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Binomial transform of Fibonacci(2n-1) ( A001519).
+20
13
1, 2, 5, 15, 50, 175, 625, 2250, 8125, 29375, 106250, 384375, 1390625, 5031250, 18203125, 65859375, 238281250, 862109375, 3119140625, 11285156250, 40830078125, 147724609375, 534472656250, 1933740234375, 6996337890625
FORMULA
G.f.: (1-3*x)/(1-5*x+5*x^2).
a(n) = (5-sqrt(5))*((5+sqrt(5))/2)^n/10 + (5+sqrt(5))*((5-sqrt(5))/2)^n/10.
a(2*n) = 5^n*Fibonacci(2*n-1), a(2*n+1) = 5^n*Lucas(2*n). - G. C. Greubel, Dec 27 2019 E.g.f.: (1/10)*exp((1/2)*(5-sqrt(5))*x)*(5 + sqrt(5) + (5 - sqrt(5))*exp(sqrt(5)*x)). - Stefano Spezia, Dec 28 2019
MAPLE
a:= n-> (<<0|1>, <-5|5>>^n. <<1, 2>>)[1, 1]:
MATHEMATICA
Table[If[EvenQ[n], 5^(n/2)*Fibonacci[n-1], 5^((n-1)/2)*LucasL[n-1]], {n, 0, 30}] (* G. C. Greubel, Dec 27 2019 *)
PROG
(Sage) [lucas_number2(n, 5, 5) for n in range(-1, 25)] # Zerinvary Lajos, Jul 08 2008 (PARI) my(x='x+O('x^30)); Vec((1-3*x)/(1-5*x+5*x^2)) \\ G. C. Greubel, Dec 27 2019 (Magma) I:=[1, 2]; [n le 2 select I[n] else 5*(Self(n-1) - Self(n-2)): n in [1..30]]; // G. C. Greubel, Dec 27 2019 (GAP) a:=[1, 2];; for n in [3..30] do a[n]:=5*(a[n-1]-a[n-2]); od; a; # G. C. Greubel, Dec 27 2019
Maximal index k of an odd Fibonacci number ( A001519) such that A001519(k) = Fibonacci(2k-1) <= n (the 'lower' odd Fibonacci Inverse).
+20
12
1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6
COMMENTS
Inverse of the odd Fibonacci sequence ( A001519), nearly, since a( A001519(n))=n except for n=0 (see A130256 for another version). a(n)+1 is the number of odd Fibonacci numbers ( A001519) <= n (for n >= 1).
FORMULA
a(n) = floor((1 + arcsinh(sqrt(5)*n/2)/log(phi))/2).
a(n) = floor((1 + arccosh(sqrt(5)*n/2)/log(phi))/2).
a(n) = floor((1 + log_phi(sqrt(5)*n))/2) for n >= 1, where phi = (1 + sqrt(5))/2.
G.f.: g(x) = 1/(1-x)*Sum_{k>=1} x^Fibonacci(2k-1).
a(n) = floor((1/2)*(1 + log_phi(sqrt(5)*n + 1))) for n >= 1.
MATHEMATICA
Table[Floor[(1 +ArcSinh[Sqrt[5]*n/2]/Log[GoldenRatio])/2], {n, 1, 100}] (* G. C. Greubel, Sep 09 2018 *)
PROG
(PARI) phi=(1+sqrt(5))/2; vector(100, n, floor((1 +asinh(sqrt(5)*n/2)/log(phi))/2)) \\ G. C. Greubel, Sep 09 2018 (Magma) phi:=(1+Sqrt(5))/2; [Floor((1 +Argsinh(Sqrt(5)*n/2)/Log(phi))/2): n in [1..100]]; // G. C. Greubel, Sep 09 2018
Minimal index k of an odd Fibonacci number A001519 such that A001519(k) = Fibonacci(2*k-1) >= n (the 'upper' odd Fibonacci Inverse).
+20
9
0, 0, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7
COMMENTS
Inverse of the odd Fibonacci sequence ( A001519), nearly, since a( A001519(n))=n except for n=1 (see A130255 for another version). a(n+1) is the number of odd Fibonacci numbers ( A001519) <= n (for n >= 0).
