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.

Showing 1-6 of 6 results.

A093005 a(n) = n * ceiling(n/2).

Original entry on oeis.org

1, 2, 6, 8, 15, 18, 28, 32, 45, 50, 66, 72, 91, 98, 120, 128, 153, 162, 190, 200, 231, 242, 276, 288, 325, 338, 378, 392, 435, 450, 496, 512, 561, 578, 630, 648, 703, 722, 780, 800, 861, 882, 946, 968, 1035, 1058, 1128, 1152, 1225, 1250, 1326, 1352, 1431, 1458
Offset: 1

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Author

Amarnath Murthy, Mar 29 2004

Keywords

Comments

Old name was: The lone multiple of n among the next n numbers.
Another old name: a(n) = n*floor((n+1)/2).
Consider the triangle
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
16 17 18 19 20 21
22 23 24 25 26 27 28
... Then sequence contains the multiple of n in the n-th row.
Interleaves A000384 and A001105. - Paul Barry, Jun 29 2006
Termwise products of the natural numbers and the natural numbers repeated.
Number of pairs (x,y) having the same parity, with x in {0,...,n} and y in {0,...,2n}. - Clark Kimberling, Jul 02 2012
Similar to generalized hexagonal numbers A000217. Other members of this family are A210977, A006578, A210978, A181995, A210981, A210982. - Omar E. Pol, Aug 09 2012
For even n, a(n) gives the sum of all the parts in the partitions of n into exactly two parts. For odd n>1, a(n) gives n plus the sum of all the parts in the partitions of n into exactly two parts. - Wesley Ivan Hurt, Nov 14 2013
Number of regions of the plane that do not contain the origin, in the arrangement of lines with polar equations rho = 1/cos(theta-k*2*Pi/n), k=0..n-1; or, by inversion, number of bounded regions in the arrangement of circles with radius 1 and centers the n-th roots of unity. - Luc Rousseau, Feb 08 2019
Numbers k such that floor(sqrt(2k)+1/2) | k. - Wesley Ivan Hurt, Dec 01 2020

Crossrefs

Programs

Formula

a(n) = n*floor((n+1)/2).
a(n) = n*A008619(n).
a(2*n-1) = n*(2*n-1), a(2*n) = 2*n^2.
From Paul Barry, Jun 29 2006: (Start)
G.f.: x*(1+x+2*x^2)/((1+x)^2*(1-x)^3).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) = n*((2*n+1) - (-1)^n)/4. (End)
a(n) = n * ceiling(n/2). - Wesley Ivan Hurt, Nov 14 2013
E.g.f.: (1/2)*x*( (x+2)*cosh(x) + (x+1)*sinh(x) ). - G. C. Greubel, Mar 14 2024
Sum_{n>=1} 1/a(n) = Pi^2/12 + 2*log(2). - Amiram Eldar, Mar 15 2024
a(n) = A183207(n) - A370980(n) + 1, by Euler's formula. - Scott R. Shannon, Jul 07 2024

Extensions

Corrected and extended by Joshua Zucker, May 08 2006
New name from Alex Ratushnyak, Apr 26 2012
New name from Wesley Ivan Hurt, Nov 14 2013

A371254 Number of vertices formed when n equally spaced points are placed around a circle and all pairs of points are joined by an interior arc whose radius equals the circle's radius.

Original entry on oeis.org

1, 2, 4, 4, 15, 7, 70, 64, 208, 220, 550, 397, 1131, 1162, 1981, 2128, 3723, 3259, 5966, 6000, 9010, 9240, 13524, 12745, 19325, 19266, 26434, 26684, 35931, 33301, 47368, 47616, 61216, 61676, 78330, 76789, 98901, 99674, 122656, 123560
Offset: 1

