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A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).
(Formerly M1769 N0700)
84
1, 2, 7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226, 1913445293767, 7141075053842, 26650854921601, 99462344632562, 371198523608647 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

Chebyshev's T(n,x) polynomials evaluated at x=2.

x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).

Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002

This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014]

a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003

a(3+6k)-1 and a(3+6k)+1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006

The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators=A005320, denominators=A001075. - Clark Kimberling, Aug 27 2008

The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008

a(n+1) is the Hankel transform of A000108(n)+A000984(n)=(n+2)*Catalan(n). - Paul Barry, Aug 11 2009

Also, numbers such that floor[a(n)^2/3] is a square: base 3 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001541. - M. F. Hasler, Jan 15 2012

Pisano period lengths:  1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12,... - R. J. Mathar, Aug 10 2012

Except for the first term, positive values of x (or y) satisfying x^2 - 4xy + y^2 + 3 = 0. - Colin Barker, Feb 04 2014

Except for the first term, positive values of x (or y) satisfying x^2 - 14xy + y^2 + 48 = 0. - Colin Barker, Feb 10 2014

A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.

M =

1, 1, 0, 0, 0, 0,...

2, 0, 3, 0, 0, 0,...

2, 0, 0, 3, 0, 0,...

2, 0, 0, 0, 3, 0,...

2, 0, 0, 0, 0, 3,...

...The triangle generated from M is:

1,

1, 1,

3, 1, 3,

11, 3, 3, 9,

41, 11, 9, 9, 27,

...The left border is A001835 and row sums are (1, 2, 7, 26, 97,...). - Gary W. Adamson, Jul 25 2016

Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2n) = 2*(a(n))^2 - 1 and a(2n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016

For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4n+2), a(3) divides a(6n+3), a(4) divides a(8n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2n+1)k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016

From Timothy L. Tiffin, Oct 21 2016: (Start)

If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms.

a(2n) == 1 mod 6 and a(2n+1) == 2 mod 6. Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476.

a(n) == 1 mod 10 if n == 0 mod 6, a(n) == 2 mod 10 if n == {1,-1} mod 6, a(n) == 7 mod 10 if n == {2,-2} mod 6, and a(n) == 6 mod 10 if n == 3 mod 6. So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)

REFERENCES

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Mcdonnell, Eugene, "Heron's Rule and Integer-Area Triangles", Vector 12.3 (January 1996) pp. 133-142

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).

P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.

LINKS

T. D. Noe, Table of n, a(n) for n = 0..200

C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.

H. Brocard, Notes élémentaires sur le problème de Peel [sic], Nouvelle Correspondance Mathématique, 4 (1878), 337-343.

Chris Caldwell, Primality Proving, Arndt's theorem.

J. B. Cosgrave and K. Dilcher, A role for generalized Fermat numbers, Math. Comp., 2016.

E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.

R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]

Kiran S. Kedlaya, The 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.

Kiran S. Kedlaya, Solutions to the 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.

Tanya Khovanova, Recursive Sequences

Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.

Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.

F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.

Index entries for sequences related to Chebyshev polynomials.

Index entries for two-way infinite sequences

Index entries for linear recurrences with constant coefficients, signature (4,-1).

FORMULA

G.f.: (1-2x)/(1-4x+x^2). E.g.f.: exp(2x)cosh(sqrt(3)x). a(n) = 4a(n-1) - a(n-2) = a(-n).

For all elements x of the sequence, 12*x^2 -12 is a square. Lim. as n-> Inf. a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002

a(n) = (S(n, 4) - S(n-2, 4))/2 = T(n, 2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U, resp. T, are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 4) = A001353(n), n>=0. See A049310 and A053120.

a(n) = A001353(n+2)-2*A001353(n+1).

a(n) = 2^(-n)*Sum_{k>=0} binomial(2n, 2k)*3^k = 2^(-n)*Sum_{k>=0} A086645(n, k)*3^k. - Philippe Deléham, Mar 01, 2004

a(n) = ((2+sqrt(3))^n + (2-sqrt(3))^n)/2; a(n) = ceiling((1/2)*(2+sqrt(3))^(n)).

