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.

A014206 a(n) = n^2 + n + 2.

Original entry on oeis.org

2, 4, 8, 14, 22, 32, 44, 58, 74, 92, 112, 134, 158, 184, 212, 242, 274, 308, 344, 382, 422, 464, 508, 554, 602, 652, 704, 758, 814, 872, 932, 994, 1058, 1124, 1192, 1262, 1334, 1408, 1484, 1562, 1642, 1724, 1808, 1894, 1982, 2072, 2164, 2258, 2354, 2452, 2552
Offset: 0

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Author

Keywords

Comments

Draw n + 1 circles in the plane; sequence gives maximal number of regions into which the plane is divided. Cf. A051890, A386480.
Number of binary (zero-one) bitonic sequences of length n + 1. - Johan Gade (jgade(AT)diku.dk), Oct 15 2003
Also the number of permutations of n + 1 which avoid the patterns 213, 312, 13452 and 34521. Example: the permutations of 4 which avoid 213, 312 (and implicitly 13452 and 34521) are 1234, 1243, 1342, 1432, 2341, 2431, 3421, 4321. - Mike Zabrocki, Jul 09 2007
If Y is a 2-subset of an n-set X then, for n >= 3, a(n-3) is equal to the number of (n-3)-subsets and (n-1)-subsets of X having exactly one element in common with Y. - Milan Janjic, Dec 28 2007
With a different offset, competition number of the complete tripartite graph K_{n, n, n}. [Kim, Sano] - Jonathan Vos Post, May 14 2009. Cf. A160450, A160457.
A related sequence is A241119. - Avi Friedlich, Apr 28 2015
From Avi Friedlich, Apr 28 2015: (Start)
This sequence, which also represents the number of Hamiltonian paths in K_2 X P_n (A200182), may be represented by interlacing recursive polynomials in arithmetic progression (discriminant =-63). For example:
a(3*k-3) = 9*k^2 - 15*k + 8,
a(3*k-2) = 9*k^2 - 9*k + 4,
a(3*k-1) = 9*k^2 - 3*k + 2,
a(3*k) = 3*(k+1)^2 - 1. (End)
a(n+1) is the area of a triangle with vertices at (n+3, n+4), ((n-1)*n/2, n*(n+1)/2),((n+1)^2, (n+2)^2) with n >= -1. - J. M. Bergot, Feb 02 2018
For prime p and any integer k, k^a(p-1) == k^2 (mod p^2). - Jianing Song, Apr 20 2019
From Bernard Schott, Jan 01 2021: (Start)
For n >= 1, a(n-1) is the number of solutions x in the interval 0 <= x <= n of the equation x^2 - [x^2] = (x - [x])^2, where [x] = floor(x). For n = 3, the a(2) = 8 solutions in the interval [0, 3] are 0, 1, 3/2, 2, 9/4, 5/2, 11/4 and 3.
This is a variant of the 4th problem proposed during the 20th British Mathematical Olympiad in 1984 (see A002061). The interval [1, n] of the Olympiad problem becomes here [0, n], and only the new solution x = 0 is added. (End)
See A386480 for the almost identical sequence 1, 2, 4, 8, 14, 22, 32, 44, 58, 74, 92, 112, 134, ... which is the maximum number of regions that can be formed in the plane by drawing n circles, and the maximum number of regions that can be formed on the sphere by drawing n great circles. - N. J. A. Sloane, Aug 01 2025

Examples

			a(0) = 0^2 + 0 + 2 = 2.
a(1) = 1^2 + 1 + 2 = 4.
a(2) = 2^2 + 2 + 2 = 8.
a(6) = 4*5/5 + 5*6/5 + 6*7/5 + 7*8/5 + 8*9/5 = 44. - _Bruno Berselli_, Oct 20 2016
		

References

  • K. E. Batcher, Sorting Networks and their Applications. Proc. AFIPS Spring Joint Comput. Conf., Vol. 32, pp. 307-314 (1968). [for bitonic sequences]
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 73, Problem 3.
  • T. H. Cormen, C. E. Leiserson and R. L. Rivest, Introduction to Algorithms. MIT Press / McGraw-Hill (1990) [for bitonic sequences]
  • Indiana School Mathematics Journal, vol. 14, no. 4, 1979, p. 4.
  • D. E. Knuth, The Art of Computer Programming, vol3: Sorting and Searching, Addison-Wesley (1973) [for bitonic sequences]
  • J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 177.
  • Derrick Niederman, Number Freak, From 1 to 200 The Hidden Language of Numbers Revealed, A Perigee Book, NY, 2009, p. 83.
  • A. M. Yaglom and I. M. Yaglom, Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 13, #44 (First published: San Francisco: Holden-Day, Inc., 1964)

Crossrefs

Cf. A014206 (dim 2), A046127 (dim 3), A059173 (dim 4), A059174 (dim 5).
A row of A059250.
Cf. A000124, A051890, A002522, A241119, A033547 (partial sums).
Cf. A002061 (central polygonal numbers).
Column 4 of A347570.

