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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A007583 a(n) = (2^(2*n + 1) + 1)/3.

Original entry on oeis.org

1, 3, 11, 43, 171, 683, 2731, 10923, 43691, 174763, 699051, 2796203, 11184811, 44739243, 178956971, 715827883, 2863311531, 11453246123, 45812984491, 183251937963, 733007751851, 2932031007403, 11728124029611, 46912496118443, 187649984473771, 750599937895083
Offset: 0

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Author

Keywords

Comments

Let u(k), v(k), w(k) be the 3 sequences defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1)=u(k)+v(k)-w(k), v(k+1)=u(k)-v(k)+w(k), w(k+1)=-u(k)+v(k)+w(k); let M(k)=Max(u(k),v(k),w(k)); then a(n)=M(2n)=M(2n-1). - Benoit Cloitre, Mar 25 2002
Also the number of words of length 2n generated by the two letters s and t that reduce to the identity 1 by using the relations ssssss=1, tt=1 and stst=1. The generators s and t along with the three relations generate the dihedral group D6=C2xD3. - Jamaine Paddyfoot (jay_paddyfoot(AT)hotmail.com) and John W. Layman, Jul 08 2002
Binomial transform of A025192. - Paul Barry, Apr 11 2003
Number of walks of length 2n+1 between two adjacent vertices in the cycle graph C_6. Example: a(1)=3 because in the cycle ABCDEF we have three walks of length 3 between A and B: ABAB, ABCB and AFAB. - Emeric Deutsch, Apr 01 2004
Numbers of the form 1 + Sum_{i=1..m} 2^(2*i-1). - Artur Jasinski, Feb 09 2007
Prime numbers of the form 1+Sum[2^(2n-1)] are in A000979. Numbers x such that 1+Sum[2^(2n-1)] is prime for n=1,2,...,x is A127936. - Artur Jasinski, Feb 09 2007
Related to A024493(6n+1), A131708(6n+3), A024495(6n+5). - Paul Curtz, Mar 27 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-6, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)*charpoly(A,2). - Milan Janjic, Feb 21 2010
Number of toothpicks in the toothpick structure of A139250 after 2^n stages. - Omar E. Pol, Feb 28 2011
Numbers whose binary representation is "10" repeated (n-1) times with "11" appended on the end, n >= 1. For example 171 = 10101011 (2). - Omar E. Pol, Nov 22 2012
a(n) is the smallest number for which A072219(a(n)) = 2*n+1. - Ramasamy Chandramouli, Dec 22 2012
An Engel expansion of 2 to the base b := 4/3 as defined in A181565, with the associated series expansion 2 = b + b^2/3 + b^3/(3*11) + b^4/(3*11*43) + .... Cf. A007051. - Peter Bala, Oct 29 2013
The positive integer solution (x,y) of 3*x - 2^n*y = 1, n>=0, with smallest x is (a(n/2), 2) if n is even and (a((n-1)/2), 1) if n is odd. - Wolfdieter Lang, Feb 15 2014
The smallest positive number that requires at least n additions and subtractions of powers of 2 to be formed. See Puzzling StackExchange link. - Alexander Cooke Jul 16 2023

References

  • H. W. Gould, Combinatorial Identities, Morgantown, 1972, (1.77), page 10.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Partial sums of A081294.
Cf. location of records in A007302.

Programs

  • GAP
    List([0..25], n-> (2^(2*n+1) + 1)/3); # G. C. Greubel, Dec 25 2019
  • Haskell
    a007583 = (`div` 3) . (+ 1) . a004171
    -- Reinhard Zumkeller, Jan 09 2013
    
  • Magma
    [(2^(2*n+1) + 1)/3: n in [0..30] ]; // Vincenzo Librandi, Apr 28 2011
    
