A176271 -id:A176271 - cp's OEIS frontend

A005917 Rhombic dodecahedral numbers: a(n) = n^4 - (n - 1)^4.

1, 15, 65, 175, 369, 671, 1105, 1695, 2465, 3439, 4641, 6095, 7825, 9855, 12209, 14911, 17985, 21455, 25345, 29679, 34481, 39775, 45585, 51935, 58849, 66351, 74465, 83215, 92625, 102719, 113521, 125055, 137345, 150415, 164289, 178991

Offset: 1

N. J. A. Sloane

Final digits of a(n), i.e., a(n) mod 10, are repeated periodically with period of length 5 {1,5,5,5,9}. There is a symmetry in this list since the sum of two numbers equally distant from the ends is equal to 10 = 1 + 9 = 5 + 5 = 2*5. Last two digits of a(n), i.e., a(n) mod 100, are repeated periodically with period of length 50. - Alexander Adamchuk, Aug 11 2006
a(n) = VarScheme(n,2) in the scheme displayed in A128195. - Peter Luschny, Feb 26 2007
If Y is a 3-subset of a 2n-set X then, for n >= 2, a(n-2) is the number of 4-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007
The numbers are the constant number found in magic squares of order n, where n is an odd number, see the comment in A006003. A Magic Square of side 1 is 1; 3 is 15; 5 is 65 and so on. - David Quentin Dauthier, Nov 07 2008
Two times the area of the triangle with vertices at (0,0), ((n - 1)^2, n^2), and (n^2, (n - 1)^2). - J. M. Bergot, Jun 25 2013
Bisection of A006003. - Omar E. Pol, Sep 01 2018
Construct an array M with M(0,n) = 2*n^2 + 4*n + 1 = A056220(n+1), M(n,0) = 2*n^2 + 1 = A058331(n) and M(n,n) = 2*n*(n+1) + 1 = A001844(n). Row(n) begins with all the increasing odd numbers from A058331(n) to A001844(n) and column(n) begins with all the decreasing odd numbers from A056220(n+1) to A001844(n). The sum of the terms in row(n) plus those in column(n) minus M(n,n) equals a(n+1). The first five rows of array M are [1, 7, 17, 31, 49, ...]; [3, 5, 15, 29, 47, ...]; [9, 11, 13, 27, 45, ...]; [19, 21, 23, 25, 43, ...]; [33, 35, 37, 39, 41, ...]. - J. M. Bergot, Jul 16 2013 [This contribution was moved here from A047926 by Petros Hadjicostas, Mar 08 2021.]
For n>=2, these are the primitive sides s of squares of type 2 described in A344332. - Bernard Schott, Jun 04 2021
(a(n) + 1) / 2 = A212133(n) is the number of cells in the n-th rhombic-dodecahedral polycube. - George Sicherman, Jan 21 2024

J. H. Conway and R. K. Guy, The Book of Numbers, p. 53.
E. Deza and M. M. Deza, Figurate Numbers, World Scientific Publishing, 2012, pp. 123-124.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Mario Defranco and Paul E. Gunnells, Hypergraph matrix models and generating functions, arXiv:2204.11361 [math.CO], 2022.
Milan Janjic, Two Enumerative Functions
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (9).
Andy Nicol, Illustration of Rhombic Dodecahedral Numbers
C. J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.
Eric Weisstein's World of Mathematics, Rhombic Dodecahedral Number.
Eric Weisstein's World of Mathematics, Nexus Number.
D. Zeitlin, A family of Galileo sequences, Amer. Math. Monthly 82 (1975), 819-822.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

(1/12)*t*(2*n^3 - 3*n^2 + n) + 2*n - 1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A063493, A063494, A063495, A063496.
Column k=3 of A047969.
Cf. A128195, A176850, A005408, A176271, A212133.
Cf. A001844, A000583, A000290.
Cf. A007588, A000292, A000332, A002817, A342354.
Cf. A031215, A008292.
Cf. A016754, A344330, A344332.

a005917 n = a005917_list !! (n-1)
a005917_list = map sum $ f 1 [1, 3 ..] where
   f x ws = us : f (x + 2) vs where (us, vs) = splitAt x ws
-- Reinhard Zumkeller, Nov 13 2014

[n^4 - (n-1)^4: n in [1..50]]; // Vincenzo Librandi, Aug 01 2011

Table[n^4-(n-1)^4,{n,40}]  (* Harvey P. Dale, Apr 01 2011 *)
#[[2]]-#[[1]]&/@Partition[Range[0,40]^4,2,1] (* More efficient than the above Mathematica program because it only has to calculate each 4th power once *) (* Harvey P. Dale, Feb 07 2015 *)
Differences[Range[0,40]^4] (* Harvey P. Dale, Aug 11 2023 *)

a(n)=n^4-(n-1)^4 \\ Charles R Greathouse IV, Jul 31 2011

A005917_list, m = [], [24, -12, 2, 1]
for _ in range(10**2):
    A005917_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

