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.

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A054569 a(n) = 4*n^2 - 6*n + 3.

Original entry on oeis.org

1, 7, 21, 43, 73, 111, 157, 211, 273, 343, 421, 507, 601, 703, 813, 931, 1057, 1191, 1333, 1483, 1641, 1807, 1981, 2163, 2353, 2551, 2757, 2971, 3193, 3423, 3661, 3907, 4161, 4423, 4693, 4971, 5257, 5551, 5853, 6163, 6481, 6807, 7141, 7483, 7833, 8191
Offset: 1

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Author

Keywords

Comments

Move in 1-7 direction in a spiral organized like A068225 etc.
Third row of A082039. - Paul Barry, Apr 02 2003
Inverse binomial transform of A036826. - Paul Barry, Jun 11 2003
Equals the "middle sequence" T(2*n,n) of the Connell sequence A001614 as a triangle. - Johannes W. Meijer, May 20 2011
Ulam's spiral (SW spoke). - Robert G. Wilson v, Oct 31 2011

Crossrefs

Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n+1) = 4*n^2 + 2*n + 1. - Paul Barry, Apr 02 2003
a(n) = 4*n^2 - 6*n+3 - 3*0^n (with leading zero). - Paul Barry, Jun 11 2003
Binomial transform of [1, 6, 8, 0, 0, 0, ...]. - Gary W. Adamson, Dec 28 2007
a(n) = 8*n + a(n-1) - 10 (with a(1)=1). - Vincenzo Librandi, Aug 07 2010
From Colin Barker, Mar 23 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(1+x)*(1+3*x)/(1-x)^3. (End)
a(n) = A000384(n) + A000384(n-1). - Bruce J. Nicholson, May 07 2017
E.g.f.: -3 + (3 - 2*x + 4*x^2)*exp(x). - G. C. Greubel, Jul 04 2019
Sum_{n>=1} 1/a(n) = A339237. - R. J. Mathar, Jan 22 2021

Extensions

Edited by Frank Ellermann, Feb 24 2002

A035608 Expansion of g.f. x*(1 + 3*x)/((1 + x)*(1 - x)^3).

Original entry on oeis.org

0, 1, 5, 10, 18, 27, 39, 52, 68, 85, 105, 126, 150, 175, 203, 232, 264, 297, 333, 370, 410, 451, 495, 540, 588, 637, 689, 742, 798, 855, 915, 976, 1040, 1105, 1173, 1242, 1314, 1387, 1463, 1540, 1620, 1701, 1785, 1870, 1958, 2047, 2139, 2232, 2328, 2425, 2525, 2626
Offset: 0

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Author

Keywords

Comments

Maximum value of Voronoi's principal quadratic form of the first type when variables restricted to {-1,0,1}. - Michael Somos, Mar 10 2004
This is the main row of a version of the "square spiral" when read alternatively from left to right (see link). See also A001107, A007742, A033954, A033991. It is easy to see that the only prime in the sequence is 5. - Emilio Apricena (emilioapricena(AT)yahoo.it), Feb 08 2009
From Mitch Phillipson, Manda Riehl, Tristan Williams, Mar 06 2009: (Start)
a(n) gives the number of elements of S_2 \wr C_k that avoid the pattern 12, using the following ordering:
In S_j, a permutation p avoids a pattern q if it has no subsequence that is order-isomorphic to q. For example, p avoids the pattern 132 if it has no subsequence abc with a < c < b. We extend this notion to S_j \wr C_n as follows. Element psi =[ alpha_1^beta_1, ... alpha_j^beta_j ] avoids tau = [ a_1 ... a_m ] (tau in S_m) if psi' = [ alpha_1*beta_1 ... alpha_j*beta_j ] avoids tau in the usual sense. For n=2, there are 5 elements of S_2 \wr C_2 that avoid the pattern 12. They are: [ 2^1,1^1 ], [ 2^2,1^1 ], [ 2^2,1^2 ], [ 2^1,1^2 ], [ 1^2,2^1 ].
For example, if psi = [2^1,1^2], then psi'=[2,2] which avoids tau=[1,2] because no subsequence ab of psi' has a < b. (End)

References

  • J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, p. 115.