  • Maple
    a[0]:=1:for n from 1 to 50 do a[n]:=4*a[n-1]-1 od: seq(a[n], n=0..23); # Zerinvary Lajos, Feb 22 2008, with correction by K. Spage, Aug 20 2014
    A007583 := proc(n)
        (2^(2*n+1)+1)/3 ;
    end proc: # R. J. Mathar, Feb 19 2015
  • Mathematica
    (* From Michael De Vlieger, Aug 22 2016 *)
    Table[(2^(2n+1) + 1)/3, {n, 0, 23}]
    Table[1 + 2Sum[4^k, {k, 0, n-1}], {n, 0, 23}]
    NestList[4# -1 &, 1, 23]
    Table[Sum[Binomial[n+k, 2k]/2^(k-n), {k, 0, n}], {n, 0, 23}]
    CoefficientList[Series[(1-2x)/(1-5x+4x^2), {x, 0, 23}], x] (* End *)
  • PARI
    a(n)=sum(k=-n\3,n\3,binomial(2*n+1,n+1+3*k))
    
  • PARI
    a=1; for(n=1,23, print1(a,", "); a=bitor(a,3*a)) \\ K. Spage, Aug 20 2014
    
  • PARI
    Vec((1-2*x)/(1-5*x+4*x^2) + O(x^30)) \\ Altug Alkan, Dec 08 2015
    
  • PARI
    apply( {A007583(n)=2<<(2*n)\/3}, [0..25]) \\ M. F. Hasler, Nov 30 2021
    
  • Sage
    [(2^(2*n+1) + 1)/3 for n in (0..25)] # G. C. Greubel, Dec 25 2019
    

Formula

a(n) = 2*A002450(n) + 1.
From Wolfdieter Lang, Apr 24 2001: (Start)
a(n) = Sum_{m = 0..n} A060920(n, m) = A002450(n+1) - 2*A002450(n).
G.f.: (1-2*x)/(1-5*x+4*x^2). (End)
a(n) = Sum_{k = 0..n} binomial(n+k, 2*k)/2^(k - n).
a(n) = 4*a(n-1) - 1, n > 0.
From Paul Barry, Mar 17 2003: (Start)
a(n) = 1 + 2*Sum_{k = 0..n-1} 4^k;
a(n) = A001045(2n+1). (End)
a(n) = A020988(n-1) + 1 = A039301(n+1) - 1 = A083584(n-1) + 2. - Ralf Stephan, Jun 14 2003
a(0) = 1; a(n+1) = a(n) * 4 - 1. - Regis Decamps (decamps(AT)users.sf.net), Feb 04 2004 (correction to lead index by K. Spage, Aug 20 2014)
a(n) = Sum_{i + j + k = n; 0 <= i, j, k <= n} (n+k)!/i!/j!/(2*k)!. - Benoit Cloitre, Mar 25 2004
a(n) = 5*a(n-1) - 4*a(n-2). - Emeric Deutsch, Apr 01 2004
a(n) = 4^n - A001045(2*n). - Paul Barry, Apr 17 2004
a(n) = 2*(A001045(n))^2 + (A001045(n+1))^2. - Paul Barry, Jul 15 2004
a(n) = left and right terms in M^n * [1 1 1] where M = the 3X3 matrix [1 1 1 / 1 3 1 / 1 1 1]. M^n * [1 1 1] = [a(n) A002450(n+1) a(n)] E.g. a(3) = 43 since M^n * [1 1 1] = [43 85 43] = [a(3) A002450(4) a(3)]. - Gary W. Adamson, Dec 18 2004
a(n) = A072197(n) - A020988(n). - Creighton Dement, Dec 31 2004
a(n) = A139250(2^n). - Omar E. Pol, Feb 28 2011
a(n) = A193652(2*n+1). - Reinhard Zumkeller, Aug 08 2011
a(n) = Sum_{k = -floor(n/3)..floor(n/3)} binomial(2*n, n+3*k)/2. - Mircea Merca, Jan 28 2012
a(n) = 2^(2*(n+1)) - A072197(n). - Vladimir Pletser, Apr 12 2014
a(n) == 2*n + 1 (mod 3). Indeed, from Regis Decamps' formula (Feb 04 2004) we have a(i+1) - a(i) == -1 (mod 3), i= 0, 1, ..., n - 1. Summing, we have a(n) - 1 == -n (mod 3), and the formula follows. - Vladimir Shevelev, May 20 2015
For n > 0 a(n) = A133494(0) + 2 * (A133494(n) + Sum_{x = 1..n - 1}Sum_{k = 0..x - 1}(binomial(x - 1, k)*(A133494(k+1) + A133494(n-x+k)))). - J. Conrad, Dec 06 2015
a(n) = Sum_{k = 0..2n} (-2)^k == 1 + Sum_{k = 1..n} 2^(2k-1). - Bob Selcoe, Aug 21 2016
E.g.f.: (1 + 2*exp(3*x))*exp(x)/3. - Ilya Gutkovskiy, Aug 21 2016
A075680(a(n)) = 1, for n > 0. - Ralf Stephan, Jun 17 2025