FORMULA
a(n) = ceiling((1+arccosh(sqrt(5)*n/2)/log(phi))/2), where phi=(1+sqrt(5))/2.
G.f.: (x/(1-x))*Sum_{k>=0} x^Fibonacci(2*k-1).
a(n) = ceiling((1/2)*(1+log_phi(sqrt(5)*n-1))) for n >= 2, where phi=(1+sqrt(5))/2.
MATHEMATICA
Join[{0, 0}, Table[Ceiling[1/2*(1 + Log[GoldenRatio, (Sqrt[5]*n - 1)])], {n, 2, 100}]] (* G. C. Greubel, Sep 12 2018 *)
PROG
(PARI) for(n=0, 100, print1(if(n==0, 0, if(n==1, 0, ceil((1/2)*(1 + log(sqrt(5)*n-1)/(log((1+sqrt(5))/2)))))), ", ")) \\ G. C. Greubel, Sep 12 2018 (Magma) [0, 0] cat [Ceiling((1/2)*(1 + Log(Sqrt(5)*n-1)/(Log((1+Sqrt(5))/2)))): n in [2..100]]; // G. C. Greubel, Sep 12 2018
1, 2, 8, 40, 208, 1088, 5696, 29824, 156160, 817664, 4281344, 22417408, 117379072, 614604800, 3218112512, 16850255872, 88229085184, 461973487616, 2418924584960, 12665653559296, 66318223015936, 347246723858432, 1818207451086848, 9520257811087360
COMMENTS
Hankel transform of 1,1,3,11,45,... (see A026375). Binomial transform of A015448. A production matrix for the sequence is M =
1, 1, 0, 0, 0, ...
1, 0, 5, 0, 0, ...
1, 0, 0, 5, 0, ...
1, 0, 0, 0, 5, ...
...
Take powers of M, extracting the upper left terms; getting
the sequence starting (1, 1, 2, 8, 40, 208, ...). (End)
The sequence is N=5 in an infinite set of INVERT transforms of powers of N prefaced with a "1". (1, 2, 8, 40, ...) is the INVERT transform of (1, 1, 5, 25, 125, ...). The first six of such sequences are shown in A006012 (N=3). - Gary W. Adamson, Jul 24 2016 The sequence is the first in an infinite set in which we perform the operation for matrix M (Cf. Jul 22 2016), but change the left border successively from (1, 1, 1, 1, ...) then to (1, 2, 2, 2, ...), then (1, 3, 3, 3, ...) ...; generally (1, N, N, N, ...). Extracting the upper left terms of each matrix operation, we obtain the infinite set beginning:
N=1 ( A154626): 1, 2, 8, 40, 208, 1088, ... N=2 ( A084120): 1, 3, 15, 81, 441, 1403, ... N=3 ( A180034): 1, 4, 22, 124, 700, 3952, ... N=4 ( A001653): 1, 5, 29, 169, 985, 5741, ... N=5 ( A000400): 1, 6, 36, 216, 1296, 7776, ... N=6 ( A015451): 1, 7, 43, 265, 1633, 10063, ... N=7 ( A180029): 1, 8, 50, 316, 1996, 12608, ... N=8 ( A180028): 1, 9, 57, 369, 1285, 15417, ... N=9 (.......): 1, 10, 64, 424, 2800, 18496, ...
N=10 ( A123361): 1, 11, 71, 481, 3241, 21851, ... N=11 (.......): 1, 12, 78, 540, 3708, 25488, ...
... Each of the sequences begins (1, (N+1), (7*N + 1),
(40*N + (N-1)^2), ... (End)
The set of infinite sequences shown (Cf. comment of Jul 27 2016), can be generated from the matrices P = [(1,N; 1,5]^n, (N=1,2,3,...) by extracting the upper left terms. Example: N=6 sequence ( A015451): (1, 7, 43, 265, ...) can be generated from the matrix P = [(1,6); (1,5)]^n. - Gary W. Adamson, Jul 28 2016
FORMULA
G.f.: (1 - 4*x) / (1 - 6*x + 4*x^2).
a(n) = (((3-sqrt(5))^n*(1+sqrt(5)) + (-1+sqrt(5))*(3+sqrt(5))^n)) / (2*sqrt(5)).
a(n) = 6*a(n-1) - 4*a(n-2) for n>1.