Programs

Formula

G.f.: 2*(x^2 - x + 1)/(1 - x)^3.
n hyperspheres divide R^k into at most C(n-1, k) + Sum_{i = 0..k} C(n, i) regions.
a(n) = A002061(n+1) + 1 for n >= 0. - Rick L. Shepherd, May 30 2005
Equals binomial transform of [2, 2, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jun 18 2008
a(n) = A003682(n+1), n > 0. - R. J. Mathar, Oct 28 2008
a(n) = a(n-1) + 2*n (with a(0) = 2). - Vincenzo Librandi, Nov 20 2010
a(0) = 2, a(1) = 4, a(2) = 8, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. - Harvey P. Dale, May 14 2011
a(n + 1) = n^2 + 3*n + 4. - Alonso del Arte, Apr 12 2015
a(n) = Sum_{i=n-2..n+2} i*(i + 1)/5. - Bruno Berselli, Oct 20 2016
Sum_{n>=0} 1/a(n) = Pi*tanh(Pi*sqrt(7)/2)/sqrt(7). - Amiram Eldar, Jan 09 2021
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(sqrt(11)*Pi/2)*sech(sqrt(7)*Pi/2).
Product_{n>=0} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)*sech(sqrt(7)*Pi/2). (End)
a(n) = 2*A000124(n). - R. J. Mathar, Mar 14 2021
E.g.f.: exp(x)*(2 + 2*x + x^2). - Stefano Spezia, Apr 30 2022

Extensions

More terms from Stefan Steinerberger, Apr 08 2006

A046127 Maximal number of regions into which space can be divided by n spheres.

Original entry on oeis.org

0, 2, 4, 8, 16, 30, 52, 84, 128, 186, 260, 352, 464, 598, 756, 940, 1152, 1394, 1668, 1976, 2320, 2702, 3124, 3588, 4096, 4650, 5252, 5904, 6608, 7366, 8180, 9052, 9984, 10978, 12036, 13160, 14352, 15614, 16948, 18356, 19840, 21402, 23044
Offset: 0

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Author

Keywords

Comments

If Y is a 2-subset of an n-set X then, for n >= 2, a(n-2) is equal to the number of 2-subsets and 4-subsets of X having exactly one element in common with Y. - Milan Janjic, Dec 28 2007

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 73, Problem 4.
  • A. M. Yaglom and I. M. Yaglom: Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 13, #45 (First published: San Francisco: Holden-Day, Inc., 1964).

Crossrefs

Cf. A014206 (dim 2), this sequence (dim 3), A059173 (dim 4), A059174 (dim 5). See also A000124, A000125. A row of A059250.
Cf. A033547.

Programs

  • Mathematica
    Join[{0},Table[n (n^2-3n+8)/3,{n,50}]]  (* Harvey P. Dale, Apr 21 2011 *)
  • Python
    def a(n): return n*(n**2 - 3*n + 8)//3 # Philip C. Ritchey, Dec 10 2017

Formula

a(n) = f(n,3) where f(n,k) = C(n-1, k) + Sum_{i=0..k} C(n, i) for hyperspheres in R^k.
a(n) = n*(n^2 - 3*n + 8)/3.
From Philip C. Ritchey, Dec 09 2017: (Start)
The above identity proved as closed form of the following summation and its corresponding recurrence relation:
a(n) = Sum_{i=1..n} (i*(i-3) + 4).
a(n) = a(n-1) + n*(n-3) + 4, a(0) = 0. (End)
From Colin Barker, Jan 28 2012: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: 2*x*(1 - 2*x + 2*x^2)/(1 - x)^4. (End)
a(n) = A033547(n-1) + 2 for n >= 1. - Jianing Song, Feb 03 2024
E.g.f.: exp(x)*x*(6 + x^2)/3. - Stefano Spezia, Feb 15 2024