Crossrefs

Partial sums of A042948.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

  • Magma
    [n^2 + n - 1 - Floor((n-1)/2): n in [0..25]]; // G. C. Greubel, Oct 29 2017
  • Maple
    A035608:=n->floor((n + 1/4)^2): seq(A035608(n), n=0..100); # Wesley Ivan Hurt, Oct 29 2017
  • Mathematica
    Table[n^2 + Floor[n/2], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Apr 12 2011 *)
    CoefficientList[Series[x (1 + 3 x)/((1 + x) (1 - x)^3), {x, 0, 60}], x] (* or *) LinearRecurrence[{2, 0, -2, 1}, {0, 1, 5, 10}, 60] (* Harvey P. Dale, Feb 21 2013 *)
  • PARI
    a(n)=n^2+n-1-(n-1)\2
    

Formula

a(n) = n^2 + n - 1 - floor((n-1)/2).
a(n) = A011848(2*n+1).
a(n) = A002378(n) - A004526(n+1). - Reinhard Zumkeller, Jan 27 2010
a(n) = 2*A006578(n) - A002378(n)/2 = A139592(n)/2. - Reinhard Zumkeller, Feb 07 2010
a(n) = A002265(n+2) + A173562(n). - Reinhard Zumkeller, Feb 21 2010
a(n) = floor((n + 1/4)^2). - Reinhard Zumkeller, Jan 27 2010
a(n) = (-1)^n*Sum_{i=0..n} (-1)^i*(2*i^2 + 3*i + 1). Omits the leading 0. - William A. Tedeschi, Aug 25 2010
a(n) = n^2 + floor(n/2), from Mathematica section. - Vladimir Joseph Stephan Orlovsky, Apr 12 2011
a(0)=0, a(1)=1, a(2)=5, a(3)=10; for n > 3, a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - Harvey P. Dale, Feb 21 2013
For n > 1: a(n) = a(n-2) + 4*n - 3; see also row sums of triangle A253146. - Reinhard Zumkeller, Dec 27 2014
a(n) = 3*A002620(n) + A002620(n+1). - R. J. Mathar, Jul 18 2015
From Amiram Eldar, Mar 20 2022: (Start)
Sum_{n>=1} 1/a(n) = 4 - 2*log(2) - Pi/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/3 - 4*(1-log(2)). (End)
E.g.f.: (x*(2*x + 3)*cosh(x) + (2*x^2 + 3*x - 1)*sinh(x))/2. - Stefano Spezia, Apr 24 2024

A054554 a(n) = 4*n^2 - 10*n + 7.

Original entry on oeis.org

1, 3, 13, 31, 57, 91, 133, 183, 241, 307, 381, 463, 553, 651, 757, 871, 993, 1123, 1261, 1407, 1561, 1723, 1893, 2071, 2257, 2451, 2653, 2863, 3081, 3307, 3541, 3783, 4033, 4291, 4557, 4831, 5113, 5403, 5701, 6007, 6321, 6643, 6973, 7311, 7657, 8011, 8373, 8743
Offset: 1

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Author

Keywords

Comments

Move in 1-3 direction in a spiral organized like A068225 etc.
Equals binomial transform of [1, 2, 8, 0, 0, 0, ...]. - Gary W. Adamson, May 03 2008
Ulam's spiral (NE spoke). - Robert G. Wilson v, Oct 31 2011
Number of ternary strings of length 2*(n-1) that have one or no 0's, one or no 1's, and an even number of 2's. For n=2, the 3 strings of length 2 are 01, 10 and 22. For n=3, the 13 strings of length 4 are the 12 permutations of 0122 and 2222. - Enrique Navarrete, Jul 25 2025

Crossrefs

Cf. A014105.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n) = 8*n + a(n-1) - 14 with n > 1, a(1)=1. - Vincenzo Librandi, Aug 07 2010
G.f.: -x*(7*x^2+1)/(x-1)^3. - Colin Barker, Sep 21 2012
For n > 2, a(n) = A014105(n) + A014105(n-1). - Bruce J. Nicholson, May 07 2017
From Leo Tavares, Feb 21 2022: (Start)
a(n) = A003215(n-2) + 2*A000217(n-1). See Hexagonal Dual Rays illustration in links.
a(n) = A227776(n-1) - 4*A000217(n-1). (End)
a(k+1) = 4k^2 - 2k + 1 in the Numberphile video. - Frank Ellermann, Mar 11 2020
E.g.f.: exp(x)*(7 - 6*x + 4*x^2) - 7. - Stefano Spezia, Apr 24 2024

Extensions

Edited by Frank Ellermann, Feb 24 2002

A087475 a(n) = n^2 + 4.

