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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A002450 a(n) = (4^n - 1)/3.

Original entry on oeis.org

0, 1, 5, 21, 85, 341, 1365, 5461, 21845, 87381, 349525, 1398101, 5592405, 22369621, 89478485, 357913941, 1431655765, 5726623061, 22906492245, 91625968981, 366503875925, 1466015503701, 5864062014805, 23456248059221, 93824992236885, 375299968947541
Offset: 0

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Keywords

Comments

For n > 0, a(n) is the degree (n-1) "numbral" power of 5 (see A048888 for the definition of numbral arithmetic). Example: a(3) = 21, since the numbral square of 5 is 5(*)5 = 101(*)101(base 2) = 101 OR 10100 = 10101(base 2) = 21, where the OR is taken bitwise. - John W. Layman, Dec 18 2001
a(n) is composite for all n > 2 and has factors x, (3*x + 2*(-1)^n) where x belongs to A001045. In binary the terms greater than 0 are 1, 101, 10101, 1010101, etc. - John McNamara, Jan 16 2002
Number of n X 2 binary arrays with path of adjacent 1's from upper left corner to right column. - R. H. Hardin, Mar 16 2002
The Collatz-function iteration started at a(n), for n >= 1, will end at 1 after 2*n+1 steps. - Labos Elemer, Sep 30 2002 [corrected by Wolfdieter Lang, Aug 16 2021]
Second binomial transform of A001045. - Paul Barry, Mar 28 2003
All members of sequence are also generalized octagonal numbers (A001082). - Matthew Vandermast, Apr 10 2003
Also sum of squares of divisors of 2^(n-1): a(n) = A001157(A000079(n-1)), for n > 0. - Paul Barry, Apr 11 2003
Binomial transform of A000244 (with leading zero). - Paul Barry, Apr 11 2003
Number of walks of length 2n between two vertices at distance 2 in the cycle graph C_6. For n = 2 we have for example 5 walks of length 4 from vertex A to C: ABABC, ABCBC, ABCDC, AFABC and AFEDC. - Herbert Kociemba, May 31 2004
Also number of walks of length 2n + 1 between two vertices at distance 3 in the cycle graph C_12. - Herbert Kociemba, Jul 05 2004
a(n+1) is the number of steps that are made when generating all n-step random walks that begin in a given point P on a two-dimensional square lattice. To make one step means to mark one vertex on the lattice (compare A080674). - Pawel P. Mazur (Pawel.Mazur(AT)pwr.wroc.pl), Mar 13 2005
a(n+1) is the sum of square divisors of 4^n. - Paul Barry, Oct 13 2005
a(n+1) is the decimal number generated by the binary bits in the n-th generation of the Rule 250 elementary cellular automaton. - Eric W. Weisstein, Apr 08 2006
a(n-1) / a(n) = percentage of wasted storage if a single image is stored as a pyramid with a each subsequent higher resolution layer containing four times as many pixels as the previous layer. n is the number of layers. - Victor Brodsky (victorbrodsky(AT)gmail.com), Jun 15 2006
k is in the sequence if and only if C(4k + 1, k) (A052203) is odd. - Paul Barry, Mar 26 2007
This sequence also gives the number of distinct 3-colorings of the odd cycle C(2*n - 1). - Keith Briggs, Jun 19 2007
All numbers of the form m*4^m + (4^m-1)/3 have the property that they are sums of two squares and also their indices are the sum of two squares. This follows from the identity m*4^m + (4^m-1)/3 = 4(4(..4(4m + 1) + 1) + 1) + 1 ..) + 1. - Artur Jasinski, Nov 12 2007
For n > 0, terms are the numbers that, in base 4, are repunits: 1_4, 11_4, 111_4, 1111_4, etc. - Artur Jasinski, Sep 30 2008
Let A be the Hessenberg matrix of order n, defined by: A[1, j] = 1, A[i, i] := 5, (i > 1), A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n) = charpoly(A,1). - Milan Janjic, Jan 27 2010
This is the sequence A(0, 1; 3, 4; 2) = A(0, 1; 4, 0; 1) of the family of sequences [a, b : c, d : k] considered by G. Detlefs, and treated as A(a, b; c, d; k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
6*a(n) + 1 is every second Mersenne number greater than or equal to M3, hence all Mersenne primes greater than M2 must be a 6*a(n) + 1 of this sequence. - Roderick MacPhee, Nov 01 2010
Smallest number having alternating bit sum n. Cf. A065359.
For n = 1, 2, ..., the last digit of a(n) is 1, 5, 1, 5, ... . - Washington Bomfim, Jan 21 2011
Rule 50 elementary cellular automaton generates this sequence. This sequence also appears in the second column of array in A173588. - Paul Muljadi, Jan 27 2011