A086645 Triangle read by rows: T(n, k) = binomial(2n, 2k).

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 15, 15, 1, 1, 28, 70, 28, 1, 1, 45, 210, 210, 45, 1, 1, 66, 495, 924, 495, 66, 1, 1, 91, 1001, 3003, 3003, 1001, 91, 1, 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1, 1, 153, 3060, 18564, 43758, 43758, 18564, 3060, 153, 1, 1, 190, 4845, 38760
Offset: 0

Views

Author

Philippe Deléham, Jul 26 2003

Keywords

Comments

Terms have the same parity as those of Pascal's triangle.
Coefficients of polynomials (1/2)*((1 + x^(1/2))^(2n) + (1 - x^(1/2))^(2n)).
Number of compositions of 2n having k parts greater than 1; example: T(3, 2) = 15 because we have 4+2, 2+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2, 3+3. - Philippe Deléham, May 18 2005
Number of binary words of length 2n - 1 having k runs of consecutive 1's; example: T(3,2) = 15 because we have 00101, 01001, 01010, 01011, 01101, 10001, 10010, 10011, 10100, 10110, 10111, 11001, 11010, 11011, 11101. - Philippe Deléham, May 18 2005
Let M_n be the n X n matrix M_n(i, j) = T(i, j-1); then for n > 0, det(M_n) = A000364(n), Euler numbers; example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385 = A000364(4). - Philippe Deléham, Sep 04 2005
Equals ConvOffsStoT transform of the hexagonal numbers, A000384: (1, 6, 15, 28, 45, ...); e.g., ConvOffs transform of (1, 6, 15, 28) = (1, 28, 70, 28, 1). - Gary W. Adamson, Apr 22 2008
From Peter Bala, Oct 23 2008: (Start)
Let C_n be the root lattice generated as a monoid by {+-2*e_i: 1 <= i <= n; +-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(C_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(C_n) [Ardila et al.]. See A127674 for (a signed version of) the corresponding array of f-vectors for these type C_n polytopes. See A008459 for the array of h-vectors for type A_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes.
The Hilbert transform of this triangle is A142992 (see A145905 for the definition of this term).
(End)
Diagonal sums: A108479. - Philippe Deléham, Sep 08 2009
Coefficients of Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
Generalized Narayana triangle for 4^n (or cosh(2x)). - Paul Barry, Sep 28 2010
Coefficients of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)*A086646(n,k). - R. J. Mathar, Mar 12 2013
Let E(y) = Sum_{n>=0} y^n/(2*n)! = cosh(sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence (2*n)! as defined in Wang and Wang. Cf. A103327. - Peter Bala, Aug 06 2013
Row 6, (1,66,495,924,495,66,1), plays a role in expansions of powers of the Dedekind eta function. See the Chan link, p. 534, and A034839. - Tom Copeland, Dec 12 2016

Examples

			From _Peter Bala_, Oct 23 2008: (Start)
The triangle begins
n\k|..0.....1.....2.....3.....4.....5.....6
===========================================
0..|..1
1..|..1.....1
2..|..1.....6.....1
3..|..1....15....15.....1
4..|..1....28....70....28.....1
5..|..1....45...210...210....45.....1
6..|..1....66...495...924...495....66.....1
...
(End)
From _Peter Bala_, Aug 06 2013: (Start)
Viewed as the generalized Riordan array (cosh(sqrt(y)), y) with respect to the sequence (2*n)! the column generating functions begin
1st col: cosh(sqrt(y)) = 1 + y/2! + y^2/4! + y^3/6! + y^4/8! + ....
2nd col: 1/2!*y*cosh(sqrt(y)) = y/2! + 6*y^2/4! + 15*y^3/6! + 28*y^4/8! + ....
3rd col: 1/4!*y^2*cosh(sqrt(y)) = y^2/4! + 15*y^3/6! + 70*y^4/8! + 210*y^5/10! + .... (End)
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 224.

