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 1-10 of 28 results. Next

A139598 A035008(n) followed by A139098(n+1).

Original entry on oeis.org

0, 8, 16, 32, 48, 72, 96, 128, 160, 200, 240, 288, 336, 392, 448, 512, 576, 648, 720, 800, 880, 968, 1056, 1152, 1248, 1352, 1456, 1568, 1680, 1800, 1920, 2048, 2176, 2312, 2448, 2592, 2736, 2888, 3040, 3200, 3360, 3528, 3696, 3872
Offset: 0

Views

Author

Omar E. Pol, May 03 2008

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 8, ... and the line from 16, in the direction 16, 48, ..., in the square spiral whose vertices are the triangular numbers A000217.
Also represents the minimum number of segments in the smooth Jordan curve which crosses every edge of an n X n square lattice exactly once. For example, the curve for a 3 X 3 lattice would have at least 32 segments. - Nikolas Novakovic, Aug 28 2022

Examples

			Array begins:
   0,   8;
  16,  32;
  48,  72;
  96, 128;
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{2,0,-2,1},{0,8,16,32},50] (* Harvey P. Dale, Sep 27 2019 *)

Formula

Array read by rows: row n gives 8*n^2 + 8*n, 8*(n+1)^2.
From Colin Barker, Jul 22 2012: (Start)
a(n) = (1 - (-1)^n + 4*n + 2*n^2).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
G.f.: 8*x/((1-x)^3*(1+x)). (End)
a(n) = 8*A002620(n+1). - R. J. Mathar, May 04 2014

A001105 a(n) = 2*n^2.

Original entry on oeis.org

0, 2, 8, 18, 32, 50, 72, 98, 128, 162, 200, 242, 288, 338, 392, 450, 512, 578, 648, 722, 800, 882, 968, 1058, 1152, 1250, 1352, 1458, 1568, 1682, 1800, 1922, 2048, 2178, 2312, 2450, 2592, 2738, 2888, 3042, 3200, 3362, 3528, 3698, 3872, 4050, 4232, 4418
Offset: 0