Sequence found by reading the line from 0, in the direction 0, 5, ... and the line from 1, in the direction 1, 21, ..., in the square spiral whose edges are the Jacobsthal numbers A001045 and whose vertices are the numbers A000975. These parallel lines are two semi-diagonals in the spiral. - Omar E. Pol, Sep 10 2011
a(n), n >= 1, is also the inverse of 3, denoted by 3^(-1), Modd(2^(2*n - 1)). For Modd n see a comment on A203571. E.g., a(2) = 5, 3 * 5 = 15 == 1 (Modd 8), because floor(15/8) = 1 is odd and -15 == 1 (mod 8). For n = 1 note that 3 * 1 = 3 == 1 (Modd 2) because floor(3/2) = 1 and -3 == 1 (mod 2). The inverse of 3 taken Modd 2^(2*n) coincides with 3^(-1) (mod 2^(2*n)) given in A007583(n), n >= 1. - Wolfdieter Lang, Mar 12 2012
If an AVL tree has a leaf at depth n, then the tree can contain no more than a(n+1) nodes total. - Mike Rosulek, Nov 20 2012
Also, this is the Lucas sequence V(5, 4). - Bruno Berselli, Jan 10 2013
Also, for n > 0, a(n) is an odd number whose Collatz trajectory contains no odd number other than n and 1. - Jayanta Basu, Mar 24 2013
Sum_{n >= 1} 1/a(n) converges to (3*(log(4/3) - QPolyGamma[0, 1, 1/4]))/log(4) = 1.263293058100271... = A321873. - K. G. Stier, Jun 23 2014
Consider n spheres in R^n: the i-th one (i=1, ..., n) has radius r(i) = 2^(1-i) and the coordinates of its center are (0, 0, ..., 0, r(i), 0, ..., 0) where r(i) is in position i. The coordinates of the intersection point in the positive orthant of these spheres are (2/a(n), 4/a(n), 8/a(n), 16/a(n), ...). For example in R^2, circles centered at (1, 0) and (0, 1/2), and with radii 1 and 1/2, meet at (2/5, 4/5). - Jean M. Morales, May 19 2015
From Peter Bala, Oct 11 2015: (Start)
a(n) gives the values of m such that binomial(4*m + 1,m) is odd. Cf. A003714, A048716, A263132.
2*a(n) = A020988(n) gives the values of m such that binomial(4*m + 2, m) is odd.
4*a(n) = A080674(n) gives the values of m such that binomial(4*m + 4, m) is odd. (End)
Collatz Conjecture Corollary: Except for powers of 2, the Collatz iteration of any positive integer must eventually reach a(n) and hence terminate at 1. - Gregory L. Simay, May 09 2016
Number of active (ON, black) cells at stage 2^n - 1 of the two-dimensional cellular automaton defined by "Rule 598", based on the 5-celled von Neumann neighborhood. - Robert Price, May 16 2016
From Luca Mariot and Enrico Formenti, Sep 26 2016: (Start)
a(n) is also the number of coprime pairs of polynomials (f, g) over GF(2) where both f and g have degree n + 1 and nonzero constant term.
a(n) is also the number of pairs of one-dimensional binary cellular automata with linear and bipermutive local rule of neighborhood size n+1 giving rise to orthogonal Latin squares of order 2^m, where m is a multiple of n. (End)
Except for 0, 1 and 5, all terms are Brazilian repunits numbers in base 4, and so belong to A125134. For n >= 3, all these terms are composite because a(n) = {(2^n-1) * (2^n + 1)}/3 and either (2^n - 1) or (2^n + 1) is a multiple of 3. - Bernard Schott, Apr 29 2017
Given the 3 X 3 matrix A = [2, 1, 1; 1, 2, 1; 1, 1, 2] and the 3 X 3 unit matrix I_3, A^n = a(n)(A - I_3) + I_3. - Nicolas Patrois, Jul 05 2017
The binary expansion of a(n) (n >= 1) consists of n 1's alternating with n - 1 0's. Example: a(4) = 85 = 1010101_2. - Emeric Deutsch, Aug 30 2017
a(n) (n >= 1) is the viabin number of the integer partition [n, n - 1, n - 2, ..., 2, 1] (for the definition of viabin number see comment in A290253). Example: a(4) = 85 = 1010101_2; consequently, the southeast border of the Ferrers board of the corresponding integer partition is ENENENEN, where E = (1, 0), N = (0, 1); this leads to the integer partition [4, 3, 2, 1]. - Emeric Deutsch, Aug 30 2017
Numbers whose binary and Gray-code representations are both palindromes (i.e., intersection of A006995 and A281379). - Amiram Eldar, May 17 2021
Starting with n = 1 the sequence satisfies {a(n) mod 6} = repeat{1, 5, 3}. - Wolfdieter Lang, Jan 14 2022
Terms >= 5 are those q for which the multiplicative order of 2 mod q is floor(log_2(q)) + 2 (and which is 1 more than the smallest possible order for any q). - Tim Seuré, Mar 09 2024
The order of 2 modulo a(n) is 2*n for n >= 2. - Joerg Arndt, Mar 09 2024

