A078608 -id:A078608 - cp's OEIS frontend

A078607 Least positive integer x such that 2*x^n > (x+1)^n.

1, 2, 3, 4, 6, 7, 9, 10, 12, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 27, 29, 30, 32, 33, 35, 36, 38, 39, 40, 42, 43, 45, 46, 48, 49, 50, 52, 53, 55, 56, 58, 59, 61, 62, 63, 65, 66, 68, 69, 71, 72, 74, 75, 76, 78, 79, 81, 82, 84, 85, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102

Offset: 0

Jon Perry, Dec 09 2002

nonn

Also, integer for which E(s) = s^n - Sum_{0 < k < s} k^n is maximal. It appears that a(n) + 2 is the least integer for which E(s) < 0. - M. F. Hasler, May 08 2020

			a(2) = 3 as 2^2 = 4, 3^2 = 9 and 4^2 = 16.
For n = 777451915729368, a(n) = 1121626023352384 = ceiling(n log 2), where n*log(2) = 1121626023352383.5 - 2.13*10^-17 and 2*floor(n log 2)^n / floor(1 + n log 2)^n = 1 - 3.2*10^-32. - _M. F. Hasler_, Nov 02 2013
a(2) is given by floor(1/(1-1/sqrt(2))). [From former A230748.]

Vincenzo Librandi, Table of n, a(n) for n = 0..20000

Cf. A224996 (the largest integer x that satisfies 2*x^n <= (x+1)^n).
Cf. A050499, A050500.
Cf. A078608, A078609. Equals A110882(n)-1 for n > 0.
Cf. A332097 (maximum of E(s), cf comments), also related to this:  A332101 (least k such that k^n <= sum of all smaller n-th powers), A030052 (least k such that k^n = sum of distinct n-th powers), A332065 (all k such that k^n is a sum of distinct n-th powers).

Table[SelectFirst[Range@ 120, 2 #^n > (# + 1)^n &], {n, 0, 71}] (* Michael De Vlieger, May 01 2016, Version 10 *)

for (n=2,50, x=2; while (2*x^n<=((x+1)^n),x++); print1(x","))

a(n)=1\(1-1/2^(1/n)) \\ Charles R Greathouse IV, Oct 31 2013

apply( A078607(n)=ceil(1/if(n>1,sqrtn(2,n)-1,!n+n/2)), [0..80]) \\ M. F. Hasler, May 08 2020

a(n) = ceiling(1/(2^(1/n)-1)) for n > 1. (For n = 1 resp. 0 this gives the integer 1 resp. infinity as argument of ceiling.) [Edited by M. F. Hasler, May 08 2020]
For most n, a(n) is the nearest integer to n/log(2), but there are exceptions, including n=777451915729368.
Following formulae merged in from former A230748, M. F. Hasler, May 14 2020:
a(n) = floor(1/(1-1/2^(1/n))).
a(n) = n/log(2) + O(1). - Charles R Greathouse IV, Oct 31 2013
a(n) = floor(1/(1-x)) with x^n = 1/2: f(n) = 1/(2^(1/n)-1) is never an integer for n > 1, whence floor(f(n)+1) = ceiling(f(n)) = a(n). - M. F. Hasler, Nov 02 2013, and Gabriel Conant, May 01 2016

Edited by Dean Hickerson, Dec 17 2002

Initial terms a(0) = 1 and a(1) = 2 added by M. F. Hasler, Nov 02 2013

A129935 Numbers n such that ceiling( 2/(2^(1/n)-1) ) is not equal to floor( 2n/log(2) ).

Original entry on oeis.org

777451915729368, 140894092055857794, 1526223088619171207, 3052446177238342414, 54545811706258836911039145, 624965662836733496131286135873807507, 1667672249427111806462471627630318921648499, 36465374036664559522628534720215805439659141

Offset: 1

Richard Stanley, Apr 30 2007 (who sent a(1))

nonn

If n belongs to this sequence and m = ceiling(2/(2^(1/n)-1)), then 0 < m/(2n) - 1/log(2) < (log(2)/3) * (1/(2n)^2) implying that m/(2n) is a convergent of 1/log(2) (note that m and 2n are not necessarily coprime). - Max Alekseyev, Jun 06 2007
From David Applegate, Jun 07 2007: (Start)
"Some background to Max Alekseyev's comments: The key point is that the Laurent series for 2/(2^(1/n)-1) about n=infinity is 2/log(2)*n - 1 + (1/6)*log(2)/n + O(1/n^3).
"Also, since 2/log(2) is irrational, 2n/log(2) is never integral, so floor(2n/log(2)) = ceiling(2n/log(2)-1).
"So the question becomes: when is 2n/log(2)-1 so close to an integer that 2/(2^(1/n)-1) is on the other side of the integer? That is why the continued fraction expansion of 2/log(2) is relevant." (End)
The appropriate generalization of ceiling(2/(2^(1/n)-1)) = ? floor(2n/log(2)) is floor(a/(b^(1/n)-1)+a/2) = ceiling(an/log(b)). When a=2, the a/2 can be hidden in floor() + 1 = ceiling(). - David Applegate, Jun 08 2007 [edited Jun 11 2007]

