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.

Showing 1-6 of 6 results.

A015441 Generalized Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 7, 13, 55, 133, 463, 1261, 4039, 11605, 35839, 105469, 320503, 953317, 2876335, 8596237, 25854247, 77431669, 232557151, 697147165, 2092490071, 6275373061, 18830313487, 56482551853, 169464432775, 508359743893, 1525146340543, 4575304803901, 13726182847159
Offset: 0

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Author

Keywords

Comments

a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x,y) = xF(n-1)(x,y) + yF(n-2)(x,y), F(0)(x,y)=0, F(1)(x,y)=1, when y=6x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 06 2002
For n>=1: number of length-(n-1) words with letters {0,1,2,3,4,5,6,7} where no two consecutive letters are nonzero, see fxtbook link below. - Joerg Arndt, Apr 08 2011
Starting with offset 1 and convolved with (1, 3, 3, 3, ...) = A003462: (1, 4, 13, 40, ...). - Gary W. Adamson, May 28 2009
a(n) is identical to its inverse binomial transform signed. Differences: A102901. - Paul Curtz, Feb 23 2010
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=2, 7*a(n-2) equals the number of 7-colored compositions of n with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
Pisano period lengths: 1, 1, 1, 2, 20, 1, 6, 2, 3, 20, 5, 2, 12, 6, 20, 4, 16, 3, 18, 20, ... - R. J. Mathar, Aug 10 2012
A015441 and A015518 are the only integer sequences (from the family of homogeneous linear recurrence relation of order 2 with positive integer coefficients with initial values a(0)=0 and a(1)=1) whose ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 14 2014
This is an autosequence of the first kind: the array of successive differences shows a main diagonal of zeros and the inverse binomial transform is identical to the sequence (with alternating signs). - Pointed out by Paul Curtz, Dec 05 2016
First two upper diagonals: A000400(n).
This is a variation on the Starhex honeycomb configuration A332243, see illustration in links. It is an alternating pattern of the 2nd iteration of the centered hexagonal numbers A003215 and centered 12-gonal 'Star' numbers A003154. - John Elias, Oct 06 2021

Examples

			G.f. = x + x^2 + 7*x^3 + 13*x^4 + 55*x^5 + 133*x^6 + 463*x^7 + 1261*x^8 + ...
		

Crossrefs

Programs

  • Magma
    I:=[0,1]; [n le 2 select I[n] else Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018
  • Maple
    A015441:=n->(1/5)*((3^n)-((-2)^n)); seq(A015441(n), n=0..30); # Wesley Ivan Hurt, Mar 14 2014
  • Mathematica
    a[n_]:=(MatrixPower[{{1,4},{1,-2}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]], {n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
    LinearRecurrence[{1,6},{0,1},30] (* Harvey P. Dale, Apr 26 2011 *)
    CoefficientList[Series[x/((1 + 2 x) (1 - 3 x)), {x, 0, 29}], x] (* Michael De Vlieger, Dec 05 2016 *)
  • PARI
    {a(n) = (3^n - (-2)^n) / 5};
    
  • Sage
    [lucas_number1(n,1,-6) for n in range(0, 27)] # Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: x/((1+2*x)*(1-3*x)).
a(n) = a(n-1) + 6*a(n-2).
a(n) = (1/5)*((3^n)-((-2)^n)). - henryk.wicke(AT)stud.uni-hannover.de
E.g.f.: (exp(3*x) - exp(-2*x))/5. - Paul Barry, Apr 20 2003
a(n+1) = Sum_{k=0..ceiling(n/2)} 6^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
a(n) = (A000244(n) - A001045(n+1)(-1)^n - A001045(n)(-1)^n)/5. - Paul Barry, Apr 27 2004
The binomial transform of [1,1,7,13,55,133,463,...] is A122117. - Philippe Deléham, Oct 19 2006
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-6)^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = 3a(n-1) + (-1)^(n+1)*A000079(n-1). - Paul Curtz, Feb 23 2010
G.f.: Q(0) -1, where Q(k) = 1 + 6*x^2 + (k+2)*x - x*(k+1 + 6*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n) = (Sum_{1<=k<=n, k odd} binomial(n,k)*5^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014
a(-n) = -(-1)^n * a(n) / 6^n for all n in Z. - Michael Somos, Mar 18 2014
From Peter Bala, Apr 01 2015: (Start)
Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} A087451(n)*x^n/n ).
For k = 0, 1, 2, ... and for n >= 1, (5^k)*a(n) | a((5^k)*n).
The expansion of exp( Sum_{n >= 1} a(5*n)/(5*a(n))*x^n/n ) has integral coefficients. Cf. A001656. (End)
From Peter Bala, Jun 27 2025: (Start)
Sum_{n >= 1} (-6)^n/(a(n)*a(n+1)) = -2, since (-6)^n/(a(n)*a(n+1)) = (-2)^n/a(n) - (-2)^(n+1)/a(n+1) for n >= 1.
The following are examples of telescoping infinite products:
Product_{n >= 0} (1 + 6^n/a(2*n+2)) = 6, since (1 + 6^(2*n-1)/a(4*n))*(1 + 6^(2*n)/a(4*n+2)) = (6 - 4^(n+1)/b(n)) / (6 - 4^n/b(n-1)), where b(n) = (2*4^n + 3*9^n)/5 = A096951(n). Similarly,
Product_{n >= 1} (1 - 6^n/a(2*n+2)) = 3/13.
Product_{n >= 0} (1 + (-6)^n/a(2*n+2)) = 6/5.
Product_{n >= 1} (1 - (-6)^n/a(2*n+2)) = 15/13.
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n. (End)