Original entry on oeis.org

4, 5, 8, 13, 20, 29, 40, 53, 68, 85, 104, 125, 148, 173, 200, 229, 260, 293, 328, 365, 404, 445, 488, 533, 580, 629, 680, 733, 788, 845, 904, 965, 1028, 1093, 1160, 1229, 1300, 1373, 1448, 1525, 1604, 1685, 1768, 1853, 1940, 2029, 2120, 2213, 2308, 2405, 2504
Offset: 0

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Author

Gary W. Adamson, Sep 09 2003

Keywords

Comments

Schroeder, p. 330, states "For positive n, these winding numbers are precisely those whose continued fraction expansion is periodic and has period length 1".
Positive X values of solutions to the equation X^3 - 4*X^2 = Y^2. To find Y values: b(n) = n*(n^2 + 4). - Mohamed Bouhamida, Nov 06 2007
From Artur Jasinski, Oct 03 2008: (Start)
General formula for cotangent recurrences type:
a(n+1) = a(n)^3 + 3*a(n) and a(1)=k is
a(n) = floor(((k + sqrt(k^2 + 4))/2)^(3^(n-1))). (End)
Given sequences of the form S(n) = N*S(n-1) + S(n-2) starting (1, N, ...), and having convergents with discriminant (N^2 + 4), S(p) == (a(n))^((p-1)/2) mod p, for n>0, p = odd prime. Example: with N = 2 we have the Pell series (1, 2, 5, 12, 29, 70, 169, ...) with P(7) = 169. Then 169 == 8^3 mod 7, with a(2) = 8. Cf. Schroeder, "Number Theory in Science and Communication", p. 90, for N = 1: F(p) == 5^((p-1)/2) mod p. - Gary W. Adamson, Feb 23 2009
The only two real solutions of the form f(x) = A*x^p with positive p that satisfy f^(n)(x) = f^[-1](x), x >= 0, n >= 1, with f^(n) the n-th derivative and f^[-1] the compositional inverse of f, are obtained for p = p1(n) = (n + sqrt(a(n)))/2 and p = p2(n) = (n - sqrt(a(n)))/2, n >= 1, and A = A(n) = (fallfac(p,n))^(-p/(p+1)), for p = p1(n) and p = p2(n), respectively. Here fallfac(x, k) := product(x - j, j = 0..k-1), the falling factorials. See the T. Koshy reference, pp. 263-264 (there is also a solution for negative p if n is even; see the corresponding comment in A002522). - Wolfdieter Lang, Oct 21 2010, Oct 28 2010
(n + sqrt(a(n)))/2 = [n;n,n,...], with the regular continued fraction with period length 1. For a simple proof see, e.g., the Schroeder reference, pp. 330-331. See also the first comment above.

Examples

			a(2) = 8, discriminant of algebraic representation of barover(2) = [2,2,2,...] = sqrt 2 - 1 = 0.41421356... = ((sqrt 8) - 2)/2. a(3) = 13, discriminant of barover(3) = [3,3,3,...] = 0.3027756... = ((sqrt 13) - 3)/2.
		

References

  • Manfred R. Schroeder, "Fractals, Chaos, Power Laws"; W.H. Freeman & Co, 1991, p. 330-331.
  • Manfred R. Schroeder, "Number Theory in Science and Communication", Springer Verlag, 5th ed., 2009. [From Gary W. Adamson, Feb 23 2009]
  • Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, New York, 2001. [From Wolfdieter Lang, Oct 21 2010]

