cp's OEIS Frontend

Let f = floor and c = ceiling. For x > 1, define four sequences as functions of x, as follows:

p1(0) = f(x), p1(n) = f(x*p1(n-1));

p2(0) = f(x), p2(n) = c(x*p2(n-1)) if n is odd and p2(n) = f(x*p1(n-1)) if n is even;

p3(0) = c(x), p3(n) = f(x*p3(n-1)) if n is odd and p3(n) = c(x*p3(n-1)) if n is even;

p4(0) = c(x), p4(n) = c(x*p4(n-1)).

The present sequence is given by a(n) = p3(n).

Following the terminology at A214986, call the four sequences power floor, power floor-ceiling, power ceiling-floor, and power ceiling sequences. In the table below, a sequence is identified with an A-numbered sequence if they appear to agree except possibly for initial terms. Notation: S(t)=sqrt(t), r = (1+S(5))/2 = golden ratio, and Limit = limit of p3(n)/p2(n).

x ......p1..... p2..... p3..... p4.......Limit

r.......A000045 A000045 A000045 A000045..r

r^2.....A001519 A001654 A061646 A001906..-1+S(5)

r^3.....A024551 A001076 A015448 A049652..-1+S(5)

r^4.....A049685 A157335 A214992 A004187..-19+9*S(5)

r^5.....A214993 A049666 A015457 A214994...(-9+5*S(5))/2

r^6.....A007805 A156085 A214995 A049660..-151+68*S(5)

1+S(2)..A024537 A000129 A001333 A048739..S(2)

2+S(2)..A007052 A214996 A214997 A007070..(1+S(2))/2

1+S(3)..A057960 A002605 A028859 A077846..(1+S(3))/2

2+S(3)..A001835 A109437 A214998 A001353..-4+3*S(3)

S(5)....A214999 A215091 A218982 A218983..1.26879683...

2+S(5)..A024551 A001076 A015448 A049652..-1+S(5)

2+S(6)..A218984 A090017 A123347 A218985..S(3/2)

2+S(7)..A218986 A015530 A126473 A218987..(1+S(7))/3

2+S(8)..A218988 A057087 A086347 A218989..(1+S(2))/2

3+S(8)..A001653 A084158 A218990 A001109..-13+10*S(2)

3+S(10).A218991 A005668 A015451 A218992..-2+S(10)

...

Properties of p1, p2, p3, p4:

(1) If x > 2, the terms of p2 and p3 interlace: p2(0) < p3(0) < p2(1) < p3(1) < p2(2) < p3(2)... Also, p1(n) <= p2(n) <= p3(n) <= p4(n) <= p1(n+1) for all x>0 and n>=0.

(2) If x > 2, the limits L(x) = limit(p/x^n) exist for the four functions p(x), and L1(x) <= L2(x) <= L3(x) <= L4 (x). See the Mathematica programs for plots of the four functions; one of them also occurs in the Odlyzko and Wilf article, along with a discussion of the special case x = 3/2.

(3) Suppose that x = u + sqrt(v) where v is a nonsquare positive integer. If u = f(x) or u = c(x), then p1, p2, p3, p4 are linear recurrence sequences. Is this true for sequences p1, p2, p3, p4 obtained from x = (u + sqrt(v))^q for every positive integer q?

(4) Suppose that x is a Pisot-Vijayaraghavan number. Must p1, p2, p3, p4 then be linearly recurrent? If x is also a quadratic irrational b + c*sqrt(d), must the four limits L(x) be in the field Q(sqrt(d))?

(5) The Odlyzko and Wilf article (page 239) raises three interesting questions about the power ceiling function; it appears that they remain open.

For positive n, a(n) equals (-1)^n times the permanent of the (4n) X (4n) tridiagonal matrix with sqrt(i)'s along the three central diagonals, where i is the imaginary unit. - John M. Campbell, Jul 12 2011

a(n) = 5^n*a(n; 3/5) = (16/5)^n*a(2*n; 3/4), and F(4*n) = 5^n*b(n; 3/5) = (16/5)^n*b(2*n; 3/4), where a(n; d) and b(n; d), n=0, 1, ..., d in C, denote the delta-Fibonacci numbers defined in comments to A014445. Two of these identities from the following relations follows: F(k+1)^n*a(n; F(k)/F(k+1)) = F(k*n+1) and F(k+1)^n*b(n; F(k)/F(k+1)) = F(k*n) (see also Witula's et al. papers). - Roman Witula, Jul 24 2012

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

Partial sums of A033888, i.e., a(n) = Sum_{k=0..n} Fibonacci(4*k). - Vladeta Jovovic, Jun 09 2002

From Paul Weisenhorn, May 17 2009: (Start)

a(n) is the solution of the 2 equations a(n)+1=A^2 and 5*a(n)+1=B^2

which are equivalent to the Pell equation (10*a(n)+3)^2-5*(A*B)^2=4.

(End)

Numbers a(n) such as a(n)+1 and 5*a(n)+1 are perfect squares. - Sture Sjöstedt, Nov 03 2011

Scaled Chebyshev U-polynomials evaluated at 3/2.

From Peter Bala, Aug 05 2019: (Start)

Let U(n;P,Q), where P and Q are integer parameters, denote the Lucas sequence of the first kind. Then, excluding the case P = -1, the sequence ( U(n;P,1) + U(2*n;P,1) )/(P + 1) is a fourth-order linear divisibility sequence with o.g.f. x*(1 - 2*(P - 1)*x + x^2)/((1 - P*x + x^2)*(1 - (P^2 - 2)*x + x^2)). This is the case P = 3. See A000027 (P = 2), A165998 (P = -2) and A238536 (P = -3).

More generally, the sequence U(n;P,1) + U(2*n;P,1) + ... + U(k*n;P,1) is a linear divisibility sequence of order 2*k. As an example, see A273625 (P = 3, k = 3 and then sequence normalized with initial term 1). (End)

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

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

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 221*a^2 = +4 with companion sequence b(n)=A078365(n+1), n>=0.

This is the m=17 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..16 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190, A004191, A078362 and A007655. The m=1..3 (signed) sequences are A049347, A056594, A010892.

For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 15'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>=2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,14}. - Milan Janjic, Jan 23 2015