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 1-4 of 4 results.

A015518 a(n) = 2*a(n-1) + 3*a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 2, 7, 20, 61, 182, 547, 1640, 4921, 14762, 44287, 132860, 398581, 1195742, 3587227, 10761680, 32285041, 96855122, 290565367, 871696100, 2615088301, 7845264902, 23535794707, 70607384120, 211822152361, 635466457082
Offset: 0

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Author

Keywords

Comments

Number of walks of length n between any two distinct vertices of the complete graph K_4. - Paul Barry and Emeric Deutsch, Apr 01 2004
For n >= 1, a(n) is the number of integers k, 1 <= k <= 3^(n-1), whose ternary representation ends in an even number of zeros (see A007417). - Philippe Deléham, Mar 31 2004
Form the digraph with matrix A=[0,1,1,1;1,0,1,1;1,1,0,1;1,0,1,1]. A015518(n) corresponds to the (1,3) term of A^n. - Paul Barry, Oct 02 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is 2. - Cino Hilliard, Sep 25 2005
(A046717(n))^2 + (2*a(n))^2 = A046717(2n). E.g., A046717(3) = 13, 2*a(3) = 14, A046717(6) = 365. 13^2 + 14^2 = 365. - Gary W. Adamson, Jun 17 2006
For n >= 2, number of ordered partitions of n-1 into parts of sizes 1 and 2 where there are two types of 1 (singletons) and three types of 2 (twins). For example, the number of possible configurations of families of n-1 male (M) and female (F) offspring considering only single births and twins, where the birth order of M/F/pair-of-twins is considered and there are three types of twins; namely, both F, both M, or one F and one M - where birth order within a pair of twins itself is disregarded. In particular, for a(3)=7, two children could be either: (1) F, then M; (2) M, then F; (3) F,F; (4) M,M; (5) F,F twins; (6) M,M twins; or (7) M,F twins (emphasizing that birth order is irrelevant here when both/all children are the same gender and when two children are within the same pair of twins). - Rick L. Shepherd, Sep 18 2004
a(n) is prime for n = {2, 3, 5, 7, 13, 23, 43, 281, 359, ...}, where only a(2) = 2 corresponds to a prime of the form (3^k - 1)/4. All prime terms, except a(2) = 2, are the primes of the form (3^k + 1)/4. Numbers k such that (3^k + 1)/4 is prime are listed in A007658. Note that all prime terms have prime indices. Prime terms are listed in A111010. - Alexander Adamchuk, Nov 19 2006
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=charpoly(A,1). - Milan Janjic, Jan 26 2010
Select an odd size subset S from {1,2,...,n}, then select an even size subset from S. - Geoffrey Critzer, Mar 02 2010
a(n) is the number of ternary sequences of length n where the numbers of (0's, 1's) are (even, odd) respectively, and, by symmetry, the number of such sequences where those numbers are (odd, even) respectively. A122983 covers (even, even), and A081251 covers (odd, odd). - Toby Gottfried, Apr 18 2010
An elephant sequence, see A175654. For the corner squares just one A[5] vector, with decimal value 341, leads to this sequence (without the leading 0). For the central square this vector leads to the companion sequence A046717 (without the first leading 1). - Johannes W. Meijer, Aug 15 2010
Let R be the commutative algebra resulting from adjoining the elements of the Klein four-group to the integers (equivalently, K = Z[x,y,z]/{x*y - z, y*z - x, x*z - y, x^2 - 1, y^2 - 1, z^2 - 1}). Then a(n) is equal to the coefficients of x, y, and z in the expansion of (x + y + z)^n. - Joseph E. Cooper III (easonrevant(AT)gmail.com), Nov 06 2010
Pisano period lengths: 1, 2, 2, 4, 4, 2, 6, 8, 2, 4, 10, 4, 6, 6, 4, 16, 16, 2, 18, 4, ... - R. J. Mathar, Aug 10 2012
The ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 09 2014
This is a divisibility sequence, also the values of Chebyshev polynomials, and also the number of ways of packing a 2 X n-1 rectangle with dominoes and unit squares. - R. K. Guy, Dec 16 2016
For n>0, gcd(a(n),a(n+1))=1. - Kengbo Lu, Jul 02 2020

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

a(n) = A080926(n-1) + 1 = (1/3)*A054878(n+1) = (1/3)*abs(A084567(n+1)).
First differences of A033113 and A039300.
Partial sums of A046717.
The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.
Cf. A046717.

