A049652 -id:A049652 - cp's OEIS frontend

A001076 Denominators of continued fraction convergents to sqrt(5).

0, 1, 4, 17, 72, 305, 1292, 5473, 23184, 98209, 416020, 1762289, 7465176, 31622993, 133957148, 567451585, 2403763488, 10182505537, 43133785636, 182717648081, 774004377960, 3278735159921, 13888945017644, 58834515230497, 249227005939632, 1055742538989025

Offset: 0

N. J. A. Sloane

a(2*n+1) with b(2*n+1) := A001077(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = -1, a(2*n) with b(2*n) := A001077(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = +1 (cf. Emerson reference).
Bisection: a(2*n+1) = T(2*n+1, sqrt(5))/sqrt(5) = A007805(n), n >= 0 and a(2*n) = 4*S(n-1,18), 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. S(n,18)=A049660(n+1). - Wolfdieter Lang, Jan 10 2003
Apart from initial terms, this is the Pisot sequence E(4,17), a(n) = floor(a(n-1)^2/a(n-2) + 1/2).
This is also the Horadam sequence (0,1,1,4), having the recurrence relation a(n) = s*a(n-1) + r*a(n-2); for n > 1, where a(0) = 0, a(1) = 1, s = 4, r = 1. a(n) / a(n-1) converges to 5^1/2 + 2 as n approaches infinity. 5^(1/2) + 2 can also be written as (2 * Phi) + 1 and Phi^2 + Phi. - Ross La Haye, Aug 18 2003
Numerators of continued fraction [4, 4, 4, ...], where the convergents to [4, 4, 4, ...] = (4/1, 17/4, 72/17, ...). Let X = the 2 X 2 matrix [0, 1; 1, 4]; then X^n = [a(n-1), a(n); a(n), a(n+1)]; e.g., X^3 = [4, 17; 17, 72]. Let C = the limit of a(n)/a(n-1) = 2 + sqrt(5) = 4.236067977...; then C^n = a(n+1) + (1/C)*a(n), where (1/C) = 0.236067977... . Example: C^3 = 76.01315556..., = 72 + 17*(0.2360679...). - Gary W. Adamson, Dec 15 2007, corrected by Greg Dresden, Sep 16 2019, corrected by Alex Mark, Jul 21 2020
Sqrt(5) = 4/2 + 4/17 + 4/(17*305) + 4/(305*5473) + 4/(5473*98209) + ... . - Gary W. Adamson, Dec 15 2007
a(p) == 20^((p-1)/2) (mod p) for odd primes p. - Gary W. Adamson, Feb 22 2009
a(n) = A167808(3*n). - Reinhard Zumkeller, Nov 12 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 4's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Moreover, a(n) is the second binomial transform of (0,1,0,5,0,25,...) (see also A033887). This fact can be proved similarly like the proof of Paul Barry's remark in A033887 by using the following scaling identity for delta-Fibonacci numbers: y^n b(n;x/y) = Sum_{k=0..n} binomial(n,k) (y-1)^(n-k) b(k;x) and the fact that b(n;2) = (1-(-1)^n) 5^floor(n/2). - Roman Witula, Jul 12 2012
Binomial transform of 0, 1, 2, 8, 24, 80, 256, ... (A063727 with offset 1). - R. J. Mathar, Feb 05 2014
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,4} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
With offset 1 is the INVERT transform of A006190: (1, 3, 10, 33, 109, 360, ...). - Gary W. Adamson, Jul 24 2015
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)
The initial 0 of this sequence is in contradiction with the fact that 0 is no valid denominator and according to all standard references, the first convergent of a continued fraction is p(0)/q(0) = b(0)/1 where b(0) is the first term of the continued fraction, given by the integer part of the number. One may artificially define q(-1) = 0 to have a recurrent relation q(n) = b(n)*q(n-1) + q(n-2), n >= 1, but then its index should be -1. - M. F. Hasler, Nov 01 2019
Number of 4-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 4-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 4 kinds of squares available. (End)
a(n) is the smallest nonnegative integer that is the sum of n, but no fewer, Fibonacci numbers including negative-index Fibonacci numbers (A039834), with that sum being a(n) = Sum_{i=0..n-1} A000045(3*i+1). a(n) is also the smallest nonnegative integer that is the sum of n, but no fewer, terms each of which is either a Fibonacci number or the negative of a Fibonacci number. (See A027941 for negatives disallowed.) - Mike Speciner, Oct 08 2023
From Enrique Navarrete, Dec 16 2023: (Start)
a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n > = 1, where P(k) = A006190(k) is the k-th 3-metallonacci number (see example below).
In general, the number of compositions with k-metallonacci number of parts is counted by the (k+1)-st metallonacci sequence (note k=1 and k=2 are the Fibonacci and the Pell numbers, respectively). (End).
a(n) is the number of tilings of a 2 X n rectangle missing the top right 1 X 1 cell, using 1 X 1 squares, dominoes and right trominoes. Compare to A110679 which is the same problem but without the missing top right cell. - Greg Dresden and Yilin Zhu, Jul 10 2025

