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 1-4 of 4 results.

A035008 Total number of possible knight moves on an (n+2) X (n+2) chessboard, if the knight is placed anywhere.

Original entry on oeis.org

0, 16, 48, 96, 160, 240, 336, 448, 576, 720, 880, 1056, 1248, 1456, 1680, 1920, 2176, 2448, 2736, 3040, 3360, 3696, 4048, 4416, 4800, 5200, 5616, 6048, 6496, 6960, 7440, 7936, 8448, 8976, 9520, 10080, 10656, 11248, 11856, 12480, 13120, 13776
Offset: 0

Views

Author

Ulrich Schimke (ulrschimke(AT)aol.com), Dec 11 1999

Keywords

Comments

16 times the triangular numbers A000217.
Centered 16-gonal numbers A069129, minus 1. Also, sequence found by reading the segment (0, 16) together with the line from 16, in the direction 16, 48, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Apr 26 2008, Nov 20 2008
For n >= 1, number of permutations of n+1 objects selected from 5 objects v, w, x, y, z with repetition allowed, containing n-1 v's. Examples: at n=1, n-1=0 (i.e., zero v's), and a(1)=16 because we have ww, wx, wy, wz, xw, xx, xy, xz, yw, yx, yy, yz, zw, zx, zy, zz; at n=2, n-1=1 (i.e., one v), and there are 3 permutations corresponding to each one in the n=1 case (e.g., the single v can be inserted in any of three places in the 2-object permutation xy, yielding vxy, xvy, and xyv), so a(2) = 3*a(1) = 3*16 = 48; at n=3, n-1=2 (i.e., two v's), and a(3) = C(4,2)*a(1) = 6*16 = 96; etc. - Zerinvary Lajos, Aug 07 2008 (this needs clarification, Joerg Arndt, Feb 23 2014)
Sequence found by reading the line from 0, in the direction 0, 16, ... and the same line from 0, in the direction 0, 48, ..., in the square spiral whose vertices are the generalized 18-gonal numbers. - Omar E. Pol, Oct 03 2011
For n > 0, a(n) is the area of the triangle with vertices at ((n-1)^2, n^2), ((n+1)^2, (n+2)^2), and ((n+3)^2, (n+2)^2). - J. M. Bergot, May 22 2014
For n > 0, a(n) is the number of self-intersecting points in star polygon {4*(n+1)/(2*n+1)}. - Bui Quang Tuan, Mar 28 2015
Equivalently: integers k such that k$ / (k/2)! and k$ / (k/2+1)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information). - Bernard Schott, Dec 02 2021

Examples

			3 X 3-Board: knight can be placed in 8 positions with 2 moves from each, so a(1) = 16.
		

Crossrefs

Cf. A033586 (King), A035005 (Queen), A035006 (Rook), A002492 (Bishop) and A049450 (Pawn).
Cf. A348692.
Subsequence of A008586 and of A349081.

Programs

Formula

a(n) = 8*n*(n+1).
G.f.: 16*x/(1-x)^3.
a(n) = A069129(n+1) - 1. - Omar E. Pol, Apr 26 2008
a(n) = binomial(n+1,2)*4^2, n >= 0. - Zerinvary Lajos, Aug 07 2008
a(n) = 8*n^2 + 8*n = 16*A000217(n) = 8*A002378(n) = 4*A046092(n) = 2*A033996(n). - Omar E. Pol, Dec 12 2008
a(n) = a(n-1) + 16*n, with a(0)=0. - Vincenzo Librandi, Nov 17 2010
E.g.f.: 8*exp(x)*x*(2 + x). - Stefano Spezia, May 19 2021
From Amiram Eldar, Feb 22 2023: (Start)
Sum_{n>=1} 1/a(n) = 1/8.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2*log(2) - 1)/8.
Product_{n>=1} (1 - 1/a(n)) = -(8/Pi)*cos(sqrt(3/2)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (8/Pi)*cos(Pi/(2*sqrt(2))). (End)

Extensions

More terms from Erich Friedman
Minor errors corrected and edited by Johannes W. Meijer, Feb 04 2010

A139098 a(n) = 8*n^2.

Original entry on oeis.org

0, 8, 32, 72, 128, 200, 288, 392, 512, 648, 800, 968, 1152, 1352, 1568, 1800, 2048, 2312, 2592, 2888, 3200, 3528, 3872, 4232, 4608, 5000, 5408, 5832, 6272, 6728, 7200, 7688, 8192, 8712, 9248, 9800, 10368, 10952, 11552, 12168, 12800, 13448, 14112, 14792, 15488, 16200
Offset: 0

Views

Author

Omar E. Pol, Apr 25 2008

Keywords

Comments

Opposite numbers to the centered 16-gonal numbers (A069129) in the square spiral whose vertices are the triangular numbers (A000217).
8 times the squares. - Omar E. Pol, Dec 09 2008
a(n-1) is the molecular topological index of the n-wheel graph W_n. - Eric W. Weisstein, Jul 11 2011
An n X n pandiagonal magic square has a(n) orientations. - Kausthub Gudipati, Sep 15 2011
Area of a square with diagonal 4n. - Wesley Ivan Hurt, Jun 19 2014
Sum of all the parts in the partitions of 4n into exactly two parts. - Wesley Ivan Hurt, Jul 23 2014
Equivalently: integers k such that k$ / (k/2-1)! and k$ / (k/2)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information). - Bernard Schott, Dec 02 2021

