A270109 -id:A270109 - cp's OEIS frontend

A008865 a(n) = n^2 - 2.

-1, 2, 7, 14, 23, 34, 47, 62, 79, 98, 119, 142, 167, 194, 223, 254, 287, 322, 359, 398, 439, 482, 527, 574, 623, 674, 727, 782, 839, 898, 959, 1022, 1087, 1154, 1223, 1294, 1367, 1442, 1519, 1598, 1679, 1762, 1847, 1934, 2023, 2114, 2207, 2302, 2399, 2498

Offset: 1

N. J. A. Sloane, R. K. Guy

For n >= 2, least m >= 1 such that f(m, n) = 0 where f(m,n) = Sum_{i=0..m} Sum_{k= 0..i} (-1)^k*(floor(i/n^k) - n*floor(i/n^(k+1))). - Benoit Cloitre, May 02 2004
For n >= 3, the a(n)-th row of Pascal's triangle always contains a triple forming an arithmetic progression. - Lekraj Beedassy, Jun 03 2004
Let C = 1 + sqrt(2) = 2.414213...; and 1/C = 0.414213... Then a(n) = (n + 1 + 1/C) * (n + 1 - C). Example: a(6) = 34 = (7 + 0.414...) * (7 - 2.414...). - Gary W. Adamson, Jul 29 2009
The sequence (n-4)^2-2, n = 7, 8, ... enumerates the number of non-isomorphic sequences of length n, with entries from {1, 2, 3} and no two adjacent entries the same, that minimally contain each of the thirteen rankings of three players (111, 121, 112, 211, 122, 212, 221, 123, 132, 213, 231, 312, 321) as embedded order isomorphic subsequences. By "minimally", we mean that the n-th symbol is necessary for complete inclusion of all thirteen words. See the arXiv paper below for proof. If n = 7, these sequences are 1213121, 1213212, 1231213, 1231231, 1231321, 1232123, and 1232132, and for each case, there are 3! = 6 isomorphs. - Anant Godbole, Feb 20 2013
a(n), n >= 0, with a(0) = -2, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 8 for b = 2*n. In general D = b^2 - 4*a*c > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
With a different offset, this is 2*n^2 - (n + 1)^2, which arises in one explanation of why Bertrand's postulate does not automatically prove Legendre's conjecture: as n gets larger, so does the range of numbers that can have primes that satisfy Bertrand's postulate yet do nothing for Legendre's conjecture. - Alonso del Arte, Nov 06 2013
x*(x + r*y)^2 + y*(y + r*x)^2 can be written as (x + y)*(x^2 + s*x*y + y^2). For r >= 0, the sequence gives the values of s: in fact, s = (r + 1)^2 - 2. - Bruno Berselli, Feb 20 2019
For n >= 2, the continued fraction expansion of sqrt(a(n)) is [n-1; {1, n-2, 1, 2n-2}]. For n=2, this collapses to [1; {2}]. - Magus K. Chu, Sep 06 2022

			G.f. = -x + 2*x^2 + 7*x^3 + 14*x^4 + 23*x^5 + 34*x^6 + 47*x^7 + 62*x^8 + 79*x^9 + ...

T. D. Noe, Table of n, a(n) for n = 1..1000
Anant Godbole and Martha Liendo, Waiting time distribution for the emergence of superpatterns, arxiv 1302.4668 [math.PR], 2013.
Eric Weisstein's World of Mathematics, Near-Square Prime.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A145067 (Zero followed by partial sums of A008865).
Cf. A000027, A013648.
Cf. A028871 (primes).
Cf. A263766 (partial products).
Cf. A270109. [Bruno Berselli, Mar 17 2016]

a008865 = (subtract 2) . (^ 2) :: Integral t => t -> t
a008865_list = scanl (+) (-1) [3, 5 ..]
-- Reinhard Zumkeller, May 06 2013

[n^2 - 2: n in [1..60]]; // Vincenzo Librandi, May 01 2014

Range[50]^2 - 2 (* Harvey P. Dale, Mar 14 2011 *)