			1 2 9 38 161 (A001077)
-,-,-,--,---, ...
0 1 4 17 72 (A001076)
G.f. = x + 4*x^2 + 17*x^3 + 72*x^4 + 305*x^5 + 1292*x^6 + 5473*x^7 + 23184*x^8 + ...
From _Enrique Navarrete_, Dec 16 2023: (Start)
From the comment on compositions with 3-metallonacci sorts of parts, A006190(k), there are A006190(1)=1 type of 1, A006190(2)=3 types of 2, A006190(3)=10 types of 3, A006190(4)=33 types of 4, A006190(5)=109 types of 5 and A006190(6)=360 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,      360;
 5+1,            2,      218;
 4+2,            2,      198;
 3+3,            1,      100;
 4+1+1,          3,       99;
 3+2+1,          6,      180;
 2+2+2,          1,       27;
 3+1+1+1,        4,       40;
 2+2+1+1,        6,       54;
 2+1+1+1+1,      5,       15;
 1+1+1+1+1+1,    1,        1;
for a total of a(6)=1292 compositions of n=6. (End)

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 23.
S. Koshkin, Non-classical linear divisibility sequences ..., Fib. Q., 57 (No. 1, 2019), 68-80. See Table 1.
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).
V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 282.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Paraskevas K. Alvanos and Konstantinos A. Draziotis, Integer Solutions of the Equation y^2 = Ax^4 + B, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.4.
Elwyn Berlekamp and Richard K. Guy, Fibonacci Plays Billiards (2003), arXiv:2002.03705 [math.HO], 2020. See p. 5.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 8.
D. W. Boyd, Some integer sequences related to the Pisot sequences, Acta Arithmetica, 34 (1979), 295-305.
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
Latham Boyle and Paul J. Steinhardt, Self-Similar One-Dimensional Quasilattices, arXiv preprint arXiv:1608.08220 [math-ph], 2016.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Thm. 1, p. 233.
M. C. Firengiz and A. Dil, Generalized Euler-Seidel method for second order recurrence relations, Notes on Number Theory and Discrete Mathematics, Vol. 20, 2014, No. 4, 21-32.
Yixing Fu, E. J. König, J. H. Wilson, Yang-Zhi Chou, and J. H. Pixley, Magic-angle semimetals, arXiv:1809.04604 [cond-mat.str-el], 2018.
Juan B. Gil and Aaron Worley, Generalized metallic means, arXiv:1901.02619 [math.NT], 2019.
Brian Hopkins and Stéphane Ouvry, Combinatorics of Multicompositions, arXiv:2008.04937 [math.CO], 2020.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 398
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Shaoxiong (Steven) Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Row n=4 of A073133, A172236 and A352361.
Cf. A000045, A001077, A015448, A175183 (Pisano periods).
Cf. A049660, A007805, A110679.
Partial sums of A033887. First differences of A049652. Bisection of A059973.
Third column of array A028412.

a:=[0,1];; for n in [3..30] do a[n]:=4*a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Mar 31 2018

I:=[0,1]; [n le 2 select I[n] else 4*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

A001076:=-1/(-1+4*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

Join[{0}, Denominator[Convergents[Sqrt[5], 30]]] (* Harvey P. Dale, Dec 10 2011 *)
a[ n_] := Fibonacci[3*n] / 2; (* Michael Somos, Feb 23 2014 *)
a[ n_] := ((2 + Sqrt[5])^n - (2 - Sqrt[5])^n) /(2 Sqrt[5]) // Simplify; (* Michael Somos, Feb 23 2014 *)
LinearRecurrence[{4, 1}, {0, 1}, 26] (* Jean-François Alcover, Sep 23 2017 *)
a[ n_] := Fibonacci[n, 4]; (* Michael Somos, Nov 02 2021 *)

a(n):=sum(4^(n-1-2*k)*binomial(n-k-1,n-2*k-1),k,0,floor((n)/2));/* Vladimir Kruchinin, Oct 02 2022 */