Views

Author

Bernd.Walter(AT)frankfurt.netsurf.de

Keywords

Comments

Number of edges of the complete bipartite graph of order 3n, K_{n,2n}. - Roberto E. Martinez II, Jan 07 2002
"If each period in the periodic system ends in a rare gas ..., the number of elements in a period can be found from the ordinal number n of the period by the formula: L = ((2n+3+(-1)^n)^2)/8..." - Nature, Jun 09 1951; Nature 411 (Jun 07 2001), p. 648. This produces the present sequence doubled up.
Let z(1) = i = sqrt(-1), z(k+1) = 1/(z(k)+2i); then a(n) = (-1)*Imag(z(n+1))/Real(z(n+1)). - Benoit Cloitre, Aug 06 2002
Maximum number of electrons in an atomic shell with total quantum number n. Partial sums of A016825. - Jeremy Gardiner, Dec 19 2004
Arithmetic mean of triangular numbers in pairs: (1+3)/2, (6+10)/2, (15+21)/2, ... . - Amarnath Murthy, Aug 05 2005
These numbers form a pattern on the Ulam spiral similar to that of the triangular numbers. - G. Roda, Oct 20 2010
Integral areas of isosceles right triangles with rational legs (legs are 2n and triangles are nondegenerate for n > 0). - Rick L. Shepherd, Sep 29 2009
Even squares divided by 2. - Omar E. Pol, Aug 18 2011
Number of stars when distributed as in the U.S.A. flag: n rows with n+1 stars and, between each pair of these, one row with n stars (i.e., n-1 of these), i.e., n*(n+1)+(n-1)*n = 2*n^2 = A001105(n). - César Eliud Lozada, Sep 17 2012
Apparently the number of Dyck paths with semilength n+3 and an odd number of peaks and the central peak having height n-3. - David Scambler, Apr 29 2013
Sum of the partition parts of 2n into exactly two parts. - Wesley Ivan Hurt, Jun 01 2013
Consider primitive Pythagorean triangles (a^2 + b^2 = c^2, gcd(a, b) = 1) with hypotenuse c (A020882) and respective odd leg a (A180620); sequence gives values c-a, sorted with duplicates removed. - K. G. Stier, Nov 04 2013
Number of roots in the root systems of type B_n and C_n (for n > 1). - Tom Edgar, Nov 05 2013
Area of a square with diagonal 2n. - Wesley Ivan Hurt, Jun 18 2014
This sequence appears also as the first and second member of the quartet [a(n), a(n), p(n), p(n)] of the square of [n, n, n+1, n+1] in the Clifford algebra Cl_2 for n >= 0. p(n) = A046092(n). See an Oct 15 2014 comment on A147973 where also a reference is given. - Wolfdieter Lang, Oct 16 2014
a(n) are the only integers m where (A000005(m) + A000203(m)) = (number of divisors of m + sum of divisors of m) is an odd number. - Richard R. Forberg, Jan 09 2015
a(n) represents the first term in a sum of consecutive integers running to a(n+1)-1 that equals (2n+1)^3. - Patrick J. McNab, Dec 24 2016
Also the number of 3-cycles in the (n+4)-triangular honeycomb obtuse knight graph. - Eric W. Weisstein, Jul 29 2017
Also the Wiener index of the n-cocktail party graph for n > 1. - Eric W. Weisstein, Sep 07 2017
Numbers represented as the palindrome 242 in number base B including B=2 (binary), 3 (ternary) and 4: 242(2)=18, 242(3)=32, 242(4)=50, ... 242(9)=200, 242(10)=242, ... - Ron Knott, Nov 14 2017
a(n) is the square of the hypotenuse of an isosceles right triangle whose sides are equal to n. - Thomas M. Green, Aug 20 2019
The sequence contains all odd powers of 2 (A004171) but no even power of 2 (A000302). - Torlach Rush, Oct 10 2019
From Bernard Schott, Aug 31 2021 and Sep 16 2021: (Start)
Apart from 0, integers such that the number of even divisors (A183063) is odd.
Proof: every n = 2^q * (2k+1), q, k >= 0, then 2*n^2 = 2^(2q+1) * (2k+1)^2; now, gcd(2, 2k+1) = 1, tau(2^(2q+1)) = 2q+2 and tau((2k+1)^2) = 2u+1 because (2k+1)^2 is square, so, tau(2*n^2) = (2q+2) * (2u+1).
The 2q+2 divisors of 2^(2q+1) are {1, 2, 2^2, 2^3, ..., 2^(2q+1)}, so 2^(2q+1) has 2q+1 even divisors {2^1, 2^2, 2^3, ..., 2^(2q+1)}.
Conclusion: these 2q+1 even divisors create with the 2u+1 odd divisors of (2k+1)^2 exactly (2q+1)*(2u+1) even divisors of 2*n^2, and (2q+1)*(2u+1) is odd. (End)
a(n) with n>0 are the numbers with period length 2 for Bulgarian and Mancala solitaire. - Paul Weisenhorn, Jan 29 2022
Number of points at L1 distance = 2 from any given point in Z^n. - Shel Kaphan, Feb 25 2023
Integer that multiplies (h^2)/(m*L^2) to give the energy of a 1-D quantum mechanical particle in a box whenever it is an integer multiple of (h^2)/(m*L^2), where h = Planck's constant, m = mass of particle, and L = length of box. - A. Timothy Royappa, Mar 14 2025

Examples

			a(3) = 18; since 2(3) = 6 has 3 partitions with exactly two parts: (5,1), (4,2), (3,3).  Adding all the parts, we get: 1 + 2 + 3 + 3 + 4 + 5 = 18. - _Wesley Ivan Hurt_, Jun 01 2013
		

References

  • Peter Atkins, Julio De Paula, and James Keeler, "Atkins' Physical Chemistry," Oxford University Press, 2023, p. 31.
  • Arthur Beiser, Concepts of Modern Physics, 2nd Ed., McGraw-Hill, 1973.
  • Martin Gardner, The Colossal Book of Mathematics, Classic Puzzles, Paradoxes and Problems, Chapter 2 entitled "The Calculus of Finite Differences," W. W. Norton and Company, New York, 2001, pages 12-13.
  • L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 44.
  • Alain M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000, p. 213.

Crossrefs

Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488.
Cf. A058331 and A247375. - Bruno Berselli, Sep 16 2014
Cf. A194715 (4-cycles in the triangular honeycomb obtuse knight graph), A290391 (5-cycles), A290392 (6-cycles). - Eric W. Weisstein, Jul 29 2017
Integers such that: this sequence (the number of even divisors is odd), A028982 (the number of odd divisors is odd), A028983 (the number of odd divisors is even), A183300 (the number of even divisors is even).