Crossrefs

Cf. A008459, A108558, A127674, A142992. - Peter Bala, Oct 23 2008
Cf. A103327 (binomial(2n+1, 2k+1)), A103328 (binomial(2n, 2k+1)), A091042 (binomial(2n+1, 2k)). -Wolfdieter Lang, Jan 06 2013
Cf. A086646 (unsigned matrix inverse), A103327.
Cf. A034839.

Programs

  • Magma
    /* As triangle: */ [[Binomial(2*n, 2*k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 14 2016
  • Maple
    A086645:=(n,k)->binomial(2*n,2*k): seq(seq(A086645(n,k),k=0..n),n=0..12);
  • Mathematica
    Table[Binomial[2 n, 2 k], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 13 2016 *)
  • Maxima
    create_list(binomial(2*n,2*k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    {T(n, k) = binomial(2*n, 2*k)};
    
  • PARI
    {T(n, k) = sum( i=0, min(k, n-k), 4^i * binomial(n, 2*i) * binomial(n - 2*i, k-i))}; /* Michael Somos, May 26 2005 */
    

Formula

T(n, k) = (2*n)!/((2*(n-k))!*(2*k)!) row sums = A081294. COLUMNS: A000012, A000384
Sum_{k>=0} T(n, k)*A000364(k) = A000795(n) = (2^n)*A005647(n).
Sum_{k>=0} T(n, k)*2^k = A001541(n). Sum_{k>=0} T(n, k)*3^k = 2^n*A001075(n). Sum_{k>=0} T(n, k)*4^k = A083884(n). - Philippe Deléham, Feb 29 2004
O.g.f.: (1 - z*(1+x))/(x^2*z^2 - 2*x*z*(1+z) + (1-z)^2) = 1 + (1 + x)*z +(1 + 6*x + x^2)*z^2 + ... . - Peter Bala, Oct 23 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A081294(n), A001541(n), A090965(n), A083884(n), A099140(n), A099141(n), A099142(n), A165224(n), A026244(n) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 08 2009
Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
From Paul Barry, Sep 28 2010: (Start)
G.f.: 1/(1-x-x*y-4*x^2*y/(1-x-x*y)) = (1-x*(1+y))/(1-2*x*(1+y)+x^2*(1-y)^2);
E.g.f.: exp((1+y)*x)*cosh(2*sqrt(y)*x);
T(n,k) = Sum_{j=0..n} C(n,j)*C(n-j,2*(k-j))*4^(k-j). (End)
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) + 2*T(n-2,k-1) - T(n-2,k) - T(n-2,k-2), with T(0,0)=T(1,0)=T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 26 2013
From Peter Bala, Sep 22 2021: (Start)
n-th row polynomial R(n,x) = (1-x)^n*T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A008459.
R(n,x) = Sum_{k = 0..n} binomial(n,2*k)*(4*x)^k*(1 + x)^(n-2*k).
R(n,x) = n*Sum_{k = 0..n} (n+k-1)!/((n-k)!*(2*k)!)*(4*x)^k*(1-x)^(n-k) for n >= 1. (End)

A090965 a(n) = 8*a(n-1) - 4*a(n-2), where a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 28, 208, 1552, 11584, 86464, 645376, 4817152, 35955712, 268377088, 2003193856, 14952042496, 111603564544, 833020346368, 6217748512768, 46409906716672, 346408259682304, 2585626450591744, 19299378566004736
Offset: 0

Views

Author

Philippe Deléham, Feb 29 2004

Keywords

Crossrefs

Cf. A001075.
Sum_{k>=0} A086645(n,k)*m^k for m = 0, 1, 2, 4 gives A000007, A081294, A001541, A083884.