numlib::fibonacci(3*n)/2 $ n = 0..30; // Zerinvary Lajos, May 09 2008

{a(n) = fibonacci(3*n) / 2}; /* Michael Somos, Aug 11 2009 */

{a(n) = imag( (2 + quadgen(20))^n )}; /* Michael Somos, Feb 23 2014 */

{a(n) = polchebyshev(n-1, 2, 2*I)/I^(n-1)}; /* Michael Somos, Nov 02 2021 */

[lucas_number1(n,4,-1) for n in range(23)] # Zerinvary Lajos, Apr 23 2009

[fibonacci(3*n)/2 for n in range(23)] # Zerinvary Lajos, May 15 2009

a(n) = 4*a(n-1) + a(n-2), n > 1. a(0)=0, a(1)=1.
G.f.: x/(1 - 4*x - x^2).
a(n) = ((2+sqrt(5))^n - (2-sqrt(5))^n)/(2*sqrt(5)).
a(n) = A014445(n)/2 = F(3n)/2.
a(n) = ((-i)^(n-1))*S(n-1, 4*i), with i^2 = -1 and S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x) = 0.
a(n) = Sum_{i=0..n} Sum_{j=0..n} Fibonacci(i+j)*n!/(i!j!(n-i-j)!)/2. - Paul Barry, Feb 06 2004
E.g.f.: exp(2*x)*sinh(sqrt(5)*x)/sqrt(5). - Vladeta Jovovic, Sep 01 2004
a(n) = F(1) + F(4) + F(7) + ... + F(3n-2), for n > 0.
Conjecture: 2a(n+1) = a(n+2) - A001077(n+1). - Creighton Dement, Nov 28 2004
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n, j)*C(j, k)*F(j)/2. - Paul Barry, Feb 14 2005
a(n) = A048876(n) - A048875(n). - Creighton Dement, Mar 19 2005
Let M = {{0, 1}, {1, 4}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = v[n][[1]]. - Roger L. Bagula, May 29 2005
a(n) = F(n, 4), the n-th Fibonacci polynomial evaluated at x=4. - T. D. Noe, Jan 19 2006
[A015448(n), a(n)] = [1,4; 1,3]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = (Sum_{k=0..n} Fibonacci(3*k-2)) + 1. - Gary Detlefs, Dec 26 2010
a(n) = (3*(-1)^n*F(n) + 5*F(n)^3)/2, n >= 0. See the general D. Jennings formula given in a comment on triangle A111125, where also the reference is given. Here the second (k=1) row [3,1] applies. - Wolfdieter Lang, Sep 01 2012
Sum_{k>=1} (-1)^(k-1)/(a(k)*a(k+1)) = (Sum_{k>=1} (-1)^(k-1)/(F_k*F_(k+1)))^3 = phi^(-3), where F_n is the n-th Fibonacci numbers (A000045) and phi is golden ratio (A001622). - Vladimir Shevelev, Feb 23 2013
G.f.: Q(0)*x/(2-4*x), where Q(k) = 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 2/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
a(-n) = -(-1)^n * a(n). - Michael Somos, Feb 23 2014
The o.g.f. A(x) = x/(1 - 4*x - x^2) satisfies A(x) + A(-x) + 8*A(x)*A(-x) = 0 or equivalently (1 + 8*A(x))*(1 + 8*A(-x)) = 1. The o.g.f. for A049660 equals -A(sqrt(x))*A(-sqrt(x)). - Peter Bala, Apr 02 2015
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)*a(n+1) = 4*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 = 4*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(n-1)*a(n+1) = a(n)^2 + (-1)^n (particular case of (5)!);
(7) a(2*n) = 2*a(n)*(2*a(n) + a(n-1));
(8) 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;
(9) a(n) - a(n-2*k+1) = alpha(k)*a(n-2*k+1) + a(n-4*k+2), where alpha(k) = (2+sqrt(5))^(2*k-1) + (2-sqrt(5))^(2*k-1);
(10) 31|Sum_{k=n..n+9} a(k), for all positive n. (End)
O.g.f.: x*exp(Sum_{n >= 1} Lucas(3*n)*x^n/n) = x + 4*x^2 + 17*x^3 + .... - Peter Bala, Oct 11 2019
a(n) = Sum_{k=0..floor(n/2)} 4^(n-2*k-1)*C(n-k-1,n-2*k-1). - Vladimir Kruchinin, Oct 02 2022
a(n) = i^(n-1)*S(n-1, -4*i), with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. - Gary Detlefs and Wolfdieter Lang, Mar 06 2023
G.f.: x/(1 - 4*x - x^2) = Sum_{n >= 0} x^(n+1) * ( Product_{k = 1..n} (m*k + 4 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024
a(n) = 4^(n-1)*hypergeom([(1-n)/2, 1-n/2], [1-n], -1/4) for n > 0. - Peter Luschny, Mar 30 2025
a(n) = a(n-1) + A110679(n-1) + A110679(n-2) = a(n-1) + Fibonacci(3*n-2). - Greg Dresden and Yilin Zhu, Jul 10 2025

A214992 Power ceiling-floor sequence of (golden ratio)^4.