Programs

  • Magma
    [Round(3^n/4): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011
    
  • Mathematica
    Table[(3^n-(-1)^n)/4,{n,0,30}] (* Alexander Adamchuk, Nov 19 2006 *)
  • Maxima
    a(n):= round(3^n/4)$ /* Dimitri Papadopoulos, Nov 28 2023 */
  • PARI
    a(n)=round(3^n/4)
    
  • Python
    for n in range(0, 20): print(int((3**n-(-1)**n)/4), end=', ') # Stefano Spezia, Nov 30 2018
    
  • Sage
    [round(3^n/4) for n in range(0,27)]
    

Formula

G.f.: x/((1+x)*(1-3*x)).
a(n) = (3^n - (-1)^n)/4 = floor(3^n/4 + 1/2).
a(n) = 3^(n-1) - a(n-1). - Emeric Deutsch, Apr 01 2004
E.g.f.: (exp(3*x) - exp(-x))/4. Second inverse binomial transform of (5^n-1)/4, A003463. Inverse binomial transform for powers of 4, A000302 (when preceded by 0). - Paul Barry, Mar 28 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k+1)*2^(2k). - Paul Barry, May 14 2003
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*4^(k-1). - Paul Barry, Apr 02 2003
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2*k)*3^k. - Paul Barry, Jul 13 2004
a(n) = U(n-1, i/sqrt(3))(-i*sqrt(3))^(n-1), i^2=-1. - Paul Barry, Nov 17 2003
G.f.: x*(1+x)^2/(1 - 6*x^2 - 8*x^3 - 3*x^4) = x(1+x)^2/characteristic polynomial(x^4*adj(K_4)(1/x)). - Paul Barry, Feb 03 2004
a(n) = sum_{k=0..3^(n-1)} A014578(k) = -(-1)^n*A014983(n) = A051068(3^(n-1)), for n > 0. - Philippe Deléham, Mar 31 2004
E.g.f.: exp(x)*sinh(2*x)/2. - Paul Barry, Oct 02 2004
a(2*n+1) = A054880(n) + 1. - M. F. Hasler, Mar 20 2008
2*a(n) + (-1)^n = A046717(n). - M. F. Hasler, Mar 20 2008
a(n) = ((1+sqrt(4))^n - (1-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008
a(n) = abs(A014983(n)). - Zerinvary Lajos, May 28 2009
a(n) = round(3^n/4). - Mircea Merca, Dec 28 2010
a(n) = Sum_{k=1,3,5,...} binomial(n,k)*2^(k-1). - Geoffrey Critzer, Mar 02 2010
From Sergei N. Gladkovskii, Jul 19 2012: (Start)
G.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - 1/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + 1/G(k+1)))))); (continued fraction).
E.g.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - (2*k+1)/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + (2*k+2)/G(k+1)))))); (continued fraction). (End)
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(4*k-1)/(x*(4*k+3) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n+1) = Sum_{k = 0..n} A238801(n,k)*2^k. - Philippe Deléham, Mar 07 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-4)^k = (-1)^(n-1)*Sum_{k=0..n-1} (-3)^k. Equals (-1)^(n-1)*Phi(n,-3), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
a(n) = 2*A006342(n-1) - n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Nov 30 2018
a(n) = 2*A033113(n-2) + n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Aug 16 2019
a(2*k) = 2*A002452(k), a(2*k+1) = A066443(k). - Yuchun Ji, Aug 14 2019
a(n+1) = 2*Sum_{k=0..n} a(k) if n odd, and 1 + 2*Sum_{k=0..n} a(k) if n even. - Kengbo Lu, May 30 2020
a(n) = F(n) + Sum_{k=1..(n-1)} a(k)*L(n-k), for F(n) and L(n) the Fibonacci and Lucas numbers. - Kengbo Lu and Greg Dresden, Jun 05 2020
From Kengbo Lu, Jun 11 2020: (Start)
a(n) = A002605(n) + Sum_{k = 1..n-2} a(k)*A002605(n-k-1).
a(n) = A006130(n-1) + Sum_{k = 1..n-1} a(k)*A006130(n-k-1). (End)
a(2n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)* 2^(2n-2i-2j-1)* 3^(i+j). - Kengbo Lu, Jul 02 2020
a(n) = 3*a(n-1) - (-1)^n. - Dimitri Papadopoulos, Nov 28 2023
G.f.: x/((1 + x)*(1 - 3*x)) = Sum_{n >= 0} x^(n+1) * Product_{k = 1..n} (k + 3*x + 1)(1 + k*x) (a telescoping series). Cf. A007482. - Peter Bala, May 08 2024
From Peter Bala, Jun 29 2025: (Start)
For n >= 1, a(n+1) = 2^n * hypergeom([1/2 - (1/2)*n, -(1/2)*n], [-n], -3).
G.f. A(x) = x*exp(Sum_{n >= 1} a(2*n)/a(n)*x^n/n) = x + 2*x^2 + 7*x^3 + 20*x^4 + ....
sqrt(A(x)/x) is the g.f. of A002426.
The following series telescope:
Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)) = -1; Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*a(n+2)*a(n+3)) = -1/98.
In general, for k >= 0, Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*...*a(n+2*k+1)) = -1/((a(1)*a(2)*...*a(2*k+1))*a(2*k+1)).
Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)) = 1/4; Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)* a(n+3)*a(n+4)) = 1/5600.
In general, for k >= 1, Sum_{n >= 1} 3^n/(a(n)*a(n+1)*...*a(n+2*k)) = 1/((a(1)*a(2)*...*a(2*k))*a(2*k)). (End)

