cp's OEIS Frontend

Chebyshev or generalized Fibonacci sequence.

This is the m=13 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..12 (nonnegative) sequences are A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913 and A004189. The m=1..3 (signed) sequences are A049347, A056594, A010892.

All positive integer solutions of Pell equation b(n)^2 - 117*a(n)^2 = +4 together with b(n+1)=A057076(n+1), n >= 0. - Wolfdieter Lang, Aug 31 2004

For positive n, a(n) equals the permanent of the tridiagonal matrix of order n with 11's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011

a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,10}. - Milan Janjic, Jan 25 2015

17*a(n+1) + 4*(a(n) + (-1)^n) is 13 times the numerator of the continued fraction [3,3,...,3,1,3,...,3,3] with n+1 3's on either side of the central 1. For example, when n=2 the continued fraction [3,3,3,1,3,3,3] = 1749/529, and 17*a(3) + 4(a(2)+1) = 17*1309 + 4*121 = 22737 = 13*1749. - Greg Dresden and Tulipa Wray, Aug 26 2025

Table starts

.....2.......5.........16............52..............169.................549

.....5......24........139...........853.............5241...............32089

....13.....115.......1202.........14042...........164014.............1905436

....34.....551......10409........231454..........5142441...........113293694

....89....2640......90157.......3815483........161243887..........6736602042

...233...12649.....780922......62897985.......5055954492........400571676322

...610...60605....6764246....1036869496.....158534446141......23818815015639

..1597..290376...58591124...17092731689....4971005036586....1416315842358249

..4181.1391275..507509767..281772661177..155870804492221...84217060496525106

.10946.6665999.4395993154.4645005493684.4887484036570530.5007720081104988709

Binomial transform of A030221. - Philippe Deléham, Nov 19 2009

Partial sums of Chebyshev sequence S(n,5) = U(n,5/2) = A004254(n) (Chebyshev's polynomials of the second kind, see A049310). - Wolfdieter Lang, Aug 31 2004

In this sequence 4*a(n)*a(n+2)+1 is a square. - Bruno Berselli, Jun 19 2012

a(n) is a divisibility sequence - m divides n implies that a(m) divides a(n). - Paul Raff, Mar 06 2009

Also number of domino tilings of the 5 X (2n-1) rectangle with upper left corner removed. For n=2 the 15 domino tilings of the 5 X 3 rectangle with upper left corner removed are:

. ._. . ._. . ._. . ._. . ._. . ._. . ._. . ._.

.|__| .| | | .|___| .|__| .|__| .| | | .| | | .|__|

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. ._. . ._. . ._. . ._. . ._. . ._. . ._.

.|__| .|__| .|__| .|__| .|__| .|__| ._| | |

|_| | | | | | |_| | |_| | |_| | | |_| | |||

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- Alois P. Heinz, Apr 14 2011

This is the r=7 member in the r-family of sequences S_r(n+1) defined in A092184 where more information can be found.

Working with an offset of 1, this sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. Case P1 = 7, P2 = 10, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 25 2014

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 165*a^2 = +4 with companion sequence b(n)=A078363(n+1), n >= 0.

This is the m=15 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..14 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190 and A004191. The m=1..3 (signed) sequences are A049347, A056594, A010892.

For positive n, a(n) equals the permanent of the tridiagonal matrix of order n with 13's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011

For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,12}. - Milan Janjic, Jan 23 2015

Riordan array (1/(1+5*x+x^2), x/(1+5*x+x^2)). - Philippe Deléham, Feb 03 2007

Chebyshev's S(n,x-5) polynomials (exponents of x in increasing order). - Philippe Deléham, Feb 22 2012

Row sums are A125905(n). - Philippe Deléham, Feb 22 2012

Diagonal sums are (-5)^n. - Philippe Deléham, Feb 22 2012

Subtriangle of triangle given by (0, -5, 1/5, -1/5, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 22 2012

Inverse of triangle in A125906. - Philippe Deléham, Feb 22 2012

Riordan array (1/(1-4*x+x^2), x/(1-4*x+x^2)).

Subtriangle of the triangle given by (0, 4, -1/4, 1/4, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Unsigned version of triangles in A124029 and in A159764.

For 1<=k<=n, T(n,k) equals the number of (n-1)-length words over {0,1,2,3,4} containing k-1 letters equal 4 and avoiding 01. - Milan Janjic, Dec 20 2016

Define {x(k)} to be an integer sequence satisfying all the following conditions:

(i) {x(k)} satisfies second-order linear recursion, that is, there exists two integers P, Q such that x(k+2) = P*x(k+1) + Q*x(k) holds for all k >= 0.

(ii) {x(k)} is (not necessarily strictly) increasing. ({A000035(k)} satisfies condition (i), but it doesn't satisfy this.)

(iii) All terms in {x(k)} do not share a common factor. ({A024023(k)} satisfies both conditions (i) and (ii), but all terms share a common factor 2.)

(iv) {x(k)} satisfies strong divisibility, that is, gcd(x(m),x(n)) = x(gcd(m,n)) holds for all m, n >= 0. ({A093131(k)} satisfies all conditions (i) to (iii), but 5 = gcd(A093131(2),A093131(3)) != A093131(gcd(2,3)) = 1.)

(v) For all positive integers n, there eventually exists some m > 0 such that n divides x(m). ({A002275(k)} satisfies all conditions (i) to (iv), but 2, 5 and 10 never divide any term.)