Original entry on oeis.org

7, 47, 323, 2213, 15169, 103969, 712615, 4884335, 33477731, 229459781, 1572740737, 10779725377, 73885336903, 506417632943, 3471038093699, 23790849022949, 163064905066945, 1117663486445665, 7660579500052711

Offset: 0

Clark Kimberling, Nov 08 2012, Jan 24 2013

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.

			a(0) = ceiling(r) = 7, where r = ((1+sqrt(5))/2)^4 = 6.8...; a(1) = floor(7*r) = 47; a(2) = ceiling(47) = 323.

Clark Kimberling, Table of n, a(n) for n = 0..250
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33, 235-240, 1991.
Index entries for linear recurrences with constant coefficients, signature (6,6,-1).
Index entries for sequences related to the Josephus Problem.

Cf. A001622, A214986, A049685, A157335, A004187.

(* Program 1.  A214992 and related sequences *)
x = GoldenRatio^4; z = 30; (* z = # terms in sequences *)
z1 = 100; (* z1 = # digits in approximations *)
f[x_] := Floor[x]; c[x_] := Ceiling[x];
p1[0] = f[x]; p2[0] = f[x]; p3[0] = c[x]; p4[0] = c[x];
p1[n_] := f[x*p1[n - 1]]
p2[n_] := If[Mod[n, 2] == 1, c[x*p2[n - 1]], f[x*p2[n - 1]]]
p3[n_] := If[Mod[n, 2] == 1, f[x*p3[n - 1]], c[x*p3[n - 1]]]
p4[n_] := c[x*p4[n - 1]]
Table[p1[n], {n, 0, z}]  (* A049685 *)
Table[p2[n], {n, 0, z}]  (* A157335 *)
Table[p3[n], {n, 0, z}]  (* A214992 *)
Table[p4[n], {n, 0, z}]  (* A004187 *)
Table[p4[n] - p1[n], {n, 0, z}]  (* A004187 *)
Table[p3[n] - p2[n], {n, 0, z}]  (* A098305 *)
(* Program 2.  Plot of power floor and power ceiling functions, p1(x) and p4(x) *)
f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
p1[x_, 0] := f[x]; p1[x_, n_] := f[x*p1[x, n - 1]];
p4[x_, 0] := c[x]; p4[x_, n_] := c[x*p4[x, n - 1]];
Plot[Evaluate[{p1[x, 10]/x^10, p4[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]
(* Program 3. Plot of power floor-ceiling and power ceiling-floor functions, p2(x) and p3(x) *)
f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
p2[x_, 0] := f[x]; p3[x_, 0] := c[x];
p2[x_, n_] := If[Mod[n, 2] == 1, c[x*p2[x, n - 1]], f[x*p2[x, n - 1]]]
p3[x_, n_] := If[Mod[n, 2] == 1, f[x*p3[x, n - 1]], c[x*p3[x, n - 1]]]
Plot[Evaluate[{p2[x, 10]/x^10, p3[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]

a(n) = floor(r*a(n-1)) if n is odd and a(n) = ceiling(r*a(n-1)) if n is even, where a(0) = ceiling(r), r = (golden ratio)^4 = (7 + sqrt(5))/2.
a(n) = 6*a(n-1) + 6*a(n-2) - a(n-3).
G.f.: (7 + 5*x - x^2)/((1 + x)*(1 - 7*x + x^2)).
a(n) = (10*(-2)^n+(10+3*sqrt(5))*(7-3*sqrt(5))^(n+2)+(10-3*sqrt(5))*(7+3*sqrt(5))^(n+2))/(90*2^n). - Bruno Berselli, Nov 14 2012
a(n) = 7*A157335(n) + 5*A157335(n-1) - A157335(n-2). - R. J. Mathar, Feb 05 2020
E.g.f.: exp(-x)*(5 + 2*exp(9*x/2)*(155*cosh(3*sqrt(5)*x/2) + 69*sqrt(5)*sinh(3*sqrt(5)*x/2)))/45. - Stefano Spezia, Oct 28 2024

A099919 a(n) = F(3) + F(6) + F(9) + ... + F(3n), F(n) = Fibonacci numbers A000045.

Original entry on oeis.org

0, 2, 10, 44, 188, 798, 3382, 14328, 60696, 257114, 1089154, 4613732, 19544084, 82790070, 350704366, 1485607536, 6293134512, 26658145586, 112925716858, 478361013020, 2026369768940, 8583840088782, 36361730124070, 154030760585064, 652484772464328, 2763969850442378

Offset: 0

Ralf Stephan, Oct 30 2004

Partial sum of the even Fibonacci numbers. - Vladimir Joseph Stephan Orlovsky, Nov 28 2010

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 25.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Project Euler, Problem 2: Even Fibonacci Numbers.
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Partial sums of A014445.
Cf. A000045, A004794, A015448, A049652.
Cf. A087635.
Case k = 3 of partial sums of fibonacci(k*n): A000071, A027941, A058038, A138134, A053606.