(End)
PROG
(Magma) [n le 2 select (n) else 6*Self(n-1)-4*Self(n-2): n in [1..25]]; // Vincenzo Librandi, May 15 2015 (PARI) Vec((1-4*x) / (1-6*x+4*x^2) + O(x^30)) \\ Colin Barker, Sep 22 2017
Sequence a(n) = 2 + 3* A001519(n+1) appearing in a certain four circle touching problem together with A246639.
+20
6
5, 8, 17, 41, 104, 269, 701, 1832, 4793, 12545, 32840, 85973, 225077, 589256, 1542689, 4038809, 10573736, 27682397, 72473453, 189737960, 496740425, 1300483313, 3404709512, 8913645221, 23336226149, 61095033224, 159948873521, 418751587337, 1096305888488, 2870166078125, 7514192345885
COMMENTS
This sequence is motivated by Kival Ngaokrajang's touching circle problem considered in A240926 and A115032. a(n), together with b(n) = A246639(n) appears as the curvature c(n) = a(n) + (4*b(n)/5)*phi (phi = (1+sqrt(5))/2, golden section) of the circle which touches i) a circle of radius 5/4 (in some length units) divided by a chord of length 2 into two unequal parts, and ii) the two touching circles in the smaller part which have curvatures A240926(n) and A240926(n+1), both also touching the circle with radius 5/4. See the illustration of Kival Ngaokrajang's link given in A240926, where the first circles in the smaller (upper) part are shown. The present circles will lie in the region between the large circle and two of these circles in the upper part. Descartes' theorem on touching circles (see the links) is applied here as c(n) = -4/5 + A(n) + A(n+1) + 2*sqrt((-4/5 )*(A(n) + A(n+1)) + A(n)*A(n+1)), with A(n) = A240926(n), n >= 0. For the proof for the first formula for a(n) given below use the formula for the curvature A240926(n) = 2 + 2*S(n, 3) - 3* S(n-1, 3) (see the W. Lang link in A240926, part II) in c(n) and compare with a(n) from c(n) = a(n) + (4*b(n)/5)*phi. This is done by using standard S-polynomial identities like the three term recurrence and the Cassini-Simson type identity S(n, x)*S(n-2, x) = -1 + S(n-1, x)^2 (here for x=3). This implies S(n, 3)*S(n-1, 3) = (-1 + S(n, 3)^2 + S(n-1, 3)^2)/3. See also the mentioned link, part III a).
a(n) appears also in the curvature for the touching circles and chord problem in the smaller part of a circle with radius 5/4 dissected by a chord of length 2, together with A246640, where details are given.
FORMULA
a(n) = 2 + 3*(S(n,3) - S(n-1,3)) = 2 + 3* A001519(n+1), n >= 0, with Chebyshev's S-polynomials (see A049310). O.g.f.: (5-12*x+5*x^2)/((1-x)*(1-3*x+x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n >=1, a(-2) = 8, a(-1) = 5, a(0) = 5.
a(n) = (2^(-1-n)*(5*2^(2+n)-3*(3-sqrt(5))^n*(-5+sqrt(5))+3*(3+sqrt(5))^n*(5+sqrt(5))))/5. - Colin Barker, Nov 02 2016
EXAMPLE
a(1) = 8 because c(n) = -4/5 + 5 + 9 + 2*sqrt((-4/5 )*(5 + 9) + 5*9) = 4*(2+(13/5)*phi). This is also 8 + (4*13/5)*phi with A246639(1) = 13.
MATHEMATICA
LinearRecurrence[{4, -4, 1}, {5, 8, 17}, 30] (* or *) CoefficientList[ Series[(5-12*x+5*x^2)/((1-x)*(1-3*x+x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)
PROG
(PARI) Vec((5-12*x+5*x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016 (Magma) I:=[5, 8, 17]; [n le 3 select I[n] else 4*Self(n-1) - 4*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017
Let T be the sequence Fibonacci(2n+1), n>=0 (cf. A001519); sequence lists the differences T(j)-T(i) for i<j.
