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 1-10 of 43 results. Next

A099367 a(n) = A054413(n-1)^2, n >= 1.

Original entry on oeis.org

0, 1, 49, 2500, 127449, 6497401, 331240000, 16886742601, 860892632649, 43888637522500, 2237459621014849, 114066552034234801, 5815156694124960000, 296458924848338725201, 15113590010571150025249, 770496631614280312562500
Offset: 0

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Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

See the comment in A099279. This is example a=7.

Crossrefs

Cf. A054413.
Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, A099366, this sequence, A099369, A099372, A099374.

Programs

  • Mathematica
    LinearRecurrence[{50,50,-1},{0,1,49},20] (* Harvey P. Dale, Jul 27 2023 *)

Formula

a(n) = A054413(n-1)^2, n >= 1. a(0)=0.
a(n) = 50*a(n-1) + 50*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=49.
a(n) = 51*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = 2*(T(n, 51/2) - (-1)^n)/53 with twice the Chebyshev polynomials of the first kind: 2*T(n, 51/2) = A099368(n).
G.f.: x*(1-x)/((1-51*x+x^2)*(1+x)) = x*(1-x)/(1-50*x-50*x^2+x^3).
a(n+1) = (1 + (-1)^n)/2 + 49*Sum_{k=1..n} k*a(n+1-k). - Michael A. Allen, Feb 21 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (7 + sqrt(53))/14 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A340096 Odd composite integers m such that A054413(m-J(m,53)) == 0 (mod m), where J(m,53) is the Jacobi symbol.

Original entry on oeis.org

25, 35, 51, 65, 91, 175, 325, 391, 455, 575, 1247, 1295, 1633, 1763, 1775, 1921, 2275, 2407, 2599, 2651, 3367, 4199, 4579, 4623, 5629, 6441, 9959, 10465, 10825, 10877, 12025, 13021, 15155, 16021, 18881, 19019, 19039, 19307, 19669
Offset: 1

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Author

Ovidiu Bagdasar, Dec 28 2020

Keywords

Comments

The generalized Lucas sequences of integer parameters (a,b) defined by U(m+2)=a*U(m+1)-b*U(m) and U(0)=0, U(1)=1, satisfy the identity
U(p-J(p,D)) == 0 (mod p) when p is prime, b=-1 and D=a^2+4.
This sequence contains the odd composite integers with U(m-J(m,D)) == 0 (mod m).
For a=7 and b=-1, we have D=53 and U(m) recovers A054413(m).
If even numbers greater than 2 that are coprime to 53 are allowed, then 10, 50, 370, 5050, ... would also be terms. - Jianing Song, Jan 09 2021

References

  • D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020.
  • D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021).
  • D. Andrica, O. Bagdasar, On generalized pseudoprimality of level k (submitted).

Crossrefs

Cf. A054413, A071904, A081264 (a=1, b=-1), A327653 (a=3,b=-1), A340095 (a=5, b=-1)
Cf. A340097 (a=3, b=1), A340098 (a=5, b=1), A340099 (a=7, b=1).

Programs

  • Mathematica
    Select[Range[3,20000, 2], CoprimeQ[#, 53] && CompositeQ[#] && Divisible[Fibonacci[#-JacobiSymbol[#, 53], 7], #] &]

A340121 Odd composite integers m such that A054413(2*m-J(m,53)) == 1 (mod m), where J(m,53) is the Jacobi symbol.

Original entry on oeis.org

25, 35, 39, 49, 51, 65, 91, 147, 175, 245, 301, 325, 343, 391, 455, 507, 575, 605, 637, 663, 741, 833, 897, 903, 935, 1127, 1205, 1225, 1247, 1295, 1505, 1595, 1633, 1715, 1763, 1775, 1911, 1921, 2107, 2275, 2401, 2407, 2499, 2599, 2651, 3025, 3143, 3185, 3311
Offset: 1

Views

Author

Ovidiu Bagdasar, Dec 28 2020

Keywords

Comments

The generalized Lucas sequences of integer parameters (a,b) defined by U(m+2)=a*U(m+1)-b*U(m) and U(0)=0, U(1)=1, satisfy U(2*p-J(p,D)) == 1 (mod p) whenever p is prime, k is a positive integer, b=-1 and D=a^2+4.
The composite integers m with the property U(k*m-J(m,D)) == U(k-1) (mod m) are called generalized Lucas pseudoprimes of level k- and parameter a. Here b=-1, a=7, D=53 and k=2, while U(m) is A054413(m).

References

  • D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020.
  • D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021).
  • D. Andrica, O. Bagdasar, On generalized pseudoprimality of level k (submitted).

Crossrefs

Cf. A054413, A071904, A340096 (a=7, b=-1, k=1).
Cf. A340118 (a=1, b=-1, k=2), A340119 (a=3, b=-1, k=2), A340120 (a=5, b=-1, k=2).

Programs

  • Mathematica
    Select[Range[3, 10000, 2], CoprimeQ[#, 53] && CompositeQ[#] && Divisible[Fibonacci[2*#-JacobiSymbol[#, 53], 7] - 1, #] &]

A340238 Odd composite integers m such that A054413(3*m-J(m,53)) == 7 (mod m), where J(m,53) is the Jacobi symbol.

Original entry on oeis.org

9, 25, 27, 51, 91, 105, 153, 185, 225, 289, 325, 425, 459, 481, 513, 747, 867, 897, 925, 945, 1001, 1189, 1299, 1469, 1633, 1785, 1921, 2241, 2245, 2599, 2601, 2651, 2769, 2907, 3051, 3277, 3825, 3897, 5681, 6225, 6507, 6777, 7225, 7361, 7803, 8023, 8227, 8701, 8721
Offset: 1

Views

Author

Ovidiu Bagdasar, Jan 01 2021

Keywords

Comments

The generalized Lucas sequences of integer parameters (a,b) defined by U(m+2)=a*U(m+1)-b*U(m) and U(0)=0, U(1)=1, satisfy U(3*p-J(p,D)) == a (mod p) whenever p is prime, k is a positive integer, b=-1 and D=a^2+4.
The composite integers m with the property U(k*m-J(m,D)) == U(k-1) (mod m) are called generalized Lucas pseudoprimes of level k- and parameter a.
Here b=-1, a=7, D=53 and k=3, while U(m) is A054413(m).

References

  • D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020.
  • D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021).
  • D. Andrica, O. Bagdasar, On generalized pseudoprimality of level k (submitted).

Crossrefs

Cf. A054413, A071904, A340096 (a=7, b=-1, k=1), A340121 (a=7, b=-1, k=2).
Cf. A340235 (a=1, b=-1, k=3), A340236 (a=3, b=-1, k=3), A340237 (a=5, b=-1, k=3).

Programs

  • Mathematica
    Select[Range[3, 10000, 2], CoprimeQ[#, 53] && CompositeQ[#] && Divisible[Fibonacci[3*#-JacobiSymbol[#, 53], 7] - 7, #] &]

A338081 Odd composite integers such that A054413(m)^2 == 1 (mod m).

Original entry on oeis.org

21, 25, 35, 49, 51, 65, 85, 91, 119, 147, 161, 175, 221, 231, 245, 325, 357, 377, 391, 399, 425, 455, 539, 559, 561, 575, 595, 629, 637, 759, 791, 833, 1001, 1105, 1127, 1225, 1247, 1295, 1309, 1495, 1547, 1633, 1763, 1775, 1921, 2001, 2015, 2261, 2275, 2407
Offset: 1

Views

Author

Ovidiu Bagdasar, Oct 08 2020

Keywords

Comments

The generalized Lucas sequence of integer parameters (a,b) is defined by
U(m+2) = a*U(m+1)-b*U(m) and U(0)=0, U(1)=1.
Whenever p is prime and b=-1,1 we have U^2(p) == 1 (mod p).
Here we define the odd composite integers for which U^2(m) == 1 (mod m) holds, for a=7, b=-1, where U(m) is A054413(m).

References

  • D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020.
  • D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021)

Crossrefs

Cf. A337231 (a=1, odd terms), A337232 (a=1, even terms), A337233 (a=2), A337234 (a=3, odd terms), A337235 (a=3, even terms), A337236 (a=4), A337237 (a=5), A338081 (a=6).

Programs

  • Mathematica
    Select[Range[3, 15000, 2], CompositeQ[#] && Divisible[Fibonacci[#, 7]*Fibonacci[#, 7] - 1, #] &]

A000129 Pell numbers: a(0) = 0, a(1) = 1; for n > 1, a(n) = 2*a(n-1) + a(n-2).

