) = (1 + (-1)^n - Lucas(n))^2. - G. C. Greubel, Mar 13 2019

A288913 a(n) = Lucas(4*n + 3).

Original entry on oeis.org

4, 29, 199, 1364, 9349, 64079, 439204, 3010349, 20633239, 141422324, 969323029, 6643838879, 45537549124, 312119004989, 2139295485799, 14662949395604, 100501350283429, 688846502588399, 4721424167835364, 32361122672259149, 221806434537978679, 1520283919093591604
Offset: 0

Views

Author

Bruno Berselli, Jun 19 2017

Keywords

Comments

a(n) mod 4 gives A101000.

Crossrefs

Cf. A033891: fourth quadrisection of A000045.
Partial sums are in A081007 (after 0).
Positive terms of A098149, and subsequence of A001350, A002878, A016897, A093960, A068397.
Quadrisection of A000032: A056854 (first), A056914 (second), A246453 (third, without 11), this sequence (fourth).

Programs

  • Magma
    [Lucas(4*n + 3): n in [0..30]]; // G. C. Greubel, Dec 22 2017
    
  • Mathematica
    LucasL[4 Range[0, 21] + 3]
    LinearRecurrence[{7,-1}, {4,29}, 30] (* G. C. Greubel, Dec 22 2017 *)
  • PARI
    Vec((4 + x)/(1 - 7*x + x^2) + O(x^30)) \\ Colin Barker, Jun 20 2017
    
  • Python
    from sympy import lucas
    def a(n):  return lucas(4*n + 3)
    print([a(n) for n in range(22)]) # Michael S. Branicky, Apr 29 2021
  • Sage
    def L():
        x, y = -1, 4
        while True:
            yield y
            x, y = y, 7*y - x
    r = L(); [next(r) for  in (0..21)] # _Peter Luschny, Jun 20 2017
    

Formula

G.f.: (4 + x)/(1 - 7*x + x^2).
a(n) = 7*a(n-1) - a(n-2) for n>1, with a(0)=4, a(1)=29.
a(n) = ((sqrt(5) + 1)^(4*n + 3) - (sqrt(5) - 1)^(4*n + 3))/(8*16^n).
a(n) = Fibonacci(4*n+4) + Fibonacci(4*n+2).
a(n) = 4*A004187(n+1) + A004187(n).
a(n) = 5*A003482(n) + 4 = 5*A081016(n) - 1.
a(n) = A002878(2*n+1) = A093960(2*n+3) = A001350(4*n+3) = A068397(4*n+3).
a(n+1)*a(n+k) - a(n)*a(n+k+1) = 15*Fibonacci(4*k). Example: for k=6, a(n+1)*a(n+6) - a(n)*a(n+7) = 15*Fibonacci(24) = 695520.

A324489 a(n) = A324488(n)/n.

Original entry on oeis.org

1, 0, 21, 31, 266, 672, 3484, 11375, 48768, 177023, 716418, 2730315, 10878520, 42485638, 169181010, 670042125, 2678678730, 10705526976, 43007270292, 173003915322, 698235680844, 2822901487191, 11439823946306, 46438021798875, 188856966693230, 769224288476860, 3137871076604544, 12817404260955810
Offset: 1

Views

Author

N. J. A. Sloane, Mar 12 2019

Keywords

Crossrefs

Programs

  • PARI
    a001350(n) = fibonacci(n+1)+fibonacci(n-1)-1-(-1)^n;
    a(n) = sumdiv(n, d, moebius(n/d)*a001350(d)^3)/n; \\ Seiichi Manyama, Apr 29 2021
    
  • PARI
    f(x) = ((1-3*x+x^2)*(1+3*x+x^2))^3*(1-x^2)^10/((1-4*x-x^2)*(1-x-x^2)^6*(1+x-x^2)^9);
    my(N=40, x='x+O('x^N)); Vec(sum(k=1, N, moebius(k)*log(f(x^k))/k)) \\ Seiichi Manyama, Apr 29 2021

Formula

From Seiichi Manyama, Apr 29 2021: (Start)
a(n) = (1/n) * Sum_{d|n} mu(n/d) * A001350(d)^3 = (1/n) * Sum_{d|n} mu(n/d) * A324487(d).
G.f.: Sum_{k>=1} mu(k) * log(f(x^k))/k , where f(x) = ((1-3*x+x^2) * (1+3*x+x^2))^3 * (1-x^2)^10/((1-4*x-x^2) * (1-x-x^2)^6 * (1+x-x^2)^9). (End)

Extensions

More terms from Seiichi Manyama, Apr 29 2021

A032189 Number of ways to partition n elements into pie slices each with an odd number of elements.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 5, 7, 10, 14, 19, 30, 41, 63, 94, 142, 211, 328, 493, 765, 1170, 1810, 2787, 4340, 6713, 10461, 16274, 25414, 39651, 62074, 97109, 152287, 238838, 375166, 589527, 927554, 1459961, 2300347, 3626242, 5721044, 9030451, 14264308, 22542397, 35646311, 56393862, 89264834, 141358275
Offset: 1

