cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

Views

Author

Jon Perry, Dec 09 2002

Keywords

Comments

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

Examples

			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.]

Links

Crossrefs

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).

Programs

  • Mathematica
    Table[SelectFirst[Range@ 120, 2 #^n > (# + 1)^n &], {n, 0, 71}] (* Michael De Vlieger, May 01 2016, Version 10 *)
  • PARI
    for (n=2,50, x=2; while (2*x^n<=((x+1)^n),x++); print1(x","))
  • PARI
    a(n)=1\(1-1/2^(1/n)) \\ Charles R Greathouse IV, Oct 31 2013
  • PARI
    apply( A078607(n)=ceil(1/if(n>1,sqrtn(2,n)-1,!n+n/2)), [0..80]) \\ M. F. Hasler, May 08 2020

Formula

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

Extensions

Edited by Dean Hickerson, Dec 17 2002
Initial terms a(0) = 1 and a(1) = 2 added by M. F. Hasler, Nov 02 2013

A078608 a(n) = ceiling(2/(2^(1/n)-1)).

Original entry on oeis.org

2, 5, 8, 11, 14, 17, 20, 23, 25, 28, 31, 34, 37, 40, 43, 46, 49, 51, 54, 57, 60, 63, 66, 69, 72, 75, 77, 80, 83, 86, 89, 92, 95, 98, 100, 103, 106, 109, 112, 115, 118, 121, 124, 126, 129, 132, 135, 138, 141, 144, 147, 150, 152, 155, 158, 161, 164, 167, 170, 173, 176, 178, 181
Offset: 1

Views

Author

Jon Perry, Dec 09 2002

Keywords

Comments

For n >= 2, a(n) is the least positive integer x such that 2*x^n > (x+2)^n. For example, a(2)=5 as 4^2=16, 5^2=25, 6^2=36 and 7^2=49.
Coincides with floor( 2*n/(log 2) ) for all n from 1 to 777451915729367 but differs at 777451915729368. See A129935.
The first few values of n for which this sequence differs from floor( 2*n/(log 2) ) were found by Dean Hickerson in 2002. - N. J. A. Sloane, Apr 30 2014
The sequence floor( log(n)/(2*log(2)) ) is mentioned by Erdős and Selfridge (1973). This sequence begins 0,0,0,1,1,1,1,... = 0 (3 times), 1 (12 times), 2 (48 times), 3 (192 times), 4 (768 times), ..., and grows too slowly to have its own entry. It is related to the game studied by Hales and Jewett (1963). - N. J. A. Sloane, Dec 02 2016

References

  • S. Golomb, "Martin Gardner and Tictacktoe," in Demaine, Demaine, and Rodgers, eds., A Lifetime of Puzzles, A K Peters, 2008, pp. 293-301.
  • 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 Dean Hickerson in 2002, and J. Buhler in 2004, and is reported in S. Golomb (2008).
  • Dean Hickerson, Email to Jon Perry and N. J. A. Sloane, Dec 16 2002. Gives first three terms of A129935: 777451915729368, 140894092055857794, 1526223088619171207, as well as five later terms. - N. J. A. Sloane, Apr 30 2014

Links

Crossrefs

Cf. A078607, A078609, A129935.

Programs

  • Haskell
    a078608 = ceiling . (2 /) . (subtract 1) . (2 **) . recip . fromIntegral
-- Reinhard Zumkeller, Mar 27 2015
  • Mathematica
    Table[(Ceiling[2/(2^(1/n)-1)]), {n, 1, 100}] (* Vincenzo Librandi, May 01 2014 *)
  • PARI
    for (n=2,50, x=2; while (2*x^n<=((x+2)^n),x++); print1(x","))

Extensions

Edited by Dean Hickerson, Dec 17 2002
Revised by N. J. A. Sloane, Jun 07 2007
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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