Original entry on oeis.org

0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428, 38613965, 93222358, 225058681, 543339720, 1311738121, 3166815962, 7645370045, 18457556052, 44560482149, 107578520350, 259717522849
Offset: 0

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Author

Keywords

Comments

Sometimes also called lambda numbers.
Also denominators of continued fraction convergents to sqrt(2): 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/A000129.
Number of lattice paths from (0,0) to the line x=n-1 consisting of U=(1,1), D=(1,-1) and H=(2,0) steps (i.e., left factors of Grand Schroeder paths); for example, a(3)=5, counting the paths H, UD, UU, DU and DD. - Emeric Deutsch, Oct 27 2002
a(2*n) with b(2*n) := A001333(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = +1 (see Emerson reference). a(2*n+1) with b(2*n+1) := A001333(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = -1.
Bisection: a(2*n+1) = T(2*n+1, sqrt(2))/sqrt(2) = A001653(n), n >= 0 and a(2*n) = 2*S(n-1,6) = 2*A001109(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - Wolfdieter Lang, Jan 10 2003
Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the denominators. - Amarnath Murthy, Mar 22 2003
This is also the Horadam sequence (0,1,1,2). Limit_{n->oo} a(n)/a(n-1) = sqrt(2) + 1 = A014176. - Ross La Haye, Aug 18 2003
Number of 132-avoiding two-stack sortable permutations.
From Herbert Kociemba, Jun 02 2004: (Start)
For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 3.
Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 1, s(n) = 2. (End)
Counts walks of length n from a vertex of a triangle to another vertex to which a loop has been added. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
Apart from initial terms, Pisot sequence P(2,5). See A008776 for definition of Pisot sequences. - David W. Wilson
Sums of antidiagonals of A038207 [Pascal's triangle squared]. - Ross La Haye, Oct 28 2004
The Pell primality test is "If N is an odd prime, then P(N)-Kronecker(2,N) is divisible by N". "Most" composite numbers fail this test, so it makes a useful pseudoprimality test. The odd composite numbers which are Pell pseudoprimes (i.e., that pass the above test) are in A099011. - Jack Brennen, Nov 13 2004
a(n) = sum of n-th row of triangle in A008288 = A094706(n) + A000079(n). - Reinhard Zumkeller, Dec 03 2004
Pell trapezoids (cf. A084158); for n > 0, A001109(n) = (a(n-1) + a(n+1))*a(n)/2; e.g., 1189 = (12+70)*29/2. - Charlie Marion, Apr 01 2006
(0!a(1), 1!a(2), 2!a(3), 3!a(4), ...) and (1,-2,-2,0,0,0,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Oct 29 2007
Let C = (sqrt(2)+1) = 2.414213562..., then for n > 1, C^n = a(n)*(1/C) + a(n+1). Example: C^3 = 14.0710678... = 5*(0.414213562...) + 12. Let X = the 2 X 2 matrix [0, 1; 1, 2]; then X^n * [1, 0] = [a(n-1), a(n); a(n), a(n+1)]. a(n) = numerator of n-th convergent to (sqrt(2)-1) = 0.414213562... = [2, 2, 2, ...], the convergents being [1/2, 2/5, 5/12, ...]. - Gary W. Adamson, Dec 21 2007
A = sqrt(2) = 2/2 + 2/5 + 2/(5*29) + 2/(29*169) + 2/(169*985) + ...; B = ((5/2) - sqrt(2)) = 2/2 + 2/(2*12) + 2/(12*70) + 2/(70*408) + 2/(408*2378) + ...; A+B = 5/2. C = 1/2 = 2/(1*5) + 2/(2*12) + 2/(5*29) + 2/(12*70) + 2/(29*169) + ... - Gary W. Adamson, Mar 16 2008
From Clark Kimberling, Aug 27 2008: (Start)
Related convergents (numerator/denominator):
lower principal convergents: A002315/A001653
upper principal convergents: A001541/A001542
intermediate convergents: A052542/A001333
lower intermediate convergents: A005319/A001541
upper intermediate convergents: A075870/A002315
principal and intermediate convergents: A143607/A002965
lower principal and intermediate convergents: A143608/A079496
upper principal and intermediate convergents: A143609/A084068. (End)
Equals row sums of triangle A143808 starting with offset 1. - Gary W. Adamson, Sep 01 2008
Binomial transform of the sequence:= 0,1,0,2,0,4,0,8,0,16,..., powers of 2 alternating with zeros. - Philippe Deléham, Oct 28 2008
a(n) is also the sum of the n-th row of the triangle formed by starting with the top two rows of Pascal's triangle and then each next row has a 1 at both ends and the interior values are the sum of the three numbers in the triangle above that position. - Patrick Costello (pat.costello(AT)eku.edu), Dec 07 2008
Starting with offset 1 = eigensequence of triangle A135387 (an infinite lower triangular matrix with (2,2,2,...) in the main diagonal and (1,1,1,...) in the subdiagonal). - Gary W. Adamson, Dec 29 2008
Starting with offset 1 = row sums of triangle A153345. - Gary W. Adamson, Dec 24 2008
From Charlie Marion, Jan 07 2009: (Start)
In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
let a(k,0) = 1, a(k,1) = 2k; for n > 0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2)
and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
let b(k,0) = 1, b(k,1) = 2k+1; for n > 0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2)
and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
for example, if k=1 and n=3, then a(1,n) = a(n+1) and
1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
(End)
Starting with offset 1 = row sums of triangle A155002, equivalent to the statement that the Fibonacci sequence convolved with the Pell sequence prefaced with a "1": (1, 1, 2, 5, 12, 29, ...) = (1, 2, 5, 12, 29, ...). - Gary W. Adamson, Jan 18 2009
It appears that P(p) == 8^((p-1)/2) (mod p), p = prime; analogous to [Schroeder, p. 90]: Fp == 5^((p-1)/2) (mod p). Example: Given P(11) = 5741, == 8^5 (mod 11). Given P(17) = 11336689, == 8^8 (mod 17) since 17 divides (8^8 - P(17)). - Gary W. Adamson, Feb 21 2009
Equals eigensequence of triangle A154325. - Gary W. Adamson, Feb 12 2009
Another combinatorial interpretation of a(n-1) arises from a simple tiling scenario. Namely, a(n-1) gives the number of ways of tiling a 1 X n rectangle with indistinguishable 1 X 2 rectangles and 1 X 1 squares that come in two varieties, say, A and B. For example, with C representing the 1 X 2 rectangle, we obtain a(4)=12 from AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB, AC, BC, CA and CB. - Martin Griffiths, Apr 25 2009
a(n+1) = 2*a(n) + a(n-1), a(1)=1, a(2)=2 was used by Theon from Smyrna. - Sture Sjöstedt, May 29 2009
The n-th Pell number counts the perfect matchings of the edge-labeled graph C_2 x P_(n-1), or equivalently, the number of domino tilings of a 2 X (n-1) cylindrical grid. - Sarah-Marie Belcastro, Jul 04 2009
As a fraction: 1/79 = 0.0126582278481... or 1/9799 = 0.000102051229...(1/119 and 1/10199 for sequence in reverse). - Mark Dols, May 18 2010
Limit_{n->oo} (a(n)/a(n-1) - a(n-1)/a(n)) tends to 2.0. Example: a(7)/a(6) - a(6)/a(7) = 169/70 - 70/169 = 2.0000845... - Gary W. Adamson, Jul 16 2010
Numbers k such that 2*k^2 +- 1 is a square. - Vincenzo Librandi, Jul 18 2010
Starting (1, 2, 5, ...) = INVERTi transform of A006190: (1, 3, 10, 33, 109, ...). - Gary W. Adamson, Aug 06 2010
[u,v] = [a(n), a(n-1)] generates all Pythagorean triples [u^2-v^2, 2uv, u^2+v^2] whose legs differ by 1. - James R. Buddenhagen, Aug 14 2010
An elephant sequence, see A175654. For the corner squares six A[5] vectors, with decimal values between 21 and 336, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A078057. - Johannes W. Meijer, Aug 15 2010
Let the 2 X 2 square matrix A=[2, 1; 1, 0] then a(n) = the (1,1) element of A^(n-1). - Carmine Suriano, Jan 14 2011
Define a t-circle to be a first-quadrant circle tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the t-circle with radius 1. Then for n > 0, define C(n) to be the next larger t-circle which is tangent to C(n - 1). C(n) has radius A001333(2n) + a(2n)*sqrt(2) and each of the coordinates of its point of intersection with C(n + 1) is a(2n + 1) + (A001333(2n + 1)*sqrt(2))/2. See similar Comments for A001109 and A001653, Sep 14 2005. - Charlie Marion, Jan 18 2012
A001333 and A000129 give the diagonal numbers described by Theon from Smyrna. - Sture Sjöstedt, Oct 20 2012
Pell numbers could also be called "silver Fibonacci numbers", since, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio, while a(n+1) = ceiling(delta*a(n)), if n is even and a(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1+sqrt(2) is the silver ratio. - Vladimir Shevelev, Feb 22 2013
a(n) is the number of compositions (ordered partitions) of n-1 into two sorts of 1's and one sort of 2's. Example: the a(3)=5 compositions of 3-1=2 are 1+1, 1+1', 1'+1, 1'+1', and 2. - Bob Selcoe, Jun 21 2013
Between every two consecutive squares of a 1 X n array there is a flap that can be folded over one of the two squares. Two flaps can be lowered over the same square in 2 ways, depending on which one is on top. The n-th Pell number counts the ways n-1 flaps can be lowered. For example, a sideway representation for the case n = 3 squares and 2 flaps is \\., .//, \./, ./., .\., where . is an empty square. - Jean M. Morales, Sep 18 2013
Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A005319(k)*(a(n-2k+1) - a(n-2k)) + a(n-4k) = A075870(k)*(a(n-2k+2) - a(n-2k+1)) - a(n-4k+2). - Charlie Marion, Nov 26 2013
An alternative formulation of the combinatorial tiling interpretation listed above: Except for n=0, a(n-1) is the number of ways of partial tiling a 1 X n board with 1 X 1 squares and 1 X 2 dominoes. - Matthew Lehman, Dec 25 2013
Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A077444(k)*a(n-2k+1) + a(n-4k+2). This formula generalizes the formula used to define this sequence. - Charlie Marion, Jan 30 2014
a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 1, 1; 1, 1, 1; 0, 1, 1], [0, 1, 1; 0, 1, 1; 1, 1, 1], [0, 1, 0; 1, 1, 1; 1, 1, 1] or [0, 0, 1; 1, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
a(n+1) counts closed walks on K2 containing two loops on the other vertex. Equivalently the (1,1) entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1;1,2). - David Neil McGrath, Oct 28 2014
For n >= 1, a(n) equals the number of ternary words of length n-1 avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)). - Tom Edgar, Jan 28 2015
A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Jan 03 2017
a(n) is the number of compositions (ordered partitions) of n-1 into two kinds of parts, n and n', when the order of the 1 does not matter, or equivalently, when the order of the 1' does not matter. Example: When the order of the 1 does not matter, the a(3)=5 compositions of 3-1=2 are 1+1, 1+1'=1+1, 1'+1', 2 and 2'. (Contrast with entry from Bob Selcoe dated Jun 21 2013.) - Gregory L. Simay, Sep 07 2017
Number of weak orderings R on {1,...,n} that are weakly single-peaked w.r.t. the total ordering 1 < ... < n and for which {1,...,n} has exactly one minimal element for the weak ordering R. - J. Devillet, Sep 28 2017
Also the number of matchings in the (n-1)-centipede graph. - Eric W. Weisstein, Sep 30 2017
Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0)=1. Let A_1(r,n) = Sum_{j=0..n} A(r,j) and let A_s(r,n) = Sum_{j=0..n} A_(s-1)(r,j). Then A_0(1,n) + A_2(3,n-4) + A_4(5,n-8) + ... + A_(2j) (2j+1, n-4j) = a(n) without the initial 0. - Gregory L. Simay, May 25 2018
(1, 2, 5, 12, 29, ...) is the fourth INVERT transform of (1, -2, 5, -12, 29, ...), as shown in A073133. - Gary W. Adamson, Jul 17 2019
Number of 2-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
Also called the 2-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence. - Michael A. Allen, Jan 23 2023
Named by Lucas (1878) after the English mathematician John Pell (1611-1685). - Amiram Eldar, Oct 02 2023
a(n) is the number of compositions of n when there are F(i) parts of size i, with i,n >= 1, F(n) the Fibonacci numbers, A000045(n) (see example below). - Enrique Navarrete, Dec 15 2023