Examples

			Apply Collatz iteration to 9: 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5 and hence 16, 8, 4, 2, 1.
Apply Collatz iteration to 27: 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5 and hence 16, 8, 4, 2, 1. [Corrected by _Sean A. Irvine_ at the suggestion of Stephen Cornelius, Mar 04 2024]
a(5) = (4^5 - 1)/3 = 341 = 11111_4 = {(2^5 - 1) * (2^5 + 1)}/3 = 31 * 33/3 = 31 * 11. - _Bernard Schott_, Apr 29 2017
		

References

  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 112.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
  • 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

Partial sums of powers of 4, A000302.
When converted to binary, this gives A094028.
Subsequence of A003714.
Primitive factors: A129735.

Programs

  • GAP
    List([0..25], n -> (4^n-1)/3); # Muniru A Asiru, Feb 18 2018
    
  • Haskell
    a002450 = (`div` 3) . a024036
    a002450_list = iterate ((+ 1) . (* 4)) 0
    -- Reinhard Zumkeller, Oct 03 2012
    
  • Magma
    [ (4^n-1)/3: n in [0..25] ]; // Klaus Brockhaus, Oct 28 2008
    
  • Magma
    [n le 2 select n-1 else 5*Self(n-1)-4*Self(n-2): n in [1..70]]; // Vincenzo Librandi, Jun 13 2015
    
  • Maple
    [seq((4^n-1)/3,n=0..40)];
    A002450:=1/(4*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation, dropping the initial zero
  • Mathematica
    Table[(4^n - 1)/3, {n, 0, 127}] (* Vladimir Joseph Stephan Orlovsky, Sep 29 2008 *)
    LinearRecurrence[{5, -4}, {0, 1}, 30] (* Harvey P. Dale, Jun 23 2013 *)
  • Maxima
    makelist((4^n-1)/3, n, 0, 30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n) = (4^n-1)/3;
    
  • PARI
    my(z='z+O('z^40)); Vec(z/((1-z)*(1-4*z))) \\ Altug Alkan, Oct 11 2015
    
  • Python
    def A002450(n): return ((1<<(n<<1))-1)//3 # Chai Wah Wu, Jan 29 2023
  • Scala
    ((List.fill(20)(4: BigInt)).scanLeft(1: BigInt)( * )).scanLeft(0: BigInt)( + ) // Alonso del Arte, Sep 17 2019
    

Formula

From Wolfdieter Lang, Apr 24 2001: (Start)
a(n+1) = Sum_{m = 0..n} A060921(n, m).
G.f.: x/((1-x)*(1-4*x)). (End)
a(n) = Sum_{k = 0..n-1} 4^k; a(n) = A001045(2*n). - Paul Barry, Mar 17 2003
E.g.f.: (exp(4*x) - exp(x))/3. - Paul Barry, Mar 28 2003
a(n) = (A007583(n) - 1)/2. - N. J. A. Sloane, May 16 2003
a(n) = A000975(2*n)/2. - N. J. A. Sloane, Sep 13 2003
a(n) = A084160(n)/2. - N. J. A. Sloane, Sep 13 2003
a(n+1) = 4*a(n) + 1, with a(0) = 0. - Philippe Deléham, Feb 25 2004
a(n) = Sum_{i = 0..n-1} C(2*n - 1 - i, i)*2^i. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
a(n+1) = Sum_{k = 0..n} binomial(n+1, k+1)*3^k. - Paul Barry, Aug 20 2004
a(n) = center term in M^n * [1 0 0], where M is the 3 X 3 matrix [1 1 1 / 1 3 1 / 1 1 1]. M^n * [1 0 0] = [A007583(n-1) a(n) A007583(n-1)]. E.g., a(4) = 85 since M^4 * [1 0 0] = [43 85 43] = [A007583(3) a(4) A007583(3)]. - Gary W. Adamson, Dec 18 2004
a(n) = Sum_{k = 0..n, j = 0..n} C(n, j)*C(j, k)*A001045(j - k). - Paul Barry, Feb 15 2005
a(n) = Sum_{k = 0..n} C(n, k)*A001045(n-k)*2^k = Sum_{k = 0..n} C(n, k)*A001045(k)*2^(n-k). - Paul Barry, Apr 22 2005
a(n) = A125118(n, 3) for n > 2. - Reinhard Zumkeller, Nov 21 2006
a(n) = Sum_{k = 0..n} 2^(n - k)*A128908(n, k), n >= 1. - Philippe Deléham, Oct 19 2008
a(n) = Sum_{k = 0..n} A106566(n, k)*A100335(k). - Philippe Deléham, Oct 30 2008
If we define f(m, j, x) = Sum_{k = j..m} binomial(m, k)*stirling2(k, j)*x^(m - k) then a(n-1) = f(2*n, 4, -2), n >= 2. - Milan Janjic, Apr 26 2009
a(n) = A014551(n) * A001045(n). - R. J. Mathar, Jul 08 2009
a(n) = 4*a(n-1) + a(n-2) - 4*a(n-3) = 5*a(n-1) - 4*a(n-2), a(0) = 0, a(1) = 1, a(2) = 5. - Wolfdieter Lang, Oct 18 2010
a(0) = 0, a(n+1) = a(n) + 2^(2*n). - Washington Bomfim, Jan 21 2011
A036555(a(n)) = 2*n. - Reinhard Zumkeller, Jan 28 2011
a(n) = Sum_{k = 1..floor((n+2)/3)} C(2*n + 1, n + 2 - 3*k). - Mircea Merca, Jun 25 2011
a(n) = Sum_{i = 1..n} binomial(2*n + 1, 2*i)/3. - Wesley Ivan Hurt, Mar 14 2015
a(n+1) = 2^(2*n) + a(n), a(0) = 0. - Ben Paul Thurston, Dec 27 2015
a(k*n)/a(n) = 1 + 4^n + ... + 4^((k-1)*n). - Gregory L. Simay, Jun 09 2016
Dirichlet g.f.: (PolyLog(s, 4) - zeta(s))/3. - Ilya Gutkovskiy, Jun 26 2016
A000120(a(n)) = n. - André Dalwigk, Mar 26 2018
a(m) divides a(m*n), in particular: a(2*n) == 0 (mod 5), a(3*n) == 0 (mod 3*7), a(5*n) == 0 (mod 11*31), etc. - M. F. Hasler, Oct 19 2018
a(n) = 4^(n-1) + a(n-1). - Bob Selcoe, Jan 01 2020
a(n) = A178415(1, n) = A347834(1, n-1), arrays, for n >= 1. - Wolfdieter Lang, Nov 29 2021
a(n) = A000225(2*n)/3. - John Keith, Jan 22 2022
a(n) = A080674(n) + 1 = A047849(n) - 1 = A163834(n) - 2 = A155701(n) - 3 = A163868(n) - 4 = A156605(n) - 7. - Ray Chandler, Jun 16 2023
From Peter Bala, Jul 23 2025: (Start)
The following are examples of telescoping products. Cf. A016153:
Product_{k = 1..2*n} 1 + 2^k/a(k+1) = a(n+1)/A007583(n) = (4^(n+1) - 1)/(2*4^n + 1).
Hence, Product_{k >= 1} 1 + 2^k/a(k+1) = 2.
Product_{k >= 1} 1 - 2^k/a(k+1) = 2/5, since 1 - 2^n/a(n+1) = b(n)/b(n-1), where b(n) = 2 - 3/(1 - 2^(n+1)).
Product_{k >= 1} 1 + (-2)^k/a(k+1) = 2/3, since 1 + (-2)^n/a(n+1) = c(n)/c(n-1), where c(n) = 2 - 1/(1 + (-2)^(n+1)).
Product_{k >= 1} 1 - (-2)^k/a(k+1) = 6/5, since 1 - (-2)^n/a(n+1) = d(n)/d(n-1), where d(n) = 2 - 1/(1 - (-2)^(n+1)). (End)