Programs

Formula

a(n) = (-1)^(n+1) * A053120(2*n, 2).
G.f.: 2*x*(1+x)/(1-x)^3.
a(n) = A100345(n, n).
Sum_{n>=1} 1/a(n) = Pi^2/12 =A072691. [Jolley eq. 319]. - Gary W. Adamson, Dec 21 2006
a(n) = A049452(n) - A033991(n). - Zerinvary Lajos, Jun 12 2007
a(n) = A016742(n)/2. - Zerinvary Lajos, Jun 20 2008
a(n) = 2 * A000290(n). - Omar E. Pol, May 14 2008
a(n) = 4*n + a(n-1) - 2, n > 0. - Vincenzo Librandi
a(n) = A002378(n-1) + A002378(n). - Joerg M. Schuetze (joerg(AT)cyberheim.de), Mar 08 2010 [Corrected by Klaus Purath, Jun 18 2020]
a(n) = A176271(n,k) + A176271(n,n-k+1), 1 <= k <= n. - Reinhard Zumkeller, Apr 13 2010
a(n) = A007607(A000290(n)). - Reinhard Zumkeller, Feb 12 2011
For n > 0, a(n) = 1/coefficient of x^2 in the Maclaurin expansion of 1/(cos(x)+n-1). - Francesco Daddi, Aug 04 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Artur Jasinski, Nov 24 2011
a(n) = A070216(n,n) for n > 0. - Reinhard Zumkeller, Nov 11 2012
a(n) = A014132(2*n-1,n) for n > 0. - Reinhard Zumkeller, Dec 12 2012
a(n) = A000217(n) + A000326(n). - Omar E. Pol, Jan 11 2013
(a(n) - A000217(k))^2 = A000217(2*n-1-k)*A000217(2*n+k) + n^2, for all k. - Charlie Marion, May 04 2013
a(n) = floor(1/(1-cos(1/n))), n > 0. - Clark Kimberling, Oct 08 2014
a(n) = A251599(3*n-1) for n > 0. - Reinhard Zumkeller, Dec 13 2014
a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+j-n+4)/3). - Wesley Ivan Hurt, Mar 12 2015
a(n) = A002061(n+1) + A165900(n). - Torlach Rush, Feb 21 2019
E.g.f.: 2*exp(x)*x*(1 + x). - Stefano Spezia, Oct 12 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/24 (A222171). - Amiram Eldar, Jul 03 2020
From Amiram Eldar, Feb 03 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sqrt(2)*sinh(Pi/sqrt(2))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(2)*sin(Pi/sqrt(2))/Pi. (End)

A069129 Centered 16-gonal numbers.

Original entry on oeis.org

1, 17, 49, 97, 161, 241, 337, 449, 577, 721, 881, 1057, 1249, 1457, 1681, 1921, 2177, 2449, 2737, 3041, 3361, 3697, 4049, 4417, 4801, 5201, 5617, 6049, 6497, 6961, 7441, 7937, 8449, 8977, 9521, 10081, 10657, 11249, 11857, 12481, 13121, 13777, 14449, 15137, 15841
Offset: 1

Views

Author

Terrel Trotter, Jr., Apr 07 2002

Keywords

Comments

Also, sequence found by reading the line from 1, in the direction 1, 17, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139098 in the same spiral. - Omar E. Pol, Apr 26 2008
The subsequence of primes begins: 17, 97, 241, 337, 449, 577, 881, 1249, 3041, 3361, 3697, 4049, 4801, 6961, 7937, 9521, 10657, 13121, 14449. See A184899: n such that the n-th centered 12-gonal number is prime. Indices of prime star numbers. - Jonathan Vos Post, Feb 27 2011
Binomial transform of [1, 16, 16, 0, 0, 0, ...] and Narayana transform (A001263) of [1, 16, 0, 0, 0, ...]. - Gary W. Adamson, Jul 28 2011
Centered hexadecagonal numbers or centered hexakaidecagonal numbers. - Omar E. Pol, Oct 03 2011
a(n) = m(n,n) for an array constructed by using the terms in A016813 as the antidiagonals; the first few antidiagonals are 1; 5,9; 13,17,21; 25,29,33,37. - J. M. Bergot, Jul 05 2013
[The first five rows begin: 1,9,21,37,57; 5,17,33,53,77; 13,29,49,73,101; 25,45,69,97,129; 41,65,93,125,161.]