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 12*x^4 + 29*x^5 + 70*x^6 + 169*x^7 + 408*x^8 + 985*x^9 + ...
From _Enrique Navarrete_, Dec 15 2023: (Start)
From the comment on compositions with Fibonacci number of parts, F(n), there are F(1)=1 type of 1, F(2)=1 type of 2, F(3)=2 types of 3, F(4)=3 types of 4, F(5)=5 types of 5 and F(6)=8 types of 6.
The following table gives the number of compositions of n=6 with Fibonacci number of parts:
Composition, number of such compositions, number of compositions of this type:
6,           1,     8;
5+1,         2,    10;
4+2,         2,     6;
3+3,         1,     4;
4+1+1,       3,     9;
3+2+1,       6,    12;
2+2+2,       1,     1;
3+1+1+1,     4,     8;
2+2+1+1,     6,     6;
2+1+1+1+1,   5,     5;
1+1+1+1+1+1, 1,     1;
for a total of a(6)=70 compositions of n=6. (End).
		

References

  • J. Austin and L. Schneider, Generalized Fibonacci sequences in Pythagorean triple preserving sequences, Fib. Q., 58:1 (2020), 340-350.
  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 76.
  • A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 941.
  • J. M. Borwein, D. H. Bailey, and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 53.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 204.
  • John Derbyshire, Prime Obsession, Joseph Henry Press, 2004, see p. 16.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.1.
  • Shaun Giberson and Thomas J. Osler, Extending Theon's Ladder to Any Square Root, Problem 3858, Elementa, No. 4 1996.
  • R. P. Grimaldi, Ternary strings with no consecutive 0's and no consecutive 1's, Congressus Numerantium, 205 (2011), 129-149.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 288.
  • Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • Paulo Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 43.
  • Paulo Ribenboim, My Numbers, My Friends: Popular Lectures on Number Theory, Springer-Verlag, NY, 2000, p. 3.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 61.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
  • Manfred R. Schroeder, "Number Theory in Science and Communication", 5th ed., Springer-Verlag, 2009, p. 90.
  • 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).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 34.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 62.

Crossrefs

Partial sums of A001333.
2nd row of A172236.
a(n) = A054456(n-1, 0), n>=1 (first column of triangle).
Cf. A175181 (Pisano periods), A214028 (Entry points), A214027 (number of zeros in a fundamental period).
A077985 is a signed version.
INVERT transform of Fibonacci numbers (A000045).
Cf. A038207.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A048739.
Cf. A073133.
Cf. A041085.
Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), this sequence (k=2), A006190 (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), A154597 (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).

Programs

  • GAP
    a := [0,1];; for n in [3..10^3] do a[n] := 2 * a[n-1] + a[n-2]; od; A000129 := a; # Muniru A Asiru, Oct 16 2017
    
  • Haskell
    a000129 n = a000129_list !! n
    a000129_list = 0 : 1 : zipWith (+) a000129_list (map (2 *) $ tail a000129_list)
    -- Reinhard Zumkeller, Jan 05 2012, Feb 05 2011
    
  • Magma
    [0] cat [n le 2 select n else 2*Self(n-1) + Self(n-2): n in [1..35]]; // Vincenzo Librandi, Aug 08 2015
    
  • Maple
    A000129 := proc(n) option remember; if n <=1 then n; else 2*procname(n-1)+procname(n-2); fi; end;
    a:= n-> (<<2|1>, <1|0>>^n)[1, 2]: seq(a(n), n=0..40); # Alois P. Heinz, Aug 01 2008
    A000129 := n -> `if`(n<2, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1)):
    seq(simplify(A000129(n)), n=0..31); # Peter Luschny, Dec 17 2015
  • Mathematica
    CoefficientList[Series[x/(1 - 2*x - x^2), {x, 0, 60}], x] (* Stefan Steinerberger, Apr 08 2006 *)
    Expand[Table[((1 + Sqrt[2])^n - (1 - Sqrt[2])^n)/(2Sqrt[2]), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *)
    LinearRecurrence[{2, 1}, {0, 1}, 60] (* Harvey P. Dale, Jan 04 2012 *)
    a[ n_] := With[ {s = Sqrt@2}, ((1 + s)^n - (1 - s)^n) / (2 s)] // Simplify; (* Michael Somos, Jun 01 2013 *)
    Table[Fibonacci[n, 2], {n, 0, 20}] (* Vladimir Reshetnikov, May 08 2016 *)
    Fibonacci[Range[0, 20], 2] (* Eric W. Weisstein, Sep 30 2017 *)
    a[ n_] := ChebyshevU[n - 1, I] / I^(n - 1); (* Michael Somos, Oct 30 2021 *)
  • Maxima
    a[0]:0$
    a[1]:1$
    a[n]:=2*a[n-1]+a[n-2]$
    A000129(n):=a[n]$
    makelist(A000129(n),n,0,30); /* Martin Ettl, Nov 03 2012 */
    
  • Maxima
    makelist((%i)^(n-1)*ultraspherical(n-1,1,-%i),n,0,24),expand; /* Emanuele Munarini, Mar 07 2018 */
    
  • PARI
    for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[2, 1]; if (a > 10^(10^3 - 6), break); write("b000129.txt", n, " ", a)); \\ Harry J. Smith, Jun 12 2009
    
  • PARI
    {a(n) = imag( (1 + quadgen( 8))^n )}; /* Michael Somos, Jun 01 2013 */
    
  • PARI
    {a(n) = if( n<0, -(-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [2, 1]}; /* Michael Somos, Jun 01 2013 */
    
  • PARI
    a(n)=([2, 1; 1, 0]^n)[2,1] \\ Charles R Greathouse IV, Mar 04 2014
    
  • PARI
    {a(n) = polchebyshev(n-1, 2, I) / I^(n-1)}; /* Michael Somos, Oct 30 2021 */
    
