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 1-7 of 7 results.

A237347 First differences of A078633.

Original entry on oeis.org

3, 3, 2, 3, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2
Offset: 1

Views

Author

Reinhard Zumkeller, Mar 18 2014

Keywords

Comments

a(n) = A078633(n+1) - A078633(n);
2 <= a(n) <= 3;
a(A049068(n)) = 2; a(A002620(n)) = 3;
A237347(n) = abs(A167752(n)) + 2.

Programs

  • Haskell
    a237347 n = a237347_list !! (n-1)
    a237347_list = zipWith (-) (tail a078633_list) a078633_list

A182834 Complement of A007590, except for initial zeros.

Original entry on oeis.org

1, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87, 88, 89, 90, 91, 92
Offset: 1

Views

Author

Clark Kimberling, Jan 07 2011

Keywords

Comments

A245575(a(n)) mod 2 = 0, or, where even terms occur in A245575; a(A078633(n)) mod 2 = 0. - Reinhard Zumkeller, Aug 05 2014

Crossrefs

Programs

  • Haskell
    a182834 n = a000196 (2 * n - 2) + n  -- Reinhard Zumkeller, Aug 05 2014
    
  • Mathematica
      a=2; b=-2;
      Table[n+Floor[(a*n+b)^(1/2)],{n,80}]
      Table[n-1+Ceiling[(n*n-b)/a],{n,50}]
  • Python
    from math import isqrt
    def A182834(n): return n+isqrt(n-1<<1) # Chai Wah Wu, Jul 28 2022

Formula

a(n) = n + floor(sqrt(2n-2)).

A117227 Minimum number of unit segments required to construct n unit cubes.

Original entry on oeis.org

12, 20, 28, 33, 41, 46, 51, 54, 62, 67, 72, 75, 83, 88, 93, 96, 101, 104, 112, 117, 122, 125, 130, 133, 138, 141, 144
Offset: 1

Views

Author

Ron R. King, Apr 22 2006

Keywords

Comments

This generalizes the 2-dimensional version which is A078633.

Examples

			a(2)=20 because 12 unit segments are required to construct each cube but 4 units are shared. I.e., a(2)=12+8=20.
		

Crossrefs

Cf. A078633.

A331207 Minimum number of matchsticks sufficient to form the edges of exactly n squares of any size >= 1.

Original entry on oeis.org

0, 4, 7, 10, 13, 12, 15, 18, 17, 20, 23, 22, 25, 28, 24, 27, 30, 29, 32, 35, 31, 34, 37, 36, 39, 42, 38, 41, 44, 43, 40, 43, 46, 45
Offset: 0

Views

Author

John King, Jan 12 2020

Keywords

Crossrefs

A331208 The minimum perimeter for exactly n matchstick squares of size >= 1.

Original entry on oeis.org

0, 4, 6, 8, 10, 8, 10, 12, 10, 12, 14, 12, 14, 16, 12, 14, 16, 14, 16, 18, 14, 16, 18, 16, 18, 20, 16, 18, 18, 18, 16, 18, 20
Offset: 0

Views

Author

John King, Jan 12 2020

Keywords

Crossrefs

A141135 Minimal number of unit edges required to construct n regular pentagons when allowing edge-sharing.

Original entry on oeis.org

5, 9, 13, 17, 21, 24, 28, 32, 36, 39, 43, 47, 50, 54, 58, 61, 65, 69, 72, 76, 80, 83, 87, 90, 94, 98, 101, 105, 109, 112
Offset: 1

Views

Author

Ralph H. Buchholz (teufel_pi(AT)yahoo.com), Jun 08 2008

Keywords

Examples

			a(6) = 24 since the first pentagon requires 5 edges, the 2nd, 3rd, 4th and 5th pentagons require an additional 4 edges each and the 6th pentagon requires 3 edges since it can share 2 edges (if one tiles via a 6-cycle). Thus 24 = 5 + 4 + 4 + 4 + 4 + 3.
		

Crossrefs

Cf. equilateral triangles A137228, squares A078633, regular hexagons A135708.
Cf. A121149.

Formula

Conjectures from Colin Barker, Apr 05 2019: (Start)
G.f.: x*(5 + 4*x + 4*x^2 - x^3 - x^5 + x^8 - x^9) / ((1 - x)^2*(1 + x + x^2)).
a(n) = a(n-1) + a(n-3) - a(n-4) for n>10.
(End)
Conjecture: if n is a term in A121149, a(n) = a(n-1) + 3, otherwise a(n) = a(n-1) + 4. - Jinyuan Wang, Apr 05 2019

Extensions

a(21)-a(30) from Jinyuan Wang, Apr 05 2019

A275868 Numbers n tracing out a spiral path in a pentagonal Z module thereby creating a ten-fold twin pattern with relations to quasicrystals.

Original entry on oeis.org

0, 1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 4, 5, 6, 5, 6, 5, 6, 7, 6, 7, 6, 7, 6, 7, 8, 7, 8, 7, 8, 7, 8, 9, 8, 9, 8, 9, 8, 9, 8, 9, 0, 9, 0, 9, 0, 9, 0, 9, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 4, 3, 4, 3, 4, 3, 4, 3
Offset: 1

Views

Author

Wolfgang Hornfeck, May 19 2017

Keywords

Comments

Interpreted as consecutive steps along directions according to a basis of vectors represented by the tenth roots of unity in the complex number plane, the sequence traces out the path of a single spiral of a ten-fold twin pattern. All points are located on a pentagonal Z module (following the ideas of Quiquandon et al.). The ten-fold twin pattern is unique in that the local structure across the twin boundaries is identically coherent to the local structure within the twin domains. The ten-fold twin pattern is enantiomorphous, depending on the sign of the irrational shift of 1/(4*tau), with tau = (1+sqrt(5))/2 the Golden Ratio, along a [110] direction of the twin domain's orthorhombic unit cell. The sequence expresses the fact that the ten-fold twin pattern has no adjustable parameters, except for an arbitrary general scaling factor.

Crossrefs

Programs

  • Mathematica
    Table[Mod[Floor[Sqrt[2*(i-1)]]+If[MemberQ[Table[2*j+Ceiling[2*Sqrt[j]],{j,1,i}],i],1,0],10],{i,1,100}]

Formula

a(n) = floor(sqrt( 2*(n-1) )) + [n in { 2*k + ceiling(2*sqrt(k)) | k in N}] mod 10. Note, that floor(sqrt( 2*n )) is A172471 (here corrected for its offset in the combined formula), while 2*k + ceiling(2*sqrt(k)) is A078633. [] denotes the Iverson bracket.
Showing 1-7 of 7 results.