A143608 A005319 and A002315 interleaved.

0, 1, 4, 7, 24, 41, 140, 239, 816, 1393, 4756, 8119, 27720, 47321, 161564, 275807, 941664, 1607521, 5488420, 9369319, 31988856, 54608393, 186444716, 318281039, 1086679440, 1855077841, 6333631924, 10812186007, 36915112104, 63018038201, 215157040700

Offset: 0

Author

Originally submitted by Clark Kimberling, Aug 27 2008. Merged with an essentially identical sequence submitted by Kenneth J Ramsey, Jun 01 2012, by N. J. A. Sloane, Aug 02 2012

Keywords

• nonn
• easy

Comments

Also, numerators of the lower principal and intermediate convergents to 2^(1/2). The lower principal and intermediate convergents to 2^(1/2), beginning with 1/1, 4/3, 7/5, 24/17, 41/29, form a strictly increasing sequence; essentially, numerators=A143608 and denominators=A079496.

Sequence a(n) such that a(2*n) = sqrt(2*A001108(2*n)) and a(2*n+1) = sqrt(A001108(2*n+1)).

For n > 0, a(n) divides A******(k+1,n+1)-A******(k,n+1) where A****** is any one of A182431, A182439, A182440, A182441 and k is any nonnegative integer.

If p is a prime of the form 8*r +/- 3 then a(p+1) == 0 (mod p); if p is a prime of the form 8*r +/- 1 then a(p-1) == 0 (mod p).

Numbers n such that sqrt(floor(n^2/2 + 1)) is an integer. The integer square roots are given by A079496. - Richard R. Forberg, Aug 01 2013

From Peter Bala, Mar 23 2018: (Start)

Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have

a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and

a(2*n) = sqrt(2)*(1 o 1 o ... o 1) (2*n terms). Cf. A084068.

This is a fourth-order divisibility sequence. Indeed, a(2*n) = sqrt(2)*U(2*n) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ). (End)

References

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

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
• Creighton Kenneth Dement, Comments on A143608 and A143609
• Vincent Granville, Successive records in mathematical sequences: surprising result, Mathematics Stack Exchange, 2019.
• Clark Kimberling, Best lower and upper approximations to irrational numbers, Elem. Math. vol. 52 iss. 3 (1997) 122-126.
• D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
• Eric Weisstein's World of Mathematics, Lehmer Number
• Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Crossrefs

Cf. A002315, A005319, A182431, A001108, A049629, A084068.

Programs

Magma

I:=[0,1,4,7]; [n le 4 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..30]]; // G. C. Greubel, Mar 27 2018

Maple

A143608 := proc(n)
    option remember;
    if n <= 3 then
        op(n+1,[0,1,4,7]) ;
    else
        6*procname(n-2)-procname(n-4) ;
    end if;
end proc: # R. J. Mathar, Jul 22 2012

Mathematica

a = -4; b = -1; Reap[While[b<2000000000, t = 4*b-a; Sow[t]; a=b; b=t; t = 2*b-a; Sow[t]; a=b; b=t]][[2,1]]
CoefficientList[Series[x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 24 2014 *)
LinearRecurrence[{0, 6, 0, -1}, {0, 1, 4, 7}, 31] (* Jean-François Alcover, Sep 21 2017 *)

PARI

a(n)=([0,1,0,0;0,0,1,0;0,0,0,1;-1,0,6,0]^n*[0;1;4;7])[1,1] \\ Charles R Greathouse IV, Jun 11 2015

PARI

concat(0, Vec(x*(1+4*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)) + O(x^50))) \\ Colin Barker, Mar 27 2016

Formula

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

a(2*n + 1) = (a(2*n) + a(2*n + 2))/4.

a(2*n) = 4*A001109(n).

a(2*n + 1) = 4*A001109(n) + A001541(n).

From Colin Barker, Jun 29 2012: (Start)

a(n) = 6*a(n-2) - a(n-4).

G.f.: x*(1 + 4*x + x^2)/((1 + 2*x - x^2)*(1 - 2*x - x^2)) = x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4). (End)

2*a(n) = A078057(n) - A123335(n-1). - R. J. Mathar, Jul 04 2012

a(2n) = A005319(n); a(2n+1) = A002315(n). - R. J. Mathar, Jul 17 2009

a(n)*a(n+1) + 1 = A001653(n+1). - Charlie Marion, Dec 11 2012

a(n) = (((-2 - sqrt(2) + (-1)^n * (-2+sqrt(2))) * ((-1+sqrt(2))^n - (1+sqrt(2))^n)))/(4*sqrt(2)). - Colin Barker, Mar 27 2016

a(n) = A084068(n) - A079496(n). - César Aguilera, Feb 14 2023