Programs

  • GAP
    a:=[1,4];; for n in [3..20] do a[n]:=8*a[n-1]-4*a[n-2]; od; a; # G. C. Greubel, Feb 03 2019
  • Magma
    m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-4*x)/(1-8*x+4*x^2) )); // G. C. Greubel, Feb 03 2019
    
  • Mathematica
    LinearRecurrence[{8,-4}, {1,4}, 20] (* G. C. Greubel, Feb 03 2019 *)
  • PARI
    my(x='x+O('x^20)); Vec((1-4*x)/(1-8*x+4*x^2)) \\ G. C. Greubel, Feb 03 2019
    
  • Sage
    [lucas_number2(n,8,4)/2 for n in range(0,21)] # Zerinvary Lajos, Jul 08 2008
    

Formula

a(n) = Sum_{k>=0} binomial(2*n, 2*k)*3^k = Sum_{k>=0} A086645(n, k)*3^k.
a(n) = 2^n*A001075(n).
G.f.: (1-4*x)/(1-8*x+4*x^2). - Philippe Deléham, Sep 07 2009
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k-4)/(x*(3*k-1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
From Peter Bala, Feb 19 2022: (Start)
a(n) = Sum_{k = 0..floor(n/2)} 4^(n-2*k)*12^k*binomial(n,2*k).
a(n) = [x^n] (4*x + sqrt(1 + 12*x^2))^n.
G.f.: A(x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. (End)

Extensions

Corrected by T. D. Noe, Nov 07 2006

A099140 a(n) = 4^n * T(n,3/2) where T is the Chebyshev polynomial of the first kind.

Original entry on oeis.org

1, 6, 56, 576, 6016, 62976, 659456, 6905856, 72318976, 757334016, 7930904576, 83053510656, 869747654656, 9108115685376, 95381425750016, 998847258034176, 10460064284409856, 109539215284371456, 1147109554861899776
Offset: 0

Views

Author

Paul Barry, Sep 30 2004

Keywords

Comments

In general, r^n * T(n,(r+2)/r) has g.f. (1-(r+2)*x)/(1-2*(r+2)*x + r^2*x^2), e.g.f. exp((r+2)*x)*cosh(2*sqrt(r+1)*x), a(n) = Sum_{k=0..n} (r+1)^k*binomial(2n,2k) and a(n) = (1+sqrt(r+1))^(2n)/2 + (1-sqrt(r+1))^(2n)/2.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{12,-16},{1,6},30] (* Harvey P. Dale, Oct 23 2012 *)
  • PARI
    a(n) = 4^n*polchebyshev(n, 1, 3/2); \\ Michel Marcus, Sep 08 2019

Formula

G.f.: (1-6*x)/(1-12*x+16*x^2);
E.g.f.: exp(6*x)*cosh(2*sqrt(5)*x);
a(n) = 4^n * T(n, 6/4) where T is the Chebyshev polynomial of the first kind;
a(n) = Sum_{k=0..n} 5^k*binomial(2n, 2k);
a(n) = (1+sqrt(5))^(2n)/2 + (1-sqrt(5))^(2n)/2.
a(n) = a(0)=1, a(1)=6, 12*a(n-1) - 16*a(n-2) for n > 1. - Philippe Deléham, Sep 08 2009

A099141 a(n) = 5^n * T(n,7/5) where T is the Chebyshev polynomial of the first kind.

Original entry on oeis.org

1, 7, 73, 847, 10033, 119287, 1419193, 16886527, 200931553, 2390878567, 28449011113, 338514191407, 4027973401873, 47928772841047, 570303484727833, 6786029465163487, 80746825394092993, 960804818888214727
Offset: 0

Views

Author

Paul Barry, Sep 30 2004

Keywords

Comments

In general, r^n * T(n,(r+2)/r) has g.f. (1-(r+2)*x)/(1-2*(r+2)*x + r^2*x^2), e.g.f. exp((r+2)*x)*cosh(2*sqrt(r+1)*x), a(n) = Sum_{k=0..n} (r+1)^k*binomial(2*n,2*k) and a(n) = (1+sqrt(r+1))^(2*n)/2 + (1-sqrt(r+1))^(2*n)/2.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{14,-25},{1,7},30] (* Harvey P. Dale, Dec 26 2014 *)