Extensions

More terms from Emeric Deutsch, Apr 01 2004
Edited by Ralf Stephan, Aug 30 2004

A066443 Number of distinct paths of length 2n+1 along edges of a unit cube between two fixed adjacent vertices.

Original entry on oeis.org

1, 7, 61, 547, 4921, 44287, 398581, 3587227, 32285041, 290565367, 2615088301, 23535794707, 211822152361, 1906399371247, 17157594341221, 154418349070987, 1389765141638881, 12507886274749927, 112570976472749341
Offset: 0

Views

Author

John W. Layman, Aug 12 2002

Keywords

Comments

All members of sequence are also hex, or central hexagonal, numbers (A003215). (If n is a hex number, 9n - 2 is always a hex number; see recurrence.) - Matthew Vandermast, Mar 30 2003
The sequence 1,1,7,61,547,... with g.f. (1-9x+6x^2)/((1-x)(1-9x)) and a(n) = A054879(n)/3 + 2*0^n/3 gives the denominators in the probability that a random walk on the cube returns to its starting corner on the 2n-th step. - Paul Barry, Mar 11 2004
Equals row sums of even row terms of triangle A158303. - Gary W. Adamson, Mar 15 2009
It appears that a(n) is the n-th record value in A120437, which gives the differences of A037314 (positive integers n such that the sum of the base 3 digits of n equals the sum of the base 9 digits of n). - John W. Layman, Dec 14 2010
Numbers in base 9 are 1, 6+1, 66+1, 666+1, 6666+1, 66666+1, etc.; that is, n 6's + 1. - Yuchun Ji, Aug 15 2019
All prime factors of a(n) are 1 mod 6. In addition, if n is not 1 mod 3 (first index being n=0), then 3 is a cubic residue modulo all prime factors of a(n). This provides a simple proof that there are infinitely many primes 1 mod 6 that have 3 as a cubic residue. - William Hu, Jul 26 2024