Crossrefs

Programs

Formula

n^2 + 4 are discriminant terms in the formula for Positive Silver Mean Constants, defined as barover(n), = (sqrt (n^2 + 4) - n)/2. Such constants barover(n) = C have the property: 1/C - C = n.
a(n) = A156701(n) / A053755(n). - Reinhard Zumkeller, Feb 13 2009
a(n) = A156798(n)/A002522(n). - Reinhard Zumkeller, Feb 16 2009
a(n) = a(n-1) + 2*n-1 (with a(0)=4). - Vincenzo Librandi, Nov 22 2010
G.f.: (4 - 7*x + 5*x^2)/(1 - x)^3. - Colin Barker, Jan 06 2012
a(n)^3 = A155965(n)^2 + A155966(n)^2. - Vincenzo Librandi, Feb 22 2012
From Amiram Eldar, Jul 13 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + 2*Pi*coth(2*Pi))/8.
Sum_{n>=0} (-1)^n/a(n) = (1 + 2*Pi*cosech(2*Pi))/8 = A371803. (End)
E.g.f.: exp(x)*(4 + x + x^2). - Stefano Spezia, Jul 08 2023
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=0} (1 - 1/a(n)) = sqrt(3)*sinh(sqrt(3)*Pi)/(2*sinh(2*Pi)).
Product_{n>=0} (1 + 1/a(n)) = sqrt(5)*sinh(sqrt(5)*Pi)/(2*sinh(2*Pi)). (End)

A054556 a(n) = 4*n^2 - 9*n + 6.

Original entry on oeis.org

1, 4, 15, 34, 61, 96, 139, 190, 249, 316, 391, 474, 565, 664, 771, 886, 1009, 1140, 1279, 1426, 1581, 1744, 1915, 2094, 2281, 2476, 2679, 2890, 3109, 3336, 3571, 3814, 4065, 4324, 4591, 4866, 5149, 5440, 5739, 6046, 6361, 6684, 7015, 7354, 7701, 8056, 8419, 8790
Offset: 1

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Comments

Move in 1-4 direction in a spiral organized like A068225 etc.
Equals binomial transform of [1, 3, 8, 0, 0, 0, ...]. - Gary W. Adamson, Apr 30 2008
Ulam's spiral (N spoke). - Robert G. Wilson v, Oct 31 2011
Also, numbers of the form m*(4*m+1)+1 for nonpositive m. - Bruno Berselli, Jan 06 2016

Crossrefs

Cf. A266883: m*(4*m+1)+1 for m = 0,-1,1,-2,2,-3,3,...
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n)^2 = Sum_{i = 0..2*(4*n-5)} (4*n^2-13*n+9+i)^2*(-1)^i = ((n-1)*(4*n-5)+1)^2. - Bruno Berselli, Apr 29 2010
From Harvey P. Dale, Aug 21 2011: (Start)
a(0)=1, a(1)=4, a(2)=15; for n > 2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: -x*(6*x^2+x+1)/(x-1)^3. (End)
From Franck Maminirina Ramaharo, Mar 09 2018: (Start)
a(n) = binomial(2*n - 2, 2) + 2*(n - 1)^2 + 1.
a(n) = A000384(n-1) + A058331(n-1).
a(n) = A130883(n-1) + A001105(n-1). (End)
E.g.f.: exp(x)*(6 - 5*x + 4*x^2) - 6. - Stefano Spezia, Apr 24 2024

Extensions

Edited by Frank Ellermann, Feb 24 2002
Incorrect formula deleted by N. J. A. Sloane, Aug 02 2009

A054567 a(n) = 4*n^2 - 7*n + 4.

Original entry on oeis.org

1, 6, 19, 40, 69, 106, 151, 204, 265, 334, 411, 496, 589, 690, 799, 916, 1041, 1174, 1315, 1464, 1621, 1786, 1959, 2140, 2329, 2526, 2731, 2944, 3165, 3394, 3631, 3876, 4129, 4390, 4659, 4936, 5221, 5514, 5815, 6124, 6441, 6766, 7099, 7440, 7789, 8146, 8511, 8884
Offset: 1

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Author

Keywords

Comments

The number 1 is placed in the middle of a sheet of squared paper and the numbers 2, 3, 4, 5, 6, etc. are written in a clockwise spiral around 1, as in A068225 etc. This sequence is read off along one of the rays from 1.
Ulam's spiral (W spoke of A054552). - Robert G. Wilson v, Oct 31 2011
Also, numbers of the form m*(4*m+1)+1 for nonnegative m. - Bruno Berselli, Jan 06 2016
The sequence forms the 1x2 diagonal of the square maze arrangement in A081344. - Jarrod G. Sage, Jul 17 2024