  • Python
    from itertools import islice
    def A000129_gen(): # generator of terms
        a, b = 0, 1
        yield from [a,b]
        while True:
            a, b = b, a+2*b
            yield b
    A000129_list = list(islice(A000129_gen(),20)) # Chai Wah Wu, Jan 11 2022
  • Sage
    [lucas_number1(n, 2, -1) for n in range(30)]  # Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: x/(1 - 2*x - x^2). - Simon Plouffe in his 1992 dissertation.
a(2n+1)=A001653(n). a(2n)=A001542(n). - Ira Gessel, Sep 27 2002
G.f.: Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (2*k + x)/(1 + 2*k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 1 + k)/(1 + k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 3 - k)/(1 - k*x) ) may all be proved using telescoping series. - Peter Bala, Jan 04 2015
a(n) = 2*a(n-1) + a(n-2), a(0)=0, a(1)=1.
a(n) = ((1 + sqrt(2))^n - (1 - sqrt(2))^n)/(2*sqrt(2)).
For initial values a(0) and a(1), a(n) = ((a(0)*sqrt(2)+a(1)-a(0))*(1+sqrt(2))^n + (a(0)*sqrt(2)-a(1)+a(0))*(1-sqrt(2))^n)/(2*sqrt(2)). - Shahreer Al Hossain, Aug 18 2019
a(n) = integer nearest a(n-1)/(sqrt(2) - 1), where a(0) = 1. - Clark Kimberling
a(n) = Sum_{i, j, k >= 0: i+j+2k = n} (i+j+k)!/(i!*j!*k!).
a(n)^2 + a(n+1)^2 = a(2n+1) (1999 Putnam examination).
a(2n) = 2*a(n)*A001333(n). - John McNamara, Oct 30 2002
a(n) = ((-i)^(n-1))*S(n-1, 2*i), with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x)=0, S(-2, x)= -1.
Binomial transform of expansion of sinh(sqrt(2)x)/sqrt(2). E.g.f.: exp(x)sinh(sqrt(2)x)/sqrt(2). - Paul Barry, May 09 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k+1)*2^k. - Paul Barry, May 13 2003
a(n-2) + a(n) = (1 + sqrt(2))^(n-1) + (1 - sqrt(2))^(n-1) = A002203(n-1). (A002203(n))^2 - 8(a(n))^2 = 4(-1)^n. - Gary W. Adamson, Jun 15 2003
Unreduced g.f.: x(1+x)/(1 - x - 3x^2 - x^3); a(n) = a(n-1) + 3*a(n-2) + a(n-2). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2k). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
Apart from initial terms, inverse binomial transform of A052955. - Paul Barry, May 23 2004
a(n)^2 + a(n+2k+1)^2 = A001653(k)*A001653(n+k); e.g., 5^2 + 70^2 = 5*985. - Charlie Marion Aug 03 2005
a(n+1) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)*(1+(-1)^(n-k))*2^k/2. - Paul Barry, Aug 28 2005
a(n) = a(n-1) + A001333(n-1) = A001333(n) - a(n-1) = A001109(n)/A001333(n) = sqrt(A001110(n)/A001333(n)^2) = ceiling(sqrt(A001108(n)/2)). - Henry Bottomley, Apr 18 2000
a(n) = F(n, 2), the n-th Fibonacci polynomial evaluated at x=2. - T. D. Noe, Jan 19 2006
Define c(2n) = -A001108(n), c(2n+1) = -A001108(n+1) and d(2n) = d(2n+1) = A001652(n); then ((-1)^n)*(c(n) + d(n)) = a(n). [Proof given by Max Alekseyev.] - Creighton Dement, Jul 21 2005
a(r+s) = a(r)*a(s+1) + a(r-1)*a(s). - Lekraj Beedassy, Sep 03 2006
a(n) = (b(n+1) + b(n-1))/n where {b(n)} is the sequence A006645. - Sergio Falcon, Nov 22 2006
From Miklos Kristof, Mar 19 2007: (Start)
Let F(n) = a(n) = Pell numbers, L(n) = A002203 = companion Pell numbers (A002203):
For a >= b and odd b, F(a+b) + F(a-b) = L(a)*F(b).
For a >= b and even b, F(a+b) + F(a-b) = F(a)*L(b).
For a >= b and odd b, F(a+b) - F(a-b) = F(a)*L(b).
For a >= b and even b, F(a+b) - F(a-b) = L(a)*F(b).
F(n+m) + (-1)^m*F(n-m) = F(n)*L(m).
F(n+m) - (-1)^m*F(n-m) = L(n)*F(m).
F(n+m+k) + (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = F(n)*L(m)*L(k).
F(n+m+k) - (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = L(n)*L(m)*F(k).
F(n+m+k) + (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = L(n)*F(m)*L(k).
F(n+m+k) - (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = 8*F(n)*F(m)*F(k). (End)
a(n+1)*a(n) = 2*Sum_{k=0..n} a(k)^2 (a similar relation holds for A001333). - Creighton Dement, Aug 28 2007
a(n+1) = Sum_{k=0..n} binomial(n+1,2k+1) * 2^k = Sum_{k=0..n} A034867(n,k) * 2^k = (1/n!) * Sum_{k=0..n} A131980(n,k) * 2^k. - Tom Copeland, Nov 30 2007
Equals row sums of unsigned triangle A133156. - Gary W. Adamson, Apr 21 2008
a(n) (n >= 3) is the determinant of the (n-1) X (n-1) tridiagonal matrix with diagonal entries 2, superdiagonal entries 1 and subdiagonal entries -1. - Emeric Deutsch, Aug 29 2008
a(n) = A000045(n) + Sum_{k=1..n-1} A000045(k)*a(n-k). - Roger L. Bagula and Gary W. Adamson, Sep 07 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
fract((1+sqrt(2))^n) = (1/2)*(1 + (-1)^n) - (-1)^n*(1+sqrt(2))^(-n) = (1/2)*(1 + (-1)^n) - (1-sqrt(2))^n.
See A001622 for a general formula concerning the fractional parts of powers of numbers x > 1, which satisfy x - x^(-1) = floor(x).
a(n) = round((1+sqrt(2))^n/(2*sqrt(2))) for n > 0. (End) [last formula corrected by Josh Inman, Mar 05 2024]
a(n) = ((4+sqrt(18))*(1+sqrt(2))^n + (4-sqrt(18))*(1-sqrt(2))^n)/4 offset 0. - Al Hakanson (hawkuu(AT)gmail.com), Aug 08 2009
If p[i] = Fibonacci(i) and if A is the Hessenberg matrix of order n defined by A[i,j] = p[j-i+1] when i<=j, A[i,j]=-1 when i=j+1, and A[i,j]=0 otherwise, then, for n >= 1, a(n) = det A. - Milan Janjic, May 08 2010
a(n) = 3*a(n-1) - a(n-2) - a(n-3), n > 2. - Gary Detlefs, Sep 09 2010
From Charlie Marion, Apr 13 2011: (Start)
a(n) = 2*(a(2k-1) + a(2k))*a(n-2k) - a(n-4k).
a(n) = 2*(a(2k) + a(2k+1))*a(n-2k-1) + a(n-4k-2). (End)
G.f.: x/(1 - 2*x - x^2) = sqrt(2)*G(0)/4; G(k) = ((-1)^k) - 1/(((sqrt(2) + 1)^(2*k)) - x*((sqrt(2) + 1)^(2*k))/(x + ((sqrt(2) - 1)^(2*k + 1))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 02 2011
In general, for n > k, a(n) = a(k+1)*a(n-k) + a(k)*a(n-k-1). See definition of Pell numbers and the formula for Sep 04 2008. - Charlie Marion, Jan 17 2012
Sum{n>=1} (-1)^(n-1)/(a(n)*a(n+1)) = sqrt(2) - 1. - Vladimir Shevelev, Feb 22 2013
From Vladimir Shevelev, Feb 24 2013: (Start)
(1) Expression a(n+1) via a(n): a(n+1) = a(n) + sqrt(2*a^2(n) + (-1)^n);
(2) a(n+1)^2 - a(n)*a(n+2) = (-1)^n;
(3) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(4) a(n)/a(n+1) = sqrt(2) - 1 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
a(-n) = -(-1)^n * a(n). - Michael Somos, Jun 01 2013
G.f.: G(0)/(2+2*x) - 1/(1+x), where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 10 2013
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 + x)/( x*(4*k+4 + x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 30 2013
a(n) = Sum_{r=0..n-1} Sum_{k=0..n-r-1} binomial(r+k,k)*binomial(k,n-k-r-1). - Peter Luschny, Nov 16 2013
a(n) = Sum_{k=1,3,5,...<=n} C(n,k)*2^((k-1)/2). - Vladimir Shevelev, Feb 06 2014
a(2n) = 2*a(n)*(a(n-1) + a(n)). - John Blythe Dobson, Mar 08 2014
a(k*n) = a(k)*a(k*n-k+1) + a(k-1)*a(k*n-k). - Charlie Marion, Mar 27 2014
a(k*n) = 2*a(k)*(a(k*n-k)+a(k*n-k-1)) + (-1)^k*a(k*n-2k). - Charlie Marion, Mar 30 2014
a(n+1) = (1+sqrt(2))*a(n) + (1-sqrt(2))^n. - Art DuPre, Apr 04 2014
a(n+1) = (1-sqrt(2))*a(n) + (1+sqrt(2))^n. - Art DuPre, Apr 04 2014
a(n) = F(n) + Sum_{k=1..n} F(k)*a(n-k), n >= 0 where F(n) the Fibonacci numbers A000045. - Ralf Stephan, May 23 2014
a(n) = round(sqrt(a(2n) + a(2n-1)))/2. - Richard R. Forberg, Jun 22 2014
a(n) = Product_{k divides n} A008555(k). - Tom Edgar, Jan 28 2015
a(n+k)^2 - A002203(k)*a(n)*a(n+k) + (-1)^k*a(n)^2 = (-1)^n*a(k)^2. - Alexander Samokrutov, Aug 06 2015
a(n) = 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1) for n >= 2. - Peter Luschny, Dec 17 2015
a(n+1) = Sum_{k=0..n} binomial(n,k)*2^floor(k/2). - Tony Foster III, May 07 2017
a(n) = exp((i*Pi*n)/2)*sinh(n*arccosh(-i))/sqrt(2). - Peter Luschny, Mar 07 2018
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)^2 - a(n-2)^2 = 2*a(n-1)*(a(n) + a(n-2)) (see A005319);
(2) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(3) Sum_{k=2..n+1} a(k)*a(k-1) = a(n+1)^2 if n is odd, else a(n+1)^2 - 1 if n is even;
(4) a(n) - a(n-2*k+1) = (A077444(k) - 1)*a(n-2*k+1) + a(n-4*k+2);
(5) Sum_{k=n..n+9} a(k) = 41*A001333(n+5). (End)
From Kai Wang, Dec 30 2019: (Start)
a(m+r)*a(n+s) - a(m+s)*a(n+r) = -(-1)^(n+s)*a(m-n)*a(r-s).
a(m+r)*a(n+s) + a(m+s)*a(n+r) = (2*A002203(m+n+r+s) - (-1)^(n+s)*A002203(m-n)*A002203(r-s))/8.
A002203(m+r)*A002203(n+s) - A002203(m+s)*A002203(n+r) = (-1)^(n+s)*8*a(m-n)*a(r-s).
A002203(m+r)*A002203(n+s) - 8*a(m+s)*a(n+r) = (-1)^(n+s)*A002203(m-n)*A002203(r-s).
A002203(m+r)*A002203(n+s) + 8*a(m+s)*a(n+r) = 2*A002203(m+n+r+s)+ (-1)^(n+s)*8*a(m-n)*a(r-s). (End)
From Kai Wang, Jan 12 2020: (Start)
a(n)^2 - a(n+1)*a(n-1) = (-1)^(n-1).
a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = (-1)^n*a(m-n).
a(m-n) = (-1)^n (a(m)*A002203(n) - A002203(m)*a(n))/2.
a(m+n) = (a(m)*A002203(n) + A002203(m)*a(n))/2.
A002203(n)^2 - A002203(n+r)*A002203(n-r) = (-1)^(n-r-1)*8*a(r)^2.
A002203(m)*A002203(n+1) - A002203(m+1)*A002203(n) = (-1)^(n-1)*8*a(m-n).
A002203(m-n) = (-1)^(n)*(A002203(m)*A002203(n) - 8*a(m)*a(n) )/2.
A002203(m+n) = (A002203(m)*A002203(n) + 8*a(m)*a(n) )/2. (End)
From Kai Wang, Mar 03 2020: (Start)
Sum_{m>=1} arctan(2/a(2*m+1)) = arctan(1/2).
Sum_{m>=2} arctan(2/a(2*m+1)) = arctan(1/12).
In general, for n > 0,
Sum_{m>=n} arctan(2/a(2*m+1)) = arctan(1/a(2*n)). (End)
a(n) = (A001333(n+3*k) + (-1)^(k-1)*A001333(n-3*k)) / (20*A041085(k-1)) for any k>=1. - Paul Curtz, Jun 23 2021
Sum_{i=0..n} a(i)*J(n-i) = (a(n+1) + a(n) - J(n+2))/2 for J(n) = A001045(n). - Greg Dresden, Jan 05 2022
From Peter Bala, Aug 20 2022: (Start)
Sum_{n >= 1} 1/(a(2*n) + 1/a(2*n)) = 1/2.
Sum_{n >= 1} 1/(a(2*n+1) - 1/a(2*n+1)) = 1/4. Both series telescope - see A075870 and A005319.
Product_{n >= 1} ( 1 + 2/a(2*n) ) = 1 + sqrt(2).
Product_{n >= 2} ( 1 - 2/a(2*n) ) = (1/3)*(1 + sqrt(2)). (End)
G.f. = 1/(1 - Sum_{k>=1} Fibonacci(k)*x^k). - Enrique Navarrete, Dec 17 2023
Sum_{n >=1} 1/a(n) = 1.84220304982752858079237158327980838... - R. J. Mathar, Feb 05 2024
a(n) = ((3^(n+1) + 1)^(n-1) mod (9^(n+1) - 2)) mod (3^(n+1) - 1). - Joseph M. Shunia, Jun 06 2024