Views

Author

Scott R. Shannon, Mar 16 2024

Keywords

Comments

Other than for n = 3, 4, and 6, all graphs so far investigated in this sequence contain some internal vertices which are created from the intersections of both 2 and 3 arcs, i.e., no graph contains only simple intersections. This is in contrast to the case where the point pairs are connected by straight lines, see A007569 and A335102, where the odd-n graphs contain only simple intersections. See the attached images.
Other patterns for the intersection arc counts are also seen. If n is divisible by 3 then a central vertex is always present that is created from the crossing of n arcs. If n is divisible by 6, then internal vertices are present that are created from the crossing of 6 arcs. For n = 15 and n = 45, internal vertices are present that are created from the crossing of 5 arcs - it is likely all graphs with n = 15+30*k, k>=0, contain such vertices.
For n = 30, the graph also contains internal vertices that are created from the crossing of 9 arcs. It is likely that all graphs with n divisible by 30 contain such vertices. As the graphs created from the straight line diagonal intersections of the regular n-gon, see A007569, also have the maximum possible line intersection count of 7 when n is divisible by 30, it is plausible that 9 is the maximum possible arc intersection count for any internal vertex, other than the central vertex when n is divisible by 3.
Assuming these patterns hold for all n, is it possible that there is a general formula for the number of vertices, analogous to that in A007569 for the intersections of chords in a regular n-gon?

Crossrefs

Cf. A371253 (regions), A371255 (edges), A371274 (k-gons), A371264 (vertex crossings), A370980 (number of circles), A371373 (complete circles), A007569, A335102, A358746, A331702, A359252.

Formula

a(n) = A371255(n) - A371253(n) + 1 by Euler's formula.

A183207 Termwise products of the natural numbers and odd integers repeated.

Original entry on oeis.org

1, 2, 9, 12, 25, 30, 49, 56, 81, 90, 121, 132, 169, 182, 225, 240, 289, 306, 361, 380, 441, 462, 529, 552, 625, 650, 729, 756, 841, 870, 961, 992, 1089, 1122, 1225, 1260, 1369, 1406, 1521, 1560, 1681, 1722, 1849, 1892, 2025
Offset: 1

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Author

Gary W. Adamson, Feb 11 2011

Keywords

Comments

There is a chessboard of n^2 squares. A pawn is standing on the lower left corner of the chessboard O (0,0) and its primary goal is to reach the upper right corner of the chessboard N (n,n). The only moves allowed are diagonal shortcuts through squares. Once a square is crossed it is destroyed so that it is impossible to cross again. The secondary goal of the pawn on its way to N is to destroy as many squares as possible. a(n) is the maximum possible number of destroyed squares, provided the pawn has reached its primary goal. - Ivan N. Ianakiev, Feb 23 2014
The sequence gives the number of curved edges created from circle intersections when a circle of radius r is drawn around each of n equally spaced points on the circumference of a circle of radius r. The number of regions in these constructions is A093005(n) and the number of vertices is A370980(n). See the attached images. - Scott R. Shannon, Jul 07 2024

Examples

			a(4) = 4*3 = 12.
		

Crossrefs

Programs

  • Magma
    I:=[1,2,9,12,25]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..60]]; // Vincenzo Librandi, Feb 23 2014
  • Mathematica
    a[n_] := n (2 Floor[(n - 1)/2] + 1); Array[a, 45] (* Robert G. Wilson v, Feb 11 2011 *)
    CoefficientList[Series[(-1 - x - 5 x^2 - x^3)/((1 + x)^2 (x - 1)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Feb 23 2014 *)
  • PARI
    a(n) = n^2-n*(1+(-1)^n)/2;
    

Formula

Termwise products of (1, 2, 3, 4, 5, 6, 7, 8...) and (1, 1, 3, 3, 5, 5, 7, 7,...).
From R. J. Mathar, Feb 12 2011: (Start)
G.f.: x*( -1-x-5*x^2-x^3 ) / ( (1+x)^2*(x-1)^3 ).
a(n) = n^2-n*(1+(-1)^n)/2. (End)
Sum_{n>=1} 1/a(n) = Pi^2/8 + log(2). - Amiram Eldar, Mar 15 2024
a(n) = A093005(n) + A370980(n) - 1, by Euler's formula. - Scott R. Shannon, Jul 07 2024

A371253 Number of regions formed when n equally spaced points are placed around a circle and all pairs of points are joined by an interior arc whose radius equals the circle's radius.

