A084534 Triangle read by rows: row #n has n+1 terms. T(n,0)=1, T(n,n)=2, T(n,m) = T(n-1,m-1) + Sum_{k=0..m} T(n-1-k,m-k).
1, 1, 2, 1, 4, 2, 1, 6, 9, 2, 1, 8, 20, 16, 2, 1, 10, 35, 50, 25, 2, 1, 12, 54, 112, 105, 36, 2, 1, 14, 77, 210, 294, 196, 49, 2, 1, 16, 104, 352, 660, 672, 336, 64, 2, 1, 18, 135, 546, 1287, 1782, 1386, 540, 81, 2, 1, 20, 170, 800, 2275, 4004, 4290, 2640, 825, 100, 2
Offset: 0
Examples
First few Chebyshev T(2*n,x) polynomials: T(2*0,x) = 1; T(2*1,x) = x^2 - 2; T(2*2,x) = x^4 - 4*x^2 + 2; T(2*3,x) = x^6 - 6*x^4 + 9*x^2 - 2; T(2*4,x) = x^8 - 8*x^6 + 20*x^4 - 16*x^2 + 2; T(2*5,x) = x^10 - 10*x^8 + 35*x^6 - 50*x^4 + 25*x^2 - 2; Triangle begins as: 1; 1, 2; 1, 4, 2; 1, 6, 9, 2; 1, 8, 20, 16, 2; 1, 10, 35, 50, 25, 2; 1, 12, 54, 112, 105, 36, 2;
References
- I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. See p. 118.
- Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990. p. 37, eq.(1.96) and p. 4. eq.(1.10).
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
- I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. [Scan of annotated copy]
Crossrefs
Programs
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Magma
A084534:= func< n,k | k eq 0 select 1 else 2*(n/k)*Binomial(2*n-k-1, k-1) >; [A084534(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 02 2022
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Maple
T := proc(n, m): if n=0 then 1 else binomial(2*n-m, m)*2*n/(2*n-m) fi: end: seq(seq(T(n,m),m=0..n),n=0..10); # Johannes W. Meijer, May 31 2018
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Mathematica
a[n_, m_] := Binomial[2n-m, m]*2n/(2n-m); a[0, 0] = 1; Table[a[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Apr 12 2016, after Wolfdieter Lang *) -
PARI
T(n,m) = if(n==0, m==0, binomial(2*n-m, m)*2*n/(2*n-m)) \\ Andrew Howroyd, Dec 18 2017
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Sage
def A084534(n,k): return 1 if (k==0) else 2*(n/k)*binomial(2*n-k-1, k-1) flatten([[A084534(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 02 2022
Formula
T(n,m) = binomial(2*n-m, m)*2*n/(2*n-m) for n > 0. - Andrew Howroyd, Dec 18 2017
Signed version from Wolfdieter Lang, Mar 07 2007: (Start)
a(n,m)=0 if n
a(n,m)=0 if n
a(n,m)=0 if nA127674(n,n-m)/2^(2*(n-m)-1) (scaled coefficients of Chebyshev's T(2*n,x), decreasing even powers). [corrected by Johannes W. Meijer, May 31 2018] (End)
Extensions
Edited by Don Reble, Nov 12 2005
A178482 Phi-antipalindromic numbers.
1, 3, 4, 7, 8, 10, 11, 18, 19, 21, 22, 25, 26, 28, 29, 47, 48, 50, 51, 54, 55, 57, 58, 65, 66, 68, 69, 72, 73, 75, 76, 123, 124, 126, 127, 130, 131, 133, 134, 141, 142, 144, 145, 148, 149, 151, 152, 170, 171, 173, 174
Offset: 1
Comments
We call m a phi-antipalindromic number if for the vector (a,...,b) (a<...=2, either a(n)+1 or a(n)-1 is in the sequence; also either a(n)+3 or a(n)-3 is in the sequence.