A002522 a(n) = n^2 + 1.

Original entry on oeis.org

1, 2, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145, 170, 197, 226, 257, 290, 325, 362, 401, 442, 485, 530, 577, 626, 677, 730, 785, 842, 901, 962, 1025, 1090, 1157, 1226, 1297, 1370, 1445, 1522, 1601, 1682, 1765, 1850, 1937, 2026, 2117, 2210, 2305, 2402, 2501
Offset: 0

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Author

Keywords

Comments

An n X n nonnegative matrix A is primitive (see A070322) iff every element of A^k is > 0 for some power k. If A is primitive then the power which should have all positive entries is <= n^2 - 2n + 2 (Wielandt).
a(n) = Phi_4(n), where Phi_k is the k-th cyclotomic polynomial.
As the positive solution to x=2n+1/x is x=n+sqrt(a(n)), the continued fraction expansion of sqrt(a(n)) is {n; 2n, 2n, 2n, 2n, ...}. - Benoit Cloitre, Dec 07 2001
a(n) is one less than the arithmetic mean of its neighbors: a(n) = (a(n-1) + a(n+1))/2 - 1. E.g., 2 = (1+5)/2 - 1, 5 = (2+10)/2 - 1. - Amarnath Murthy, Jul 29 2003
Equivalently, the continued fraction expansion of sqrt(a(n)) is (n;2n,2n,2n,...). - Franz Vrabec, Jan 23 2006
Number of {12,1*2*,21}-avoiding signed permutations in the hyperoctahedral group.
The number of squares of side 1 which can be drawn without lifting the pencil, starting at one corner of an n X n grid and never visiting an edge twice is n^2-2n+2. - Sébastien Dumortier, Jun 16 2005
Also, numbers m such that m^3 - m^2 is a square, (n*(1 + n^2))^2. - Zak Seidov
1 + 2/2 + 2/5 + 2/10 + ... = Pi*coth Pi [Jolley], see A113319. - Gary W. Adamson, Dec 21 2006
For n >= 1, a(n-1) is the minimal number of choices from an n-set such that at least one particular element has been chosen at least n times or each of the n elements has been chosen at least once. Some games define "matches" this way; e.g., in the classic Parker Brothers, now Hasbro, board game Risk, a(2)=5 is the number of cards of three available types (suits) required to guarantee at least one match of three different types or of three of the same type (ignoring any jokers or wildcards). - Rick L. Shepherd, Nov 18 2007
Positive X values of solutions to the equation X^3 + (X - 1)^2 + X - 2 = Y^2. To prove that X = n^2 + 1: Y^2 = X^3 + (X - 1)^2 + X - 2 = X^3 + X^2 - X - 1 = (X - 1)(X^2 + 2X + 1) = (X - 1)*(X + 1)^2 it means: (X - 1) must be a perfect square, so X = n^2 + 1 and Y = n(n^2 + 2). - Mohamed Bouhamida, Nov 29 2007
{a(k): 0 <= k < 4} = divisors of 10. - Reinhard Zumkeller, Jun 17 2009
Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n)^2/4 + 1), n=1, 2, 3, ... . - Johannes W. Meijer, Jun 12 2010
For n > 0, continued fraction [n,n] = n/a(n); e.g., [5,5] = 5/26. - Gary W. Adamson, Jul 15 2010
The only real solution of the form f(x) = A*x^p with negative p which satisfies f^(m)(x) = f^[-1](x), x >= 0, m >= 1, with f^(m) the m-th derivative and f^[-1] the compositional inverse of f, is obtained for m=2*n, p=p(n)= -(sqrt(a(n))-n) and A=A(n)=(fallfac(p(n),2*n))^(-p(n)/(p(n)+1)), with fallfac(x,k):=Product_{j=0..k-1} (x-j) (falling factorials). See the T. Koshy reference, pp. 263-4 (there are also two solutions for positive p, see the corresponding comment in A087475). - Wolfdieter Lang, Oct 21 2010
n + sqrt(a(n)) = [2*n;2*n,2*n,...] with the regular continued fraction with period 1. This is the even case. For the general case see A087475 with the Schroeder reference and comments. For the odd case see A078370.
a(n-1) counts configurations of non-attacking bishops on a 2 X n strip [Chaiken et al., Ann. Combin. 14 (2010) 419]. - R. J. Mathar, Jun 16 2011
Also numbers k such that 4*k-4 is a square. Hence this sequence is the union of A053755 and A069894. - Arkadiusz Wesolowski, Aug 02 2011
a(n) is also the Moore lower bound on the order, A191595(n), of an (n,5)-cage. - Jason Kimberley, Oct 17 2011
Left edge of the triangle in A195437: a(n+1) = A195437(n,0). - Reinhard Zumkeller, Nov 23 2011
If h (5,17,37,65,101,...) is prime is relatively prime to 6, then h^2-1 is divisible by 24. - Vincenzo Librandi, Apr 14 2014
The identity (4*n^2+2)^2 - (n^2+1)*(4*n)^2 = 4 can be written as A005899(n)^2 - a(n)*A008586(n)^2 = 4. - Vincenzo Librandi, Jun 15 2014
a(n) is also the number of permutations simultaneously avoiding 213 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014
a(n-1) is the maximum number of stages in the Gale-Shapley algorithm for finding a stable matching between two sets of n elements given an ordering of preferences for each element (see Gura et al.). - Melvin Peralta, Feb 07 2016
Because of Fermat's little theorem, a(n) is never divisible by 3. - Altug Alkan, Apr 08 2016
For n > 0, if a(n) points are placed inside an n X n square, it will always be the case that at least two of the points will be a distance of sqrt(2) units apart or less. - Melvin Peralta, Jan 21 2017
Also the limit as q->1^- of the unimodal polynomial (1-q^(n*k+1))/(1-q) after making the simplification k=n. The unimodal polynomial is from O'Hara's proof of unimodality of q-binomials after making the restriction to partitions of size <= 1. See G_1(n,k) from arXiv:1711.11252. As the size restriction s increases, G_s->G_infinity=G: the q-binomials. Then substituting k=n and q=1 yields the central binomial coefficients: A000984. - Bryan T. Ek, Apr 11 2018
a(n) is the smallest number congruent to both 1 (mod n) and 2 (mod n+1). - David James Sycamore, Apr 04 2019
a(n) is the number of permutations of 1,2,...,n+1 with exactly one reduced decomposition. - Richard Stanley, Dec 22 2022
From Klaus Purath, Apr 03 2025: (Start)
The odd prime factors of these terms are always of the form 4*k + 1.
All a(n) = D satisfy the Pell equation (k*x)^2 - D*y^2 = -1. The values for k and the solutions x, y can be calculated using the following algorithm: k = n, x(0) = 1, x(1) = 4*D - 1, y(0) = 1, y(1) = 4*D - 3. The two recurrences are of the form (4*D - 2, -1). The solutions x, y of the Pell equations for n = {1 ... 14} are in OEIS.
It follows from the above that this sequence is a subsequence of A031396. (End)

