A128127 -id:A128127 - cp's OEIS frontend

A128218 First differences of A128217.

1, 3, 1, 3, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 15, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

John W. Layman, Feb 19 2007

nonn

a(A130883(n-1)) = 2*n-1 and a(m) != 2*n-1 for m < A130883(n-1). - Reinhard Zumkeller, Jun 20 2015

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A128127.

Cf. A130883, A152271 (run lengths after initial term).

a128218 n = a128218_list !! (n-1)
a128218_list = zipWith (-) (tail a128217_list) a128217_list
-- Reinhard Zumkeller, Jun 20 2015

nsrQ[n_]:=Module[{sr=Sqrt[n]},Abs[First[sr-Nearest[{Floor[sr], Ceiling[ sr]}, sr]]]<1/4];Differences[Select[Range[0,250],nsrQ]] (* Harvey P. Dale, May 02 2012 *)

default(realprecision, 10000);
is_A128217(n) = ((abs(sqrt(n)-sqrtint(n))<(1/4)) || (abs(sqrt(n)-(1+sqrtint(n)))<(1/4)));
k=0; n=0; prevm=0; while(k<20000, n++; if(is_A128217(n), k++; write("b128218.txt", k, " ", (n-prevm)); prevm = n)); \\ Antti Karttunen, Jan 16 2025

Let A(1)={1}. Then, for k=2,3,4,..., form A(k) by appending to A(k-1) the term k-1 followed by k-1 1's, if k is even, or by appending to A(k-1) the term k followed by k-1 1's, if k is odd. {a(n)} appears to be the limit of {A(k)} as k->infinity.

Offset changed by Reinhard Zumkeller, Jun 20 2015

A128153 The number of regular pentagons found by constructing n equally-spaced points on each side of the pentagon and drawing lines parallel to the pentagon sides, as well as lines connecting vertices.

Original entry on oeis.org

1, 9, 20, 37, 58, 85, 116, 153, 194, 241, 292, 349, 410

Offset: 0

Noah Priluck (npriluck(AT)gmail.com), May 02 2007

Similar to constructions for A002717 (dividing a triangle), A000330 (dividing a square) and sequences pending for dividing other polygons.

Use 1 midpoint (resp. 2 points) on each side placed to divide each side into 2 (resp. 3) equally-sized segments or so on, do the same construction for every side of the pentagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least 1 side of the polygon. Also connect all vertices of the pentagon with lines that are parallel to at least 1 side of the pentagon.

			With 0 points, there is only 1 pentagon. With 1 point (a midpoint on each side), 9 regular pentagons are found. With 2 points, 20 regular pentagons are found in total.

Michel Marcus, Figure with 1 midpoint on each side
Noah Priluck, On Counting Regular Polygons Formed by Special Families of Parallel Lines, Geombinatorics Quarterly, Vol XVII (4), 2008, pp. 166-171. (Note that there is no document to download; see A128127 for PDF file.)

Conjecture: a(n) = (10*n^2 + 16*n + 9 -(-1)^n)/4 for n > 0.
From Chai Wah Wu, Oct 21 2017: (Start)
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 4 (conjectured).
G.f.: (-x^4 + x^3 - 2*x^2 - 7*x - 1)/((x - 1)^3*(x + 1)) (conjectured). (End)

Edited by Michel Marcus, Jul 10 2013

a(4)-a(12) from Giovanni Resta, Aug 20 2017

A128828 The number of regular hexagons found by constructing n equally-spaced points on each side of the hexagon and drawing lines parallel to the hexagon side.

Original entry on oeis.org

1, 2, 15, 28, 65, 120

Offset: 0

Noah Priluck (npriluck(AT)gmail.com), May 08 2007

Use 1 midpoint (resp. 2 points) on each side placed to divide each side into 2 (resp. 3) equally-sized segments (or so on), do the same construction for every side of the hexagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least 1 side of the hexagon.

Similar constructions to sequences A002717 (dividing a triangle), A000330 (dividing a square) and other sequences pending for similar constructions in other polygons.

			With 1 point (a midpoint on each side), 2 regular hexagons are found. With 3 points on each side, 15 regular hexagons are found in total and so on.

Noah Priluck, On Counting Regular Polygons Formed by Special Families of Parallel Lines, Geombinatorics Quarterly, Vol XVII (4), 2008, pp. 166-171. (note there is no document to download, see A128127 for pdf file).

Cf. A128127, A128153.

a(n) = (11*n^4 + 78*n^3 + 1413*n^2 - 2322*n + 324)/324 when n = 3k,
  (-13*n^4 + 670*n^3 - 3219*n^2 + 9934*n - 6724)/324 when n = 1 + 3k,
  (5*n^4-46*n^3 + 1515*n^2 - 6046*n + 7940)/108 when n = 2 + 3k (conjecture).

Edited by Michel Marcus, Jul 15 2013

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A128218 First differences of A128217.

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A128153 The number of regular pentagons found by constructing n equally-spaced points on each side of the pentagon and drawing lines parallel to the pentagon sides, as well as lines connecting vertices.

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A128828 The number of regular hexagons found by constructing n equally-spaced points on each side of the hexagon and drawing lines parallel to the hexagon side.

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