Examples

			From _Michael B. Porter_, Aug 22 2016: (Start)
Give coordinates (a,b,c) to the vertices of the cube, where a, b, and c are either 0 or 1. For n = 1, the a(1) = 7 paths of length 2n + 1 = 3 from (0,0,0) to (0,0,1) are:
(0,0,0) -> (0,0,1) -> (0,0,0) -> (0,0,1)
(0,0,0) -> (0,0,1) -> (0,1,1) -> (0,0,1)
(0,0,0) -> (0,0,1) -> (1,0,1) -> (0,0,1)
(0,0,0) -> (0,1,0) -> (0,0,0) -> (0,0,1)
(0,0,0) -> (0,1,0) -> (0,1,1) -> (0,0,1)
(0,0,0) -> (1,0,0) -> (0,0,0) -> (0,0,1)
(0,0,0) -> (1,0,0) -> (1,0,1) -> (0,0,1) (End)
		

Crossrefs

Cf. A158303, A037314, A120437, A083234 (binomial transform), A083233 (inverse binomial transform), A054879 (recurrent walks), A125857 (walks ending on face diagonal), A054880 (walks ending on space diagonal).

Programs

  • Magma
    [(3^(2*n+1)+1)/4: n in [0..20]]; // Vincenzo Librandi, Jun 16 2011
    
  • Maple
    seq((3^(2*n+1) + 1)/4, n=0..18); # Zerinvary Lajos, Jun 16 2007
  • Mathematica
    NestList[9 # - 2 &, 1, 18] (* or *)
    Table[(3^(2 n + 1) + 1)/4, {n, 0, 18}] (* or *)
    CoefficientList[Series[(1 - 3 x)/((1 - x) (1 - 9 x)), {x, 0, 18}], x] (* Michael De Vlieger, Aug 22 2016 *)
  • PARI
    a(n)=3^(2*n+1)\/4 \\ Charles R Greathouse IV, Jul 02 2013
    
  • PARI
    Vec((1-3*x)/((1-x)*(1-9*x)) + O(x^50)) \\ Altug Alkan, Nov 13 2015

Formula

a(n) = (3^(2*n+1)+1)/4. - Vladeta Jovovic, Dec 22 2002
a(n) = 9*a(n-1) - 2. - Matthew Vandermast, Mar 30 2003
From Paul Barry, Apr 21 2003: (Start)
G.f.: (1-3*x)/((1-x)*(1-9*x)).
E.g.f.: (3*exp(9*x) + exp(x))/4. (End)
a(n) = (-1)^n times the (i, i)-th element of M^n (for any i), where M = ((1, 1, 1, -2), (1, 1, -2, 1), (1, -2, 1, 1), (-2, 1, 1, 1)). - Simone Severini, Nov 25 2004
a(n) = Sum_{k=0..n} binomial(2*n+1, 2*k)*4^(n-k). - Paul Barry, Jan 22 2005
a(n) = A054880(n) + 1.
a(n) = A057660(3^n). - Henry Bottomley, Nov 08 2015
a(n) = Sum_{k=0..2n} (-3)^k == 1 + Sum_{k=1..n} 2*3^(2k-1). - Bob Selcoe, Aug 21 2016
a(n) = 3^(2*n+1) * a(-1-n) for all n in Z. - Michael Somos, Jul 02 2017
a(n) = 6*A002452(n) + 1. - Yuchun Ji, Aug 15 2019

Extensions

Corrected by Vladeta Jovovic, Dec 22 2002

A081251 Numbers n such that A081249(m)/m^2 has a local maximum for m = n.