Examples

			G.f. = 1 + 2*x + 5*x^2 + 10*x^3 + 17*x^4 + 26*x^5 + 37*x^6 + 50*x^7 + 65*x^8 + ...
		

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 120).
  • E. Gura and M. Maschler, Insights into Game Theory: An Alternative Mathematical Experience, Cambridge, 2008; p. 26.
  • Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, New York, 2001.

Crossrefs

Left edge of A055096.
Cf. A059100, A117950, A087475, A117951, A114949, A117619 (sequences of form n^2 + K).
a(n+1) = A101220(n, n+1, 3).
Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), A005843 (g=4), this sequence (g=5), A051890 (g=6), A188377 (g=7). - Jason Kimberley, Oct 30 2011
Cf. A002496 (primes).
Cf. A254858.
Subsequence of A031396.

Programs

Formula

O.g.f.: (1-x+2*x^2)/((1-x)^3). - Eric Werley, Jun 27 2011
Sequences of the form a(n) = n^2 + K with offset 0 have o.g.f. (K - 2*K*x + K*x^2 + x + x^2)/(1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a*(n-3). - R. J. Mathar, Apr 28 2008
For n > 0: a(n-1) = A143053(A000290(n)) - 1. - Reinhard Zumkeller, Jul 20 2008
A143053(a(n)) = A000290(n+1). - Reinhard Zumkeller, Jul 20 2008
a(n)*a(n-2) = (n-1)^4 + 4. - Reinhard Zumkeller, Feb 12 2009
a(n) = A156798(n)/A087475(n). - Reinhard Zumkeller, Feb 16 2009
From Reinhard Zumkeller, Mar 08 2010: (Start)
a(n) = A170949(A002061(n+1));
A170949(a(n)) = A132411(n+1);
A170950(a(n)) = A002061(n+1). (End)
For n > 1, a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n - 2)^2 + (a(n) + n - 1 + a(n) + n)^2 = (n+1) *(6*n^4 + 18*n^3 + 26*n^2 + 19*n + 6) / 6 = (a(n) + n)^2 + ... + (a(n) + 2*n)^2. - Charlie Marion, Jan 10 2011
From Eric Werley, Jun 27 2011: (Start)
a(n) = 2*a(n-1) - a(n-2) + 2.
a(n) = a(n-1) + 2*n - 1. (End)
a(n) = (n-1)^2 + 2(n-1) + 2 = 122 read in base n-1 (for n > 3). - Jason Kimberley, Oct 20 2011
a(n)*a(n+1) = a(n*(n+1) + 1) so a(1)*a(2) = a(3). More generally, a(n)*a(n+k) = a(n*(n+k) + 1) + k^2 - 1. - Jon Perry, Aug 01 2012
a(n) = (n!)^2* [x^n] BesselI(0, 2*sqrt(x))*(1+x). - Peter Luschny, Aug 25 2012
a(n) = A070216(n,1) for n > 0. - Reinhard Zumkeller, Nov 11 2012
E.g.f.: exp(x)*(1 + x + x^2). - Geoffrey Critzer, Aug 30 2013
a(n) = A254858(n-2,3) for n > 2. - Reinhard Zumkeller, Feb 09 2015
Sum_{n>=0} (-1)^n / a(n) = (1+Pi/sinh(Pi))/2 = 0.636014527491... = A367976 . - Vaclav Kotesovec, Feb 14 2015
Sum_{n>=0} 1/a(n) = (1 + Pi*coth(Pi))/2 = 2.076674... = A113319. - Vaclav Kotesovec, Apr 10 2016
4*a(n) = A001105(n-1) + A001105(n+1). - Bruno Berselli, Jul 03 2017
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi)*sinh(sqrt(2)*Pi).
Product_{n>=1} (1 - 1/a(n)) = Pi*csch(Pi). (End)
Sum_{n>=0} a(n)/n! = 3*e. - Davide Rotondo, Feb 16 2025

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A006190 a(n) = 3*a(n-1) + a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, 42837, 141481, 467280, 1543321, 5097243, 16835050, 55602393, 183642229, 606529080, 2003229469, 6616217487, 21851881930, 72171863277, 238367471761, 787274278560, 2600190307441, 8587845200883, 28363725910090
Offset: 0

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Author

Keywords

Comments

Denominators of continued fraction convergents to (3+sqrt(13))/2. - Benoit Cloitre, Jun 14 2003
a(n) and A006497(n) occur in pairs: (a,b): (1,3), (3,11), (10,36), (33,119), (109,393), ... such that b^2 - 13a^2 = 4(-1)^n. - Gary W. Adamson, Jun 15 2003
Form the 4-node graph with matrix A = [1,1,1,1; 1,1,1,0; 1,1,0,1; 1,0,1,1]. Then this sequence counts the walks of length n from the vertex with degree 5 to one (any) of the other vertices. - Paul Barry, Oct 02 2004
a(n+1) is the binomial transform of A006138. - Paul Barry, May 21 2006
a(n+1) is the diagonal sum of the exponential Riordan array (exp(3x),x). - Paul Barry, Jun 03 2006
Number of paths in the right half-plane from (0,0) to the line x=n-1, consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0). Example: a(3)=10 because we have hh, H, UD, DU, hU, Uh, UU, hD, Dh and DD. - Emeric Deutsch, Sep 03 2007
Equals INVERT transform of A000129. Example: a(5) = 109 = (29, 12, 5, 2, 1) dot (1, 1, 3, 10, 33) = (29 + 12 + 15 + 20 + 33). - Gary W. Adamson, Aug 06 2010
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 3's along the main diagonal, 1's along the superdiagonal and subdiagonal, and 0's everywhere else. - John M. Campbell, Jul 08 2011
These numbers could also be called "bronze Fibonacci numbers". Indeed, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio; analogously, for Pell numbers (A000129), or "silver Fibonacci numbers", P(n+1) = ceiling(delta*a(n)), if n is even and P(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1 + sqrt(2) is the silver ratio. Here, for n >= 1, we have a(n+1) = ceiling(c*a(n)), if n is even and a(n+1) = floor(c*a(n)), if n is odd, where c = (3 + sqrt(13))/2 is the bronze ratio (cf. comment in A098316). - Vladimir Shevelev, Feb 23 2013
Let p(n,x) denote the Fibonacci polynomial, defined by p(1,x) = 1, p(2,x) = x, p(n,x) = x*p(n-1,x) + p(n-2,x). Let q(n,x) be the numerator polynomial of the rational function p(n, x + 1 + 1/x). Then q(n,1) = a(n). - Clark Kimberling, Nov 04 2013
The (1,1)-entry of the matrix A^n where A = [0,1,0; 1,2,1; 1,1,2]. - David Neil McGrath, Jul 18 2014
a(n+1) counts closed walks on K2, containing three loops on the other vertex. Equivalently the (1,1)-entry of A^(n+1) where the adjacency matrix of digraph is A = (0,1; 1,3). - David Neil McGrath, Oct 29 2014
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,2,3} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
With offset 1 is the INVERTi transform of A001076. - Gary W. Adamson, Jul 24 2015
Apart from the initial 0, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = 1 - 3 S. See A291219. - Clark Kimberling, Sep 02 2017
From Rogério Serôdio, Mar 30 2018: (Start)
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)).
gcd(a(n),a(n+k)) = a(gcd(n, k)) for all positive integers n and k. (End)
Numbers of straight-chain fatty acids involving oxo and/or hydroxy groups, if cis-/trans isomerism and stereoisomerism are neglected. - Stefan Schuster, Apr 04 2018
Number of 3-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
From Michael A. Allen, Jan 25 2023: (Start)
Also called the 3-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 3 kinds of squares available. (End)
a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n >= 1, P(k) = A000129(k) is the k-th Pell number (see example below). - Enrique Navarrete, Dec 15 2023