Views

Author

Keywords

Comments

a(n) is also the total number of cyclic compositions of n into odd parts assuming that two compositions are equivalent if one can be obtained from the other by a cyclic shift. For example, a(5)=3 because 5 has the following three cyclic compositions into odd parts: 5, 1+3+1, 1+1+1+1+1. - Petros Hadjicostas, Dec 27 2016

Crossrefs

Programs

  • Mathematica
    a1350[n_] := Sum[Binomial[k - 1, 2k - n] n/(n - k), {k, 0, n - 1}];
    a[n_] := 1/n Sum[EulerPhi[n/d] a1350[d], {d, Divisors[n]}];
    Array[a, 50] (* Jean-François Alcover, Jul 29 2018, after Petros Hadjicostas *)
  • PARI
    N=66;  x='x+O('x^N);
    B(x)=x/(1-x^2);
    A=sum(k=1,N,eulerphi(k)/k*log(1/(1-B(x^k))));
    Vec(A)
    /* Joerg Arndt, Aug 06 2012 */
    
  • Python
    from sympy import totient, lucas, divisors
    def A032189(n): return sum(totient(n//k)*(lucas(k)-((k&1^1)<<1)) for k in divisors(n,generator=True))//n # Chai Wah Wu, Sep 23 2023

Formula

a(n) = A000358(n)-(1+(-1)^n)/2.
"CIK" (necklace, indistinct, unlabeled) transform of 1, 0, 1, 0...(odds)
G.f.: Sum_{k>=1} phi(k)/k * log( 1/(1-B(x^k)) ) where B(x) = x/(1-x^2). [Joerg Arndt, Aug 06 2012]
a(n) = (1/n)*Sum_{d divides n} phi(n/d)*A001350(d). - Petros Hadjicostas, Dec 27 2016

A324485 a(n) = A324484(n)/n.

Original entry on oeis.org

1, 0, 5, 6, 24, 40, 120, 250, 640, 1452, 3600, 8510, 20880, 50460, 124024, 303750, 750120, 1853120, 4600200, 11437548, 28527320, 71281800, 178526880, 447893250, 1125750120, 2833844040, 7144449920, 18036271740, 45591631800, 115381449692, 292329067800, 741410192250
Offset: 1

Views

Author

N. J. A. Sloane, Mar 12 2019

Keywords

Crossrefs

Programs

  • PARI
    a001350(n) = fibonacci(n+1)+fibonacci(n-1)-1-(-1)^n;
    a(n) = sumdiv(n, d, moebius(n/d)*a001350(d)^2)/n; \\ Seiichi Manyama, Apr 29 2021
    
  • PARI
    f(x) = ((1-x-x^2)*(1+x-x^2))^2/((1-3*x+x^2)*(1-x)^2*(1+x)^4);
    my(N=40, x='x+O('x^N)); Vec(sum(k=1, N, moebius(k)*log(f(x^k))/k)) \\ Seiichi Manyama, Apr 29 2021

Formula

From Seiichi Manyama, Apr 29 2021: (Start)
a(n) = (1/n) * Sum_{d|n} mu(n/d) * A001350(d)^2 = (1/n) * Sum_{d|n} mu(n/d) * A152152(d).
G.f.: Sum_{k>=1} mu(k) * log(f(x^k))/k , where f(x) = ((1-x-x^2) * (1+x-x^2))^2/((1-3*x+x^2) * (1-x)^2 * (1+x)^4). (End)

Extensions

More terms from Seiichi Manyama, Apr 29 2021

A294203 Number of partitions of n into distinct Lucas parts (A000204) greater than 1.

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 0, 2, 0, 0, 1, 2, 0, 0, 2, 1, 0, 0, 3, 0, 0, 2, 2, 0, 0, 3, 0, 0, 1, 3, 0, 0, 3, 2, 0, 0, 4, 0, 0, 2, 3, 0, 0, 3, 1, 0, 0, 4, 0, 0, 3, 3, 0, 0, 5, 0, 0, 2, 4, 0, 0, 4, 2, 0, 0, 5, 0, 0, 3, 3, 0, 0, 4, 0, 0, 1, 4, 0, 0, 4, 3, 0, 0, 6, 0, 0, 3, 5, 0, 0, 5, 2, 0, 0, 6, 0, 0, 4, 4
Offset: 0