{for(n=1, 47, print1(n^2-2, ","))} \\ Klaus Brockhaus, Oct 17 2008

For n > 1: a(n) = A143053(A000290(n)), A143054(a(n)) = A000290(n). - Reinhard Zumkeller, Jul 20 2008
G.f.: (x-5*x^2+2*x^3)/(-1+3*x-3*x^2+x^3). - Klaus Brockhaus, Oct 17 2008
E.g.f.: (x^2 + x -2)*exp(x) + 2. - G. C. Greubel, Aug 19 2017
a(n+1) = A101986(n) - A101986(n-1) = A160805(n) - A160805(n-1). - Reinhard Zumkeller, May 26 2009
For n > 1, a(n) = floor(n^5/(n^3 + n + 1)). - Gary Detlefs, Feb 10 2010
a(n) = a(n-1) + 2*n - 1 for n > 1, a(1) = -1. - Vincenzo Librandi, Nov 18 2010
Right edge of the triangle in A195437: a(n) = A195437(n-2, n-2). - Reinhard Zumkeller, Nov 23 2011
a(n)*a(n-1) + 2 = (a(n) - n)^2 = A028552(n-2)^2. - Bruno Berselli, Dec 07 2011
a(n+1) = A000096(n) + A000096(n-1) for all n in Z. - Michael Somos, Nov 11 2015
From Amiram Eldar, Jul 13 2020: (Start)
Sum_{n>=1} 1/a(n) = (1 - sqrt(2)*Pi*cot(sqrt(2)*Pi))/4.
Sum_{n>=1} (-1)^n/a(n) = (1 - sqrt(2)*Pi*cosec(sqrt(2)*Pi))/4. (End)
Assume offset 0. Then a(n) = 2*LaguerreL(2, 1 - n). - Peter Luschny, May 09 2021
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=1} (1 - 1/a(n)) = sqrt(2/3)*sin(sqrt(3)*Pi)/sin(sqrt(2)*Pi).
Product_{n>=2} (1 + 1/a(n)) = -Pi/(sqrt(2)*sin(sqrt(2)*Pi)). (End)

A130883 a(n) = 2*n^2 - n + 1.

Original entry on oeis.org

1, 2, 7, 16, 29, 46, 67, 92, 121, 154, 191, 232, 277, 326, 379, 436, 497, 562, 631, 704, 781, 862, 947, 1036, 1129, 1226, 1327, 1432, 1541, 1654, 1771, 1892, 2017, 2146, 2279, 2416, 2557, 2702, 2851, 3004, 3161, 3322, 3487, 3656, 3829, 4006, 4187, 4372, 4561

Offset: 0

Mohammad K. Azarian, Jul 26 2007

Maximum number of regions determined by n bent lines (or angular sectors). See Concrete Mathematics reference.
A "bent line" may also be regarded as a "long-legged letter V", meaning a letter V with both line segments extended to infinity.  See A117625 for the analogous sequence for a long-legged Z. - N. J. A. Sloane, Jun 18 2025
a(n)*Pi is the total length of half circle spiral after n rotations. It is formed as irregular spiral with two center points. At the 2nd stage, there are two alternatives: (1) select 2nd half circle radius, r2  = 2, the sequence will be A014105 or (2) select r2 = 0, the sequence will be A130883. See illustration in links. - Kival Ngaokrajang, Jan 19 2014
A128218(a(n)) = 2*n+1 and A128218(m) != 2*n+1 for m < a(n). - Reinhard Zumkeller, Jun 20 2015

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, Reading, MA, 1994, pp. 7-8, and Problem 1.18, pages 19 and 500.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Dmitry Efimov, Hafnian of two-parameter matrices, arXiv:2101.09722 [math.CO], 2021.
Guo-Niu Han, Enumeration of Standard Puzzles
Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Kival Ngaokrajang, Illustration of irregular spirals (center points: 1, 2)
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.
N. J. A. Sloane, Illustration for a(3) = 16
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000124, A000384, A014105, A084849, A128218, A128918, A152947, A270109.
See also A117625.
A row of the array in A386478.