Examples

			From _Enrique Navarrete_, Dec 15 2023: (Start)
From the comment on compositions with Pell number of parts, A000129(k), there are A000129(1)=1 type of 1, A000129(2)=2 types of 2, A000129(3)=5 types of 3, A000129(4)=12 types of 4, A000129(5)=29 types of 5 and A000129(6)=70 types of 6.
The following table gives the number of compositions of n=6:
Composition, number of such compositions, number of compositions of this type:
  6,              1,     70;
  5+1,            2,     58;
  4+2,            2,     48;
  3+3,            1,     25;
  4+1+1,          3,     36;
  3+2+1,          6,     60;
  2+2+2,          1,      8;
  3+1+1+1,        4,     20;
  2+2+1+1,        6,     24;
  2+1+1+1+1,      5,     10;
  1+1+1+1+1+1,    1,      1;
for a total of a(6)=360 compositions of n=6. (End).
		

References

  • H. L. Abbott and D. Hanson, A lattice path problem, Ars Combin., 6 (1978), 163-178.
  • A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 128.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • L.-N. Machaut, Query 3436, L'Intermédiaire des Mathématiciens, 16 (1909), 62-63. - N. J. A. Sloane, Mar 08 2022

Crossrefs

Row sums of Pascal's rhombus (A059317). Also row sums of triangle A054456(n, m).
Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), A000129 (k=2), this sequence (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), A154597 (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).
Cf. A006497, A052906, A175182 (Pisano periods), A201001 (prime subsequence), A092936 (squares).
Cf. A243399.

Programs

  • GAP
    a:=[0,1];; for n in [3..30] do a[n]:=3*a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Mar 31 2018
  • Haskell
    a006190 n = a006190_list !! n
    a006190_list = 0 : 1 : zipWith (+) (map (* 3) $ tail a006190_list) a006190_list
    -- Reinhard Zumkeller, Feb 19 2011
    
  • Magma
    [ n eq 1 select 0 else n eq 2 select 1 else 3*Self(n-1)+Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 19 2011
    
  • Maple
    a[0]:=0: a[1]:=1: for n from 2 to 35 do a[n]:= 3*a[n-1]+a[n-2] end do: seq(a[n],n=0..30); # Emeric Deutsch, Sep 03 2007
    A006190:=-1/(-1+3*z+z**2); # Simon Plouffe in his 1992 dissertation, without the leading 0
    seq(combinat[fibonacci](n,3),n=0..30); # R. J. Mathar, Dec 07 2011
  • Mathematica
    a[n_] := (MatrixPower[{{1, 3}, {1, 2}}, n].{{1}, {1}})[[2, 1]]; Table[ a[n], {n, -1, 24}] (* Robert G. Wilson v, Jan 13 2005 *)
    LinearRecurrence[{3,1},{0,1},30] (* or *) CoefficientList[Series[x/ (1-3x-x^2), {x,0,30}], x] (* Harvey P. Dale, Apr 20 2011 *)
    Table[If[n==0, a1=1; a0=0, a2=a1; a1=a0; a0=3*a1+a2], {n, 0, 30}] (* Jean-François Alcover, Apr 30 2013 *)
    Table[Fibonacci[n, 3], {n, 0, 30}] (* Vladimir Reshetnikov, May 08 2016 *)
  • PARI
    a(n)=if(n<1,0,contfracpnqn(vector(n,i,2+(i>1)))[2,1])
    
  • PARI
    a(n)=([1,3;1,2]^n)[2,1] \\ Charles R Greathouse IV, Mar 06 2014
    
  • PARI
    concat([0],Vec(x/(1-3*x-x^2)+O(x^30))) \\ Joerg Arndt, Apr 30 2013
    
  • Sage
    [lucas_number1(n,3,-1) for n in range(0, 30)] # Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: x/(1 - 3*x - x^2).
From Benoit Cloitre, Jun 14 2003: (Start)
a(3*n) = 2*A041019(5*n-1), a(3*n+1) = A041019(5*n), a(3*n+2) = A041019(5*n+3).
a(2*n) = 3*A004190(n-1); a(3*n) = 10*A041613(n-1) for n >= 1. (End)
From Gary W. Adamson, Jun 15 2003: (Start)
a(n-1) + a(n+1) = A006497(n).
A006497(n)^2 - 13*a(n)^2 = 4(-1)^n. (End)
a(n) = U(n-1, (3/2)i)(-i)^(n-1), i^2 = -1. - Paul Barry, Nov 19 2003
a(n) = Sum_{k=0..n-1} binomial(n-k-1,k)*3^(n-2*k-1). - Paul Barry, Oct 02 2004
a(n) = F(n, 3), the n-th Fibonacci polynomial evaluated at x=3.
Let M = {{0, 1}, {1, 3}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = Abs[v[n][[1]]]. - Roger L. Bagula, May 29 2005 [Or a(n) = [M^(n+1)]{1,1}. - _L. Edson Jeffery, Aug 27 2013]
From Paul Barry, May 21 2006: (Start)
a(n+1) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(k-j).
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(n-j-k).
a(n+1) = Sum_{k=0..floor(n/2)} C(n-k,k)*3^(n-2*k).
a(n) = Sum_{k=0..n} C(k,n-k)*3^(2*k-n). (End)
E.g.f.: exp(3*x/2)*sinh(sqrt(13)*x/2)/(sqrt(13)/2). - Paul Barry, Jun 03 2006
a(n) = (ap^n - am^n)/(ap - am), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2.
Let C = (3 + sqrt(13))/2 = exp arcsinh(3/2) = 3.3027756377... Then C^n, n > 0 = a(n)*(1/C) + a(n+1). Let X = the 2 X 2 matrix [0, 1; 1, 3]. Then X^n = [a(n-1), a(n); a(n), a(n+1)]. - Gary W. Adamson, Dec 21 2007
1/3 = 3/(1*10) + 3/(3*33) + 3/(10*109) + 3/(33*360) + 3/(109*1189) + ... . - Gary W. Adamson, Mar 16 2008
a(n) = ((3 + sqrt(13))^n - (3 - sqrt(13))^n)/(2^n*sqrt(13)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
a(p) == 13^((p-1)/2) mod p, for odd primes p. - Gary W. Adamson, Feb 22 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
Limit_{k->oo} a(n+k)/a(k) = (A006497(n) + a(n)*sqrt(13))/2.
Limit_{n->oo} A006497(n)/a(n) = sqrt(13). (End)
Sum_{k>=1} (-1)^(k-1)/(a(k)*a(k+1)) = (sqrt(13)-3)/2. - Vladimir Shevelev, Feb 23 2013
From Vladimir Shevelev, Feb 24 2013: (Start)
(1) Expression a(n+1) via a(n): a(n+1) = (3*a(n) + sqrt(13*a(n)^2 + 4*(-1)^n))/2;
(2) a^2(n+1) - a(n)*a(n+2) = (-1)^n;
(3) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(4) a(n)/a(n+1) = (sqrt(13)-3)/2 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
a(n) = sqrt(13*(A006497(n))^2 + (-1)^(n-1)*52)/13. - Vladimir Shevelev, Mar 13 2013
Sum_{n >= 1} 1/( a(2*n) + 1/a(2*n) ) = 1/3; Sum_{n >= 1} 1/( a(2*n + 1) - 1/a(2*n + 1) ) = 1/9. - Peter Bala, Mar 26 2015
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)*a(n+1) = 3*Sum_{k=1..n} a(k)^2;
(2) a(n)^2 + a(n+1)^2 = a(2*n+1);
(3) a(n)^2 - a(n-2)^2 = 3*a(n-1)*(a(n) + a(n-2));
(4) a(m*(p+1)) = a(m*p)*a(m+1) + a(m*p-1)*a(m);
(5) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(6) a(2*n) = a(n)*(3*a(n) + 2*a(n-1));
(7) 3*Sum_{k=2..n+1} a(k)*a(k-1) is equal to a(n+1)^2 if n odd, and is equal to a(n+1)^2 - 1 if n is even;
(8) a(n) = alpha(k)*a(n-2*k+1) + a(n-4*k+2), where alpha(k) = ap^(2*k-1) + am^(2*k-1), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2;
(9) 131|Sum_{k=n..n+9} a(k), for all positive n. (End)
From Kai Wang, Feb 10 2020: (Start)
a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2 - Catalan's identity.
arctan(1/a(2n)) - arctan(1/a(2n+2)) = arctan(a(2)/a(2n+1)).
arctan(1/a(2n)) = Sum_{m>=n} arctan(a(2)/a(2m+1)).
The same formula holds for Fibonacci numbers and Pell numbers. (End)
a(n+2) = 3^(n+1) + Sum_{k=0..n} a(k)*3^(n-k). - Greg Dresden and Gavron Campbell, Feb 22 2022
G.f. = 1/(1-Sum_{k>=1} P(k)*x^k), P(k)=A000129(k) (with a(0)=1). - Enrique Navarrete, Dec 17 2023
G.f.: x/(1 - 3*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 3 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024
Sum_{n>=0} a(n)/phi^(3*n) = 1, where phi = A001622 is the golden ratio. - Diego Rattaggi, Apr 07 2025