[(Fibonacci(3*n+2) - 1)/2: n in [0..30]]; // G. C. Greubel, Jan 17 2018

CoefficientList[Series[2 x/((1 - x) (1 - 4 x - x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 15 2014 *)
LinearRecurrence[{5, -3, -1}, {0, 2, 10}, 30] (* G. C. Greubel, Jan 17 2018 *)
Accumulate[Fibonacci[3Range[0, 19]]] (* Alonso del Arte, Dec 23 2018 *)

a(n) = sum(i=1, n, fibonacci(3*i)); \\ Michel Marcus, Mar 15 2014

a(n) = fibonacci(3*n+2)\2 \\ Charles R Greathouse IV, Jun 11 2015

a(n) = (Fibonacci(3*n + 2) - 1)/2 = (A015448(n+1)-1)/2.
G.f.: 2*x/((1 - x)*(1 - 4*x - x^2)).
a(n) = (F(3n + 2) - 1)/2 = 2 * A049652(n).
a(n) = Sum_{0 <= j <= i <= n} binomial(i, j)*F(i + j). - Benoit Cloitre, May 21 2005
From Gary Detlefs, Dec 08 2010: (Start)
a(n) = 4*a(n - 1) + a(n - 2) + 2, n > 1.
a(n) = 5*a(n - 1) - 3*a(n - 2) - a(n - 3), n > 2.
a(n) = (Fibonacci(3*n + 3) + Fibonacci(3*n) - 2)/4. (End)
a(n) = (-10 + (5 - 3*sqrt(5))*(2 - sqrt(5))^n + (2 + sqrt(5))^n*(5 + 3*sqrt(5)))/20. - Colin Barker, Nov 26 2016
E.g.f.: exp(x)*(exp(x)*(5*cosh(sqrt(5)*x) + 3*sqrt(5)*sinh(sqrt(5)*x)) - 5)/10. - Stefano Spezia, Jun 03 2024

a(0) = 0 prepended by Joerg Arndt, Mar 13 2014

A049651 a(n) = (F(3*n+1) - 1)/2, where F=A000045 (the Fibonacci sequence).

Original entry on oeis.org

0, 1, 6, 27, 116, 493, 2090, 8855, 37512, 158905, 673134, 2851443, 12078908, 51167077, 216747218, 918155951, 3889371024, 16475640049, 69791931222, 295643364939, 1252365390980, 5305104928861, 22472785106426, 95196245354567, 403257766524696, 1708227311453353, 7236167012338110

Offset: 0

Clark Kimberling

This is the sequence A(0,1;4,1;2) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010

For n>0, a(n) is the least number whose greedy Fibonacci-union-Lucas representation (as at A214973), has n terms. - Clark Kimberling, Oct 23 2012

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 24.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Hans Koch, Golden mean renormalization for the almost Mathieu operator and related skew products, arXiv:1907.06804 [math-ph], 2019.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Cf. A000045, A033887, A077259, A214973.

Pairwise sums of A049652.

[(Fibonacci(3*n+1) - 1)/2: n in [0..30]]; // G. C. Greubel, Dec 05 2017

(Fibonacci[Range[1,5!,3]]-1)/2 (* Vladimir Joseph Stephan Orlovsky, May 18 2010 *)
LinearRecurrence[{5, -3, -1}, {0, 1, 6}, 50] (* G. C. Greubel, Dec 05 2017 *)

vector(30,n,n--;(fibonacci(3*n+1) -1)/2) \\ G. C. Greubel, Dec 05 2017

[(fibonacci(3*n+1)-1)/2 for n in (0..30)] # G. C. Greubel, Apr 19 2019

From Ralf Stephan, Jan 23 2003: (Start)
a(n) = 4*a(n-1) + a(n-2) + 2, a(0)=0, a(1)=1.
G.f.: x*(1+x)/((1-x)*(1-4*x-x^2)).
a(n) is asymptotic to -1/2+(sqrt(5)+5)/20*(sqrt(5)+2)^n. (End)
a(n+1) = F(2) + F(5) + F(8) + ... + F(3n+2).
a(n) = 5*a(n-1) - 3*a(n-2) - a(n-3), a(0)=0, a(1)=1, a(2)= 6. Observation by G. Detlefs. See the W. Lang link. - Wolfdieter Lang, Oct 18 2010
a(2*n) = A077259(2*n); a(2*n+1) = A294262(2*n+1). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019
E.g.f.: exp(x)*(exp(x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x)) - 5)/10. - Stefano Spezia, May 24 2024

A214984 Array: T(m,n) = (F(m) + F(2m) + ... + F(nm))/F(m), by antidiagonals, where F = A000045 (Fibonacci numbers).