a130883 = a128918 . (* 2)  -- Reinhard Zumkeller, Oct 27 2013

[2*n^2 - n + 1 : n in [0..50]]; // Wesley Ivan Hurt, Mar 25 2020

a[n_]:=2*n^2-n+1; (* or *) Array[ -#*(1-#*2)+1&,5!,0] (* Vladimir Joseph Stephan Orlovsky, Dec 21 2008 *)
LinearRecurrence[{3,-3,1},{1,2,7},50] (* Harvey P. Dale, Jul 20 2011 *)

a(n)=2*n^2-n+1 \\ Charles R Greathouse IV, Sep 24 2015

def A130883(n): return n*(2*n - 1) + 1 # Chai Wah Wu, May 24 2022

a(n) = a(n-1) + 4*n - 3 for n > 0, a(0)=1. - Vincenzo Librandi, Nov 23 2010
a(n) = A000124(2*n) - 2*n. - Geoffrey Critzer, Mar 30 2011
O.g.f.: (4*x^2-x+1)/(1-x)^3. - Geoffrey Critzer, Mar 30 2011
a(n) = 2*a(n-1) - a(n-2) + 4. - Eric Werley, Jun 27 2011
a(0)=1, a(1)=2, a(2)=7; for n > 2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jul 20 2011
a(n) = A128918(2*n). - Reinhard Zumkeller, Oct 27 2013
a(n) = 1 + A000384(n). - Omar E. Pol, Apr 27 2017
E.g.f.: (2*x^2 + x + 1)*exp(x). - G. C. Greubel, Jul 14 2017
a(n) = A152947(2*n+1). - Franck Maminirina Ramaharo, Jan 10 2018

A027444 a(n) = n^3 + n^2 + n.

Original entry on oeis.org

0, 3, 14, 39, 84, 155, 258, 399, 584, 819, 1110, 1463, 1884, 2379, 2954, 3615, 4368, 5219, 6174, 7239, 8420, 9723, 11154, 12719, 14424, 16275, 18278, 20439, 22764, 25259, 27930, 30783, 33824, 37059, 40494, 44135, 47988, 52059, 56354, 60879, 65640, 70643, 75894

Offset: 0

Patrick De Geest and Mark Milhet (mm992395(AT)shellus.com)

For n>1, a(n) is the volume of a truncated square pyramid with height n and base lengths n+2 and n-1. - Wesley Ivan Hurt, Apr 05 2016

			For n = 4, 4^3 + 4^2 + 4 = 64 + 16 + 4 = 84.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. Applegate, M. LeBrun, and N. J. A. Sloane, Dismal Arithmetic, J. Int. Seq. 14 (2011) # 11.9.8.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Eric Weisstein's World of Mathematics, Magic Circles.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Column k=3 of A228275.

Cf. A270109.

[n^3 + n^2 + n: n in [0..50]]; // Vincenzo Librandi, Jun 09 2011

A027444:=n->n^3+n^2+n: seq(A027444(n), n=0..100); # Wesley Ivan Hurt, Apr 05 2016

Table[n^3 + n^2 + n, {n, 0, 50}] (* Harvey P. Dale, Dec 13 2013 *)

O.g.f.: x*(3 + 2*x + x^2)/(1 - x)^4. - R. J. Mathar, Feb 04 2008

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - Wesley Ivan Hurt, Apr 05 2016

A270867 a(n) = n^3 + 2n^2 + 4n + 1.

Original entry on oeis.org

1, 8, 25, 58, 113, 196, 313, 470, 673, 928, 1241, 1618, 2065, 2588, 3193, 3886, 4673, 5560, 6553, 7658, 8881, 10228, 11705, 13318, 15073, 16976, 19033, 21250, 23633, 26188, 28921, 31838, 34945, 38248, 41753, 45466, 49393, 53540, 57913, 62518, 67361, 72448

Offset: 0

Vincenzo Librandi, Apr 01 2016

Numbers of the type (m+1)^3 - (m-1)*m. Similar sequences are: A069778 with the closed form (m+1)^3 - m*(m+1), A152015 with (m+1)^3 - (m+1)*(m+2).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Andrew Misseldine, Counting Schur Rings over Cyclic Groups, arXiv preprint arXiv:1508.03757 [math.RA], 2015 (page 19, 4th row; page 21, 3rd row).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A069778, A016957, A152015, A270109.