A059173 Maximal number of regions into which 4-space can be divided by n hyperspheres.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 62, 114, 198, 326, 512, 772, 1124, 1588, 2186, 2942, 3882, 5034, 6428, 8096, 10072, 12392, 15094, 18218, 21806, 25902, 30552, 35804, 41708, 48316, 55682, 63862, 72914, 82898, 93876, 105912, 119072, 133424, 149038
Offset: 0

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Author

N. J. A. Sloane, Feb 15 2001

Keywords

Comments

n hyperspheres divide R^k into at most C(n-1, k) + Sum_{i=0..k} C(n, i) regions.
From Raphie Frank Nov 24 2012: (Start)
Define the gross polygonal sum, GPS(n), of an n-gon as the maximal number of combined points (p), intersections (i), connections (c = edges (e) + diagonals (d)) and areas (a) of a fully connected n-gon, plus the area outside the n-gon. The gross polygonal sum (p + i + c + a + 1) is equal to this sequence and, for all n > 0, then individual components of this sum can be calculated from the first 5 entries in row (n-1) of Pascal's triangle.
For example, the gross polygonal sum of a 7-gon (the heptagon):
Let row 6 of Pascal's triangle = {1, 6, 15, 20, 15, 6, 1} = A B C D E F G.
Points = 1 + 6 = A + B = 7 [A000027(n)].
Intersections = 20 + 15 = D + E = 35 [A000332(n+2)].
Connections = 6 + 15 = B + C = 21 [A000217(n)].
Areas inside = 15 + 20 + 15 = C + D + E = 50 [A006522(n+1)].
Areas outside = 1 = A = 1 [A000012(n)].
Then, GPS(7) = 7 + 35 + 21 + 50 + 1 = 2(A + B + C + D + E) = 114 = a(7). In general, a(n) = GPS(n). (End)

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 73, Problem 4.

Crossrefs

Cf. A014206 (dim 2), A046127 (dim 3), A059173 (dim 4), A059174 (dim 5).
A row of A059250.

Programs

  • Mathematica
    LinearRecurrence[{5,-10,10,-5,1},2^Range[0,5],50] (* Paolo Xausa, Dec 29 2023 *)

Formula

a(0) = 1; a(n) = 2 * A000127(n), for n >= 1.
G.f.: -(x^5 + x^4 - 2*x^3 + 4*x^2 - 3*x + 1)/(x-1)^5. - Colin Barker, Oct 06 2012
E.g.f.: exp(x)*(2 + x^2 + x^4/12) - 1. - Stefano Spezia, May 19 2024

A059174 Maximal number of regions into which 5-space can be divided by n hyperspheres.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 126, 240, 438, 764, 1276, 2048, 3172, 4760, 6946, 9888, 13770, 18804, 25232, 33328, 43400, 55792, 70886, 89104, 110910, 136812, 167364, 203168, 244876, 293192, 348874, 412736, 485650, 568548, 662424, 768336, 887408, 1020832, 1169870
Offset: 0

Views

Author

N. J. A. Sloane, Feb 15 2001

Keywords

Comments

n hyperspheres divide R^k into at most binomial(n-1, k) + Sum_{i=0..k} binomial(n, i) regions.

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 73, Problem 4.

Crossrefs

Cf. A014206 (dim 2), A046127 (dim 3), A059173 (dim 4), this sequence (dim 5).
Fifth row (k=5) of A059250.
Cf. A006261.

Programs

  • GAP
    Concatenation([1], List([1..40], n-> Binomial(n-1,5) + Sum([0..5], i-> Binomial(n,i)))); # Muniru A Asiru, Dec 18 2018
    
  • Magma
    [1] cat [(n^5-5*n^4+25*n^3+5*n^2+94*n+120)/60: n in [0..40]]; // Vincenzo Librandi, Dec 21 2018
  • Maple
    seq(coeff(series((x^6+3*x^4-6*x^3+7*x^2-4*x+1)/(1-x)^6,x,n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Dec 18 2018
  • Mathematica
    Join[{1}, Table[((n^5 - 5 n^4 + 25 n^3 + 5 n^2 + 94 n + 120) / 60), {n, 0, 50}]] (* Vincenzo Librandi, Dec 21 2018 *)
  • PARI
    a(n) = binomial(n-1, 5) + sum(i=0, 5, binomial(n, i)); \\ Michel Marcus, Jan 29 2016
    

Formula

a(n) = binomial(n-1, 5) + Sum_{i=0..5} binomial(n, i).
G.f.: (x^6 + 3*x^4 - 6*x^3 + 7*x^2 - 4*x + 1)/(x - 1)^6. - Colin Barker, Oct 06 2012
a(n) = 2*A006261(n-1), for n > 0. - Günter Rote, Dec 18 2018, by elementary manipulations.
E.g.f.: 1 + (1/60)*(120*x + 20*x^3 + x^5)*exp(x). - Franck Maminirina Ramaharo, Dec 21 2018

A178522 Triangle read by rows: T(n,k) is the number of nodes at level k in the Fibonacci tree of order n (n>=0, 0<=k<=n-1).