A001047 a(n) = 3^n - 2^n.

Original entry on oeis.org

0, 1, 5, 19, 65, 211, 665, 2059, 6305, 19171, 58025, 175099, 527345, 1586131, 4766585, 14316139, 42981185, 129009091, 387158345, 1161737179, 3485735825, 10458256051, 31376865305, 94134790219, 282412759265, 847255055011, 2541798719465, 7625463267259, 22876524019505
Offset: 0

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Author

Keywords

Comments

a(n+1) is the sum of the elements in the n-th row of triangle pertaining to A036561. - Amarnath Murthy, Jan 02 2002
Number of 2 X n binary arrays with a path of adjacent 1's and no path of adjacent 0's from top row to bottom row. - R. H. Hardin, Mar 21 2002
With offset 1, partial sums of A027649. - Paul Barry, Jun 24 2003
Number of distinct lines through the origin in the n-dimensional lattice of side length 2. A049691 has the values for the 2-dimensional lattice of side length n. - Joshua Zucker, Nov 19 2003
a(n+1)/(n+1)=(3*3^n-2*2^n)/(n+1) is the second binomial transform of the harmonic sequence 1/(n+1). - Paul Barry, Apr 19 2005
a(n+1) is the sum of n-th row of A036561. - Reinhard Zumkeller, May 14 2006
The sequence gives the sum of the lengths of the segments in Cantor's dust generating sequence up to the i-th step. Measurement unit = length of the segment of i-th step. - Giorgio Balzarotti, Nov 18 2006
Let T be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xTy if x is a proper subset of y. Then a(n) = |T|. - Ross La Haye, Dec 22 2006
From Alexander Adamchuk, Jan 04 2007: (Start)
a(n) is prime for n in A057468.
p divides a(p) - 1 for prime p.
Quotients (3^p - 2^p - 1)/p, where p = prime(n), are listed in A127071.
Numbers k such that k divides 3^k - 2^k - 1 are listed in A127072.
Pseudoprimes in A127072(n) include all powers of primes {2,3,7} and some composite numbers that are listed in A127073, which includes all Carmichael numbers A002997.
Numbers n such that n^2 divides 3^n - 2^n - 1 are listed in A127074.
5 divides a(2n).
5^2 divides a(2*5n).
5^3 divides a(2*5^2n).
5^4 divides a(2*5^3n).
7^2 divides a(6*7n).
13 divides a(4n).
13^2 divides a(4*13n).
19 divides a(3n).
19^2 divides a(3*19n).
23^2 divides a(11n).
23^3 divides a(11*23n).
23^4 divides a(11*23^2n).
29 divides a(7n).
p divides a((p-1)n) for prime p>3.
p divides a((p-1)/2) for prime p in A097934. Also primes p such that 6 is a square mod p, except {2,3}, A038876(n).
p^(k+1) divides a(p^k*(p-1)/2*n) for prime p in A097934.
p^(k+1) divides a(p^k*(p-1)*n) for prime p>3.
Note the exception that for p = 23, p^(k+2) divides a(p^k*(p-1)/2*n).
There are no more such exceptions for primes p up to 600000. (End)
a(n) divides a(q*(n+1)-1), for all q integer. Leonardo Sarasua, Apr 15 2024
Final digits of terms follow sequence 1,5,9,5. - Enoch Haga, Nov 26 2007
This is also the second column sequence of the Sheffer triangle A143494 (2-restricted Stirling2 numbers). See the e.g.f. given below. - Wolfdieter Lang, Oct 08 2011
Partial sums give A000392. - Jon Perry, Apr 05 2014
For n >= 1, this is also row 2 of A281890: when consecutive positive integers are written as a product of primes in nondecreasing order, "3" occurs in n-th position a(n) times out of every 6^n. - Peter Munn, May 17 2017
a(n) is the number of ternary sequences of length n which include the digit 2. For example, a(2)=5 since the sequences are 02,20,12,21,22. - Enrique Navarrete, Apr 05 2021
a(n-1) is the number of ways we can form disjoint unions of two nonempty subsets of [n] such that the union contains n. For example, for n = 3, a(2) = 5 since the disjoint unions are {1}U{3}, {1}U{2,3}, {2}U{3}, {2}U{1,3}, and {1,2}U{3}. Cf. A000392 if we drop the requirement that the union contains n. - Enrique Navarrete, Aug 24 2021
Configures as a composite Koch Snowflake Fractal (see illustration in links) based on the five-fold division of the Cantor Square/Cantor Dust Fractal of (9^n-4^n)/5 see my illustration in (A016153). - John Elias, Oct 13 2021
Number of pairs (A,B) where B is a subset of {1,2,...,n} and A is a proper subset of B. - Jianing Song, Jun 18 2022
From Manfred Boergens, Mar 29 2023: (Start)
With regard to the comments by Ross La Haye and Jianing Song: Omitting "proper" gives A000244.
Number of pairs (A,B) where B is a nonempty subset of {1,2,...,n} and A is a nonempty subset of B. For nonempty proper subsets see a(n+1) in A028243. (End)
a(n) is the number of n-digit numbers whose smallest decimal digit is 7. - Stefano Spezia, Nov 15 2023
a(n-1) is the number of all possible player-reduced binary games observed by each player in an nx2 game assuming the individual strategies of k < n - 1 players are fixed and the remaining n - k - 1 player will play as one, either maintaining their status quo strategies or jointly adopting an alternative strategy. - Ambrosio Valencia-Romero, Apr 11 2024