Crossrefs

Programs

Formula

a(n) = 8*A000290(n) = 4*A001105(n) = 2*A016742(n). - Omar E. Pol, Dec 13 2008
G.f.: -8*x*(1+x)/(x-1)^3. - R. J. Mathar, Nov 27 2015
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/48 (A245058).
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/96.
Product_{n>=1} (1 + 1/a(n)) = sqrt(8)*sinh(Pi/sqrt(8))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(8)*sin(Pi/sqrt(8))/Pi. (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Wesley Ivan Hurt, Dec 03 2021
From Elmo R. Oliveira, Dec 01 2024: (Start)
E.g.f.: 8*x*(1 + x)*exp(x).
a(n) = n*A008590(n) = A001105(2*n). (End)

A349079 Numbers k such that there exists m, 1 <= m <= k with the property that A000178(k) / m! is a square, where A000178(k) = k$ = 1!*2!*...*k! is the superfactorial of k.

Original entry on oeis.org

1, 2, 4, 8, 12, 14, 16, 18, 20, 24, 28, 32, 34, 36, 40, 44, 48, 52, 56, 60, 62, 64, 68, 72, 76, 80, 84, 88, 92, 96, 98, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 142, 144, 148, 152, 156, 160, 164, 168, 172, 176, 180, 184, 188, 192, 194, 196, 200, 204, 208, 212, 216, 220, 224, 228
Offset: 1

Views

Author

Bernard Schott, Nov 07 2021

Keywords

Comments

If k is a term, then A348692(k) lists integers m such that k$ / m! is a square; and for each k, there exist only one (A349080) or two (A349081) such integers m.
See A348692 for further information, links and references about Olympiads.
Except for 1, all terms are even, and, when k is such an even term, corresponding m belong(s) to {k/2 - 2, k/2 - 1, k/2, k/2 + 1, k/2 + 2}.
This sequence is the union of {1} and of three infinite and disjoint subsequences:
-> A008586, so every positive multiple of 4 is a term and in this case, for k=4*q, (k$)/(k/2)! = ( 2^(k/4) * Product_{j=1..k/2} ((2j-1)!) )^2 (see example 4).
-> A060626, so every k = 4*q^2 - 2 (q >= 1) is a term (see examples 2 and 14).
-> 2*A055792 = {k = 2q^2 with q>1 in A001541} = {18, 578, ...} (see example 18).

Examples

			2 is a term as 2$ / 2! = 1^2.
4 is a term as 4$ / 2! = 12^2.
14 is a term as 14$ / 8! = 1309248519599593818685440000000^2 and also 14$ / 9! = 436416173199864606228480000000^2.
18 is a term as 18$ / 7! = 29230177671473293820176594405114531928195727360000000000000^2.
		

Crossrefs

Programs

  • Mathematica
    supfact[n_] := supfact[n] = BarnesG[n + 2]; fact[n_] := fact[n] = n!; q[k_] := AnyTrue[Range[k], IntegerQ @ Sqrt[supfact[k]/fact[#]] &]; Select[Range[230], q] (* Amiram Eldar, Nov 08 2021 *)
  • PARI
    f(n) = prod(k=2, n, k!); \\ A000178
    isok(k) = my(sf=f(k)); for (m=1, k, if (issquare(sf/m!), return(1))); \\ Michel Marcus, Nov 08 2021

A349766 Numbers of the form 2*t^2-4 when t > 1 is a term in A001541.

Original entry on oeis.org

14, 574, 19598, 665854, 22619534, 768398398, 26102926094, 886731088894, 30122754096398, 1023286908188734, 34761632124320654, 1180872205318713598, 40114893348711941774, 1362725501650887306814, 46292552162781456489998, 1572584048032918633353214, 53421565080956452077519374
Offset: 1

Views

Author

Bernard Schott, Dec 04 2021

Keywords

Comments

Equivalently: integers k such that k$ / (k/2+1)! and k$ / (k/2+2)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information).
The 3 subsequences of A349081 are A035008, A139098 and this one.

Examples

			A001541(1) = 3, then for t = 3, 2*t^2-4 = 14; also for k = 14, 14$ / 8! = 1309248519599593818685440000000^2 and 14$ / 9! = 436416173199864606228480000000^2. Hence, 14 is a term.
		

Crossrefs

Programs

  • Maple
    with(orthopoly):
    sequence = (2*T(n,3)^2-4, n=1..20);
  • Mathematica
    (2*#^2 - 4) & /@ LinearRecurrence[{6, -1}, {3, 17}, 17] (* Amiram Eldar, Dec 04 2021 *)
    LinearRecurrence[{35, -35, 1},{14, 574, 19598},17] (* Ray Chandler, Mar 01 2024 *)
  • PARI
    a(n) = my(t=subst(polchebyshev(n), 'x, 3)); 2*t^2-4; \\ Michel Marcus, Dec 04 2021

Formula

a(n) = 2*(cosh(2*n*arcsinh(1)))^2 - 4.
a(n) = 16*A001110(n) - 2. - Hugo Pfoertner, Dec 04 2021
Showing 1-4 of 4 results.