Examples

			a(5) = 161 because 8*5^2 - 8*5 + 1 = 200 - 40 + 1 = 161.
		

Crossrefs

Programs

Formula

a(n) = 8*n^2 - 8*n + 1.
a(n) = A035008(n-1) + 1. - Omar E. Pol, Apr 26 2008
a(n) = 16*n + a(n-1) - 16 with n > 1, a(1)=1. - Vincenzo Librandi, Aug 08 2010
G.f.: -x*(1+14*x+x^2) / (x-1)^3. - R. J. Mathar, Feb 04 2011
E.g.f.: (8*x^2 + 1)*exp(x). - G. C. Greubel, Jul 18 2017
a(n) = A056220(2n-1). - Bruce J. Nicholson, Aug 31 2017
Sum_{n>=1} 1/a(n) = Pi * tan(Pi/(2*sqrt(2))) / (4*sqrt(2)). - Vaclav Kotesovec, Jul 23 2019
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} a(n)/n! = 9*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 9/e - 1. (End)
Product_{n>=2} (a(n) - 1) / (a(n) + 1) = Pi/4. - Dimitris Valianatos, Jun 27 2020
a(n) = A016754(n-1) + 8*A000217(n-1). - Leo Tavares, Jul 19 2021

A077221 a(0) = 0 and then alternately even and odd numbers in increasing order such that the sum of any two successive terms is a square.

Original entry on oeis.org

0, 1, 8, 17, 32, 49, 72, 97, 128, 161, 200, 241, 288, 337, 392, 449, 512, 577, 648, 721, 800, 881, 968, 1057, 1152, 1249, 1352, 1457, 1568, 1681, 1800, 1921, 2048, 2177, 2312, 2449, 2592, 2737, 2888, 3041, 3200, 3361, 3528, 3697, 3872, 4049, 4232
Offset: 0

Views

Author

Amarnath Murthy, Nov 03 2002

Keywords

Comments

This sequence arises from reading the line from 0, in the direction 0, 1, ... and the same line from 0, in the direction 0, 8, ..., in the square spiral whose vertices are the triangular numbers A000217. Cf. A139591, etc. - Omar E. Pol, May 03 2008
The general formula for alternating sums of powers of odd integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,0)-(-1)^k*P(n,2*k))/2. Here n=2, thus a(k) = |(P(2,0)-(-1)^k*P(2,2*k))/2|. - Peter Luschny, Jul 12 2009
Axis perpendicular to A046092 in the square spiral whose vertices are the triangular numbers A000217. See the comment above. - Omar E. Pol, Sep 14 2011
Column 8 of A195040. - Omar E. Pol, Sep 28 2011
Concentric octagonal numbers. A139098 and A069129 interleaved. - Omar E. Pol, Sep 17 2011
Subsequence of A194274. - Bruno Berselli, Sep 22 2011
Partial sums of A047522. - Reinhard Zumkeller, Jan 07 2012
Alternating sum of the first n odd squares in decreasing order, n >= 1. Also number of "ON" cells at n-th stage in simple 2-dimensional cellular automaton. The rules are: on the infinite square grid, start with all cells OFF, so a(0) = 0. Turn a single cell to the ON state, so a(1) = 1. At each subsequent step, the neighbor cells of each cell of the old generation are turned ON, and the cells of the old generation are turned OFF. Here "neighbor" refers to the eight adjacent cells of each ON cell. See example. - Omar E. Pol, Feb 16 2014

Examples

			From _Omar E. Pol_, Feb 16 2014: (Start)
Illustration of initial terms as a cellular automaton:
.
.                                   O O O O O O O
.                     O O O O O     O           O
.           O O O     O       O     O   O O O   O
.     O     O   O     O   O   O     O   O   O   O
.           O O O     O       O     O   O O O   O
.                     O O O O O     O           O
.                                   O O O O O O O
.
.     1       8           17              32
.
(End)
		

Crossrefs

Programs

Formula

a(2n) = 8*n^2, a(2n+1) = 8*n(n+1) + 1.
From Ralf Stephan, Mar 31 2003: (Start)
a(n) = 2*n^2 + 4*n + 1 [+1 if n is odd] with a(0)=1.
G.f.: x*(x^2+6*x+1)/(1-x)^3/(1+x). (End)
Row sums of triangle A131925; binomial transform of (1, 7, 2, 4, -8, 16, -32, ...). - Gary W. Adamson, Jul 29 2007
a(n) = a(-n); a(n+1) = A195605(n) - (-1)^n. - Bruno Berselli, Sep 22 2011
a(n) = 2*n^2 + ((-1)^n-1)/2. - Omar E. Pol, Sep 28 2011
Sum_{n>=1} 1/a(n) = Pi^2/48 + tan(Pi/(2*sqrt(2)))*Pi /(4*sqrt(2)). - Amiram Eldar, Jan 16 2023

Extensions

Extended by Ralf Stephan, Mar 31 2003

A016802 a(n) = (4*n)^2.