A102900 a(n) = 3*a(n-1) + 4*a(n-2), a(0)=a(1)=1.

Original entry on oeis.org

1, 1, 7, 25, 103, 409, 1639, 6553, 26215, 104857, 419431, 1677721, 6710887, 26843545, 107374183, 429496729, 1717986919, 6871947673, 27487790695, 109951162777, 439804651111, 1759218604441, 7036874417767, 28147497671065
Offset: 0

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Author

Paul Barry, Jan 17 2005

Keywords

Comments

Binomial transform of A102901.
Hankel transform is = 1,6,0,0,0,0,0,0,0,0,0,0,... - Philippe Deléham, Nov 02 2008

References

  • Maria Paola Bonacina and Nachum Dershowitz, Canonical Inference for Implicational Systems, in Automated Reasoning, Lecture Notes in Computer Science, Volume 5195/2008, Springer-Verlag.

Crossrefs

Cf. A001045, A004171, A046717, A086901, A102901, A247666 (which appears to be the run length transform of this sequence).

Programs

  • Haskell
    a102900 n = a102900_list !! n
    a102900_list = 1 : 1 : zipWith (+)
                   (map (* 4) a102900_list) (map (* 3) $ tail a102900_list)
    -- Reinhard Zumkeller, Feb 13 2015
    
  • Magma
    [n le 2 select 1 else 3*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 28 2015
    
  • Mathematica
    a[n_]:=(MatrixPower[{{2,2},{3,1}},n].{{2},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
    LinearRecurrence[{3, 4}, {1, 1}, 30] (* Vincenzo Librandi, Dec 28 2015 *)
  • PARI
    a(n)=([0,1; 4,3]^n*[1;1])[1,1] \\ Charles R Greathouse IV, Mar 28 2016
    
  • SageMath
    A102900=BinaryRecurrenceSequence(3,4,1,1)
    [A102900(n) for n in range(51)] # G. C. Greubel, Dec 09 2022

Formula

G.f.: (1-2*x)/(1-3*x-4*x^2).
a(n) = (2*4^n + 3*(-1)^n)/5.
a(n) = ceiling(4^n/5) + floor(4^n/5) = (ceiling(4^n/5))^2 - (floor(4^n/5))^2.
a(n) + a(n+1) = 2^(2*n+1) = A004171(n).
a(n) = Sum_{k=0..n} binomial(2*n-k, 2*k)*2^k. - Paul Barry, Jan 20 2005
a(n) = upper left term in the 2 X 2 matrix [1,3; 2,2]^n. - Gary W. Adamson, Mar 14 2008
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(8*4^k-3*(-1)^k)/(x*(8*4^k-3*(-1)^k) + (2*4^k+3*(-1)^k)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
a(n) = 2^(2*n-1) - a(n-1), a(1)=1. - Ben Paul Thurston, Dec 27 2015; corrected by Klaus Purath, Aug 02 2020
From Klaus Purath, Aug 02 2020: (Start)
a(n) = 4*a(n-1) + 3*(-1)^n.
a(n) = 6*4^(n-2) + a(n-2), n>=2. (End)

A347237 Möbius transform of A347236.