Conjecture: this is the sequence of numbers k for which f(k) is an integer, where f(x) is the change-of-base function defined at A214969 using b=phi and c=b^2. - Clark Kimberling, Oct 17 2012
There is a 21-state automaton accepting the Zeckendorf representations of those n in this sequence. - Jeffrey Shallit, May 03 2023
Kimberling's conjecture has been proven by Ingrid Vukusic and myself. Along the way we prove an alternate characterization of the sequence: they are the positive integers whose base-phi expansion consists only of even exponents of phi. - Jeffrey Shallit, Aug 28 2025
Alternatively, this sequence consists of those numbers k such that either k or k-1 can be written as the (possibly empty) sum of distinct Lucas numbers L_i where i>=2 and i is even. - Jeffrey Shallit, Aug 28 2025
Examples
The vectors of exponents of 4 and 5 are (-2,0,2) and (-4,-1,3) correspondingly (cf.A104605). Therefore by definition 4 is a phi-antipalindromic number, while 5 is not. Let n=38. Then k=5. Thus a(38)=A005248(5)+a(6)=123+10=133. The vector of exponents of phi in the base-phi expansion of 133 is (-10,-4,-2,2,4,10).
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..3071 from R. J. Mathar)
- Jeffrey Shallit, Proving Properties of phi-Representations with the Walnut Theorem-Prover, arXiv:2305.02672 [math.NT], 2023.
Crossrefs
Programs
-
Mathematica
phiAPQ[1] = True; phiAPQ[n_] := Module[{d = RealDigits[n, GoldenRatio, 2*Ceiling[Log[GoldenRatio, n]]]}, e = d[[2]] - Flatten @ Position[d[[1]], 1]; Reverse[e] == -e]; Select[Range[200], phiAPQ] (* Amiram Eldar, Apr 23 2020 *)
Formula
For k>=1, a(2^k)=A005248(k); if 2^k
A180664 Golden Triangle sums: a(n) = a(n-1) + A001654(n+1) with a(0)=0.
0, 2, 8, 23, 63, 167, 440, 1154, 3024, 7919, 20735, 54287, 142128, 372098, 974168, 2550407, 6677055, 17480759, 45765224, 119814914, 313679520, 821223647, 2149991423, 5628750623, 14736260448, 38580030722, 101003831720
Offset: 0
Comments
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).
Programs
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Magma
[(1/10)*((-1)^n - 15 + 2*Lucas(2*n+4)): n in [0..40]]; // G. C. Greubel, Jan 21 2022
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Maple
nmax:=26: with(combinat): for n from 0 to nmax+1 do A001654(n):=fibonacci(n) * fibonacci(n+1) od: a(0):=0: for n from 1 to nmax do a(n) := a(n-1)+A001654(n+1) od: seq(a(n),n=0..nmax);
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Mathematica
Table[Sum[Fibonacci[i+2]*Fibonacci[i+3], {i,0,n-1}], {n,0,40}] (* Rigoberto Florez, Jul 07 2020 *) LinearRecurrence[{3,0,-3,1},{0,2,8,23},30] (* Harvey P. Dale, Mar 30 2023 *) -
Sage
[(1/10)*((-1)^n - 15 + 2*lucas_number2(2*n+4,1,-1)) for n in (0..40)] # G. C. Greubel, Jan 21 2022
Formula
a(n+1) = Sum_{k=0..n} A180662(2*n-k+2, k+2).
a(n) = (-15 + (-1)^n + (6-2*A)*A^(-n-1) + (6-2*B)*B^(-n-1))/10 with A=(3+sqrt(5))/2 and B=(3-sqrt(5))/2.
G.f.: (2*x+2*x^2-x^3)/(1-3*x-x^4+3*x^3).
a(n) = Sum_{i=0..n-1} F(i+2)*F(i+3), where F(i) = A000045(i). - Rigoberto Florez, Jul 07 2020
a(n) = (1/10)*((-1)^n - 15 + 2*Lucas(2*n+4)). - G. C. Greubel, Jan 21 2022
A198633 Total number of round trips, each of length 2*n on the graph P_3 (o-o-o).
3, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648
Offset: 0
Comments
See the array and triangle A198632 for the general case for the graph P_N (there N is n and the length is l=2*k).
Examples
With the graph P_3 as 1-2-3: n=0: 3, from the length 0 walks starting at 1, 2 and 3. n=2: 8, from the walks of length 4, namely 12121, 12321, 21212, 23232, 21232, 23212, 32323 and 32123.
Links
- Index entries for linear recurrences with constant coefficients, signature (2).
Programs
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Mathematica
Join[{3},NestList[2#&,4,30]] (* Harvey P. Dale, Nov 07 2020 *) -
PARI
a(n)=if(n,2<
Formula
a(n) = w(3,2*n), n>=0, with w(3,l) the total number of closed walks on the graph P_3 (the simple path with 3 points (vertices) and 2 lines (or edges)).