Original entry on oeis.org

0, 16, 64, 144, 256, 400, 576, 784, 1024, 1296, 1600, 1936, 2304, 2704, 3136, 3600, 4096, 4624, 5184, 5776, 6400, 7056, 7744, 8464, 9216, 10000, 10816, 11664, 12544, 13456, 14400, 15376, 16384, 17424, 18496, 19600, 20736, 21904, 23104, 24336, 25600, 26896, 28224
Offset: 0

Views

Author

Keywords

Comments

A bisection of A016742. Sequence arises from reading the line from 0, in the direction 0, 16, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
Also, sequence found by reading the line from 0, in the direction 0, 16, ... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Sep 10 2011

Crossrefs

Programs

Formula

a(n) = 16*n^2 = 16*A000290(n). - Omar E. Pol, Dec 11 2008
a(n) = 8*A001105(n) = 4*A016742(n) = 2*A139098(n). - Omar E. Pol, Dec 13 2008
a(n) = a(n-1) + 16*(2*n-1) (with a(0)=0). - Vincenzo Librandi, Nov 20 2010
From Amiram Eldar, Jan 25 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/96.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/192.
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/4)/(Pi/4).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/4)/(Pi/4) = 2*sqrt(2)/Pi (A112628). (End)
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 16*x*(1 + x)/(1-x)^3.
E.g.f.: 16*x*(1 + x)*exp(x).
a(n) = n*A008598(n) = A195146(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A051062 a(n) = 16*n + 8.

Original entry on oeis.org

8, 24, 40, 56, 72, 88, 104, 120, 136, 152, 168, 184, 200, 216, 232, 248, 264, 280, 296, 312, 328, 344, 360, 376, 392, 408, 424, 440, 456, 472, 488, 504, 520, 536, 552, 568, 584, 600, 616, 632, 648, 664, 680, 696, 712, 728, 744, 760, 776, 792, 808, 824, 840
Offset: 0

Views

Author

Keywords

Comments

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0(97).
n such that 32 is the largest power of 2 dividing A003629(k)^n-1 for any k. - Benoit Cloitre, Mar 23 2002
Continued fraction expansion of tanh(1/8). - Benoit Cloitre, Dec 17 2002
If Y and Z are 2-blocks of a (4n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 28 2007
General form: (q*n+x)*q x=+1; q=2=A016825, q=3=A017197, q=4=A119413, ... x=-1; q=3=A017233, q=4=A098502, ... x=+2; q=4=A051062, ... - Vladimir Joseph Stephan Orlovsky, Feb 16 2009
a(n)*n+1 = (4n+1)^2 and a(n)*(n+1)+1 = (4n+3)^2 are both perfect squares. - Carmine Suriano, Jun 01 2014
For all positive integers n, there are infinitely many positive integers k such that k*n + 1 and k*(n+1) + 1 are both perfect squares. Except for 8, all the numbers of this sequence are the smallest integers k which are solutions for getting two perfect squares. Example: a(1) = 24 and 24 * 1 + 1 = 25 = 5^2, then 24 * (1+1) + 1 = 49 = 7^2. [Reference AMM] - Bernard Schott, Sep 24 2017
Numbers k such that 3^k + 1 is divisible by 17*193. - Bruno Berselli, Aug 22 2018

References

  • Letter from Gary W. Adamson concerning Prouhet-Thue-Morse sequence, Nov 11 1999.