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 86-87.
  • 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

a(n) = row sums of A091913, row 2 of A047969, column 1 of A090888 and column 1 of A038719.
Cf. partitions: A241766, A241759.
A diagonal of A262307.

Programs

  • Haskell
    a001047 n = a001047_list !! n
    a001047_list = map fst $ iterate (\(u, v) -> (3 * u + v, 2 * v)) (0, 1)
    -- Reinhard Zumkeller, Jun 09 2013
  • Magma
    [3^n - 2^n: n in [0..30]]; // Vincenzo Librandi, Jul 17 2011
    
  • Maple
    seq(3^n - 2^n, n=0..40); # Giorgio Balzarotti, Nov 18 2006
    A001047:=1/(3*z-1)/(2*z-1); # Simon Plouffe in his 1992 dissertation, dropping the initial zero
  • Mathematica
    Table[ 3^n - 2^n, {n, 0, 25} ]
    LinearRecurrence[{5, -6}, {0, 1}, 25] (* Harvey P. Dale, Aug 18 2011 *)
    Numerator@NestList[(3#+1)/2&,1/2,100] (* Zak Seidov, Oct 03 2011 *)
  • PARI
    {a(n) = 3^n - 2^n};
    
  • Python
    [3**n - 2**n for n in range(25)] # Ross La Haye, Aug 19 2005; corrected by David Radcliffe, Jun 26 2016
    
  • Sage
    [lucas_number1(n, 5, 6) for n in range(26)]  # Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: x/((1-2*x)*(1-3*x)).
a(n) = 5*a(n-1) - 6*a(n-2).
a(n) = 3*a(n-1) + 2^(n-1). - Jon Perry, Aug 23 2002
Starting 0, 0, 1, 5, 19, ... this is 3^n/3 - 2^n/2 + 0^n/6, the binomial transform of A086218. - Paul Barry, Aug 18 2003
a(n) = A083323(n)-1 = A056182(n)/2 = (A002783(n)-1)/2 = (A003063(n+2)-A003063(n+1))/2. - Ralf Stephan, Jan 12 2004
Binomial transform of A000225. - Ross La Haye, Feb 07 2005
a(n) = Sum_{k=0..n-1} binomial(n, k)*2^k. - Ross La Haye, Aug 20 2005
a(n) = 2^(2n) - A083324(n). - Ross La Haye, Sep 10 2005
a(n) = A112626(n, 1). - Ross La Haye, Jan 11 2006
E.g.f.: exp(3*x) - exp(2*x). - Mohammad K. Azarian, Jan 14 2009
a(n) = A217764(n,1). - Ross La Haye, Mar 27 2013
a(n) = 2*a(n-1) + 3^(n-1). - Toby Gottfried, Mar 28 2013
a(n) = A000244(n) - A000079(n). - Omar E. Pol, Mar 28 2013
a(n) = Sum_{k=0..2} Stirling1(2,k)*(k+1)^n = c_2^{(-n)}, poly-Cauchy numbers. - Takao Komatsu, Mar 28 2013
a(n) = A227048(n,A098294(n)). - Reinhard Zumkeller, Jun 30 2013
a(n+1) = Sum_{k=0..n} 2^k*3^(n-k). - J. M. Bergot, Mar 27 2018
Sum_{n>=1} 1/a(n) = A329064. - Amiram Eldar, Nov 20 2020
a(n) = (1/2)*Sum_{k=0..n} binomial(n, k)*(2^(n-k) + 2^k - 2).
a(n) = A001117(n) + 2*A000918(n) + 1. - Ambrosio Valencia-Romero, Mar 08 2022
a(n) = A000225(n) + A028243(n+1). - Ambrosio Valencia-Romero, Mar 09 2022
From Peter Bala, Jun 27 2025: (Start)
exp(Sum_{n >=1} a(2*n)/a(n)*x^n/n) = Sum_{n >= 0} a(n+1)*x^n.
exp(Sum_{n >=1} a(3*n)/a(n)*x^n/n) = 1 + 19*x + 247*x^2 + ... is the g.f. of A019443.
exp(Sum_{n >=1} a(4*n)/a(n)*x^n/n) = 1 + 65*x + 2743*x^2 + ... is the g.f. of A383754.
The following are all examples of telescoping series:
Sum_{n >= 1} 6^n/(a(n)*a(n+1)) = 2, since 6^n/(a(n)*a(n+1)) = b(n) - b(n+1), where b(n) = 2^n/a(n);
Sum_{n >= 1} 18^n/(a(n)*a(n+1)*a(n+2)) = 22/75, since 18^n/(a(n)*a(n+1)*a(n+2)) = c(n) - c(n+1), where c(n) = (5*6^n - 2*4^n)/(15*a(n)*a(n+1));
Sum_{n >= 1} 54^n/(a(n)*a(n+1)*a(n+2)*a(n+3)) = 634/48735 since 54^n/(a(n)*a(n+1)*a(n+2)*a(n+3)) = d(n) - d(n+1), where d(n) = (57*18^n - 38*12^n + 8*8^n)/(513*a(n)*a(n+1)*a(n+2)).
Sum_{n >= 1} 6^n/(a(n)*a(n+2)) = 14/25; Sum_{n >= 1} (-6)^n/(a(n)*a(n+2)) = -6/25.
Sum_{n >= 1} 6^n/(a(n)*a(n+3)) = 306/1805.
Sum_{n >= 1} 6^n/(a(n)*a(n+4)) = 4282/80275; Sum_{n >= 1} (-6)^n/(a(n)*a(n+4)) = -1698/80275. (End)

