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Numbers such that A166986(n)/2 - A167389(n) = 1.
2

%I #36 Nov 22 2022 23:11:02

%S 50,59,111,120,181,242,251,303,312,373,434,495,504,556,565,626,687,

%T 696,748,757,818,879,940,949,1001,1010,1071,1132,1141,1193,1202,1254,

%U 1263,1324,1385,1394,1446,1455,1516,1577,1638,1647,1699,1708,1769,1830,1839

%N Numbers such that A166986(n)/2 - A167389(n) = 1.

%C It is conjectured that the successive differences of this sequence, A172470, are limited to three numbers: 9, 52 and 61 where it is noted that 61 - 52 = 9, (52 + 2)/9 = 6, (61 + 2)/9 = 7 and we have lcm(9, 52, 61) = 28548 = 13^4 - 13 and 6 + 7 = 13.

%C From _Travis Scott_, Oct 16 2022: (Start)

%C Given sequences S(n), T(n) such that S'(n), T'(n) both ~ r for some real number r, if S(n) - T(n) converges to c then I(n) = floor(S(n)) - floor(T(n)) - floor(c) converges to the indicator [(r*n) mod 1 < c mod 1]. Take a new sequence 1(k) from the k-th n indicated by I(n). If S(n) - T(n) - floor(c) is [I] nonincreasing and [II] < 1 for all n > m, it is easy to see that the first differences of 1(k) for all values > m are capped at max(a, b) by the earliest unordered pair of positive and negative residues {r*a, r*b} == {-x, y} (mod 1)_[-1/2, 1/2) satisfying x + y < c mod 1, since they are jointly sufficient to map any interval [0, c mod 1 <= c_n < 1] back to itself modulo 1.

%C Rearranging terms, an equivalent statement of A172468 is 1(k) indicated by [((n + 2)/log(2)) mod 1 < c_n] with [III] c_n = (n + 2 - Im(W(n,-log(2)/2))/(2*Pi))/log(2) - 5/2. By the asymptotics of W_n [see for example Corless et al., s. 4] we have Im(W(n,z)) ~ 2*Pi*n + arg(z) - Pi/2 as n->oo, taking the entire negative real axis to [IV] Im(W(n,-r)) ~ 2*Pi*n + Pi/2 and [ibid., from Eq. 4.20] Im(W(n,-r)) - 2*Pi*n = C strictly increasing from n >= 0 for the slice [V] (-1/e, 0). [Empirically this holds to -0.93568951... due to the fact that W_n is a discrete assembly of branches.]

%C The conjectured closure of {9,52,61} then follows from plugging [IV] into [III] to get c = lim_{n->oo} c_n = 7/log(16) - 5/2 and by inspection of (n/log(2)) mod 1 for n <= 61, noting that -log(2)/2 satisfies [V] so that c_n satisfies [I] and [II]. Some basic pigeonholing further restricts the differences to three admissible runs [{52,9}, {61,61,9}, {61,61,61,9}] and issuing the sequence simplifies to evaluating ((n + 2)/log(2)) mod 1 once per run, comparing it to two constants, and measuring a low watermark for c_n as infrequently as possible.

%C Asymptotically, with {a, b} = {9/log(2), 52/log(2)} mod 1_[-1/2, 1/2), the three differences shuffle [0, c] by sending [c+a, c] to [0, -a] at n+9, [0, b] to [c-b, c] at n+52, and (b, c+a) to (-a, c-b) at n+61. 1254, 9892, 111768, 137237, 3194660, 11530771, 47096480, 208252803, 2084612060, 2581695828, 8931808997, 29473399808, 36320596745... is the subsequence of terms that transgress the asymptote, in that c < ((n + 2)/log(2)) mod 1 < c_n and a run that would be {61,61,61,9} splits to {52,9}, {61,61,9}.

%C Note finally that A172468 runs naturally from 52, 61, ... and that the offset to 50, 59, ... is an artifact of defining A166986 in terms of (n+2)/log(2) instead of n/log(2). (End)

%H R. M. Corless et al. <a href="https://doi.org/10.1007/BF02124750">On the Lambert W function.</a>, Adv. Comput. Math. 5 (1996), pp. 329-359.

%H Stephen Crowley, <a href="http://vixra.org/abs/1203.0004">A Mysterious Three Term Integer Sequence Related to a Lambert W Function Solution to a Certain Transcendental Equation</a> [Dead link ?]

%F a(n) ~ c*n, where c = 4*log(2)/(7 - 10*log(2)) = 40.4590949.... - _Travis Scott_, Oct 16 2022

%p [ListTools[SearchAll](1, [seq(round(evalf(floor((n+2)/ln(2))-2-(argument(exp(-(ln(2)+LambertW(n, -ln(sqrt(2))))/ln(2)))*ln(2)+Im(LambertW(n,-ln(sqrt(2)))))/(2*Pi*ln(2)))), n = 1 .. 10000)])]

%t Select[Range@1888,Floor[(#+2)/Log@2]==Floor[Im@LambertW[#,-Log@2/2]/Log@4/Pi+7/2]&] (* Or, accelerated per comment: *) Module[{x=0,m,z=1},Flatten@Table[m=Mod[x/Log@2,1];If[m<145/2-201/Log@16||(m<z&&m<(z=145/2-(Im@LambertW[x,-Log@2/2]/2/Pi-x+50)/Log@2)),{x+=52},Table[x+=61,2+Boole[m<189-131/Log@2]]]~Join~{x+=9},18]]-2 (* _Travis Scott_, Oct 16 2022 *)

%Y Cf. A166986, A167389.

%K nonn

%O 1,1

%A _Stephen Crowley_, Feb 03 2010