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 1, 2, 4, 2, 1, 2, 4, 6, 2, 1, 2, 4, 8, 8, 2, 1, 2, 4, 8, 14, 10, 2, 1, 2, 4, 8, 16, 22, 12, 2, 1, 2, 4, 8, 16, 30, 32, 14, 2, 1, 2, 4, 8, 16, 32, 52, 44, 16, 2, 1, 2, 4, 8, 16, 32, 62, 84, 58, 18, 2, 1, 2, 4, 8, 16, 32, 64, 114, 128, 74, 20, 2, 1, 2, 4, 8, 16, 32, 64, 126
Offset: 0

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Author

Emeric Deutsch, Jun 15 2010

Keywords

Comments

A Fibonacci tree of order n (n>=2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node.
Sum of entries in row n is A001595(n).
Sum_{k=0..n-1} k*T(n,k) = A178523(n).

Examples

			Triangle starts:
1,
1,
1,2,
1,2,2,
1,2,4,2,
1,2,4,6,2,
1,2,4,8,8,2,
1,2,4,8,14,10,2,
1,2,4,8,16,22,12,2,
1,2,4,8,16,30,32,14,2,
...
		

References

  • D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.

Crossrefs

Cf. A001595, A059214, A178523, A067331, A002940. See A059250 for another version.

Programs

  • Maple
    G := (1-t*z+t*z^2)/((1-z)*(1-t*z-t*z^2)): Gser := simplify(series(G, z = 0, 17)): for n from 0 to 15 do P[n] := sort(coeff(Gser, z, n)) end do: 1; for n to 13 do seq(coeff(P[n], t, k), k = 0 .. n-1) end do; # yields sequence in triangular form

Formula

G.f.: G(t,z)=(1-tz+tz^2)/[(1-z)(1-tz-tz^2)].
T(k,n) = T(k-1,n-1)+T(k-1,n) with T(0,0)=1, T(k,0)=1 for k>0, T(0,n)=2 for n>0. - Frank M Jackson, Aug 30 2011

A059214 Square array T(k,n) = C(n-1,k) + Sum_{i=0..k} C(n,i) read by antidiagonals (k >= 1, n >= 1).

Original entry on oeis.org

2, 2, 4, 2, 4, 6, 2, 4, 8, 8, 2, 4, 8, 14, 10, 2, 4, 8, 16, 22, 12, 2, 4, 8, 16, 30, 32, 14, 2, 4, 8, 16, 32, 52, 44, 16, 2, 4, 8, 16, 32, 62, 84, 58, 18, 2, 4, 8, 16, 32, 64, 114, 128, 74, 20, 2, 4, 8, 16, 32, 64, 126, 198, 186, 92, 22, 2, 4, 8, 16, 32, 64
Offset: 1

Views

Author

N. J. A. Sloane, Feb 15 2001

Keywords

Comments

For k > 1, gives maximal number of regions into which k-space can be divided by n hyperspheres.
The maximum number of subsets of a set of n points in k-space that can be formed by intersecting it with a hyperplane. - Günter Rote, Dec 18 2018

Examples

			Array begins
   2 4 6  8 10 12 ...
   2 4 8 14 22 32 ...
   2 4 8 16 30 52 ...
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 73, Problem 4.

Crossrefs

Cf. A014206 (dim 2), A046127 (dim 3), A059173 (dim 4), A059174 (dim 5).
Equals twice A216274.
Apart from left border, same as A059250. A178522 is probably the best version.

Programs

  • Mathematica
    A059214[k_,n_]:=Binomial[n-1,k]+Sum[Binomial[n,i],{i,0,k}];
    Table[A059214[k-n+1,n],{k,10},{n,k}] (* Paolo Xausa, Dec 29 2023 *)

Formula

T(k,n) = C(n-1, k) + Sum_{i=0..k} C(n, i).
Showing 1-6 of 6 results.