Crossrefs

Programs

Formula

a(n) = A118413(n+1,4) for n>3. - Reinhard Zumkeller, Apr 27 2006
a(n) = 32*n - a(n-1) for n>0, a(0)=8. - Vincenzo Librandi, Aug 06 2010
A003484(a(n)) = 8; A209675(a(n)) = 9. - Reinhard Zumkeller, Mar 11 2012
A007814(a(n)) = 3; A037227(a(n)) = 7. - Reinhard Zumkeller, Jun 30 2012
a(-1 - n) = - a(n). - Michael Somos, Jun 02 2014
Sum_{n>=0} (-1)^n/a(n) = Pi/32 (A244978). - Amiram Eldar, Feb 28 2023
From Elmo R. Oliveira, Apr 16 2024: (Start)
G.f.: 8*(1+x)/(1-x)^2.
E.g.f.: 8*exp(x)*(1 + 2*x).
a(n) = 8*A005408(n) = A008598(n) + 8 = A139098(n+1) - A139098(n).
a(n) = 4*A016825(n) = 2*A017113(n) = 2*a(n-1) - a(n-2) for n >= 2. (End)
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=0} (1 - (-1)^n/a(n)) = sqrt(2)*sin(7*Pi/32).
Product_{n>=0} (1 + (-1)^n/a(n)) = sqrt(2)*cos(7*Pi/32). (End)

A244630 a(n) = 17*n^2.

Original entry on oeis.org

0, 17, 68, 153, 272, 425, 612, 833, 1088, 1377, 1700, 2057, 2448, 2873, 3332, 3825, 4352, 4913, 5508, 6137, 6800, 7497, 8228, 8993, 9792, 10625, 11492, 12393, 13328, 14297, 15300, 16337, 17408, 18513, 19652, 20825, 22032, 23273, 24548, 25857, 27200, 28577, 29988
Offset: 0

Views

Author

Vincenzo Librandi, Jul 03 2014

Keywords

Comments

First bisection of A195047. - Bruno Berselli, Jul 03 2014
Norms of purely imaginary numbers in Z[sqrt(-17)] (for example, 3*sqrt(-17) has norm 153). - Alonso del Arte, Jun 23 2018

Crossrefs

Cf. similar sequences of the type k*n^2: A000290 (k = 1), A001105 (k = 2), A033428 (k = 3), A016742 (k = 4), A033429 (k = 5), A033581 (k = 6), A033582 (k = 7), A139098 (k = 8), A016766 (k = 9), A033583 (k = 10), A033584 (k = 11), A135453 (k = 12), A152742 (k = 13), A144555 (k = 14), A064761 (k = 15), A016802 (k = 16), this sequence (k = 17), A195321 (k = 18), A244631 (k = 19), A195322 (k = 20), A064762 (k = 21), A195323 (k = 22), A244632 (k = 23), A195824 (k = 24), A016850 (k = 25), A244633 (k = 26), A244634 (k = 27), A064763 (k = 28), A244635 (k = 29), A244636 (k = 30).

Programs

Formula

G.f.: 17*x*(1 + x)/(1 - x)^3. [corrected by Bruno Berselli, Jul 03 2014]
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = 17*A000290(n). - Omar E. Pol, Jul 03 2014
a(n) = a(-n). - Muniru A Asiru, Jun 29 2018
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 17*x*(1 + x)*exp(x).
a(n) = n*A008599(n) = A195047(2*n). (End)

A195321 a(n) = 18*n^2.

Original entry on oeis.org

0, 18, 72, 162, 288, 450, 648, 882, 1152, 1458, 1800, 2178, 2592, 3042, 3528, 4050, 4608, 5202, 5832, 6498, 7200, 7938, 8712, 9522, 10368, 11250, 12168, 13122, 14112, 15138, 16200, 17298, 18432, 19602, 20808, 22050, 23328, 24642, 25992, 27378, 28800, 30258, 31752
Offset: 0

Views

Author

Omar E. Pol, Sep 16 2011

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 18, ..., in the square spiral whose vertices are the generalized hendecagonal numbers A195160. Semi-axis opposite to A195316 in the same spiral.
Area of a square with diagonal 6n. - Wesley Ivan Hurt, Jun 19 2014
Number of identical tessellation tiles that are composed of 48 equilateral edge joined triangles that can be formed into a order n hexagon. The example tiles shown in the link below are tessellated with eight sphinx tiles. See A291582. - Craig Knecht, Sep 02 2017

Crossrefs

Programs

Formula

a(n) = 18*A000290(n) = 9*A001105(n) = 6*A033428(n) = 3*A033581(n) = 2*A016766(n).
G.f.: 18*x*(1+x)/(1-x)^3. - Wesley Ivan Hurt, Jun 20 2014
From Elmo R. Oliveira, Dec 01 2024: (Start)
E.g.f.: 18*x*(1 + x)*exp(x).
a(n) = n*A008600(n) = A195147(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A195322 a(n) = 20*n^2.