O.g.f. for w(3,l) (with zeros for odd l): y*(d/dy)S(3,y)/S(3,y) with y=1/x and Chebyshev S-polynomials (coefficients A049310). See A198632, also for a rewritten form.
Empirical g.f.: (3-2*x)/(1-2*x). - Colin Barker, Jan 02 2012
This g.f. follows from the Chebyshev o.g.f. given above with x -> sqrt(x). Therefore a(0) = 3 and a(n) = 2^(n+1), n >= 1. - Wolfdieter Lang, Feb 18 2013.
A032170 "CHK" (necklace, identity, unlabeled) transform of 1, 2, 3, 4, ...
1, 2, 5, 10, 24, 50, 120, 270, 640, 1500, 3600, 8610, 20880, 50700, 124024, 304290, 750120, 1854400, 4600200, 11440548, 28527320, 71289000, 178526880, 447910470, 1125750120, 2833885800, 7144449920, 18036373140, 45591631800, 115381697740, 292329067800
Offset: 1
Keywords
Comments
Apparently, for n > 2, the same as A072337. - Ralf Stephan, Feb 01 2004
a(n) is the number of prime period-n periodic orbits of Arnold's cat map. - Bruce Boghosian, Apr 26 2009
From Petros Hadjicostas, Nov 17 2017: (Start)
A first proof of the g.f., given below, can be obtained using the first of Vladeta Jovovic's formulae. If b(n) = A004146(n), then B(x) = Sum_{n >= 1} b(n)*x^n = x*(1 + x)/((1 - x)*(1 - 3*x + x^2)) (see the documentation for sequence A004146). From Jovovich's first formula, A(x) = Sum_{n >= 1} a(n)*x^n = Sum_{n >= 1} (1/n)*Sum_{d | n} mu(d)*b(n/d)*x^n. Letting m = n/d, we get A(x) = Sum_{d >= 1} (mu(d)/d)*Sum_{m >= 1} b(m)*(x^d)^m/m = Sum_{d >= 1} (mu(d)/d)*f(x^d), where f(y) = Sum_{m >= 1} b(m)*y^m/m = int(B(w)/w, w = 0..y) = int((1 + w)/((1 - w)*(1 - 3*w + w^2)), w = 0..y) = log((1 - y)^2/(1 - 3*y + y^2)) for |y| < (3 - sqrt(5))/2.
A second proof of the g.f. can be obtained using C. G. Bower's definition of the CHK transform of a sequence (e(n): n>=1) with g.f. E(x) (see the links below). If (c_k(n): n >= 1) = CHK_k(e(n): n >= 1), then (c_k(n): n >= 1) = (1/k)*(MOEBIUS*AIK)k (e_n: n >= 1) = (1/k)*Sum{d | gcd(n,k)} mu(d)*AIK_{k/d}(e(n/d): n multiple of d), where the * between MOEBIUS and AIK denotes Dirichlet convolution and (d_k(n): n >= 1) = AIK_k(e(n): n >= 1) has g.f. E(x)^k. (There is a typo in the given definition of CHK in the link.)
If C(x) is the g.f. of CHK(e(n): n >= 1) = Sum_{k = 1..n} CHK_k(e(n): n >= 1), then C(x) = Sum_{n>=1} Sum_{k = 1..n} c_k(n)*x^n = Sum_{k >= 1} (1/k) Sum_{n >= k} Sum_{d | gcd(n,k)} mu(d)*d_{k/d}(n/d)*x^n. Letting m = n/d and s = k/d and using the fact that E(0) = 0, we get C(x) = Sum_{d >= 1} (mu(d)/d)*Sum_{s >= 1} (1/s)*Sum_{m >= s} d_s(m)*(x^d)^m = Sum_{d >= 1} (mu(d)/d)*Sum_{s >= 1} E(x^d)^s. Thus, C(x) = -Sum_{d >= 1} (mu(d)/d)*log(1 - E(x^d)).
For the sequence (e(n): n >= 1) = (n: n >= 1), we have E(x) = Sum_{n>=1} n*x^n = x/(1 - x)^2, and thus A(x) = C(x) = -Sum_{d >= 1} (mu(d)/d)*log(1 - x/(1-x)^2), from which we can easily get the g.f. given in the formula section.