Extensions

Edited by Charles R Greathouse IV, Mar 24 2010

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)

A036561 Nicomachus triangle read by rows, T(n, k) = 2^(n - k)*3^k, for 0 <= k <= n.

Original entry on oeis.org

1, 2, 3, 4, 6, 9, 8, 12, 18, 27, 16, 24, 36, 54, 81, 32, 48, 72, 108, 162, 243, 64, 96, 144, 216, 324, 486, 729, 128, 192, 288, 432, 648, 972, 1458, 2187, 256, 384, 576, 864, 1296, 1944, 2916, 4374, 6561, 512, 768, 1152, 1728, 2592, 3888, 5832, 8748, 13122, 19683
Offset: 0

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Keywords

Comments

The triangle pertaining to this sequence has the property that every row, every column and every diagonal contains a nontrivial geometric progression. More interestingly every line joining any two elements contains a nontrivial geometric progression. - Amarnath Murthy, Jan 02 2002
Kappraff states (pp. 148-149): "I shall refer to this as Nicomachus' table since an identical table of numbers appeared in the Arithmetic of Nicomachus of Gerasa (circa 150 A.D.)" The table was rediscovered during the Italian Renaissance by Leon Battista Alberti, who incorporated the numbers in dimensions of his buildings and in a system of musical proportions. Kappraff states "Therefore a room could exhibit a 4:6 or 6:9 ratio but not 4:9. This ensured that ratios of these lengths would embody musical ratios". - Gary W. Adamson, Aug 18 2003
After Nichomachus and Alberti several Renaissance authors described this table. See for instance Pierre de la Ramée in 1569 (facsimile of a page of his Arithmetic Treatise in Latin in the links section). - Olivier Gérard, Jul 04 2013
The triangle sums, see A180662 for their definitions, link Nicomachus's table with eleven different sequences, see the crossrefs. It is remarkable that these eleven sequences can be described with simple elegant formulas. The mirror of this triangle is A175840. - Johannes W. Meijer, Sep 22 2010
The diagonal sums Sum_{k} T(n - k, k) give A167762(n + 2). - Michael Somos, May 28 2012
Where d(n) is the divisor count function, then d(T(i,j)) = A003991, the rows of which sum to the tetrahedral numbers A000292(n+1). For example, the sum of the divisors of row 4 of this triangle (i = 4), gives d(16) + d(24) + d(36) + d(54) + d(81) = 5 + 8 + 9 + 8 + 5 = 35 = A000292(5). In fact, where p and q are distinct primes, the aforementioned relationship to the divisor function and tetrahedral numbers can be extended to any triangle of numbers in which the i-th row is of form {p^(i-j)*q^j, 0<=j<=i}; i >= 0 (e.g., A003593, A003595). - Raphie Frank, Nov 18 2012, corrected Dec 07 2012
Sequence (or tree) generated by these rules: 1 is in S, and if x is in S, then 2*x and 3*x are in S, and duplicates are deleted as they occur; see A232559. - Clark Kimberling, Nov 28 2013
Partial sums of rows produce Stirling numbers of the 2nd kind: A000392(n+2) = Sum_{m=1..(n^2+n)/2} a(m). - Fred Daniel Kline, Sep 22 2014
A permutation of A003586. - L. Edson Jeffery, Sep 22 2014
Form a word of length i by choosing a (possibly empty) word on alphabet {0,1} then concatenating a word of length j on alphabet {2,3,4}. T(i,j) is the number of such words. - Geoffrey Critzer, Jun 23 2016
Form of Zorach additive triangle (see A035312) where each number is sum of west and northwest numbers, with the additional condition that each number is GCD of the two numbers immediately below it. - Michel Lagneau, Dec 27 2018

Examples

			The start of the sequence as a triangular array read by rows:
   1
   2   3
   4   6   9
   8  12  18  27
  16  24  36  54  81
  32  48  72 108 162 243
  ...
The start of the sequence as a table T(n,k) n, k > 0:
    1    2    4    8   16   32 ...
    3    6   12   24   48   96 ...
    9   18   36   72  144  288 ...
   27   54  108  216  432  864 ...
   81  162  324  648 1296 2592 ...
  243  486  972 1944 3888 7776 ...
  ...
- _Boris Putievskiy_, Jan 08 2013
		

References

  • Jay Kappraff, Beyond Measure, World Scientific, 2002, p. 148.
  • Flora R. Levin, The Manual of Harmonics of Nicomachus the Pythagorean, Phanes Press, 1994, p. 114.