Crossrefs

Cf. A266883: m*(4*m+1)+1 for m = 0,-1,1,-2,2,-3,3,...
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n) = 8*n+a(n-1)-11 for n>1, a(1)=1. - Vincenzo Librandi, Aug 07 2010
a(n) = A204674(n-1) / n. - Reinhard Zumkeller, Jan 18 2012
From Colin Barker, Oct 25 2014: (Start)
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3).
G.f.: -x*(4*x^2+3*x+1) / (x-1)^3. (End)
E.g.f.: exp(x)*(4 - 3*x + 4*x^2) - 4. - Stefano Spezia, Apr 24 2024
a(n) = A016742(n-1) + n. - Jarrod G. Sage, Jul 17 2024

Extensions

Edited by Frank Ellermann, Feb 24 2002
Typo fixed by Charles R Greathouse IV, Oct 28 2009

A008527 Coordination sequence for body-centered tetragonal lattice.

Original entry on oeis.org

1, 10, 34, 74, 130, 202, 290, 394, 514, 650, 802, 970, 1154, 1354, 1570, 1802, 2050, 2314, 2594, 2890, 3202, 3530, 3874, 4234, 4610, 5002, 5410, 5834, 6274, 6730, 7202, 7690, 8194, 8714, 9250, 9802, 10370, 10954, 11554, 12170, 12802, 13450, 14114, 14794, 15490, 16202, 16930, 17674
Offset: 0

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Author

Keywords

Comments

Also sequence found by reading the segment (1, 10) together with the line from 10, in the direction 10, 34, ..., in the square spiral whose vertices are the generalized hexagonal numbers A000217. - Omar E. Pol, Nov 02 2012

Crossrefs

Apart from leading term, same as A108100.
Cf. A206399.
Cf. A016754 (SE), A054554 (NE), A054569 (SW), A053755 (NW), A033951 (S), A054552 (E), A054556 (N), A054567 (W) (Ulam spiral spokes).
A143839 (SSE) + A143855 (ESE) = A143838 (SSW) + A143856 (ENE) = A143854 (WSW) + A143861 (NNE) = A143859 (WNW) + A143860 (NNW) = even bisection = a(2n) = A010021(n).

Programs

  • GAP
    Concatenation([1], List([1..40], n-> 2*(1+4*n^2) )); # G. C. Greubel, Nov 09 2019
  • Magma
    [1] cat [2*(1 + 4*n^2): n in [1..50]]; // G. C. Greubel, Nov 09 2019
    
  • Maple
    1, seq(8*k^2+2, k=1..50);
  • Mathematica
    a[0]:= 1; a[n_]:= 8n^2 +2; Table[a[n], {n,0,50}] (* Alonso del Arte, Sep 06 2011 *)
    LinearRecurrence[{3,-3,1},{1,10,34,74},50] (* Harvey P. Dale, Feb 13 2022 *)
  • PARI
    vector(51, n, if(n==1,1, 2*(1+(2*n-2)^2)) ) \\ G. C. Greubel, Nov 09 2019
    
  • Sage
    [1]+[2*(1+4*n^2) for n in (1..40)] # G. C. Greubel, Nov 09 2019
    

Formula

a(0) = 1; a(n) = 8*n^2+2 for n>0.
From Gary W. Adamson, Dec 27 2007: (Start)
a(n) = (2n+1)^2 + (2n-1)^2 for n>0.
Binomial transform of [1, 9, 15, 1, -1, 1, -1, 1, ...]. (End)
From Colin Barker, Apr 13 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3.
G.f.: (1+x)*(1+6*x+x^2)/(1-x)^3. (End)
From Bruce J. Nicholson, Jul 31 2019: (Start) Assume n>0.
a(n) = A016754(n) + A016754(n-1).
a(n) = 2 * A053755(n).
a(n) = A054554(n+1) + A054569(n+1).
a(n) = A033951(n) + A054552(n).
a(n) = A054556(n+1) + A054567(n+1). (End)
E.g.f.: -1 + 2*exp(x)*(1 + 2*x)^2. - Stefano Spezia, Aug 02 2019
Sum_{n>=0} 1/a(n) = 3/4+1/8*Pi*coth(Pi/2) = 1.178172.... - R. J. Mathar, May 07 2024

A080335 Diagonal in square spiral or maze arrangement of natural numbers.