Apparently, for this sequence and for sequences A032165, A032166, A032167, the author assumes that C(0) = 0 (i.e., he assumes the CHK transform has no constant term), while for sequences A032164, A108529, and possibly others, he assumes that the CHK transform starts with the constant term 1 (i.e., he assumes C(x) = 1 - Sum_{d >= 1} (mu(d)/d)*log(1 - E(x^d))). (End)
From Petros Hadjicostas, Jul 13 2020: (Start)
We elaborate further on Michel Marcus's claim below. Consider his sequence (b(n): n >= 1) with b(1) = 3 and b(n) = a(n) for n >= 2.
Using the identity -Sum_{k >= 1} (mu(k)/k)*log(1 - x^k) = x for |x| < 1 and the g.f. of (a(n): n >= 1) below, we see that Sum_{n >= 1} b(n)*x^n = 3*x - a(1)*x + Sum_{n >= 1} a(n)*x^n = 2*x + Sum_{k >= 1} (mu(k)/k)*(2*log(1 - x^k) - log(1 - 3*x^k + x^(2*k))) = -Sum_{k >= 1} (mu(k)/k)*log(1 - 3*x^k + x^(2*k)).
Following Kam Cheong Au (2020), let d(w,N) be the dimension of the Q-span of weight w and level N of colored multiple zeta values (CMZV). Here Q are the rational numbers.
Deligne's bound says that d(w,N) <= D(w,N), where 1 + Sum_{w >= 1} D(w,N)*t^w = (1 - a*t + b*t^2)^(-1) when N >= 3, where a = phi(N)/2 + omega(N) and b = omega(N) - 1 (with omega(N) being the number of distinct primes of N).
For N = 6, a = phi(6)/2 + omega(6) = 2/2 + 2 = 3 and b = omega(6) - 1 = 1. It follows that D(w, N=6) = A001906(w+1) = Fibonacci(2*(w+1)).
For some reason, Kam Cheong Au (2020) assumes Deligne's bound is tight, i.e., d(w,N) = D(w,N). He sets Sum_{w >= 1} c(w,N)*t^w = log(1 + Sum_{w >= 1} d(w,N)*t^w) = log(1 + Sum_{w >= 1} D(w,N)*t^w) = -log(1 - a*t + b*t^2) for N >= 3.
For N = 6, we get that c(w, N=6) = A005248(w)/w.
He defines d*(w,N) = Sum_{k | w} (mu(k)/k)*c(w/k,N) to be the "number of primitive constants of weight w and level N". (Using the terminology of A113788, we may perhaps call d*(w,N) the number of irreducible colored multiple zeta values at weight w and level N.)
Using standard techniques of the theory of g.f.'s, we can prove that Sum_{w >= 1} d*(w,N)*t^w = Sum_{s >= 1} (mu(s)/s) Sum_{k >= 1} c(k,N)*(t^s)^k = -Sum_{s >= 1} (mu(s)/s)*log(1 - a*t^s + b*t^(2*s)).
For N = 6, we saw that a = 3 and b = 1, and hence d*(w, N=6) = b(w) for w >= 1 (as claimed by Michel Marcus below). See Table 1 on p. 6 in Kam Cheong Au (2020). (End)
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..2400
- Kam Cheong Au, Evaluation of one-dimensional polylogarithmic integral, with applications to infinite series, arXiv:2007.03957 [math.NT], 2020. See line N = 6 in Table 1 (p. 6).
- C. G. Bower, Transforms (2).
- Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
- Eric Weisstein's World of Mathematics, Arnold's cat map.
- Wikipedia, Arnold's cat map.
- Index entries for sequences related to Lyndon words
Crossrefs
Programs
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Mathematica
Table[DivisorSum[n, MoebiusMu[n/#] (LucasL[2 #] - 2) &]/n, {n, 31}] (* Michael De Vlieger, Nov 18 2017 *)
Formula
a(n) = (1/n)*Sum_{d | n} mu(n/d)*A004146(d). - Vladeta Jovovic, Feb 15 2003
Inverse EULER transform of Fibonacci(2*n). - Vladeta Jovovic, May 04 2006
G.f.: Sum_{n >= 1} (mu(n)/n)*f(x^n), where f(y) = log((1 - y)^2/(1 - 3*y + y^2)). - Petros Hadjicostas, Nov 17 2017
It appears that the sequence b(1) = 3, b(n) = a(n) for n >= 2 is related to the rational sequence (c(w, N=6): w >= 1) = (A005248(w)/w: w >= 1) whose g.f. is log(1/(1 - a*t + b*t^2)), where a = phi(N)/2 + omega(N) and b = omega(N) - 1 when N = 6, where phi is A000010 and omega is A001221. See Kam Cheong Au (2020). - Michel Marcus, Jul 13 2020 [Edited by Petros Hadjicostas, Jul 13 2020]
A067902 a(n) = 14*a(n-1) - a(n-2); a(0) = 2, a(1) = 14.