Magma
```
[n^3+2*n^2+4*n+1: n in [0..50]];
```

A270867:=n->n^3+2*n^2+4*n+1: seq(A270867(n), n=0..100); # Wesley Ivan Hurt, Apr 01 2016

Table[n^3 + 2 n^2 + 4 n + 1, {n, 0, 40}]

x='x+O('x^99); Vec((1+4*x-x^2+2*x^3)/(1-x)^4) \\ Altug Alkan, Apr 01 2016

for i in range(0,100):print(i**3+2*i**2+4*i+1) # Soumil Mandal, Apr 02 2016

O.g.f.: (1 + 4*x - x^2 + 2*x^3)/(1 - x)^4.
E.g.f.: (1 + 7*x + 5*x^2 + x^3)*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = -A270109(-n-1). - Bruno Berselli, Apr 01 2016
a(n+2) - 2*a(n+1) + a(n) = A016957(n+1). - Wesley Ivan Hurt, Apr 02 2016

A085490 Number of pairs with two different elements which can be obtained by selecting unique elements from two sets with n+1 and n^2 elements respectively and n common elements.

Original entry on oeis.org

0, 1, 10, 33, 76, 145, 246, 385, 568, 801, 1090, 1441, 1860, 2353, 2926, 3585, 4336, 5185, 6138, 7201, 8380, 9681, 11110, 12673, 14376, 16225, 18226, 20385, 22708, 25201, 27870, 30721, 33760, 36993, 40426, 44065, 47916, 51985, 56278, 60801, 65560, 70561, 75810, 81313

Offset: 0

Polina S. Dolmatova (polinasport(AT)mail.ru), Aug 15 2003

			a(2) = 10 because we can write a(2) = 2^3 + 2^2 - 2 = 10.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A270109.

[n^3+n^2-n: n in [0..50]]; // Vincenzo Librandi, Jun 22 2017

a:=n->sum(n*k, k=0..n):seq(a(n)+sum(n*k, k=2..n), n=0..30); # Zerinvary Lajos, Jun 10 2008
a:=n->sum(-2+sum(2+sum(2, j=1..n),j=1..n),j=1..n):seq(a(n)/2,n=0..40);# Zerinvary Lajos, Dec 06 2008
seq(n^3+n^2-n, n=0..100); # Robert Israel, Dec 05 2014

LinearRecurrence[{4, -6, 4, -1}, {0, 1, 10, 33}, 60] (* Vincenzo Librandi, Jun 22 2017 *)

a(n) = n^3 + n^2 - n = n*A028387(n-1).
a(n) = A081437(n-1), n>0. - R. J. Mathar, Sep 12 2008
G.f.: x*(1+6*x-x^2)/(1-x)^4. - Robert Israel, Dec 05 2014
E.g.f.: x*(1+4*x+x^2)*exp(x). - Robert Israel, Dec 05 2014
For q a prime power, a(q) is the number of pairs of commuting nilpotent 2*2 matrices with coefficients in GL(q). (Proof: the zero matrix commutes with all q^2 nilpotent matrices, each of the remaining q^2-1 nilpotent matrices commutes with exactly q nilpotent matrices.) - Mark Wildon, Jun 18 2017

cp's OEIS Frontend

A008865 a(n) = n^2 - 2.

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A130883 a(n) = 2*n^2 - n + 1.

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A027444 a(n) = n^3 + n^2 + n.

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A270867 a(n) = n^3 + 2*n^2 + 4*n + 1.

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A085490 Number of pairs with two different elements which can be obtained by selecting unique elements from two sets with n+1 and n^2 elements respectively and n common elements.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A270867 a(n) = n^3 + 2n^2 + 4n + 1.