|
| |
|
|
A071111
|
|
a(n) is the least integer x such that there exists an integer in the open interval (x/(i+1), x/i) for i= n-1, n-2 ..., 3, 2, 1.
|
|
2
| |
|
|
3, 5, 7, 13, 17, 26, 31, 43, 57, 65, 82, 101, 111, 133, 157, 183, 197, 226, 257, 290, 307, 343, 381, 421, 463, 485, 530, 577, 626, 677, 703, 757, 813, 871, 931, 993, 1025, 1090, 1157, 1226, 1297, 1370, 1407, 1483, 1561, 1641, 1723, 1807, 1893, 1937, 2026, 2117
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 2,1
|
|
|
COMMENTS
| a(10) = 57, a(100) = 8191, a(1000) = 937993.
a(n) is the least integer that for k=1, 2, ..., (n-1) can be expressed as: a(n)=p*k + b for some positive integers p and b such that p>1 and p>b>0.
This is the same sequence (apart from the initial term) as A001000. The identity of these two sequences was first proved by Rustem Aidagulov and a detailed version of the proof can be found in the Alekseyev link below.
Comments from Christopher Carl Heckman (checkman(AT)mathi17.la.asu.edu), May 23 2004: "This problem was given in Crux Mathematicorum, Vol. 23 #6 (October 1997) as Problem #2272. A solution, which includes a general formula, can be found in Crux Mathematicorum, Vol. 24 #7 (November 1998): a(n) = floor (((n + x_n) / 2)^2 + 1), where x_n = floor (n + 1 - 2 sqrt (n - 1)).
"This formula was found by Florian Herzig (then a student at Cambridge, UK), who also proved that the proposer's conjecture that a(n) = cases (1 + (n-m)^2, if m^2 <= n - 2, 1 + (n-m)^2 + (n - m), otherwise) where m = floor ((1 + sqrt (4 n - 7)) / 2) also is true although 'the proof of this fact is quite challenging'.
"The problem was also solved by Peter Tingley (then an undergraduate student at the University of Waterloo, Waterloo, Ontario), who gave the alternate formula: a(n) = n y_n + floor ((n - y_n)^2 / 4 + 1), where y_n = floor (n - 2 sqrt(n - 1) + 1), which 'is readily seen to be the same as the one obtained by Herzig.'"
|
|
|
LINKS
| T. D. Noe, Table of n, a(n) for n=2..1000
Max Alekseyev, Proof that A001000 and A071111 are essentially the same sequence
|
|
|
FORMULA
| For 2<=n<=200, a(n) = n^2 - n*c(n) + floor(c(n)^2/4) + 1, where c(n) = floor(sqrt(4n-5)). Is this true for all n>=2?
|
|
|
EXAMPLE
| a(4)=7 because 2 lies in (7/4,7/3), 3 lies in (7/3,7/2) and 4, 5 and 6 lie in (7/2,7) and for x<7 the definition doesn't hold.
|
|
|
MATHEMATICA
| test[x_, n_] := Module[{k}, For[k=n, x<=k(k-1), k--, If[Ceiling[x/(k-1)]-Floor[x/k]<2, Return[False]]]; True]; a[n_] := For[x=1, True, x++, If[test[x, n], Return[x]]]
|
|
|
CROSSREFS
| Sequence in context: A104294 A171566 A137270 * A177070 A038929 A070806
Adjacent sequences: A071108 A071109 A071110 * A071112 A071113 A071114
|
|
|
KEYWORD
| nonn,nice
|
|
|
AUTHOR
| Fernando Delgado, Paul Monasterios and Adolfo Rodriguez (misterioso53(AT)hotmail.com), May 27 2002
|
|
|
EXTENSIONS
| Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com) and Robert G. Wilson v (rgwv(AT)rgwv.com), Jun 04 2002
|
| |
|
|