Crossrefs

Cf. A001047 (row sums), A000400 (central terms), A013620, A007318.
Triangle sums (see the comments): A001047 (Row1); A015441 (Row2); A005061 (Kn1, Kn4); A016133 (Kn2, Kn3); A016153 (Fi1, Fi2); A016140 (Ca1, Ca4); A180844 (Ca2, Ca3); A180845 (Gi1, Gi4); A180846 (Gi2, Gi3); A180847 (Ze1, Ze4); A016185 (Ze2, Ze3). - Johannes W. Meijer, Sep 22 2010, Sep 10 2011
Antidiagonal cumulative sum: A000392; square arrays cumulative sum: A160869. Antidiagonal products: 6^A000217; antidiagonal cumulative products: 6^A000292; square arrays products: 6^A005449; square array cumulative products: 6^A006002.

Programs

  • Haskell
    a036561 n k = a036561_tabf !! n !! k
    a036561_row n = a036561_tabf !! n
    a036561_tabf = iterate (\xs@(x:_) -> x * 2 : map (* 3) xs) [1]
    -- Reinhard Zumkeller, Jun 08 2013
    
  • Magma
    /* As triangle: */ [[(2^(i-j)*3^j)/3: j in [1..i]]: i in [1..10]]; // Vincenzo Librandi, Oct 17 2014
  • Maple
    A036561 := proc(n,k): 2^(n-k)*3^k end:
    seq(seq(A036561(n,k),k=0..n),n=0..9);
    T := proc(n,k) option remember: if k=0 then 2^n elif k>=1 then procname(n,k-1) + procname(n-1,k-1) fi: end: seq(seq(T(n,k),k=0..n),n=0..9);
    # Johannes W. Meijer, Sep 22 2010, Sep 10 2011
  • Mathematica
    Flatten[Table[ 2^(i-j) 3^j, {i, 0, 12}, {j, 0, i} ]] (* Flatten added by Harvey P. Dale, Jun 07 2011 *)
  • PARI
    for(i=0,9,for(j=0,i,print1(3^j<<(i-j)", "))) \\ Charles R Greathouse IV, Dec 22 2011
    
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, 2^(n - k) * 3^k)} /* Michael Somos, May 28 2012 */
    

Formula

T(n,k) = A013620(n,k)/A007318(n,k). - Reinhard Zumkeller, May 14 2006
T(n,k) = T(n,k-1) + T(n-1,k-1) for n>=1 and 1<=k<=n with T(n,0) = 2^n for n>=0. - Johannes W. Meijer, Sep 22 2010
T(n,k) = 2^(k-1)*3^(n-1), n, k > 0 read by antidiagonals. - Boris Putievskiy, Jan 08 2013
a(n) = 2^(A004736(n)-1)*3^(A002260(n)-1), n > 0, or a(n) = 2^(j-1)*3^(i-1) n > 0, where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor[(-1+sqrt(8*n-7))/2]. - Boris Putievskiy, Jan 08 2013
G.f.: 1/((1-2x)(1-3yx)). - Geoffrey Critzer, Jun 23 2016
T(n,k) = (-1)^n * Sum_{q=0..n} (-1)^q * C(k+3*q, q) * C(n+2*q, n-q). - Marko Riedel, Jul 01 2024

A180845 a(n) = (16^n-3^n)/13.

Original entry on oeis.org

0, 1, 19, 313, 5035, 80641, 1290499, 20648713, 330381595, 5286112081, 84577812979, 1353245066713, 21651921244555, 346430740444321, 5542891848703459, 88686269584038313, 1418980313358961915
Offset: 0

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Author

Johannes W. Meijer, Sep 21 2010

Keywords

Comments

The a(n+1) appear in several triangle sums of Nicomachus' table A036561, i.e Gi1(4*n), Gi1(4*n+1)/2, Gi1(4*n+2)/4, Gi1(4*n+3)/8 and Gi4(n). See A180662 for information about these giraffe and other chess sums.

Crossrefs

Programs

Formula

a(n) = (16^n-3^n)/13
G.f.: x/((16*x-1)*(3*x-1))

A118004 a(n) = 9^n - 4^n.

Original entry on oeis.org

0, 5, 65, 665, 6305, 58025, 527345, 4766585, 42981185, 387158345, 3485735825, 31376865305, 282412759265, 2541798719465, 22876524019505, 205890058352825, 1853015893884545, 16677164519797385, 150094566577522385, 1350851442795085145, 12157664359545301025, 109418984733465848105, 984770884591425188465
Offset: 0

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Author

Eric W. Weisstein, Apr 09 2006

Keywords

Crossrefs

Programs

Formula

a(n) = 5*A016153(n).
a(n) = 13*a(n-1) - 36*a(n-2), n>=2. - Vincenzo Librandi, Mar 16 2011
G.f.: 5*x / ( (1-4*x)*(1-9*x) ). - R. J. Mathar, Mar 18 2011
From G. C. Greubel, Nov 11 2024: (Start)
a(n) = A001047(n)*A007689(n).
E.g.f.: 2*exp(13*x/2)*sinh(5*x/2). (End)

Extensions

More terms added by G. C. Greubel, Nov 11 2024

A180844 a(n) = (27^n - 2^n)/25.