a(n) = cosh( n * log( 2 + sqrt(3))).

a(n) = sum{k=0..floor(n/2); C(n, 2k)2^(n-2k)3^k }. - Paul Barry, May 08 2003

a(n+2) = 2*a(n+1) + 3*Sum_{k>=0} a(n-k)*2^k. - Philippe Deléham, Mar 03 2004

a(n) = 2*a(n-1)+3*A001353(n-1). - Lekraj Beedassy, Jul 21 2006

a(n) = left term of M^n * [1,0] where M = the 2 X 2 matrix [2,3; 1,2]. Right term = A001353(n). Example: a(4) = 97 since M^4 * [1,0] = [A001075(4), A001353(4)] = [97, 56]. - Gary W. Adamson, Dec 27 2006

Binomial transform of A026150: (1, 1, 4, 10, 28, 76,...). - Gary W. Adamson, Nov 23 2007

First differences of A001571. - N. J. A. Sloane, Nov 03 2009

Sequence satisfies -3 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008

a(n) = Sum_{k, 0<=k<=n} A201730(n,k)*2^k. - Philippe Deléham, Dec 06 2011

G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k-4)/(x*(3*k-1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013

a(n) = Sum_{k = 0..n} A238731(n,k). - Philippe Deléham, Mar 05 2014

EXAMPLE

2^6 -1 = 63 does not divide a(2^4) = 708158977, therefore 63 is composite. 2^5 -1 = 31 divides a(2^3) = 18817, therefore 31 is prime.

G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 97*x^4 + 362*x^5 + 1351*x^6 + 5042*x^7 + ...

MAPLE

A001075:=-(-1+2*z)/(1-4*z+z**2); # Simon Plouffe in his 1992 dissertation

MATHEMATICA

Table[ Ceiling[(1/2)*(2 + Sqrt[3])^n], {n, 0, 24}]

f[x_] := (1-2*x) / (1-4*x+x^2); CoefficientList[ Series[ f[x], {x, 0, 24}], x] (* Jean-François Alcover, Dec 21 2011, after Simon Plouffe *)

a=1; b=1; Join[{1, 2}, Table[c=(b=b+a)+(a=a+b*2), {n, 0, 30}]] (* Vladimir Joseph Stephan Orlovsky, Jan 29 2012 *)

LinearRecurrence[{4, -1}, {1, 2}, 30] (* Harvey P. Dale, Aug 22 2015 *)

Round@Table[LucasL[2n, Sqrt[2]]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)

PROG

(PARI) {a(n) = subst(poltchebi(abs(n)), x, 2)};

(PARI) {a(n) = real((2 + quadgen(12))^abs(n))};

(PARI) {a(n) = polsym(1 - 4*x + x^2, abs(n))[1 + abs(n)]/2};

(PARI) a(n)=polchebyshev(n, 1, 2) \\ Charles R Greathouse IV, Nov 07 2016

(Sage) [lucas_number2(n, 4, 1)/2 for n in xrange(0, 25)] # Zerinvary Lajos, May 14 2009

(Haskell)

a001075 n = a001075_list !! n

a001075_list =

   1 : 2 : zipWith (-) (map (4 *) $ tail a001075_list) a001075_list

-- Reinhard Zumkeller, Aug 11 2011

(Sage)

def a(n):

    Q = QuadraticField(3, 't')

    u = Q.units()[0]

    return (u^n).lift().coeffs()[0]  # Ralf Stephan, Jun 19 2014

CROSSREFS

a(n) = sqrt(1+3*A001353(n)) (cf. Richardson comment).

Bisections are A011943 and A094347.

Cf. A065918, A071954, A001571, A001834, A003500, A016064, A082840, A026150, A277434.

Sequence in context: A055988 A275013 A278351 * A113436 A126223 A273320

Adjacent sequences:  A001072 A001073 A001074 * A001076 A001077 A001078

KEYWORD

nonn,easy,nice

AUTHOR

N. J. A. Sloane

EXTENSIONS

More terms from James A. Sellers, Jul 10 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

STATUS

approved

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Last modified December 5 11:38 EST 2016. Contains 278764 sequences.