Original entry on oeis.org

2, 6, 20, 60, 182, 546, 1640, 4920, 14762, 44286, 132860, 398580, 1195742, 3587226, 10761680, 32285040, 96855122, 290565366, 871696100, 2615088300, 7845264902, 23535794706, 70607384120, 211822152360, 635466457082, 1906399371246
Offset: 1

Views

Author

Klaus Brockhaus, Mar 17 2003

Keywords

Comments

The limit of the local maxima, lim A081249(n)/n^2 = 1/6. For local minima cf. A081250.
Also the number of different 4- and 3-colorings for the vertices of all triangulated planar polygons on a base with n+2 vertices, if the colors of the two base vertices are fixed. - Patrick Labarque, Mar 23 2010
From Toby Gottfried, Apr 18 2010: (Start)
a(n) = the number of ternary sequences of length n+1 where the numbers of (0's, 1's) are both odd.
A015518 covers the (odd, even) and (even, odd) cases, and A122983 covers (even, even). (End)

Examples

			6 is a term since A081249(5)/5^2 = 4/25 = 0.160, A081249(6)/6^2 = 7/36 = 0.194, A081249(7)/7^2 = 9/49 = 0.184.
		

Crossrefs

Programs

  • GAP
    List([1..30], n-> (9*3^(n-1) -(-1)^n -2)/4); # G. C. Greubel, Jul 14 2019
  • Magma
    [Floor(3^(n+1)/4) : n in [1..30]]; // Vincenzo Librandi, Jun 25 2011
    
  • Maple
    seq(floor(3^(n+1)/4), n=1..30). # Mircea Merca, Dec 27 2010
  • Mathematica
    a[n_]:= Floor[3^(n+1)/4]; Array[a, 30]
    Table[(9*3^(n-1) -(-1)^n -2)/4, {n, 1, 30}] (* G. C. Greubel, Jul 14 2019 *)
  • PARI
    vector(30, n, (9*3^(n-1) -(-1)^n -2)/4) \\ G. C. Greubel, Jul 14 2019
    
  • Sage
    [(9*3^(n-1) -(-1)^n -2)/4 for n in (1..30)] # G. C. Greubel, Jul 14 2019
    

Formula

G.f.: 2/((1-x)*(1+x)*(1-3*x)).
a(n) = a(n-2) + 2*3^(n) for n > 1.
a(n+2) - a(n) = A008776(n).
a(n) = 2*A033113(n+1).
a(2*n+1) = A054880(n+1).
a(n) = floor(3^(n+1)/4). - Mircea Merca, Dec 26 2010
From G. C. Greubel, Jul 14 2019: (Start)
a(n) = (9*3^(n-1) -(-1)^n -2)/4.
E.g.f.: (3*exp(3*x) - 2*exp(x) - exp(-x))/4. (End)

A172253 Numbers k such that the squarefree kernel of 9^k*(9^k - 1) is 3*(9^k - 1)/4.

Original entry on oeis.org

1, 3, 7, 9, 11, 13, 17, 19, 23, 27, 29, 31, 33, 37, 41, 43, 47, 49, 51, 53, 57, 59, 61, 67, 69, 71, 73, 77, 79, 81, 83, 87, 89, 91, 93, 97, 99, 101, 103, 107, 109, 111, 113, 119, 121, 123, 127, 129, 131, 133, 137, 139, 141, 143, 149, 151, 153, 157, 159, 161
Offset: 1

Views

Author

Artur Jasinski, Jan 29 2010

Keywords

Comments

From Artur Jasinski: (Start)
The maximal value of the squarefree kernel of a*b*9^k for every number 9^k and every a,b such that a + b = 9^k and gcd(a,b,3)=1 is never less than 3*(9^k - 1)/4 and is exactly equal to 3*(9^k - 1)/4 for exponents k in this sequence.
Conjecture: This sequence is infinite. (End)

Crossrefs

Programs

  • PARI
    rad(n) = factorback(factor(n)[, 1]); \\ A007947
    isok(k) = rad(9^k*(9^k - 1)) == 3*(9^k - 1)/4; \\ Michel Marcus, Dec 24 2022

Extensions

Edited by Jon E. Schoenfield, Dec 23 2022
More terms from Sean A. Irvine, Jun 15 2024
Showing 1-4 of 4 results.