A084068 a(1) = 1, a(2) = 2; a(2*k) = 2*a(2*k-1) - a(2*k-2), a(2*k+1) = 4*a(2*k) - a(2*k-1).

1, 2, 7, 12, 41, 70, 239, 408, 1393, 2378, 8119, 13860, 47321, 80782, 275807, 470832, 1607521, 2744210, 9369319, 15994428, 54608393, 93222358, 318281039, 543339720, 1855077841, 3166815962, 10812186007, 18457556052, 63018038201, 107578520350

Offset: 1

Author

Benoit Cloitre, May 10 2003

Keywords

• nonn
• easy

Comments

The upper principal and intermediate convergents to 2^(1/2), beginning with 2/1, 3/2, 10/7, 17/12, 58/41, form a strictly decreasing sequence; essentially, numerators=A143609 and denominators=A084068. - Clark Kimberling, Aug 27 2008

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. We have

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

a(2*n) = (1/sqrt(2))*(1 o 1 o ... o 1) (2*n terms). Cf. A049629, A108412 and A143608.

This is a fourth-order divisibility sequence. Indeed, a(2*n) = U(2*n)/sqrt(2) 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 ).

It appears that this sequence consists of those numbers m such that 2*m^2 = floor( m*sqrt(2) * ceiling(m*sqrt(2)) ). Cf. A084069. (End)

Conjecture: a(n) is the earliest occurrence of n in A348295, which is to say, a(n) is the least m such that Sum_{k=1..m} (-1)^(floor(k*(sqrt(2)-1))) = Sum_{k=1..m} (-1)^A097508(k) = n. This has been confirmed for the first 32 terms by Chai Wah Wu, Oct 21 2021. - Jianing Song, Jul 16 2022

References

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

Links

• Indranil Ghosh, Table of n, a(n) for n = 1..2608
• Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (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

Bisections are A001542 and A002315.

Cf. A084069, A084070, A133566, A079496, A001541, A001653, A008844, A055792, A049629, A108412, A143608.

Programs

Maple

a := proc (n) if `mod`(n, 2) = 1 then (1/2)*(sqrt(2) + 1)^n - (1/2)*(sqrt(2) - 1)^n else (1/2)*((sqrt(2) + 1)^n - (sqrt(2) - 1)^n)/sqrt(2) end if;
end proc:
seq(simplify(a(n)), n = 1..30); # Peter Bala, Mar 25 2018

Mathematica

a[n_] := ((Sqrt[2]+1)^n - (Sqrt[2]-1)^n) ((-1)^n(Sqrt[2]-2) + (Sqrt[2]+2))/8;
Table[Simplify[a[n]], {n, 30}] (* after Paul Barry, Peter Luschny, Mar 29 2018 *)

PARI

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^(n-1)*[1;2;7;12])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

Formula

"A Diofloortin equation": n such that 2*n^2=floor(n*sqrt(2)*ceiling(n*sqrt(2))).

a(n)*a(n+3) = -2 + a(n+1)*a(n+2).

From Paul Barry, Jun 06 2006: (Start)

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

a(n) = ((sqrt(2)+1)^n-(sqrt(2)-1)^n)*((sqrt(2)/8-1/4)*(-1)^n+sqrt(2)/8+1/4);

a(n) = Sum_{k=0..floor(n/2)} 2^k*(C(n,2*k)-C(n-1,2*k+1)*(1+(-1)^n)/2). (End)

A000129(n+1) = A079496(n) + a(n). - Gary W. Adamson, Sep 18 2007

Equals A133566 * A000129, where A000129 = the Pell sequence. - Gary W. Adamson, Sep 18 2007

From Peter Bala, Mar 23 2018: (Start)

a(2*n + 2) = a(2*n + 1) + sqrt( (1 + a(2*n + 1)^2)/2 ).

a(2*n + 1) = 2*a(2*n) + sqrt( (1 + 2*a(2*n)^2) ).

More generally,

a(2*n+2*m+1) = sqrt(2)*a(2*n) o a(2*m+1), where o is the binary operation defined above, that is,

a(2*n+2*m+1) = sqrt(2)*a(2*n)*sqrt(1 + a(2*m+1)^2) + a(2*m+1)*sqrt(1 + 2*a(2*n)^2).

sqrt(2)*a(2*(n + m)) = (sqrt(2)*a(2*n)) o (sqrt(2)*a(2*m)), that is,

a(2*n+2*m) = a(2*n)*sqrt(1 + 2*a(2*m)^2) + a(2*m)*sqrt(1 + 2*a(2*n)^2).

sqrt(1 + 2*a(2*n)^2) = A001541(n).

1 + 2*a(2*n)^2 = A055792(n+1).

a(2*n) - a(2*n-1) = A001653(n).

(1 + a(2*n+1)^2)/2 = A008844(n). (End)

a(n) = A000129(n) for even n and A001333(n) for odd n. - R. J. Mathar, Oct 15 2021