Original entry on oeis.org

1, 0, 1, 6, 1, 0, 3, 6, 17, 0, 1, 6, 3, 0, 1, 42, 1, 0, 3, 6, 3, 0, 5, 6, 37, 0, 49, 18, 1, 0, 5, 78, 1, 0, 3, 102, 3, 0, 3, 6, 1, 0, 3, 6, 17, 0, 5, 42, 89, 0, 1, 18, 5, 0, 1, 18, 3, 0, 1, 6, 5, 0, 51, 330, 3, 0, 3, 6, 5, 0, 1, 102, 5, 0, 37, 18, 3, 0, 3, 42, 353, 0, 5, 18, 1, 0, 1, 6, 7, 0, 9, 30, 5, 0, 3, 78, 3, 0, 17
Offset: 1

Views

Author

Antti Karttunen, Aug 24 2021

Keywords

Comments

Dirichlet convolution of A003972 (prime shifted phi) with A061019.
Dirichlet convolution of A003961 with A158523.
Multiplicative because A003972 and A061019 are, and also because A347236 is.
From Antti Karttunen, Aug 25 2021: (Start)
All terms are nonnegative because sequence is multiplicative and a(p^k) >= 0 for all primes p and k >= 0.
Proof: For any prime p, sequence a(p^k), k>=0, is obtained as an ordinary convolution of sequences (-p)^k and the first differences of q^k, where q = A151800(p). (E.g., for powers of 2, the sequences convolved are A122803 and A025192, giving A102901.) This convolution is an alternating sum, with the terms 1*(q-1)*q^(k-1), -(p)*(q-1)*q^(k-2), (p^2)*(q-1)*q^(k-3), -(p^3)*(q-1)*q^(k-4), ..., (p^(k-1))*(q-1), -(p^k), for odd k, with sum of each consecutive pair being nonnegative because q >= p+1, while with an even exponent k, the leftover term p^k at the end is also positive, thus the whole sum is nonnegative also in that case.
(End)

Crossrefs

Programs

  • PARI
    A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A061019(n) = (((-1)^bigomega(n))*n);
    A347237(n) = sumdiv(n,d,A061019(d)*eulerphi(A003961(n/d)));
    \\ Or alternatively as:
    A158523(n) = { my(f=factor(n)); prod(i=1, #f~, my(p=f[i,1], e=f[i,2]); ((-1)^e)*(p+1)*(p^(e-1))); };
    A347237(n) = sumdiv(n,d,A003961(n/d)*A158523(d));

Formula

a(n) = Sum_{d|n} A008683(n/d) * A347236(d).
a(n) = Sum_{d|n} A003972(n/d) * A061019(d).
a(n) = Sum_{d|n} A003961(n/d) * A158523(d).
For all n >= 1, a(A000040(n)) = A001223(n) - 1.
For all n >= 0, a(2^n) = A102901(n).
For all n >= 0, a(3^n) = A120612(n).
Multiplicative with a(p^e) = (-p)^e + (A151800(p)-1)*(A151800(p)^e-(-p)^e)/(A151800(p)+p). - Sebastian Karlsson, Sep 02 2021

A140796 a(n)=a(n-1)+6a(n-2), n>2.

Original entry on oeis.org

1, 5, 14, 44, 128, 392, 1160, 3512, 10472, 31544, 94376, 283640, 849896, 2551736, 7651112, 22961528, 68868200, 206637368, 619846568, 1859670776, 5578750184, 16736774840, 50209275944, 150629924984, 451885580648, 1355665130552
Offset: 0

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Author

Paul Curtz, Jul 15 2008

Keywords

Comments

The binomial transform is A037481.
The recurrence of the definition is also satisfied by A087451, A102901 and A140725.

Crossrefs

Programs

  • Mathematica
    Join[{1},LinearRecurrence[{1,6},{5,14},30]] (* Harvey P. Dale, Nov 20 2011 *)

Formula

a(n+1)-3a(n) = (-1)^n*A000079(n-1), n>0.
d(n+1)-3d(n) = (-1)^(n+1)*A000079(n-1), n>0, where d(n) is the sequence of pair sums d(n)= a(n)+a(n+1)=6, 19, 58, 172,...
O.g.f.: (1+x)(3x+1)/((2x+1)(1-3x)). - R. J. Mathar, Jul 29 2008
a(n) = (-1)^(n+1)*2^n/10+8*3^n/5, n>0. - R. J. Mathar, Jul 29 2008
a(n) = A140725(n)+A140725(n+1). - Philippe Deléham, Nov 17 2013

Extensions

Edited and extended by R. J. Mathar, Jul 29 2008

A232015 Expansion of (1-2*x)/((1+2*x)*(1-3*x)).