S. W. Golomb and A. W. Hales, "Hypercube Tic-Tac-Toe", in "More Games of No Chance", ed. R. J. Nowakowski, MSRI Publications 42, Cambridge University Press, 2002, pp. 167-182. Here it is stated that the first counterexample is at n=6847196937, an error due to faulty multiprecision arithmetic. The correct value was found by J. Buhler in 2004 and is reported in S. Golomb, "Martin Gardner and Tictacktoe," in Demaine, Demaine, and Rodgers, eds., A Lifetime of Puzzles, A K Peters, 2008, pp. 293-301.
Dean Hickerson, Email to Jon Perry and N. J. A. Sloane, Dec 16 2002. Gives first three terms: 777451915729368, 140894092055857794, 1526223088619171207, as well as five later terms. - N. J. A. Sloane, Apr 30 2014

Max Alekseyev and Robert Gerbicz, Table of n, a(n) for n = 1..100
K. O'Bryant, The sequence of fractional parts of roots, arXiv preprint arXiv:1410.2927 [math.NT], 2014-2015.
Max Alekseyev and others, Integer Parts [in Russian]
Art of Problem Solving, Logarithmic identity
S. W. Golomb and A. W. Hales, Hypercube Tic-Tac-Toe, More Games of No Chance, MSRI Publications, Vol. 42, 2002.
N. J. A. Sloane, Two Sequences that Agree for a Long Time (Vugraph from a talk about the OEIS)

Cf. A078608 for the sequence ceiling( 2/(2^(1/n)-1) ).

Cf. A016730, A120754, A120755.

(* Mma 9.0.1 code from Bill Gosper, Mar 15 2013. He comments: "This reproduces the hundred values in the b-file, and probably works up to around half a billion digits. When Mathematica gets fixed, change 999999999 to infinity." *)
$MaxExtraPrecision = 999999999; For[{lo = {0, 1}, hi = {1, 0}, nu = {0, 0}, n = 0}, nu[[2]] < 10^386, nu = lo + hi; For[{k = nu[[2]]}, Floor[k*2/Log[2]] != Ceiling[2/(2^(1/k) - 1)], k += nu[[2]], Print[{++n, k}]];
  If[nu[[1]]*Log[2] > 2*nu[[2]], hi = nu, lo = nu]]

prec=1500;default(realprecision,prec);c=contfrac(log(2)/2);default(realprecision,prec*2+50); i=0;for(n=2,#c-1, cand=contfracpnqn(vecextract(c,2^n-1))[1,1];forstep(m=cand,c[n+1]*cand,cand, if(ceil(2/(2^(1/m)-1)) != floor(2*m/log(2)), i++;print(i" "m), break))) /* Phil Carmody, Mar 20 2013 */

More terms from Max Alekseyev, Jun 06 2007

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 08 2007

A078609 Least positive integer x such that 2*x^n>(x+3)^n.

Original entry on oeis.org

8, 12, 16, 21, 25, 29, 34, 38, 42, 47, 51, 55, 60, 64, 68, 73, 77, 81, 86, 90, 94, 99, 103, 107, 112, 116, 120, 125, 129, 133, 138, 142, 146, 150, 155, 159, 163, 168, 172, 176, 181, 185, 189, 194, 198, 202, 207, 211, 215, 220, 224, 228, 233, 237, 241, 246, 250

Offset: 2

Jon Perry, Dec 09 2002

nonn

			a(2)=8 as 7^2=49, 8^2=64, 10^2=100 and 11^2=121.

Cf. A078607, A078608.

for (n=2,50, x=2; while (2*x^n<=((x+3)^n),x++); print1(x","))

a(n) = ceiling(3/(2^(1/n)-1)). For most n, a(n) = floor(3n/log(2)-1/2), but there are exceptions, starting with n=32 and n=52113.

Edited by Dean Hickerson, Dec 17 2002

A258703 a(n) = floor(n/sqrt(2) - 1/2).

Original entry on oeis.org

0, 0, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 9, 10, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 17, 18, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25, 26, 27, 27, 28, 29, 29, 30, 31, 32, 32, 33, 34, 34, 35, 36, 36, 37, 38, 39, 39, 40, 41, 41, 42, 43, 44, 44, 45, 46, 46, 47, 48, 48, 49, 50, 51, 51, 52, 53, 53