2, 14, 194, 2702, 37634, 524174, 7300802, 101687054, 1416317954, 19726764302, 274758382274, 3826890587534, 53301709843202, 742397047217294, 10340256951198914, 144021200269567502, 2005956546822746114, 27939370455248878094, 389145229826661547202, 5420093847118012782734
Offset: 0
Comments
Solves for x in x^2 - 3*y^2 = 4. [Complete nonnegative solutions are in A003500 and A052530. - Wolfdieter Lang, Sep 05 2021]
For n>0, a(n)+2 is the number of dimer tilings of a 4 X 2n Klein bottle (cf. A103999).
This is the Lucas sequence V(14,1). In addition to the comment above: If x = a(n) then y(n) = (a(n+1) - a(n-1))/24, n >= 1. - Klaus Purath, Aug 17 2021
Examples
G.f. = 2 + 14*x + 194*x^2 + 2702*x^3 + 37634*x^4 + 524174*x^5 + ...
Links
Programs
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GAP
m:=7;; a:=[2,14];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
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Magma
[Floor((2+Sqrt(3))^(2*n)+(1+Sqrt(3))^(-n)): n in [0..19]]; // Vincenzo Librandi, Mar 31 2011
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Maple
a := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(14) fi: 14*a(n-1)-a(n-2): end: for n from 0 to 30 do printf(`%d,`,a(n)) od: seq( simplify(2*ChebyshevT(n, 7)), n=0..20); # G. C. Greubel, Dec 23 2019
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Mathematica
a[0]=2; a[1]=14; a[n_]:= 14a[n-1] -a[n-2]; Table[a[n], {n,0,20}] (* Robert G. Wilson v, Jan 30 2004 *) 2*ChebyshevT[Range[21] -1, 7] (* G. C. Greubel, Dec 23 2019 *) -
PARI
vector( 21, n, 2*polchebyshev(n-1, 1, 7) ) \\ G. C. Greubel, Dec 23 2019
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Sage
[lucas_number2(n,14,1) for n in range(0,20)] # Zerinvary Lajos, Jun 26 2008
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Sage
[2*chebyshev_T(n,7) for n in (0..20)] # G. C. Greubel, Dec 23 2019
Formula
G.f.: 2*(1-7*x)/(1-14*x+x^2). - N. J. A. Sloane, Nov 22 2006
a(n) = p^n + q^n, where p = 7 + 4*sqrt(3) and q = 7 - 4*sqrt(3). - Tanya Khovanova, Feb 06 2007
a(n) = 2*A011943(n+1). - R. J. Mathar, Sep 27 2014
From Peter Bala, Oct 16 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 7 - 4*sqrt(3). This sequence gives the partial denominators in the simple continued fraction expansion of 1 + F(alpha) = 2.07140228197873197080... = 2 + 1/(14 + 1/(194 + 1/(2702 + ...))). Cf. A005248.
12*Sum_{n >= 1} 1/(a(n) - 16/a(n)) = 1.
16*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 12/a(n)) = 1.
Series acceleration formula for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/12 - 16*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 16)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(7-4*sqrt(3)))^2 - 1 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499.
(End)
From Klaus Purath, Aug 17 2021: (Start)
a(n) = (a(n-1)*a(n-2) + 2688)/a(n-3), n >= 3.
a(n) = (a(n-1)^2 + 192)/a(n-2), n >= 2.
a(2*n+1) = 14*A302332(n). (End)
a(n) = A003500(2*n) = S(2*n,4) - S(2*n-2, 4) = 2*T(2*n,2), for n >= 0, with Chebyshev S and T. S(n, 4) = A001353(n+1) and T(n, 2) = A001075(n). - Wolfdieter Lang, Sep 06 2021
A100545 Expansion of (7-2*x) / (1-3*x+x^2).