Formula

G.f.: (1-7*x)/(1-14*x+25*x^2);
e.g.f.: exp(7*x)*cosh(2*sqrt(6)*x);
a(n) = 5^n * T(n, 7/5) where T is the Chebyshev polynomial of the first kind;
a(n) = Sum_{k=0..n} 6^k * binomial(2n, 2k);
a(n) = (1+sqrt(6))^(2n)/2 + (1-sqrt(6))^(2n)/2.
a(0)=1, a(1)=7, a(n) = 14*a(n-1) - 25*a(n-2) for n > 1. - Philippe Deléham, Sep 08 2009

A099142 a(n) = 6^n * T(n, 4/3) where T is the Chebyshev polynomial of the first kind.

Original entry on oeis.org

1, 8, 92, 1184, 15632, 207488, 2757056, 36643328, 487039232, 6473467904, 86042074112, 1143628341248, 15200538791936, 202038000386048, 2685388609667072, 35692849740775424, 474411605904392192
Offset: 0

Views

Author

Paul Barry, Sep 30 2004

Keywords

Comments

In general, r^n * T(n,(r+2)/r) has g.f. (1-(r+2)*x)/(1-2*(r+2)*x + r^2*x^2), e.g.f. exp((r+2)*x)*cosh(2*sqrt(r+1)*x), a(n) = Sum_{k=0..n} (r+1)^k*binomial(2n,2k) and a(n) = (1+sqrt(r+1))^(2*n)/2 + (1-sqrt(r+1))^(2*n)/2.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{16,-36},{1,8},20] (* Harvey P. Dale, Mar 09 2018 *)
  • PARI
    a(n) = 6^n*polchebyshev(n, 1, 4/3); \\ Michel Marcus, Sep 08 2019

Formula

G.f.: (1-8*x)/(1-16*x+36*x^2);
E.g.f.: exp(8*x)*cosh(2*sqrt(7)*x).
a(n) = 6^n * T(n, 8/6) where T is the Chebyshev polynomial of the first kind.
a(n) = Sum_{k=0..n} 7^k * binomial(2n, 2k).
a(n) = (1+sqrt(7))^(2*n)/2 + (1-sqrt(7))^(2*n)/2.
a(0)=1, a(1)=8, a(n) = 16*a(n-1) - 36*a(n-2) for n > 1. - Philippe Deléham, Sep 08 2009

A333988 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of (1-(k+1)*x) / (1-2*(k+1)*x+((k-1)*x)^2).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 8, 1, 1, 4, 17, 32, 1, 1, 5, 28, 99, 128, 1, 1, 6, 41, 208, 577, 512, 1, 1, 7, 56, 365, 1552, 3363, 2048, 1, 1, 8, 73, 576, 3281, 11584, 19601, 8192, 1, 1, 9, 92, 847, 6016, 29525, 86464, 114243, 32768, 1, 1, 10, 113, 1184, 10033, 62976, 265721, 645376, 665857, 131072, 1
Offset: 0

Views

Author

Seiichi Manyama, Sep 04 2020

Keywords

Examples

			Square array begins:
  1,   1,    1,     1,     1,     1, ...
  1,   2,    3,     4,     5,     6, ...
  1,   8,   17,    28,    41,    56, ...
  1,  32,   99,   208,   365,   576, ...
  1, 128,  577,  1552,  3281,  6016, ...
  1, 512, 3363, 11584, 29525, 62976, ...
		

Crossrefs

Main diagonal gives A333990.

Programs

  • Mathematica
    T[n_, 0] := 1; T[n_, k_] := Sum[k^j * Binomial[2*n, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Sep 04 2020 *)
  • PARI
    {T(n, k) = sum(j=0, n, k^j*binomial(2*n, 2*j))}

Formula

T(n,k) = Sum_{j=0..n} k^j * binomial(2*n,2*j).
T(0,k) = 1, T(1,k) = k+1 and T(n,k) = 2 * (k+1) * T(n-1,k) - (k-1)^2 * T(n-2,k) for n>1.