Offset: 1

N. J. A. Sloane, Jun 09 2015

nonn

From Michel Dekking, Aug 11 2022: (Start)
By definition, (a(n)) is an inhomogeneous Beatty sequence. The associated Sturmian word  is s(alpha, rho) = (floor((n + 1)*alpha + rho) - floor(n*alpha + rho), n= 0, 1, 2,...) = 1,0,1,1,1,0,1,1,0,1,1,...,  with slope alpha = sqrt(2)/2, and intercept rho = -1/2.
Also, s(alpha, rho) = s(alpha,rho+1) - 1. Since 0 < alpha < 1 and 0 < rho +1 < 1, with algebraic conjugates
alpha* = -sqrt(2)/2, and (rho +1)* = 1/2,
Yasutomi's criterion gives that s(alpha, rho) is fixed point of a morphism.
The morphism can be found following the ideas of Chapter 2 in Lothaire's book and Section 4 of my paper "Substitution invariant Sturmian words and binary trees" (cf. A006340).
For a better fit with the literature we will determine the morphism that fixes the binary complement s(1-alpha, 1-(1+rho) ) = 0,1,0,0,0,1,0,0,1,0,0....
Let psi_1 and psi_8 be the elementary Sturmian morphisms given by
    psi_1(0)=01 , psi_1(1)=0;  psi_8(0)=01, psi(8)1=1.
Let psi := psi_1 psi_8. Then psi is given by
    psi(0)=010 , psi(1)=0.
We see that psi  fixes the Octanacci sequence A324772.
That psi is the right morphism can be proved by checking that  (x,y)  =  (1-alpha, -rho) is fixed point of the composition T_1 T_8 of the fractional linear maps
    T_1(x,y) = ((1-x)/(2-x), (1-y)/(2-x)),
    T_8(x,y) = ((1/(2-x), y/(2-x))).
Conclusion, taking the binary complement of psi: the Sturmian word equals A104521. (End)

Michel Dekking, Table of n, a(n) for n = 1..10000
Michel Dekking, Substitution invariant Sturmian words and binary trees, arXiv:1705.08607 [math.CO], (2017).
Michel Dekking, Substitution invariant Sturmian words and binary trees, Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A17.
M. Lothaire, Algebraic combinatorics on words, Cambridge University Press. Online publication date: April 2013; Print publication year: 2002.
K. O'Bryant, The sequence of fractional parts of roots, arXiv preprint, arXiv:1410.2927 [math.NT], 2014-2015.

Cf. A078608, A129935, A104521.

Cf. A049472, A001951.

a258703 = floor . (/ 2) . subtract 1 . (* sqrt 2) . fromIntegral
-- Reinhard Zumkeller, Jun 09 2015

[Floor(n/Sqrt(2) - 1/2): n in [1..80]]; // Vincenzo Librandi, Jun 09 2015

Table[Floor[n/Sqrt[2] - 1/2], {n, 1, 100}] (* Vincenzo Librandi, Jun 09 2015 *)

vector(100, n, n--; floor(n/sqrt(2) - 1/2)) \\ G. C. Greubel, Sep 30 2018

a(n) = floor(1/(exp(sqrt(2)/n)-1)) for all positive integers n [O'Bryant].

a(n) = floor((n*sqrt(2) - 1) / 2). - Reinhard Zumkeller, Jun 09 2015

Offset changed from 0 to 1 Michel Dekking, Aug 11 2022

A342831 a(n) is the smallest positive integer k such that the n-dimensional cube [0,k]^n contains at least as many internal lattice points as external lattice points.

Original entry on oeis.org

3, 6, 9, 12, 15, 18, 21, 24, 26, 29, 32, 35, 38, 41, 44, 47, 50, 52, 55, 58, 61, 64, 67, 70, 73, 76, 78, 81, 84, 87, 90, 93, 96, 99, 101, 104, 107, 110, 113, 116, 119, 122, 125, 127, 130, 133, 136, 139, 142, 145, 148, 151, 153, 156, 159, 162, 165, 168, 171, 174, 177, 179, 182

Offset: 1

Gary Yane, Mar 23 2021

nonn

			a(2) > 5 because the number of internal lattice points = 4^2 = 16 < 20 = 6^2 - 16 = the number of external lattice points, therefore a(2)=6 because the number of internal lattice points = 5^2 = 25 > 24 = 7^2 - 25 = number of external lattice points.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A078608.

a:= n-> ceil(1+2/(2^(1/n)-1)):
seq(a(n), n=1..65);  # Alois P. Heinz, Apr 20 2021

a[1] = 3; a[n_] := Floor[2^(1/n + 1)/(2^(1/n) - 1)]; Array[a, 100] (* Amiram Eldar, Mar 31 2021 *)

a(1) = 3 and a(n) = floor(2^(1/n+1)/(2^(1/n)-1)) for n > 1.

a(n) = A078608(n) + 1.

cp's OEIS Frontend

A078607 Least positive integer x such that 2*x^n > (x+1)^n.

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A129935 Numbers n such that ceiling( 2/(2^(1/n)-1) ) is not equal to floor( 2n/log(2) ).

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A078609 Least positive integer x such that 2*x^n>(x+3)^n.

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A258703 a(n) = floor(n/sqrt(2) - 1/2).

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A342831 a(n) is the smallest positive integer k such that the n-dimensional cube [0,k]^n contains at least as many internal lattice points as external lattice points.

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