Original entry on oeis.org

1, 2, 1, 4, 4, 1, 7, 12, 5, 1, 12, 33, 22, 8, 1, 20, 88, 94, 56, 12, 1, 33, 232, 399, 385, 134, 19, 1, 54, 609, 1691, 2640, 1487, 342, 30, 1, 88, 1596, 7164, 18096, 16492, 6138, 872, 48, 1, 143, 4180, 30348, 124033, 182900, 110143, 25319, 2256, 77, 1

Offset: 1

Clark Kimberling, Oct 28 2012

col 1: A001612 (except for initial term)
row 1: A000071
row 2: A027941
row 3: A049652
row 4: A092521
row 6: A049664
row 8: A156093 without minus signs

			Northwest corner:
1...2....4.....7......12......20
1...4....12....33.....88......232
1...5....22....94.....399.....1691
1...8....56....385....2640....18096
1...12...134...1487...16492...182900

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A214978, A214985, A214986.

F[n_] := Fibonacci[n]; L[n_] := LucasL[n];
t[m_, n_] := (1/F[m])*Sum[F[m*k], {k, 1, n}]
TableForm[Table[t[m, n], {m, 1, 10}, {n, 1, 10}]]
Flatten[Table[t[k, n + 1 - k], {n, 1, 12}, {k, 1, n}]]

For odd-numbered rows (m odd):
T(m,n) = (F(m*n+m) + F(m*n) - F(m))/(F(m)*L(m)).
For even-numbered rows (m even):
T(m,n) = (F(m*n+m) - F(m*n) - F(m))/(F(m)*(L(m)-2)).

A214986 Power ceiling array for the golden ratio, by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 7, 8, 5, 1, 1, 12, 21, 22, 7, 1, 1, 20, 55, 94, 48, 12, 1, 1, 33, 144, 399, 329, 134, 18, 1, 1, 54, 377, 1691, 2255, 1487, 323, 30, 1, 1, 88, 987, 7164, 15456, 16492, 5796, 872, 47, 1, 1, 143, 2584, 30348, 105937, 182900

Offset: 1

Clark Kimberling, Oct 28 2012

row 0: A000012 ... row 6: A049660
row 1: A000071 ... row 8: A049668
row 2: A001906 ... col 0: A000012
row 3: A049652 ... col 1: A169986
row 4: A004187
For x>1, define c(x,0) = 1 and c(x,n) = ceiling(x*c(x,n-1)) for n>0.  Row m of A214986 is the sequence c(r^m,n), where r = golden ratio = (1 + sqrt(5))/2.  The name of the array corresponds to the power ceiling function f(x) = limit of c(x,n)/x^n as n increases without bound; f(x) generalizes the case for x = 3/2 as described under "Power Ceilings" at MathWorld.  For a graph of f(x), see the Mathematica program at A083286.
The term "power ceiling sequence" extends to sequences generated by recurrences P(n) = ceiling(x*P(n-1)) + g(n), and "power ceiling functions" f(x) to the limit of P(n)/x^n in case x>1 and g(n)/x^n -> 0.
Suppose that h is a nonnegative integer and g(n) is a constant.  If x is a positive integer power of the golden ratio r, then f(x), in many cases, lies in the field Q(sqrt(5)).  Examples matching rows of A214986, using g(n) = 0, follow:
...
x ... P ........ f(x)
r ... A000071 .. (5 + 2*sqrt(5))/2 = 1.8944... (A010532)
r^2 . A001906 .. (5 + 3*sqrt(5))/10 = 1.7082...(A176015)
r^3 . A049652 .. (25 + 11*sqrt(5))/40 = 1.2399...
r^4 . A004187 .. (15 + 7*sqrt(5))/10 = 1.0219...
...
If k is odd, then f(r^k) = r^k((b(k) + c(k))/d(k)), where
b(k) = L(j)^2 + L(j-1)^2, where j=[(k+1)/2], L=A000032 (Lucas numbers); c(k) = (L(k)+2)*sqrt(5); d(k) = 10*F(k)*L(k), where F=A000045 (Fibonacci numbers).  If k is even, then f(r^k) = r^k/(F(k)*sqrt(5)).