Original entry on oeis.org

1, -1, 5, -1, 29, 23, 197, 335, 1517, 3527, 12629, 33791, 109565, 312311, 969701, 2843567, 8661773, 25723175, 77693813, 232032863, 698195741, 2090392919, 6279567365, 18821924879, 56499329069, 169430878343, 508426852757, 1525012122815, 4575573239357
Offset: 0

Views

Author

Philippe Deléham, Nov 17 2013

Keywords

Crossrefs

Programs

  • Mathematica
    Table[(3^n + 4 (-2)^n)/5, {n, 0, 30}] (* Bruno Berselli, Nov 18 2013 *)
    CoefficientList[Series[(1-2x)/((1+2x)(1-3x)),{x,0,40}],x] (* or *) LinearRecurrence[ {1,6},{1,-1},30] (* Harvey P. Dale, Apr 20 2017 *)
  • PARI
    Vec((1-2*x)/((1+2*x)*(1-3*x))+O(x^20)) \\ Edward Jiang, Sep 06 2014

Formula

G.f.: (1 - 2*x) / (1 - x - 6*x^2).
a(n) = a(n-1) + 6*a(n-2) for n>1, a(0)=1, a(1)=-1.
a(n) = sum_{k=0..n} A108561(n,k)*2^k.
a(n) = A102901(n) - A015441(n).
From Bruno Berselli, Nov 18 2013: (Start)
a(n) = (3^n + 4*(-2)^n)/5.
a(n+1) + a(n) = 4*A015441(n).
a(n+1) - a(n) = -2*(-1)^n*A165405(n).
Sum(a(i), i=0..n) = A091001(n+1). (End)

A208459 Triangle T_x = T(n,k) given by (0, 1/x, 1-1/x, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (x, 1/x-1, -1/x, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938, for x = 0.

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 0, 1, 0, -1, 0, 1, 0, 1, 2, 0, 1, 0, 2, 0, -3, 0, 1, 0, 3, -1, 0, 5, 0, 1, 0, 4, -2, 3, 2, -8, 0, 1, 0, 5, -3, 7, -2, -5, 13, 0, 1, 0, 6, -4, 12, -8, 2, 12, -21, 0, 1, 0, 7, -5, 18, -16, 15, 3, -25, 34
Offset: 0

Views

Author

Philippe Deléham, Feb 27 2012

Keywords

Comments

Triangle T_x : T_1 = A103631, T_2 = A208343, T_3 = A208345.

Examples

			Triangle begins :
1
0, 0
0, 1, 1
0, 1, 0, -1
0, 1, 0, 1, 2
0, 1, 0, 2, 0, -3
0, 1, 0, 3, -1, 0, 5
0, 1, 0, 4, -2, 3, 2, -8
0, 1, 0, 5, -3, 7, -2, -5, 13
0, 1, 0, 6, -4, 12, -8, 2, 12, -21
0, 1, 0, 7, -5, 18, -16, 15, 3, -25, 34
		

Crossrefs

Cf. A103631, A208343, A208345, A000045 (Fibonacci)

Formula

T(n,k) = T(n-1,k) - T(n-1,k-1) + T(n-2,k-1) + T(n-2,k-2) with T(0,0) = 1 T(1,0) = 0, T(1,1) = 0, T(n,k) = 0 if k<0 or if k>n.
G.f.: (1-x+y*x)/(1-x+y*x- y^2*x^2-y*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = 12*A015548(n-1), 6*A085939(n-1), A106434(n), A000007(n), A000007(n), A077957(n), (-1)^n*A102901(n) for x = -4, -3, -2, -1, 0, 1, 2 respectively.
Sm_{k, 0<=k<=n} T(n,k)*x^(n-k) = A000007(n), A034834(n-1), A077957(n), A052533(n), (-1)^n*A086344(n) for x = -1, 0, 1, 2, 3 respectively.
Showing 1-6 of 6 results.