Views

Author

Ilya Gutkovskiy, Oct 24 2017

Keywords

Comments

Convolution of the sequences A003263 and A033999.
Positions of 0: 1, 2, 5, 6, 8, 9, 12, 13, ... = A287775(n) - 1 (conjecture).
From Michel Dekking, Dec 30 2017: (Start)
Proof of the 'positions of 0' conjecture: let (z(n))=1,2,5,6,8,9,12,... be the positions of 0. The crucial observation is that if a number n is the sum of distinct Lucas parts greater than 1, then n+1 is a sum of Lucas parts. This implies that (z(2n))=2,6,9,13,... is the sequence of numbers A054770 that are not a sum of Lucas numbers. We see there that Ian Agol proved that b(n):=A054770(n)=floor(phi*n)+2n-1. But then the sequence of first differences (b(n+1)-b(n)) equals the Fibonacci word on the alphabet {4,3}, yielding that (z(2n)-z(2n-1)) equals the Fibonacci word on {3,2}, and we already know that z(2n+1)-z(2n)=1 for all n. On the other hand, A287775 has the same first difference sequence given by A108103. Since A287775(1)=2, the conjecture follows. (End)
Positions of 1: 0, 3, 4, 10, 15, 28, 44, 75, ... = A001350(n+1) - 1 (conjecture).

Examples

			a(7) = 2 because we have [7] and [4, 3].
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[Product[1 + x^LucasL[k], {k, 2, 15}], {x, 0, 100}], x]

Formula

G.f.: Product_{k>=2} (1 + x^Lucas(k)).

A324488 Inflation orbit counts b^{(3)}_n for Danzer's F-type tiling and other 3D cut and project patterns with tau-inflation.

Original entry on oeis.org

1, 0, 63, 124, 1330, 4032, 24388, 91000, 438912, 1770230, 7880598, 32763780, 141420760, 594798932, 2537715150, 10720674000, 45537538410, 192699485568, 817138135548, 3460078306440, 14662949297724, 62103832718202, 263115950765038, 1114512523173000, 4721424167330750
Offset: 1

Views

Author

N. J. A. Sloane, Mar 12 2019

Keywords

Crossrefs

Programs

  • PARI
    a001350(n) = fibonacci(n+1)+fibonacci(n-1)-1-(-1)^n;
    a(n) = sumdiv(n, d, moebius(n/d)*a001350(d)^3); \\ Seiichi Manyama, Apr 29 2021

Formula

a(n) = Sum_{d|n} mu(n/d) * A001350(d)^3 = Sum_{d|n} mu(n/d) * A324487(d). - Seiichi Manyama, Apr 29 2021

Extensions

More terms from Seiichi Manyama, Apr 29 2021

A365857 Number of cyclic compositions of 2*n into odd parts.

Original entry on oeis.org

1, 2, 4, 7, 14, 30, 63, 142, 328, 765, 1810, 4340, 10461, 25414, 62074, 152287, 375166, 927554, 2300347, 5721044, 14264308, 35646311, 89264834, 223959710, 562878429, 1416953362, 3572233420, 9018211989, 22795835726, 57690911720, 146164582455, 370705552702, 941109975022, 2391391374017, 6081865318124
Offset: 1

Views

Author

Joshua P. Bowman, Sep 20 2023

Keywords

Comments

Even bisection of A032189.
Also the number of cyclic compositions into an even number of odd parts; because such a sum must be even, alternating terms are zero and have been removed.
Also the number of dual classes of cyclic n-color compositions of n. A cyclic composition is a sum of positive integers in which the order of the parts is considered up to cyclic permutation. In other words, it is the collection of components remaining in the cycle graph C_n on n vertices when one or more edges are removed, and rotations are considered equivalent. In an n-color composition, each part of size k is assigned one of k "colors" which may be represented graphically by marking one vertex in the part. (See A032198 for the number of cyclic n-color compositions.) The dual of a cyclic n-color composition is obtained by switching the roles of edges and vertices in C_n, then removing each edge that came from a previously marked vertex while marking each vertex that came from a previously removed edge. Each cyclic n-color composition of n either belongs to a dual pair or is self-dual. (See A365859 for the number of self-dual cyclic n-color compositions.)

Crossrefs

Programs

  • PARI
    N=99;  x='x+O('x^N); B(x)=x/(1-x^2);
    A=Vec(sum(k=1, N, eulerphi(k)/k*log(1/(1-B(x^k)))));
    vector(#A\2,n,A[2*n]) \\ Joerg Arndt, Sep 22 2023
    
  • Python
    from sympy import totient, lucas, divisors
    def A365857(n): return sum(totient((n<<1)//k)*(lucas(k)-((k&1^1)<<1)) for k in divisors(n<<1,generator=True))//n>>1 # Chai Wah Wu, Sep 23 2023

Formula

G.f.: (1/2)*(Sum_{k>=1} phi(k)/k * log((1-2*x^k+x^(2*k))/(1-3*x^k+x^(2*k))) + Sum_{m>=1} phi(2*m)/(2*m) * log((1+x^m-x^(2*m))/(1-x^m-x^(2*m)))).
a(n) = (1/(2*n)) * Sum_{k divides 2*n} phi(k)*A001350((2*n)/k).
a(n) = (A032198(n) + A365859(n))/2.
Previous Showing 11-20 of 26 results. Next