			Northwest corner:
1...1....1.....1......1.......1
1...2....4.....7......12......20
1...3....8.....21.....55......144
1...5....22....94.....399.....1691
1...7....48....329....2255....15456
1...19...134...1487...16492...182900

Clark Kimberling, Antidiagonals n = 1..35, flattened
Eric Weisstein's World of Mathematics, Power Ceilings

Cf. A000045, A214978, A214984, A214987.

r = GoldenRatio;
s[x_, 0] := 1; s[x_, n_] := Ceiling[x*s[x, n - 1]];
t = TableForm[Table[s[r^m, n], {m, 0, 10}, {n, 0, 10}]  ]
u = Flatten[Table[s[r^m, n - m], {n, 0, 10}, {m, 0, n}]]

The odd-numbered rows of A214986 are even-numbered rows of A213978; the even-numbered rows of A214986 are odd-numbered rows of A214984.

A181661 Upper Beatty array of the golden ratio, (1+sqrt(5))/2.

Original entry on oeis.org

1, 2, 2, 6, 5, 3, 23, 17, 7, 4, 95, 68, 24, 10, 5, 400, 284, 95, 35, 13, 6, 1692, 1199, 396, 141, 46, 15, 7, 7165, 5075, 1671, 590, 186, 53, 18, 8, 30349, 21494, 7072, 2492, 778, 214, 64, 20, 9, 128558, 91046, 29951, 10549, 3286, 896, 259, 71, 23, 10, 544578

Offset: 1

Clark Kimberling, Nov 18 2010

(row 1)=-1+A049652.
(column 1)=A000027.
(column 2)=A001950=(u(n)), or simply u.
(column 3)=u(u(n))+l(l(n)), or simply uu+ll.
(column 4)=u(uu+ll)+l(ul+lu),
whereas Column 4 of the lower Beatty array
is u(ul+lu)+l(uu+ll).
U(n,k)-L(n,k)=n for n>=1, k>=0.

			Northwest corner of the array:
  1     2     6    23    95    400 ...
  2     5    17    68   284   1199 ...
  3     7    24    95   396   1671 ...
  4    10    35   141   590   2492 ...

Cf. A181886, A000201, A001950, A000045.

Here we introduce Beatty arrays.  Suppose that
(u(1),u(2),...) and (l(1),l(2),...) are the Beatty
sequences of positive real numbers r and s=r/(1-r), where
r=1, let
U(n,0)=n, U(n,1)=u(1), L(n,0)=0, L(n,1)=l(1),
and for k>=2 let x=floor(r*u(k-1)), y=floor(r*l(k-1)),
a=x+u(k-1), b=x, c=y+l(k-1), d=y,
U(n,k)=a+d, L(n,k)=b+c.  We call U and L the upper and
lower Beatty arrays of r (and of s).  Note that
U(n,k)-L(n,k)=U(n,1)-L(n,1) for all n>=1 and k>=1.

A082574 a(1)=1, a(n) = ceiling(r(3)a(n-1)) where r(3) = (1/2)(3 + sqrt(13)) is the positive root of X^2 = 3*X + 1.

Original entry on oeis.org

1, 4, 14, 47, 156, 516, 1705, 5632, 18602, 61439, 202920, 670200, 2213521, 7310764, 24145814, 79748207, 263390436, 869919516, 2873148985, 9489366472, 31341248402, 103513111679, 341880583440, 1129154862000, 3729345169441, 12317190370324

Offset: 1

Benoit Cloitre, May 06 2003

nonn

More generally the sequence a(1)=1, a(n) = ceiling(r(z)*a(n-1)) where r(z) = (1/2)*(z + sqrt(z^2 + 4)) is the positive root of X^2 = z*X + 1 satisfies the linear recurrence: for n > 3, a(n) = (z+1)*a(n-1) - (z-1)*a(n-2) - a(n-3) and the closed-form formula: a(n) = floor(t(z)*r(z)^n) where t(z) = (1/(2*z))*(1+(z+2)/sqrt(z^2+4)) is the positive root of z*(z^2 + 4)*X^2 = (z^2 + 4)*X + 1.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Ignas Gasparavičius, Andrius Grigutis, and Juozas Petkelis, Picturesque convolution-like recurrences and partial sums' generation, arXiv:2507.23619 [math.NT], 2025. See p. 24.
Index entries for linear recurrences with constant coefficients, signature (4,-2,-1).

Cf. A000071, A048739, A049652.