7, 19, 50, 131, 343, 898, 2351, 6155, 16114, 42187, 110447, 289154, 757015, 1981891, 5188658, 13584083, 35563591, 93106690, 243756479, 638162747, 1670731762, 4374032539, 11451365855, 29980065026, 78488829223, 205486422643, 537970438706, 1408424893475, 3687304241719, 9653487831682, 25273159253327
Offset: 0
Comments
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Mark W. Coffey, James L. Hindmarsh, Matthew C. Lettington, John Pryce, On Higher Dimensional Interlacing Fibonacci Sequences, Continued Fractions and Chebyshev Polynomials, arXiv:1502.03085 [math.NT], 2015 (see p. 31).
- Tanya Khovanova, Recursive Sequences
- Index entries for linear recurrences with constant coefficients, signature (3,-1).
Programs
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GAP
List([0..30], n-> Fibonacci(2*n+4) +Lucas(1,-1,2*n+3)[2] ); # G. C. Greubel, Jan 17 2020
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Magma
[Fibonacci(2*n+4) +Lucas(2*n+3): n in [0..30]]; // G. C. Greubel, Jan 17 2020
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Maple
F := proc(n) combinat[fibonacci](n) ; end: A100545 := proc(n) 4*F(2*(n+1)) + F(2*n+1)+F(2*n+3) ; end: for n from 0 to 30 do printf("%d,",A100545(n)) ; od ; # R. J. Mathar, Oct 26 2006
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Mathematica
Table[Fibonacci[2*(n+2)] + LucasL[2*n+3], {n,0,30}] (* G. C. Greubel, Jan 17 2020 *) -
PARI
Vec((7-2*x)/(1-3*x+x^2) + O(x^30)) \\ Michel Marcus, Feb 11 2015
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Sage
[fibonacci(2*n+4) +lucas_number2(2*n+3,1,-1) for n in (0..30)] # G. C. Greubel, Jan 17 2020
Formula
a(n-1) = 4*Fibonacci(2*n) + Fibonacci(2*n-1) + Fibonacci(2*n+1).
a(n) + a(n+1) = A055849(n+2).
a(n) = 3*a(n-1) - a(n-2) with a(0)=7 and a(1)=19. - Philippe Deléham, Nov 16 2008
a(n) = (2^(-1-n)*((3-sqrt(5))^n*(-17+7*sqrt(5)) + (3+sqrt(5))^n*(17+7*sqrt(5)))) / sqrt(5). - Colin Barker, Oct 14 2015
From G. C. Greubel, Jan 17 2020: (Start)
a(n) = Fibonacci(2*n+4) + Lucas(2*n+3).
E.g.f.: 2*exp(3*t/2)*(cosh(sqrt(5)*t/2) + (4/sqrt(5))*sinh(sqrt(5)*t/2)). (End)
Extensions
Corrected and extended by T. D. Noe and R. J. Mathar, Oct 26 2006
A153173 a(n) = L(5*n)/L(n) where L(n) = Lucas number A000204(n).
11, 41, 341, 2161, 15251, 103361, 711491, 4868641, 33391061, 228811001, 1568437211, 10749853441, 73681573691, 505018447961, 3461454668501, 23725145626561, 162614613425891, 1114577020834241, 7639424866266611
Offset: 1
Comments
All numbers in this sequence are congruent to 1 mod 10.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- L. Carlitz, Problem B-185, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 8, No. 3 (1970), p. 325; Lucas Ratio I, Solution to Problem B-185 by C. B. A. Peck, ibid., Vol. 9, No. 1 (1971), p. 109.
- L. Carlitz, Problem B-186, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 8, No. 3 (1970), p. 326; Lucas Ratio II, Solution to Problem B-186 by John Wessner, ibid., Vol. 9, No. 1 (1971), pp. 109-110.
- Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).