A270473 Expansion of g.f. (1-5*x)/(1-9*x).

Original entry on oeis.org

1, 4, 36, 324, 2916, 26244, 236196, 2125764, 19131876, 172186884, 1549681956, 13947137604, 125524238436, 1129718145924, 10167463313316, 91507169819844, 823564528378596, 7412080755407364, 66708726798666276, 600378541187996484, 5403406870691968356, 48630661836227715204
Offset: 0

Views

Author

Colin Barker, Mar 17 2016

Keywords

Comments

Also squares that can be expressed as the sum of two powers of three (3^x + 3^y), except a(0). - Karl-Heinz Hofmann, Sep 03 2022

Crossrefs

Cf. A001019 (powers of 9), A083884 (partial sums).
Cf. A067403: (1-4*x)/(1-9*x); A102518: (1-6*x)/(1-9*x).

Programs

  • Mathematica
    Join[{1},NestList[9#&,4,20]] (* Harvey P. Dale, Oct 23 2022 *)
  • PARI
    Vec((1-5*x)/(1-9*x) + O(x^30))

Formula

G.f.: (1-5*x)/(1-9*x).
a(n) = 9*a(n-1) for n>1.
a(n) = 4*9^(n-1) for n>0.
E.g.f.: (4*exp(9*x) + 5)/9. - Stefano Spezia, Jul 09 2024

A164907 a(n) = (3*3^n-(-1)^n)/2.

Original entry on oeis.org

1, 5, 13, 41, 121, 365, 1093, 3281, 9841, 29525, 88573, 265721, 797161, 2391485, 7174453, 21523361, 64570081, 193710245, 581130733, 1743392201, 5230176601, 15690529805, 47071589413, 141214768241, 423644304721, 1270932914165
Offset: 0

Views

Author

Klaus Brockhaus, Aug 31 2009

Keywords

Comments

Interleaving of A096053 and A083884 without initial term 1.
Partial sums are (essentially) in A080926.
First differences are (essentially) in A105723.
a(n)+a(n+1) = A008776(n+1) = A099856(n+1) = A110593(n+2).
Binomial transform of A056450. Inverse binomial transform of A164908.

Crossrefs

Equals A046717 without initial term 1 and A080925 without initial term 0. Equals A084182 / 2 from second term onward.

Programs

Formula

a(n) = 2*a(n-1)+3*a(n-2) for n > 1; a(0) = 1, a(1) = 5.
G.f.: (1+3*x)/((1+x)*(1-3*x)).
a(n) = 3*a(n-1)+2*(-1)^n. - Carmine Suriano, Mar 21 2014

A165224 a(0)=1, a(1)=9, a(n) = 18*a(n-1) - 49*a(n-2) for n > 1.

Original entry on oeis.org

1, 9, 113, 1593, 23137, 338409, 4957649, 72655641, 1064876737, 15607654857, 228758827313, 3352883803641, 49142725927201, 720277760311209, 10557006115168913, 154732499817791193, 2267891697076964737
Offset: 0

Views

Author

Philippe Deléham, Sep 08 2009

Keywords

Comments

a(n)/a(n-1) tends to 9 + 4*sqrt(2) = 14.65685424... - Klaus Brockhaus, Sep 25 2009

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{18,-49},{1,9},20] (* Harvey P. Dale, Sep 30 2016 *)

Formula

G.f.: (1-9x)/(1-18x+49x^2);
e.g.f.: exp(9x)*cosh(4*sqrt(2)x);
a(n) = Sum_{k=0..n} 8^k*binomial(2n,2k) = Sum_{k=0..n} 8^k*A086645(n,k);
a(n) = 7^n*T(n,9/7) where T is the Chebyshev polynomial of the first kind;
a(n) = (1+sqrt(8))^(2n)/2 + (1-sqrt(8))^(2n)/2.
a(n) = ((9-4*sqrt(2))^n + (9+4*sqrt(2))^n)/2. - Klaus Brockhaus, Sep 25 2009
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