+20
4
1, 3, 4, 8, 11, 12, 21, 29, 32, 33, 55, 76, 84, 87, 88, 144, 199, 220, 228, 231, 232, 377, 521, 576, 597, 605, 608, 609, 987, 1364, 1508, 1563, 1584, 1592, 1595, 1596, 2584, 3571, 3948, 4092, 4147, 4168, 4176, 4179, 4180, 6765, 9349, 10336, 10713, 10857, 10912, 10933
MATHEMATICA
Module[{nn=31, fbs}, fbs=Fibonacci[Range[1, nn, 2]]; Sort[Flatten[Table[ fbs[[n]]- Take[fbs, n-1], {n, nn/2}]]]] (* Harvey P. Dale, Aug 30 2015 *)
a(n) = 1 + L(n) + F(2*n-1) with {L(n)}_{n>=0} the Lucas numbers ( A000032) and F(2*n-1)_{n>=0} the bisected Fibonacci numbers ( A001519).
(Formerly M2537)
+20
3
4, 3, 6, 10, 21, 46, 108, 263, 658, 1674, 4305, 11146, 28980, 75547, 197262, 515594, 1348477, 3528150, 9233244, 24167167, 63261114, 165604618, 433534041, 1134967250, 2971318756, 7778909811, 20365282518, 53316730378, 139584573093, 365436446014, 956723886540
COMMENTS
The Name field was changed. The former Name-field entry was:
a(n) = Fibonacci(2n)*(1 + Fibonacci(n-1))/Fibonacci(n) for n even; a(n) = 2 + Fibonacci(2n)*(1 + Fibonacci(n-1))/Fibonacci(n) for n odd.
The new entry in the Name field is based on Equation (1) of the Grieg paper cited in the Links section. This formula is simpler than the former formulas given in the Name and Formula fields of this sequence. We make 6 comments:
(Comment I) The former formula in the Name section is consistent with a(0)=4 since 0 is a removable singularity of the former formula in the Name section. To see this we use the continuous Binet form, F(x) = (alpha^x - beta^x)/sqrt(5) with alpha = (sqrt(5)+1)/2 and beta = 1-alpha. Hence, for all nonzero x, F(2x)/F(x) = (alpha^(2x) - beta^(2x))/(alpha^x - beta^x) = alpha^x + beta^x. It immediately follows that lim_{x->0} F(2x)/F(x) = alpha^0 + beta^0 = 2. Consequently, lim_{x->0} Fibonacci(2x)*(1 + Fibonacci(x-1))/Fibonacci(x) = 2(1+Fibonacci(-1)) = 4 as required.
(Comment II) As pointed out in references [2,3] of the Grieg paper, a(n) = F(2n) C(n), with C(n) = Sum_{m>=0} 1/F(n*b), b=2^m, and {F(n)}_{n>=0}, the Fibonacci numbers ( A000045). In other words, the a(n) = C(n) F(2n) are the numerators of the sequence of C(n) = a(n) / F(2n), which are the values of infinite sums of reciprocals of subsets of the Fibonacci numbers. (Comment III) The probable reason that the elegant formulas in the Grieg papers were overlooked is that the Grieg papers are difficult to read because of their nonstandard notation and terminology. For example, the generalized Fibonacci and Lucas numbers are indicated by P with subscripts rather than by F. Similarly, Grieg refers to the sequence of odd-indexed Fibonacci numbers as the "bisected" Fibonacci numbers ( A001519), a term which rarely appears in modern books on the Fibonacci numbers. (Comment IV) The defining recursion for a(n) is best given in factored form using annihilating operators (Equation (12) in the Grieg paper). Let E be the translation operator so that E(f)(n) = f(n+1). Then, for example, (E^2-E-1)(L(n)) = L(n+2) - L(n+1) - L(n) = 0 ( A000032). Thus the operator E^2-E-1 annihilates the Lucas sequence. Similarly, (E^2-3E+1) annihilates the bisected Fibonacci numbers ( A001519) and (E-1) annihilates the constant sequence ( A000012). It follows that the product of these annihilators annihilates the sum of the sequences, which is a(n). So the defining recursion may be obtained by expanding (E-1)(E^2-3E+1)(E^2-E-1). It follows that the sequence a(n) satisfies a(n) - 5 a(n-1) + 7 a(n-2) - a(n-3) - 3 a(n-4) + a(n-5) = 0. For example, 1*108 - 5*46 + 7*21 - 1*10 - 3*6 + 1*3 = 0. Note that the derivation of this recursion, based on annihilators, is