Then it's easy to show that the only solutions to {x(k)} are x(k) = A172236(n,k) or x(k) = T(n,k), i.e., x(0) = 0, x(1) = 1, P >= 1, Q = 1 or P >= 2, Q = -1.

The case n = 0 is not included since it gives the period-4 signed sequence 0, 1, 0, -1, 0, 1, 0, -1, ..., the g.f. of which is the inverse of the 4th cyclotomic polynomial.

The case n = 1 is not included since it gives the period-6 signed sequence 0, 1, 1, 0, -1, -1, ..., the g.f. of which is the inverse of the 6th cyclotomic polynomial.

The congruence property: let p be an odd prime which is not divisible by n^2 - 4, then T(n,(p-1)/2) == 1/2(((n-2)/p) - ((n+2)/p)) (mod p), T(n,(p+1)/2) == 1/2(((n-2)/p) + ((n+2)/p)) (mod p). Here ((n-2)/p) is the Legendre symbol. Or equivalently:

((n-2)/p)...((n+2)/p)...T(n,(p-1)/2) mod p...T(n,(p+1)/2) mod p

.....1...........1...............0....................1

....-1..........-1...............0...................-1

.....1..........-1...............1....................0

....-1...........1..............-1....................0

To prove this, rewrite (n +- sqrt(n^2-4))/2 as ((sqrt(n+2) +- sqrt(n-2))/2)^2.

Let E(n,m) be the smallest number l such that m divides T(n,l), we have: E(n,p) divides (p - ((n^2-4)/p))/2 for odd primes p that are not divisible by n^2 - 4. E(n,p) = p for odd primes p that are divisible by n^2 - 4. E(n,2) = 2 for even n and 3 for odd n.

E(n,p^e) <= p^(e-1)*E(n,p) for all primes p. If p^2 does not divide T(n,E(n,p)), then E(n,p^e) = p^(e-1)*E(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 - 4, p^2 is never divisible by T(n,p), so E(n,p^e) = p^e as described above. E(n,m_1*m_2) = lcm(E(n,m_1),E(n,m_2)) if gcd(m_1,m_2) = 1.

Given n, the largest possible value of E(n,m)/m is 1 for even n and 3/2 for odd n. It can be obtained by the value of E(n,2) described above.

Let pi(n,m) be the Pisano period of T(n,k) modulo m, i.e, the smallest number l such that T(n,k+1) == T(n,k) (mod m) holds for all k >= 0, we have: for odd primes p that are not divisible by n^2 - 4, pi(n,p) divides p + ((n-2)/p) if ((n+2)/p) = -1 and (p - ((n-2)/p))/2 if ((n+2)/p) = 1. pi(n,p) = p for odd primes p that are divisible by n - 2 and 2p for odd primes p that are divisible by n + 2. pi(n,2) = 2 even n and 3 for odd n.

pi(n,p^e) <= p^(e-1)*pi(n,p) for all primes p. If p^2 does not divide T(n,E(n,p)), then pi(n,p^e) = p^(e-1)*pi(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 - 4, p^2 is never divisible by T(n,p), so pi(n,p^e) = p^e or 2p^e as described above. pi(n,m_1*m_2) = lcm(pi(n,m_1),pi(n,m_2)) if gcd(m_1,m_2) = 1.

Given n, the largest possible value of pi(n,m)/m is:

Parity of n...n + 2 is a power of 2 or 3...max{pi(n,m)/m}.....obtained by

....even..........yes (even exponent)............1...........pi(n,2^e) = 2^e

....even...........yes (odd exponent)...........4/3............pi(n,3) = 4

....even...................no....................2.............pi(n,p) = 2p (p >= 3 is any prime factor of n + 2)

.....odd..................yes....................2..........pi(n,3^e) = 2*3^e

.....odd...................no....................3..........pi(n,2p^e) = 6p^e (p >= 5 is any prime factor of n + 2)

The largest possible value of pi(n,m)/m is obtained by infinitely many m except for the case n = 10, in which we have pi(10,3) = 6, pi(10,7) = 8, pi(10,21) = 24 and pi(10,m)/m <= 14/13 for all other m. [Corrected by Jianing Song, Nov 04 2018]

Let z(n,m) be the number of zeros in a period of T(n,k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: for odd primes p that are not divisible by n^2 - 4, z(n,p) = 2 if ((n+2)/p) = -1; 1 or 2 if ((n+2)/p) = 1. z(n,p) = 1 for odd primes p that are divisible by n - 2 and 2 for odd primes p that are divisible by n + 2. z(n,2) = 1.

For all odd primes p, z(n,p) = 2 if and only if pi(n,p) is even, z(n,p) = 1 if and only if pi(n,p) is odd. For all odd primes p, if E(n,p) is even then z(n,p) = 2 (the converse is not necessarily true). [Comment revised by Jianing Song, Jul 06 2019]

z(n,p^e) = z(n,p) for all odd primes p. z(n,4) = 1 if n == 2, 3 (mod 4) and 2 if n == 0, 1 (mod 4). z(n,2^e) = 1 for even n and 2 for odd n, e >= 3.

By induction it is easy to show that T(n,k) = T(n,m+1)*T(n,k-m) - T(n,m)*T(k-m-1). Let k = 2m we have T(n,2m) = T(n,m)*(T(n,m+1)-T(m-1)); let k = 2m+1 we have T(n,2m+1) = T(n,m+1)^2 - T(n,m)^2 = (T(n,m+1)+T(n,m))*(T(n,m+1)-T(n,m)). So T(n,k) is composite if n >= 3, k >= 3. - Jianing Song, Jul 06 2019