Original entry on oeis.org

0, 1, 29, 787, 21257, 573955, 15496817, 418414123, 11297181449, 305023899379, 8235645283745, 222362422662139, 6003785411879801, 162102206120758723, 4376759565260493713, 118172508262033346635, 3190657723074900391913
Offset: 0

Views

Author

Johannes W. Meijer, Sep 21 2010

Keywords

Comments

The a(n+1) appear in several triangle sums of Nicomachus's table A036561, i.e., Ca2(3*n), Ca2(3*n+1)/3, Ca2(3*n+2)/9 and Ca3(n). See A180662 for information about these camel sums and other chess sums.

Crossrefs

Programs

  • Mathematica
    (#[[1]]-#[[2]])/25&/@Partition[Riffle[27^Range[0,20],2^Range[0,20]],2]  (* Harvey P. Dale, Jan 22 2011 *)
  • PARI
    a(n) = (27^n - 2^n)/25 \\ Iain Fox, Dec 12 2017
    
  • PARI
    first(n) = Vec(x/((27*x-1)*(2*x-1)) + O(x^n), -n) \\ Iain Fox, Dec 12 2017

Formula

a(n) = (27^n - 2^n)/25.
G.f.: x/((27*x-1)*(2*x-1)).

A180846 a(n) = (81^n - 2^n)/79.

Original entry on oeis.org

0, 1, 83, 6727, 544895, 44136511, 3575057423, 289579651327, 23455951757615, 1899932092367071, 153894499481733263, 12465454458020395327, 1009701811099652023535, 81785846699071813910431, 6624653582624816926753103
Offset: 0

Views

Author

Johannes W. Meijer, Sep 21 2010

Keywords

Comments

The a(n+1) appear in several triangle sums of Nicomachus's table A036561, i.e., Gi2(4*n), Gi2(4*n+1)/2, Gi2(4*n+2)/4, Gi2(4*n+3)/8 and Gi3(n). See A180662 for information about these giraffe and other chess sums.

Crossrefs

Programs

Formula

a(n) = (81^n - 2^n)/79.
G.f.: x/((81*x-1)*(2*x-1)).

A180847 a(n) = (27^n-4^n)/23.

Original entry on oeis.org

0, 1, 31, 853, 23095, 623821, 16844191, 454797253, 12279542215, 331547705341, 8951788306351, 241698285320053, 6525853707835735, 176198050128342061, 4757347353532344511, 128448378545641737253, 3468106220733400647655
Offset: 0

Views

Author

Johannes W. Meijer, Sep 21 2010

Keywords

Comments

For n>0, a(n) appears in several triangle sums of Nicomachus' table A036561, i.e. Ze1(2*n), Ze1(2*n+1)/2; Ze4(3*n), Ze4(3*n+1)/3 and Ze4(3*n+2)/9. See A180662 for information about these zebra and other chess sums.

Crossrefs

Programs

  • Mathematica
    Table[(27^n-4^n)/23,{n,0,20}] (* or *) LinearRecurrence[{31,-108},{0,1},20]  (* Harvey P. Dale, Sep 01 2011 *)
  • PARI
    a(n)=(27^n-4^n)/23 \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = (27^n-4^n)/23.
G.f.: x/((27*x-1)*(4*x-1)).
a(0)=0, a(1)=1, a(n) = 31*a(n-1)-108*a(n-2). - Harvey P. Dale, Sep 01 2011

A099459 Expansion of 1/(1 - 7*x + 9*x^2).

Original entry on oeis.org

1, 7, 40, 217, 1159, 6160, 32689, 173383, 919480, 4875913, 25856071, 137109280, 727060321, 3855438727, 20444528200, 108412748857, 574888488199, 3048504677680, 16165536349969, 85722212350663, 454565659304920
Offset: 0

Views

Author

Paul Barry, Oct 16 2004

Keywords

Comments

Associated to the knot 9_48 by the modified Chebyshev transform A(x) -> (1/(1+x^2)^2)*A(x/(1+x^2)). See A099460 and A099461.

Crossrefs

Programs

  • Magma
    [n le 2 select 7^(n-1) else 7*Self(n-1) -9*Self(n-2): n in [1..31]]; // G. C. Greubel, Nov 18 2021
  • Mathematica
    LinearRecurrence[{7,-9},{1,7},30] (* Harvey P. Dale, Jan 06 2012 *)
  • Sage
    [lucas_number1(n,7,9) for n in range(1, 22)] # Zerinvary Lajos, Apr 23 2009
    

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-9)^k*7^(n-2*k).
a(n) = Sum{k=0..n} binomial(2*n-k+1, k) * 3^k. - Paul Barry, Jan 17 2005
a(n) = 7*a(n-1) - 9*a(n-2), n >= 2. - Vincenzo Librandi, Mar 18 2011
a(n) = ((7 + sqrt(13))^(n+1) - (7 - sqrt(13))^(n+1))/(2^(n+1)*sqrt(13)). - Rolf Pleisch, May 19 2011
a(n) = 3^(n-1)*ChebyshevU(n-1, 7/6). - G. C. Greubel, Nov 18 2021
From Peter Bala, Jul 23 2025: (Start)
The following products telescope:
Product_{k >= 1} 1 + 3^k/a(k) = (1 + sqrt(13))/2.
Product_{k >= 1} 1 - 3^k/a(k) = (1 + sqrt(13))/14,
Product_{k >= 1} 1 + (-3)^k/a(k) = (13 + sqrt(13))/26.
Product_{k >= 1} 1 - (-3)^k/a(k) = (13 + sqrt(13))/14. (End)
Showing 1-10 of 13 results. Next