more straightforward than the derivation in the Grieg paper based on central difference operators. (Comment V) Using standard techniques, we may obtain the g.f. of the a(n) from the defining recursion, of order 5. Since a(n) = 1 + F(2n-1) + L(n), it follows that the g.f. for the a(n) is the product of the g.f. for the three summands 1, F(2n-1), L(n). The g.f. for the constant sequence, 1, is C(x)=1/(1-x) ( A000012); the g.f. for the Lucas sequence is L(x)=(2-x)/(1-x-x^2)( A000032); and the g.f. for the bisected Fibonacci numbers F(2n-1) is B(x)=(1 - 2x)/(1 - 3x + x^2) (sequence A001519). It follows that the g.f. for the a(n) is the sum of these g.f.s: C(x) + L(x) + B(x) = 1/(1-x) + (1-2x)/(1-3x+x^2) + (2-x)/(1-x-x^2). This is a partial fraction decomposition of a rational function, that is an equivalent g.f. for the a(n), A(x) = (4 - 17x + 19x^2 - 3x^3 - 2x^4)/(1 - 5x + 7x^2 - x^3 - 3x^4 + x^5). (Comment VI) We can now comment on the g.f., P(x), conjectured by Simon Plouffe and mentioned in the MAPLE section below. Expanding P(x) as a series, we obtain the following g.f. of the a(n): 3 + 6x + 10x^2 + 21x^3 ... . In fact, it is easily checked that P(x) = 1/x (A(x)-4). This settles the Plouffe conjecture as follows: (i) A(x) is the rational function g.f. of a(n)_{n>=0}, (ii) P(x) is the rational function g.f. of a(n)_{n>=1}, and (iii) C(x)+L(x)+B(x) is the partial fraction g.f. of a(n)_{n>=0}. (End) Let phi = (1+sqrt(5))/2, p(n) = phi^n - (-phi)^(-n) and FL(n) = 1 + (p(n-1) + p(n+1) + p(2*n-1))/sqrt(5). FL is a doubly infinite sequence with FL(n) = A005522(n) and FL(-n) = A006172(n) for n >= 0. - Peter Luschny, Mar 09 2015
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
a(n) = ((sqrt(5)+5)/10)*(((3+sqrt(5))/2)^(n-1)) + ((5-sqrt(5))/10)*(((3-sqrt(5))/2)^(n-1)) + ((1+sqrt(5))/2)^n + ((1-sqrt(5))/2)^n + 1. - Tim Monahan, Jul 23 2011 a(n) = 1 + L(n) + F(2*n-1), with L(n) the Lucas numbers, sequence A000032 and F(2n-1) the bisected Fibonacci numbers, sequence A001519. - Russell Jay Hendel, Mar 02 2015 a(n) = 5*a(n-1) - 7*a(n-2) + a(n-3) + 3*a(n-4) - a(n-5). - Russell Jay Hendel, Mar 02 2015 Sum_{n>=1} (a(2*n-1)/Fibonacci(4*n-2) - 1/phi) = A079586, where phi is the golden ratio ( A001622) (Greig, 1977). - Amiram Eldar, Jan 29 2022
MAPLE
with(combinat): a:=proc(n) if n mod 2 = 0 then fibonacci(2*n)*(1+fibonacci(n-1))/fibonacci(n) else 2+fibonacci(2*n)*(1+fibonacci(n-1))/fibonacci(n) fi end: seq(a(n), n=1..32); # Emeric Deutsch, Apr 01 2005 A005522:=-(-3+9*z-z**2-10*z**3+4*z**4)/(z-1)/(z**2-3*z+1)/(z**2+z-1); # conjectured by Simon Plouffe in his 1992 dissertation
FL := proc(n) local a, p; a := (1+sqrt(5))/2; p := m -> a^m - (-a)^(-m);
1 + (p(n-1) + p(n+1) + p(2*n-1))/sqrt(5) end: A005522 := n -> FL(n):
MATHEMATICA
Series[1/(1-x)+ (2-x)/(1-x-x^2)+(1-2*x)/(1-3*x+x^2), {x, 0, 30}] (* Russell Jay Hendel, Mar 02 2015 *) LinearRecurrence[{5, -7, 1, 3, -1}, {4, 3, 6, 10, 21}, 31] (* Russell Jay Hendel, Mar 02 2015 *) CoefficientList[ Series[(4 - 17x + 19x^2 - 3x^3 - 2x^4)/(1 - 5x + 7x^2 - x^3 - 3x^4 + x^5), {x, 0, 30}], x] (* Russell Jay Hendel, Mar 02 2015 *)
PROG
(PARI)
\ps {31};
C(x) = 1/(1-x);
L(x) = (2-x)/(1-x-x^2);
B(x) = (1-2*x)/(1-3*x+x^2);
A(x) = C(x)+L(x)+B(x);
Ser(A(x), x)
Vec(Ser(A(x), x))
(Magma) [1 + Lucas(n) + Fibonacci(2*n - 1): n in [0..30]]; // Vincenzo Librandi, Mar 08 2015 (PARI) {lucas(n) = fibonacci(n+1) + fibonacci(n-1)};
for(n=0, 30, print1(1 + fibonacci(2*n-1) + lucas(n), ", ")) \\ G. C. Greubel, Jan 01 2018
Sequence a(n) = 3 + 5* A001519(n+1) appearing in a certain three circle touching problem, together with A246638.