Extensions

Second formula corrected by Johannes W. Meijer, Jun 02 2010

A073133 Table by antidiagonals of T(n,k) = n*T(n,k-1) + T(n,k-2) starting with T(n,1) = 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 5, 3, 1, 4, 10, 12, 5, 1, 5, 17, 33, 29, 8, 1, 6, 26, 72, 109, 70, 13, 1, 7, 37, 135, 305, 360, 169, 21, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 1, 10, 82, 528, 2549, 8658, 18901, 23184, 12970, 2378, 89
Offset: 1

Views

Author

Henry Bottomley, Jul 16 2002

Keywords

Comments

Columns of the array are generated from Fibonacci polynomials f(x). They are: (1), (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 +1), ... If column headings start 0, 1, 2, ... then the terms in the n-th column are generated from the n-th degree Fibonacci polynomial. For example, column 5 (8, 70, 360, ...) is generated from f(x), x = 1,2,3,...; fifth-degree polynomial x^5 + 4x^3 + 3x; e.g., f(2) = 70 = 2^5 + 4*8 + 3*2. - Gary W. Adamson, Apr 02 2006
The ratio of two consecutive entries of the sequence in the n-th row approaches (n + sqrt(n^2 + 4))/2. Example: The sequence beginning (1, 3, 10, 33, ...) tends to 3.302775... = (3 + sqrt(13))/2. - Gary W. Adamson, Aug 12 2013
As to the array sequences, (n+1)-th sequence is the INVERT transform of the n-th sequence. - Gary W. Adamson, Aug 20 2013
The array can be extended infinitely above the Fibonacci row by taking successive INVERTi transforms, resulting in:
...
1, -2, 5, -12, 29, -70, ...
1, -1, 2, -3, 5, -8, ...
l, 0, 1, 0, 1, 0, ...
1, 1, 2, 3, 5, 8, ...
1, 2, 5, 12, 29, 70, ...
...
This results in an infinite array in which sequences above the (1, 0, 1, 0, ...) are reflections of the sequences below, except for the alternate signs. Any sequence in the (+ sign) row starting (1, n, ...) is the (2*n-th) INVERT transform of the same sequence but with alternate signs. Example: (1, 2, 5, 12, ...) is the (2*2) = fourth INVERT transform of (1, -2, 5, -12, ...) by inspection. Conjecture: This "reflection" principle will result from taking successive INVERT transforms of any aerated sequence starting 1, ... and with positive signs. Likewise, the rows above the aerated sequence are successive INVERTi transforms of the aerated sequence. - Gary W. Adamson, Jul 14 2019
From Michael A. Allen, Feb 21 2023: (Start)
Row n is the n-metallonacci sequence.
T(n,k) is the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

Examples

			Table begins:
  1, 1,  2,  3,   5,    8,   13, ...
  1, 2,  5, 12,  29,   70,  169, ...
  1, 3, 10, 33, 109,  360, 1189, ...
  1, 4, 17, 72, 305, 1292, 5473, ... etc.
		

Crossrefs

Programs

  • GAP
    T:= function(n,k)
        if k<0 then return 0;
        elif k=1 then return 1;
        else return n*T(n,k-1) + T(n,k-2);
        fi;
      end;
    Flat(List([1..15], n-> List([1..n], k-> T(n-k+1,k) ))); # G. C. Greubel, Aug 12 2019
  • Maple
    A073133 := proc(n,k)
        option remember;
        if k <= 1 then
            k;
        else
            n*procname(n,k-1)+procname(n,k-2) ;
        end if;
    end proc:
    seq(seq( A073133(d-k,k),k=1..d-1),d=2..13) ; # R. J. Mathar, Aug 16 2019
  • Mathematica
    T[n_, 1]:= 1; T[n_, k_]:= T[n, k] = If[k<0, 0, n*T[n, k-1] + T[n, k-2]]; Table[T[n-k+1, k], {n, 15}, {k, n}]//Flatten (* G. C. Greubel, Aug 12 2019 *)
  • PARI
    T(n,k) = if(k==1, 1, k<0, 0, n*T(n,k-1)+T(n,k-2));
    for(n=1,15, for(k=1,n, print1(T(n-k+1,k), ", "))) \\ G. C. Greubel, Aug 12 2019
    
  • Sage
    def T(n, k):
        if (k<0): return 0
        elif (k==1): return 1
        else: return n*T(n, k-1) + T(n, k-2)
    [[T(n-k+1, k) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Aug 12 2019
    

Formula

T(n, k) = A073134(n, k) + 2*A073135(n, k-2) = Sum_{j=0..k-1} abs(A049310(k-1, j)*n^j).
T(n,k) = [[0,1; 1,n]^{k+1}]{1,1}, n,k in {1,2,...}. - _L. Edson Jeffery, Sep 23 2012
G.f. for row n: x/(1-n*x-x^2). - L. Edson Jeffery, Aug 28 2013

A078370 a(n) = 4*(n+1)*n + 5.

Original entry on oeis.org

5, 13, 29, 53, 85, 125, 173, 229, 293, 365, 445, 533, 629, 733, 845, 965, 1093, 1229, 1373, 1525, 1685, 1853, 2029, 2213, 2405, 2605, 2813, 3029, 3253, 3485, 3725, 3973, 4229, 4493, 4765, 5045, 5333, 5629, 5933, 6245, 6565, 6893, 7229, 7573, 7925, 8285, 8653, 9029
Offset: 0

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

This is the generic form of D in the (nontrivially) solvable Pell equation x^2 - D*y^2 = -4. See A078356, A078357.
1/5 + 1/13 + 1/29 + ... = (Pi/8)*tanh Pi [Jolley]. - Gary W. Adamson, Dec 21 2006
Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n+1)^2 + 4), n = 1, 2, 3, ... . - Johannes W. Meijer, Jun 12 2010
(2*n + 1 + sqrt(a(n)))/2 = [2*n + 1; 2*n + 1, 2*n + 1, ...], n >= 0, with the regular continued fraction with period length 1. This is the odd case. See A087475 for the general case with the Schroeder reference and comments. For the even case see A002522.
Primes in the sequence are in A005473. - Russ Cox, Aug 26 2019
The continued fraction expansion of sqrt(a(n)) is [2n+1; {n, 1, 1, n, 4n+2}]. For n=0, this collapses to [2; {4}]. - Magus K. Chu, Aug 27 2022
Discriminant of the binary quadratic forms y^2 - x*y - A002061(n+1)*x^2. - Klaus Purath, Nov 10 2022
From Klaus Purath, Apr 08 2025: (Start)
There are no squares in this sequence. The prime factors of these terms are always of the form 4*k + 1.
All a(n) = D satisfy the Pell equation (k*x)^2 - D*(m*y)^2 = -1 for any integer n where m = (D - 3)/2. The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*m^2 - 1), x(0) = 1, x(1) = 4*D*m^2 - 1, y(0) = 1, y(1) = 4*D*m^2 - 3. The two recurrences are of the form (4*D*m^2 - 2, -1).
It follows from the above that this sequence belongs to A031396. (End)

References

  • L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 176.

Crossrefs

Subsequence of A077426 (D values (not a square) for which Pell x^2 - D*y^2 = -4 is solvable in positive integers).
Subsequence of A031396.

Programs

  • Magma
    [4*n^2+4*n+5 : n in [0..80]]; // Wesley Ivan Hurt, Aug 29 2022
  • Mathematica
    Table[4 n (n + 1) + 5, {n, 0, 45}] (* or *)
    Table[8 Binomial[n + 1, 2] + 5, {n, 0, 45}] (* or *)
    CoefficientList[Series[(5 - 2 x + 5 x^2)/(1 - x)^3, {x, 0, 45}], x] (* Michael De Vlieger, Jan 04 2017 *)
  • PARI
    a(n)=4*n^2+4*n+5 \\ Charles R Greathouse IV, Sep 24 2015
    
  • Python
    a= lambda n: 4*n**2+4*n+5 # Indranil Ghosh, Jan 04 2017
    
  • Scala
    (1 to 99 by 2).map(n => n * n + 4) // Alonso del Arte, May 29 2019
    

Formula

a(n) = (2*n + 1)^2 + 4.
a(n) = 4*(n+1)*n + 5 = 8*binomial(n+1, 2) + 5, hence subsequence of A004770 (5 (mod 8) numbers). [Typo fixed by Zak Seidov, Feb 26 2012]
G.f.: (5 - 2*x + 5*x^2)/(1 - x)^3.
a(n) = 8*n + a(n-1), with a(0) = 5. - Vincenzo Librandi, Aug 08 2010
a(n) = A016754(n) + 4. - Leo Tavares, Feb 22 2023
From Elmo R. Oliveira, Oct 31 2024: (Start)
E.g.f.: (5 + 8*x + 4*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

Extensions

More terms from Max Alekseyev, Mar 03 2010
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