Original entry on oeis.org

1, 5, 9, 17, 25, 37, 49, 65, 81, 101, 121, 145, 169, 197, 225, 257, 289, 325, 361, 401, 441, 485, 529, 577, 625, 677, 729, 785, 841, 901, 961, 1025, 1089, 1157, 1225, 1297, 1369, 1445, 1521, 1601, 1681, 1765, 1849, 1937, 2025, 2117, 2209, 2305, 2401, 2501
Offset: 0

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Author

Paul Barry, Mar 19 2003

Keywords

Comments

Interleaves the odd squares A016754 with (1+4n^2), A053755.
Squares of positive integers (plus 1 if n is odd). - Wesley Ivan Hurt, Oct 10 2013
a(n) is the maximum total number of queens that can coexist without attacking each other on an [n+3] X [n+3] chessboard, when the lone queen is in the most vulnerable position on the board. Specifically, the lone queen will placed in any center position, facing an opponent's "army" of size a(n)-1 == A137932(n+2). - Bob Selcoe, Feb 12 2015
a(n) is also the edge chromatic number of the complement of the (n+2) X (n+2) rook graph. - Eric W. Weisstein, Jan 31 2024

Crossrefs

Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n) = (3 + 4*n + 2*n^2 - (-1)^n)/2.
a(2*n) = A016754(n), a(2*n+1) = A053755(n+1).
E.g.f.: exp(x)*(2 + 3*x + x^2) - cosh(x). The sequence 1,1,5,9,... is given by n^2+(1+(-1)^n)/2 with e.g.f. exp(1+x+x^2)*exp(x)-sinh(x). - Paul Barry, Sep 02 2003 and Sep 19 2003
a(0)=1, a(1)=5, a(2)=9, a(3)=17, a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - Harvey P. Dale, Jan 29 2012
a(n)+(-1)^n = A137928(n+1). - Philippe Deléham, Feb 17 2012
G.f.: (1 + 3*x - x^2 + x^3)/((1-x)^3*(1+x)). - Colin Barker, Mar 18 2012
a(n) = A000035(n) + A000290(n+1). - Wesley Ivan Hurt, Oct 10 2013
From Bob Selcoe, Feb 12 2015: (Start)
a(n) = A137932(n+2) + 1.
a(n) = (n+1)^2 when n is even; a(n) = (n+1)^2 + 1 when n is odd.
a(n) = A002378(n+2) - A047238(n+3) + 1.
(End)
Sum_{n>=0} 1/a(n) = Pi*coth(Pi/2)/4 + Pi^2/8 - 1/2. - Amiram Eldar, Jul 07 2022

A137932 Terms in an n X n spiral that do not lie on its principal diagonals.

Original entry on oeis.org

0, 0, 0, 4, 8, 16, 24, 36, 48, 64, 80, 100, 120, 144, 168, 196, 224, 256, 288, 324, 360, 400, 440, 484, 528, 576, 624, 676, 728, 784, 840, 900, 960, 1024, 1088, 1156, 1224, 1296, 1368, 1444, 1520, 1600, 1680, 1764, 1848, 1936, 2024, 2116, 2208, 2304, 2400, 2500, 2600, 2704, 2808
Offset: 0

Views

Author

William A. Tedeschi, Feb 29 2008

Keywords

Comments

The count of terms not on the principal diagonals is always even.
The last digit is the repeating pattern 0,0,0,4,8,6,4,6,8,4, which is palindromic if the leading 0's are removed, 4864684.
The sum of the last digits is 40, which is the count of the pattern times 4.
A 4 X 4 spiral is the only spiral, aside from a 0 X 0, whose count of terms that do not lie on its principal diagonals equal the count of terms that do [A137932(4) = A042948(4)] making the 4 X 4 the "perfect spiral".
Yet another property is mod(a(n), A042948(n)) = 0 iff n is even. This is a large family that includes the 4 X 4 spiral.
a(n) is the maximum number of queens of one color that can coexist without attacking one queen of the opponent's color on an [n+1] X [n+1] chessboard, when the lone queen is in the most vulnerable position on the board, i.e., on a center square. - Bob Selcoe, Feb 12 2015
Also the circumference of the (n-1) X (n-1) grid graph for n > 2. - Eric W. Weisstein, Mar 25 2018
Also the crossing number of the complete bipartite graph K_{5,n}. - Eric W. Weisstein, Sep 11 2018

