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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).
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%I #59 Feb 13 2024 06:59:22

%S 1,2,7,12,41,70,239,408,1393,2378,8119,13860,47321,80782,275807,

%T 470832,1607521,2744210,9369319,15994428,54608393,93222358,318281039,

%U 543339720,1855077841,3166815962,10812186007,18457556052,63018038201,107578520350

%N 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).

%C 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

%C From _Peter Bala_, Mar 23 2018: (Start)

%C 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

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

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

%C 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 ).

%C 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)

%C 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

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

%H Indranil Ghosh, <a href="/A084068/b084068.txt">Table of n, a(n) for n = 1..2608</a>

%H Clark Kimberling, <a href="http://dx.doi.org/10.1007/s000170050020">Best lower and upper approximates to irrational numbers</a>, Elemente der Mathematik, 52 (1997) 122-126.

%H D. H. Lehmer, <a href="http://www.jstor.org/stable/1968235">An extended theory of Lucas' functions</a>, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.

%H E. W. Weisstein, <a href="http://mathworld.wolfram.com/LehmerNumber.html">MathWorld: Lehmer Number</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,0,-1).

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

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

%F From _Paul Barry_, Jun 06 2006: (Start)

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

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

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

%F A000129(n+1) = A079496(n) + a(n). - _Gary W. Adamson_, Sep 18 2007

%F Equals A133566 * A000129, where A000129 = the Pell sequence. - _Gary W. Adamson_, Sep 18 2007

%F From _Peter Bala_, Mar 23 2018: (Start)

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

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

%F More generally,

%F 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,

%F 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).

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

%F 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).

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

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

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

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

%F (End)

%F a(n) = A000129(n) for even n and A001333(n) for odd n. - _R. J. Mathar_, Oct 15 2021

%p 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;

%p end proc:

%p seq(simplify(a(n)), n = 1..30); # _Peter Bala_, Mar 25 2018

%t a[n_] := ((Sqrt[2]+1)^n - (Sqrt[2]-1)^n) ((-1)^n(Sqrt[2]-2) + (Sqrt[2]+2))/8;

%t Table[Simplify[a[n]], {n, 30}] (* after _Paul Barry_, _Peter Luschny_, Mar 29 2018 *)

%o (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

%Y Bisections are A001542 and A002315.

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

%K nonn,easy

%O 1,2

%A _Benoit Cloitre_, May 10 2003