Programs
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Magma
I:=[11, 41, 341, 2161, 15251]; [n le 5 select I[n] else 5*Self(n-1)+15*Self(n-2)-15*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..30]]; // G. C. Greubel, Dec 21 2017
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Mathematica
Table[LucasL[5*n]/LucasL[n], {n, 1, 50}] CoefficientList[Series[x*(11-14*x-29*x^2+6*x^3+x^4)/((1-x)*(x^2-7*x+1)*(x^2+3*x+1)), {x,0,50}], x] (* G. C. Greubel, Dec 21 2017 *) a[ n_] := 1 + 5*Fibonacci[n]*Fibonacci[3*n]; (* Michael Somos, Apr 23 2022 *) -
PARI
{L(n)=fibonacci(n-1)+fibonacci(n+1)}; a(n) = L(5*n)/L(n) \\ Charles R Greathouse IV, Jun 11 2015 -
PARI
my(x='x+O('x^30)); Vec(x*(11-14*x-29*x^2+6*x^3+x^4 )/((1-x)*(x^2-7*x +1)*(x^2+3*x+1))) \\ G. C. Greubel, Dec 21 2017 -
PARI
{a(n) = 1 + 5*fibonacci(n)*fibonacci(3*n)}; /* Michael Somos, Apr 23 2022 */
Formula
From R. J. Mathar, Oct 22 2010: (Start)
a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: -x*(11-14*x-29*x^2+6*x^3+x^4)/( (x-1)*(x^2-7*x+1)*(x^2+3*x+1) ).
From Amiram Eldar, Feb 02 2022: (Start)
a(n) = Lucas(2*n)^2 - (-1)^n*Lucas(2*n) - 1 (Carlitz, Problem B-185).
a(n) = (Lucas(2*n) - 3*(-1)^n)^2 + (-1)^n*(5*Fibonacci(n))^2 (Carlitz, Problem B-186). (End)
a(n) = a(-n) = 1 + 10*A085695(n) = 5 + L(n-1)*L(n)^2*L(n+1) for all n in Z. - Michael Somos, Apr 23 2022
A195693 Decimal expansion of arctan(1/phi), where phi = (1+sqrt(5))/2 (the golden ratio).
5, 5, 3, 5, 7, 4, 3, 5, 8, 8, 9, 7, 0, 4, 5, 2, 5, 1, 5, 0, 8, 5, 3, 2, 7, 3, 0, 0, 8, 9, 2, 6, 8, 5, 2, 0, 0, 3, 5, 0, 2, 3, 8, 2, 2, 7, 0, 0, 7, 1, 6, 3, 2, 3, 3, 3, 8, 2, 6, 9, 6, 0, 3, 7, 1, 6, 8, 5, 5, 1, 6, 9, 4, 8, 8, 6, 8, 1, 3, 9, 7, 0, 0, 6, 7, 0, 8, 5, 6, 4, 3, 4, 3, 0, 8, 5, 3, 2, 0, 7
Offset: 0
Comments
Radian measure of half the smaller angle in the golden rhombus. - Eric W. Weisstein, Dec 11 2018
The angle between the diagonal and the longer side of a golden rectangle. - Amiram Eldar, May 18 2021
Examples
arctan(1/phi) = 0.5535743588970452515085327300892685200... . tan(0.5535743588970452515085327300...) = 1/(golden ratio). cot(0.5535743588970452515085327300...) = (golden ratio).
Links
- Paul S. Bruckman, Problem H-549, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 37, No. 1 (1999), p. 91; Resurrection, Solution to Problem H-549 by Charles K. Cook, ibid., Vol. 38, No. 2 (2000), pp. 191-192.
- Hei-Chi Chan, Machin-type formulas expressing Pi in terms of phi, The Fibonacci Quarterly, Vol. 46/47, No. 1 (2008/2009), pp. 32-37.
- Verner E. Hoggatt, Jr. and I. D. Bruggles, A Primer on the Fibonacci Sequence, Part V, The Fibonacci Quarterly, Vol. 2, No. 1 (1964), pp. 59-65.
- Eric Weisstein's World of Mathematics, Golden Rhombus.
- Index entries for transcendental numbers
Programs
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Mathematica
(See also A195692.) RealDigits[ArcCot[GoldenRatio], 10, 100][[1]] (* or *) RealDigits[(Pi - ArcTan[4/3])/4, 10, 100][[1]] (* Eric W. Weisstein, Dec 11 2018 *)
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PARI
atan(2)/2 \\ Michel Marcus, Feb 05 2022
Formula
Equals Pi/2 - A195723. - Amiram Eldar, May 18 2021
Equals arctan(2)/2. - Christoph B. Kassir, Dec 04 2021
From Amiram Eldar, Jan 11 2022: (Start)
Equals arccot(phi).
Equals (Pi - arctan(phi^5))/3.
Equals (Pi - arctan(4/3))/4.