Original entry on oeis.org

1, 1, 6, 5, 26, 18, 99, 89, 270, 271, 650, 516, 1288, 1303, 2250, 2337, 4047, 3636, 6404, 6401, 9597, 9769, 14261, 13632, 20251, 20125, 27594, 27749, 37324, 35040, 49043, 49185, 63228, 63547, 80676, 79380, 101640, 102259, 125853, 126561
Offset: 1

Views

Author

Scott R. Shannon, Mar 16 2024

Keywords

Comments

See A371254 for further information.

Crossrefs

Cf. A371254 (vertices), A371255 (edges), A371274 (k-gons), A370980 (number of circles), A371374 (complete circles), A006533, A358782, A359046, A359253, A007678.

Formula

a(n) = A371255(n) - A371254(n) + 1 by Euler's formula.

A374825 Place n equally spaced points on the circumference of a circle of radius r and then connect each pair of points with straight lines whose intersections create A007569(n) - n additional points. Draw a circle of radius r around each of the A007569(n) points. The sequence gives the total number of vertices formed from all circle intersections.

Original entry on oeis.org

0, 1, 4, 13, 71, 313, 1625, 3073, 17443, 28601, 115094, 95965, 527463, 587441
Offset: 1

Views

Author

Scott R. Shannon, Jul 21 2024

Keywords

Crossrefs

Cf. A374826 (regions), A374827 (edges), A374828 (k-gons), A007569 (total circles), A370980, A374338.

Formula

a(n) = A374827(n) - A374826(n) + 1, by Euler's formula.

A111869 Least number k such that C(2k,k) is divisible by n^2.

Original entry on oeis.org

1, 3, 5, 15, 13, 5, 25, 63, 41, 13, 61, 15, 85, 25, 14, 255, 145, 41, 181, 23, 25, 61, 265, 95, 313, 85, 365, 27, 421, 14, 481, 1023, 61, 145, 39, 53, 685, 181, 86, 63, 841, 25, 925, 61, 44, 265, 1105, 383, 1201, 313, 145, 85, 1405, 365, 63, 95, 181, 421, 1741, 23, 1861
Offset: 1

Views

Author

Robert G. Wilson v, Nov 23 2005

Keywords

Comments

From David A. Corneth, Aug 15 2025: (Start)
Conjecture 1: a(n) = (n^2 - 1)/2 + 1 for odd prime n.
Conjecture 2: Let q be the largest prime factor of n. Let e be the multiplicity of q in the factorization of n. Then a(n) >= (q^(2*e) - 1)/2 + 1. for n != 2.
These conjectures hold for n = 1..4002.
Conjecture 3: a(2^k) = 4^k - 1 for k >= 1.
This conjecture holds for k = 1..11. (End)

Examples

			From _David A. Corneth_, Aug 15 2025: (Start)
a(4) = 15 as 4^2 = 16 | binomial(2*15, 15) = binomial(30, 15) and for any k < 15 we have 16 does not divide binomial(2*k, k). We don't really need to compute binomial(30, 15) and not the previous binomial(2*k, k) but just find how many factors 2 they have. binomial(30, 15) = 30! / (15!)^2.
We find the 2-adic valuation of 30! as follows: let b(0) = 30 and let b(n + 1) = floor(b(n) / 2). Then the 2-adic valuation of 30! is Sum{k = 1..floor(log(30)/log(2))} b(k) = b(1) + b(2) + b(3) + b(4) = 15 + 7 + 3 + 1 = 26.
Similar for 15! it is 7 + 3 + 1 = 11. 26 - 2*11 = 4 >= 4 so a(4) <= 15 and checking the others gives a(4) = 15. (End)
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{k = 1, m = n^2}, While[ Mod[ Binomial[2k, k], m] != 0, k++ ]; k]; Array[f, 61]
  • PARI
    See Corneth link

Formula

a(n) = A073078(n^2).
Showing 1-6 of 6 results.