+20
3
8, 13, 28, 68, 173, 448, 1168, 3053, 7988, 20908, 54733, 143288, 375128, 982093, 2571148, 6731348, 17622893, 46137328, 120789088, 316229933, 827900708, 2167472188, 5674515853, 14856075368, 38893710248, 101825055373, 266581455868, 697919312228, 1827176480813, 4783610130208, 12523653909808
COMMENTS
See the comments under A246638. The curvature c(n) defined there is c(n) = A246638(n) + (4*a(n)/5)*phi with phi = (1+sqrt(5))/2, the golden section. It lives in the quadratic number field Q(sqrt(5)). Descartes' theorem on touching circles gives c(n) = -4/5 + A(n) + A(n+1) + 2*sqrt((-4/5 )*(A(n) + A(n+1)) + A(n)*A(n+1)), with A(n) = A240926(n), n >= 0. For the proof of the first formula given below one compares this a(n) with the a(n) in c(n) given above. This uses standard Chebyshev S-polynomial identities with x = 3, like the three term recurrence and the Cassini-Simson type identity S(n, x)*S(n-2, x) = -1 + S(n-1, x)^2 (here for x=3). This implies S(n, 3)*S(n-1, 3) = (-1 + S(n, 3)^2 + S(n-1, 3)^2)/3. See also the W. Lang link in A240926, part III a).
FORMULA
a(n) = 3 + 5*(S(n, 3) - S(n-1, 3)) = 3 + 5* A001519(n+1), n >= 0, with Chebyshev S-polynomials (see A049310) with S(-1, x) = 0. O.g.f.: (8-19*x+8*x^2)/((1-x)*(1-3*x+x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n >=1, a(-2) = 13, a(-1) = 8, a(0) = 8.
a(n) = 2^(-1-n)*(3*2^(1+n)-(3-sqrt(5))^n*(-5+sqrt(5))+(3+sqrt(5))^n*(5+sqrt(5))). - Colin Barker, Nov 02 2016
MATHEMATICA
CoefficientList[Series[(8-19*x+8*x^2)/((1-x)*(1-3*x+x^2)), {x, 0, 50}], x] (* or *) LinearRecurrence[{4, -4, 1}, {8, 13, 28}, 30] (* G. C. Greubel, Dec 20 2017 *)
PROG
(PARI) Vec((8-19*x+8*x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016 (Magma) I:=[8, 13, 28]; [n le 3 select I[n] else 4*Self(n-1) -4*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017
Sequence a(n) = 1 + A001519(n+1) appearing in a certain touching problem for three circles and a chord, together with A246638.