Original entry on oeis.org

0, 20, 80, 180, 320, 500, 720, 980, 1280, 1620, 2000, 2420, 2880, 3380, 3920, 4500, 5120, 5780, 6480, 7220, 8000, 8820, 9680, 10580, 11520, 12500, 13520, 14580, 15680, 16820, 18000, 19220, 20480, 21780, 23120, 24500, 25920, 27380, 28880, 30420, 32000, 33620, 35280
Offset: 0

Views

Author

Omar E. Pol, Sep 16 2011

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 20, ..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. Semiaxis opposite to A195317 in the same spiral.
a(n) is the sum of all the integers less than 10*n which are not multiple of 2 or 5. a(2) = (1 + 3 + 7 + 9) + (11 + 13 + 17 + 19) = 20 + 60 = 80 = 20 * 2^2. (Link Crux Mathematicorum). - Bernard Schott, May 15 2017
Number of terms less than 10^k (k=0, 1, 2, ...): 1, 1, 3, 8, 23, 71, 224, 708, 2237, 7072, 22361, 70711, ... - Muniru A Asiru, Feb 01 2018

Examples

			From _Muniru A Asiru_, Feb 01 2018: (Start)
n=0, a(0) = 20*0^2 = 0.
n=1, a(1) = 20*1^2 = 20.
n=1, a(2) = 20*2^2 = 80.
n=1, a(3) = 20*3^2 = 180.
n=1, a(4) = 20*4^2 = 320.
...
(End)
		

Crossrefs

Programs

Formula

a(n) = 20*A000290(n) = 10*A001105(n) = 5*A016742(n) = 4*A033429(n) = 2*A033583(n).
a(0)=0, a(1)=20, a(2)=80; for n > 2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jan 18 2013
a(n) = A010014(n) - A005899(n) for n > 0. - R. J. Cano, Sep 29 2015
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 20*x*(1 + x)/(1-x)^3.
E.g.f.: 20*x*(1 + x)*exp(x).
a(n) = n*A008602(n) = A195148(2*n). (End)

A157914 a(n) = 8*n^2 - 1.

Original entry on oeis.org

7, 31, 71, 127, 199, 287, 391, 511, 647, 799, 967, 1151, 1351, 1567, 1799, 2047, 2311, 2591, 2887, 3199, 3527, 3871, 4231, 4607, 4999, 5407, 5831, 6271, 6727, 7199, 7687, 8191, 8711, 9247, 9799, 10367, 10951, 11551, 12167, 12799, 13447, 14111, 14791
Offset: 1

Views

Author

Vincenzo Librandi, Mar 09 2009

Keywords

Comments

The identity (8*n^2 - 1)^2 - (16*n^2 - 4)*(2*n)^2 = 1 can be written as a(n)^2 - A158443(n)*A005843(n)^2 = 1.
Sequence found by reading the line from 7, in the direction 7, 31, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 03 2011
Bisection of A195605 (odd part). - Bruno Berselli, Sep 21 2011
The identity (8*n^2 - 1)^2 - (64*n^2 - 16)*(n)^2 = 1 can be written as a(n)^2 - A157913(n)*(n)^2 = 1. - Vincenzo Librandi, Feb 09 2012

Crossrefs

Programs

  • Magma
    I:=[7, 31, 71]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]];
    
  • Mathematica
    Table[8n^2-1,{n,50}]
    LinearRecurrence[{3,-3,1},{7,31,71},50] (* Harvey P. Dale, Jul 16 2025 *)
  • PARI
    a(n)=8*n^2-1 \\ Charles R Greathouse IV, Sep 03 2011

Formula

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f: x*(7+10*x-x^2)/(1-x)^3.
a(n) = A139098(n) - 1. - Omar E. Pol, Sep 03 2011
E.g.f.: (8*x^2 + 8*x - 1)*exp(x) + 1. - G. C. Greubel, Jul 15 2017
From Amiram Eldar, Feb 04 2021: (Start)
Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(8))*cot(Pi/sqrt(8)))/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(8))*csc(Pi/sqrt(8)) - 1)/2.
Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(8))*csc(Pi/sqrt(8)).
Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(8))/sqrt(2). (End)
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