Equals Sum_{k>=1} ((-1)^(k+1) * arctan(1/Fibonacci(2*k))) (Bruckman, 1999). (End)
Equals Sum_{k>=1} arctan(1/Lucas(2*k)) (Hoggatt and Bruggles, 1964). - Amiram Eldar, Feb 05 2022
A005247 a(n) = 3*a(n-2) - a(n-4), a(0)=2, a(1)=1, a(2)=3, a(3)=2. Alternates Lucas (A000032) and Fibonacci (A000045) sequences for even and odd n.
2, 1, 3, 2, 7, 5, 18, 13, 47, 34, 123, 89, 322, 233, 843, 610, 2207, 1597, 5778, 4181, 15127, 10946, 39603, 28657, 103682, 75025, 271443, 196418, 710647, 514229, 1860498, 1346269, 4870847, 3524578, 12752043, 9227465, 33385282, 24157817
Offset: 0
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- T. D. Noe, Table of n, a(n) for n = 0..500
- T. Crilly, Double sequences of positive integers, Math. Gaz., 69 (1985), 263-271.
- R. K. Guy, Letter to N. J. A. Sloane, Feb 1986
- 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
- Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1).
Programs
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Haskell
a005247 n = a005247_list !! n a005247_list = f a000032_list a000045_list where f (x::xs) (:y:ys) = x : y : f xs ys -- Reinhard Zumkeller, Dec 27 2012
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Magma
I:=[2,1,3,2]; [n le 4 select I[n] else 3*Self(n-2) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 21 2017
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Maple
with(combinat): A005247 := n-> if n mod 2 = 1 then fibonacci(n) else fibonacci(n+1)+fibonacci(n-1); fi; A005247:=-(z+1)*(3*z**2-z-1)/(z**2-z-1)/(z**2+z-1); # Simon Plouffe in his 1992 dissertation. Gives sequence with an additional leading 1.
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Mathematica
CoefficientList[Series[(2 + x - 3x^2 - x^3)/(1 - 3x^2 + x^4), {x, 0, 40}], x] LinearRecurrence[{0,3,0,-1},{2,1,3,2},50] (* Harvey P. Dale, Oct 10 2012 *) -
PARI
a(n)=if(n%2,fibonacci(n),fibonacci(n+1)+fibonacci(n-1))
Formula
a(0)=2, a(1)=1, a(2)=3, a(n) = (1+a(n-1)a(n-2))/a(n-3), n >= 3. a(-n) = a(n).
G.f.: (2+x-3*x^2-x^3)/((1-x-x^2)*(1+x-x^2))
a(n) = F(n) if n odd, a(n) = L(n) if n even. a(n) = F(n+1)+(-1)^nF(n-1). - Mario Catalani (mario.catalani(AT)unito.it), Sep 20 2002
a(n) = ((5+sqrt(5))/10)*(((1+sqrt(5))/2)^n+((-1+sqrt(5))/2)^n)+((5-sqrt(5))/10)*(((1-sqrt(5))/2)^n+((-1-sqrt(5))/2)^n). With additional leading 1: a(n)=((sqrt(5))/5)*(((1+sqrt(5))/2)^n-((1-sqrt(5))/2)^n)+((5+3*sqrt(5))/10)*((-1+sqrt(5))/2)^n+((5-3*sqrt(5))/10)*((-1-sqrt(5))/2)^n. - Tim Monahan, Jul 25 2011
From Peter Bala, Jan 11 2013: (Start)
Let phi = 1/2*(sqrt(5) - 1). This sequence is the simple continued fraction expansion of the real number 1 + product {n >= 0} (1 + sqrt(5)*phi^(4*n+1))/(1 + sqrt(5)*phi^(4*n+3)) = 2.77616 23282 02325 23857 ... = 2 + 1/(1 + 1/(3 + 1/(2 + 1/(7 + ...)))). Cf. A005248.
Furthermore, for k = 0,1,2,... the simple continued fraction expansion of 1 + product {n >= 0} (1 + 1/5^k*sqrt(5)*phi^(4*n+1))/(1 + 1/5^k*sqrt(5)*phi^(4*n+3)) equals [2; 1*5^k, 3, 2*5^k, 7, 5*5^k, 18, 13*5^k, 47, ...]. (End)
a(n) = hypergeom([(1-n)/2, n mod 2 - n/2], [1 - n], -4) for n > 2. - Peter Luschny, Sep 03 2019
E.g.f.: 2*cosh(x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Mar 15 2022
Extensions
Additional comments from Michael Somos, May 01 2000
Comments