I:=[1,4,14]; [n le 3 select I[n] else 4*Self(n-1)-2*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 12 2017

a:=n->sum(fibonacci(i,3), i=0..n): seq(a(n), n=1..30); # Zerinvary Lajos, Mar 20 2008

LinearRecurrence[{4, -2, -1}, {1, 4, 14}, 30] (* Vincenzo Librandi, Sep 12 2017 *)
Table[Sum[Fibonacci[k, 3], {k,0,n}], {n,1,30}] (* G. C. Greubel, May 31 2019 *)

Vec(1/((1-x)*(1-3*x-x^2)) + O(x^30)) \\ Michel Marcus, Sep 12 2017

(1/((1-x)*(1-3*x-x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 31 2019

a(1)=1, a(2)=4, a(3)=14, a(n) = 4*a(n-1) - 2*a(n-2) - a(n-3).
a(n) = floor(t(3)*r(3)^n) where t(3) = (1/6)*(1 + 5/sqrt(13)) is the positive root of 39*X^2 = 13*X + 1.
G.f.: 1/((1-x)*(1-3*x-x^2)). Partial sums of A006190. - Paul Barry, Jul 10 2004

A181886 Lower Beatty array of the golden ratio, (1+sqrt(5))/2.

Original entry on oeis.org

0, 1, 0, 5, 3, 0, 22, 15, 4, 0, 94, 66, 21, 6, 0, 399, 282, 92, 31, 8, 0, 1691, 1197, 393, 137, 41, 9, 0, 7164, 5073, 1668, 586, 181, 47, 11, 0, 30348, 21492, 7069, 2488, 773, 208, 57, 12, 0, 128557, 91044

Offset: 1

Clark Kimberling, Nov 19 2010

Upper and lower Beatty arrays are defined at A181661.
(row 1)=A049652.
(column 1)=A000004.
(column 2)=A000201 (lower Wythoff sequence).
(column 3)=u(l(n))+l(u(n)), or simply ul+lu.
(column 4)=u(ul+lu)+l(uu+ll),
whereas column 4 of the upper Beatty array is
u(uu+ll)+l(ul+lu).
U(n,k)-L(n,k)=n for n>=1, k>=0.

			Northwest corner of the array:
0....1....5....22....94....399...
0....3...15....66...282...1197...
0....4...21....92...393...1668...
0....6...31...137...586...2488...

Cf. A181661, A000201, A001950, A000045.

A214985 Array: T(m,n) = (F(n) + F(2n) + ... + F(nm))/F(n), by antidiagonals; transpose of A214984.

Original entry on oeis.org

1, 1, 2, 1, 4, 4, 1, 5, 12, 7, 1, 8, 22, 33, 12, 1, 12, 56, 94, 88, 20, 1, 19, 134, 385, 399, 232, 33, 1, 30, 342, 1487, 2640, 1691, 609, 54, 1, 48, 872, 6138, 16492, 18096, 7164, 1596, 88, 1, 77, 2256, 25319, 110143, 182900, 124033, 30348, 4180, 143

Offset: 1

Clark Kimberling, Oct 28 2012

row 1: A001612 (except for initial term)
col 1: A000071
col 2: A027941
col 3: A049652
col 4: A092521
col 6: A049664
col 8: A156093 without minus signs

			Northwest corner:
1....1.....1......1.......1
2....4.....5......8.......12
4....12....22.....56......134
7....33....94.....385.....1487
12...88....399....2640....16492
20...232...1691...18096...182900

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A214978, A214984, A214986.

F[n_] := Fibonacci[n]; L[n_] := LucasL[n];
t[m_, n_] := (1/F[n])*Sum[F[k*n], {k, 1, m}]
TableForm[Table[t[m, n], {m, 1, 10}, {n, 1, 10}]]
Flatten[Table[t[k, n + 1 - k], {n, 1, 12}, {k, 1, n}]]

For odd-numbered columns (m odd):
T(m,n) = (F(m*n+m) + F(m*n) - F(m))/(F(m)*L(m)).
For even-numbered columns (m even):
T(m,n) = (F(m*n+m) - F(m*n) - F(m))/(F(m)*(L(m)-1)).

cp's OEIS Frontend

A001076 Denominators of continued fraction convergents to sqrt(5).

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A214992 Power ceiling-floor sequence of (golden ratio)^4.

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A099919 a(n) = F(3) + F(6) + F(9) + ... + F(3n), F(n) = Fibonacci numbers A000045.

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A049651 a(n) = (F(3*n+1) - 1)/2, where F=A000045 (the Fibonacci sequence).

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A214984 Array: T(m,n) = (F(m) + F(2*m) + ... + F(n*m))/F(m), by antidiagonals, where F = A000045 (Fibonacci numbers).

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A214986 Power ceiling array for the golden ratio, by antidiagonals.

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A214984 Array: T(m,n) = (F(m) + F(2m) + ... + F(nm))/F(m), by antidiagonals, where F = A000045 (Fibonacci numbers).

A082574 a(1)=1, a(n) = ceiling(r(3)a(n-1)) where r(3) = (1/2)(3 + sqrt(13)) is the positive root of X^2 = 3*X + 1.

A214985 Array: T(m,n) = (F(n) + F(2n) + ... + F(nm))/F(n), by antidiagonals; transpose of A214984.