+20
3
2, 3, 6, 14, 35, 90, 234, 611, 1598, 4182, 10947, 28658, 75026, 196419, 514230, 1346270, 3524579, 9227466, 24157818, 63245987, 165580142, 433494438, 1134903171, 2971215074, 7778742050, 20365011075, 53316291174, 139583862446, 365435296163, 956722026042, 2504730781962
COMMENTS
This sequence is motivated by Kival Ngaokrajang's touching circle problem considered in A240926 and A115032. a(n), together with b(n) = A246638(n), appears in a curvature c(n) = b(n) + 4*a(n)*phi, with phi = (1+sqrt(5))/2, the golden section. This is an integer in the real quadratic field Q(sqrt(5)). c(n) is the curvature of the circle which touches i) a chord of length 2 (in some length units) of a circle of radius 5/4 which is divided by this chord in two unequal parts, and ii) the two touching circles in the smaller part which have curvatures A240926(n) and A240926(n+1). These two touching circles touch also the circle with radius 5/4 and the chord. See the illustration of Kival Ngaokrajang's link given in A240926, where the first circles in the smaller (upper) part are shown. c(n) is an integer in the real quadratic field Q(sqrt(5)). From Descartes' theorem on touching circles (see the links) one has here: c(n) = A(n) + A(n+1) + 2*sqrt(A(n)*A(n+1)),
with A(n) = A240926(n), n >= 0. In this application the chord has curvature 0. For the proof for the first formula for a(n) given below use the formula for the curvature A240926(n) = 2 + 2*S(n, 3) - 3* S(n-1, 3) (see the W. Lang link found in A240926, part II) in c(n) from Descartes' formula and compare it with a(n) from c(n) = A246638(n) + 4*a(n)*(1+sqrt(5))/2. This can be done by using standard S-polynomial identities like the three term recurrence and the Cassini-Simson type identity (see a comment on A246638) which implies the formula S(n, 3)*S(n-1, 3) = (-1 + S(n, 3)^2 + S(n-1, 3)^2)/3. See also the above mentioned W. Lang link part III b).
FORMULA
a(n) = 1 + S(n, 3) - S(n-1, 3) = 1 + A001519(n+1), n>=0, with Chebyshev's S-polynomials (see A049310), and S(-1, x) = 0. O.g.f.: (2-5*x+2*x^2)/((1-x)*(1-3*x+x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n >= 1, with a(-2) = 3, a(-1) = 2 and a(0) = 2.
a(n) = 1+(2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/sqrt(5). - Colin Barker, Nov 02 2016
EXAMPLE
a(1) = 3 because c(1) = 0 + 5 + 9 + 2*sqrt(5*9) = 8 + 12*phi which is indeed 8 + 4*3*phi, with 8 = A246638(1).
MATHEMATICA
LinearRecurrence[{4, -4, 1}, {2, 3, 6}, 30] (* or *) CoefficientList[ Series[ (2-5*x+ 2*x^2)/ ((1-x)*(1-3*x+x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)
PROG
(PARI) Vec((2-5*x+2*x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016 (Magma) I:=[2, 3, 6]; [n le 3 select I[n] else 4*Self(n-1) - 4*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017
1, 1, 2, 2, 2, 5, 4, 4, 5, 13, 8, 8, 10, 13, 34, 16, 16, 20, 26, 34, 89, 32, 32, 40, 52, 68, 89, 233, 64, 64, 80, 104, 136, 178, 233, 610, 128, 128, 160, 208, 272, 356, 466, 610, 1597, 256, 256, 320, 416, 544, 712, 932, 1220, 1597, 4181
COMMENTS
Row sums = A061667: (1, 3, 9, 26, 73, 201, ...).
FORMULA
Let M = an infinite lower triangular matrix with A011782: (1, 1, 2, 4, 8, 16, ...) in every column; and Q = an infinite lower triangular matrix with odd-indexed Fibonacci numbers, A001519: (1, 2, 5, 13, 34, 89, ...) as the main column and the rest zeros.
EXAMPLE
First few rows of the triangle =
1;
1, 2;
2, 2, 5;
4, 4, 5, 13;
8, 8, 10, 13, 34;
16, 16, 20, 26, 34, 89;
32, 32, 40, 52, 68, 89, 233;
64, 64, 80, 104, 136, 178, 233, 610;
128, 128, 160, 208, 272, 356, 466, 610, 1597;
256, 256, 320, 416, 544, 712, 932, 1220, 1597, 4181;
...
Row 3 = (4, 4, 5, 13) = termwise products of (4, 2, 1, 1) and (1, 2, 5, 13). Row 3 sum of terms = 26 = (1, 1, 2, 4) convolved with (1, 2, 5, 13).
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