Examples

			a(0) = 0^2 - (2(0) - mod(0,2)) = 0.
a(3) = 3^2 - (2(3) - mod(3,2)) = 4.
		

Crossrefs

Cf. A042948.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n) = n^2 - (2*n - mod(n,2)) = n^2 - A042948(n).
a(n) = 2*A007590(n-1). - Enrique Pérez Herrero, Jul 04 2012
G.f.: -4*x^3 / ( (1+x)*(x-1)^3 ). a(n) = 4*A002620(n-1). - R. J. Mathar, Jul 06 2012
From Bob Selcoe, Feb 12 2015: (Start)
a(n) = (n-1)^2 when n is odd; a(n) = (n-1)^2 - 1 when n is even.
a(n) = A002378(n) - A047238(n+1). (End)
From Amiram Eldar, Mar 20 2022: (Start)
Sum_{n>=3} 1/a(n) = Pi^2/24 + 1/4.
Sum_{n>=3} (-1)^(n+1)/a(n) = Pi^2/24 - 1/4. (End)
E.g.f.: x*(x - 1)*cosh(x) + (x^2 - x + 1)*sinh(x). - Stefano Spezia, Oct 17 2022

A156859 The main column of a version of the square spiral.

Original entry on oeis.org

0, 3, 7, 14, 22, 33, 45, 60, 76, 95, 115, 138, 162, 189, 217, 248, 280, 315, 351, 390, 430, 473, 517, 564, 612, 663, 715, 770, 826, 885, 945, 1008, 1072, 1139, 1207, 1278, 1350, 1425, 1501, 1580, 1660, 1743, 1827, 1914, 2002, 2093, 2185, 2280, 2376, 2475, 2575
Offset: 0

Views

Author

Emilio Apricena (emilioapricena(AT)yahoo.it), Feb 17 2009

Keywords

Comments

This spiral is sometimes called an Ulam spiral, but square spiral is a better name. - N. J. A. Sloane, Jul 27 2018
It is easy to see that the only two primes in the sequence are 3, 7. Therefore the primes of the version of Ulam spiral are divided into four parts (see also A035608): northeast (NE), northwest (NW), southwest (SW), and southeast (SE).
Number of pairs (x,y) having x and y of opposite parity with x in {0,...,n} and y in {0,...,2n}. - Clark Kimberling, Jul 02 2012
Partial Sums of A014601(n). - Wesley Ivan Hurt, Oct 11 2013

Crossrefs

Cf. A000290, A000384, A004526, A014601 (first differences), A115258.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n) = n^2 + n + floor((n+1)/2) = A002378(n) + A004526(n+1) = A002620(n+1) + 3*A002620(n).
From R. J. Mathar, Feb 20 2009: (Start)
G.f.: x*(3+x)/((1+x)*(1-x)^3).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). (End)
a(n-1) = floor(n/(e^(1/n)-1)). - Richard R. Forberg, Jun 19 2013
a(n) = A000290(n+1) + A004526(-n-1). - Wesley Ivan Hurt, Jul 15 2013
a(n) + a(n+1) = A014105(n+1). - R. J. Mathar, Jul 15 2013
a(n) = floor(A000384(n+1)/2). - Bruno Berselli, Nov 11 2013
E.g.f.: (x*(5 + 2*x)*cosh(x) + (1 + 5*x + 2*x^2)*sinh(x))/2. - Stefano Spezia, Apr 24 2024
Sum_{n>=1} 1/a(n) = 4/9 + 2*log(2) - Pi/3. - Amiram Eldar, Apr 26 2024

Extensions

More terms added by Wesley Ivan Hurt, Oct 11 2013
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