a(n) = (2*n - 1)*(2*n^2 - 2*n + 1).
Sum_{i=1..n} a(i) = n^4 = A000583(n). First differences of A000583.
G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
More generally, g.f. for n^m - (n - 1)^m is Euler(m, x)/(1 - x)^m, where Euler(m, x) is Eulerian polynomial of degree m (cf. A008292). E.g.f.: x*(exp(y/(1 - x)) - exp(x*y/(1 - x)))/(exp(x*y/(1 - x))-x*exp(y/(1 - x))). - Vladeta Jovovic, May 08 2002
a(n) = sum of the next (2*n - 1) odd numbers; i.e., group the odd numbers so that the n-th group contains (2*n - 1) elements like this: (1), (3, 5, 7), (9, 11, 13, 15, 17), (19, 21, 23, 25, 27, 29, 31), ... E.g., a(3) = 65 because 9 + 11 + 13 + 15 + 17 = 65. - Xavier Acloque, Oct 11 2003
a(n) = 2*n - 1 + 12*Sum_{i = 1..n} (i - 1)^2. - Xavier Acloque, Oct 16 2003
a(n) = (4*binomial(n,2) + 1)*sqrt(8*binomial(n,2) + 1). - Paul Barry, Mar 14 2004
Binomial transform of [1, 14, 36, 24, 0, 0, 0, ...], if the offset is 0. - Gary W. Adamson, Dec 20 2007
Sum_{i=1..n-1}(a(i) + a(i+1)) = 8*Sum_{i=1..n}(i^3 + i) = 16*A002817(n-1) for n > 1. - Bruno Berselli, Mar 04 2011
a(n+1) = a(n) + 2*(6*n^2 + 1) = a(n) + A005914(n). - Vincenzo Librandi, Mar 16 2011
a(n) = -a(-n+1). a(n) = (1/6)*(A181475(n) - A181475(n-2)). - Bruno Berselli, Sep 26 2011
a(n) = A045975(2*n-1,n) = A204558(2*n-1)/(2*n - 1). - Reinhard Zumkeller, Jan 18 2012
a(n+1) = Sum_{k=0..2*n+1} (A176850(n,k) - A176850(n-1,k))*(2*k + 1), n >= 1. - L. Edson Jeffery, Nov 02 2012
a(n) = A005408(n-1) * A001844(n-1) = (2*(n - 1) + 1) * (2*(n - 1)*n + 1) = A000290(n-1)*12 + 2 + a(n-1). - Bruce J. Nicholson, May 17 2017
a(n) = A007588(n) + A007588(n-1) = A000292(2n-1) + A000292(2n-2) + A000292(2n-3) = A002817(2n-1) - A002817(2n-2). - Bruce J. Nicholson, Oct 22 2017
a(n) = A005898(n-1) + 6*A000330(n-1) (cf. Deza, Deza, 2012, p. 123, Section 2.6.2). - Felix Fröhlich, Oct 01 2018
a(n) = A300758(n-1) + A005408(n-1). - Bruce J. Nicholson, Apr 23 2020
G.f.: polylog(-4, x)*(1-x)/x. See the Simon Plouffe formula above (with expanded numerator), and the g.f. of the rows of A008292 by Vladeta Jovovic, Sep 02 2002. - Wolfdieter Lang, May 10 2021

A000466 a(n) = 4*n^2 - 1.

Original entry on oeis.org

-1, 3, 15, 35, 63, 99, 143, 195, 255, 323, 399, 483, 575, 675, 783, 899, 1023, 1155, 1295, 1443, 1599, 1763, 1935, 2115, 2303, 2499, 2703, 2915, 3135, 3363, 3599, 3843, 4095, 4355, 4623, 4899, 5183, 5475, 5775, 6083, 6399, 6723, 7055, 7395

Offset: 0

Chan Siu Kee (skchan5(AT)hkein.ie.cuhk.hk)

Sum_{n>=1} (-1)^n*a(n)/n! = 1 - 1/e = A068996. - Gerald McGarvey, Nov 06 2007
Sequence arises from reading the line from -1, in the direction -1, 15, ... and the same line from 3, in the direction 3, 35, ..., in the square spiral whose nonnegative vertices are the squares A000290. - Omar E. Pol, May 24 2008
a(n) is the product of the consecutive odd integers 2n-1 and 2n+1 (cf. A005408). - Doug Bell, Mar 08 2009
For n>0: a(n) = A176271(2*n,n); cf. A016754, A053755. - Reinhard Zumkeller, Apr 13 2010
a(n+1) gives the curvature c(n) of the n-th circle touching the two equal semicircles of the symmetric arbelos (1/2, 1/2) and the (n-1)-st circle, with input c(0) = 3 =  A059100(1) (referring to the second circle of the Pappus chain), for n >= 0. - Wolfdieter Lang and Kival Ngaokrajang, Jul 03 2015
After 3, a(n) is pseudoprime to base 2n. For example: (2*2)^(a(2)-1) == 1 (mod a(2)), in fact 4^14 = 15*17895697+1. - Bruno Berselli, Sep 24 2015
Numbers m such that m+1 and (m+1)/4 are squares. - Bruno Berselli, Mar 03 2016
After -1, the least common multiple of 2*m+1 and 2*m-1. - Colin Barker, Feb 11 2017
This sequence contains all products of the twin prime pairs (see A037074). - Charles Kusniec, Oct 03 2019

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
L. B. W. Jolley, Summation of Series, Dover, 2nd ed., 1961.
Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), pp. 980-981.
A. Languasco and A. Zaccagnini, Manuale di Crittografia, Ulrico Hoepli Editore (2015), p. 259.

Vincenzo Librandi, Table of n, a(n) for n = 0..900
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Kival Ngaokrajang, Illustration of the Pappus chain (downwards direction).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A001539, A016286, A016742.
Factor of A160466. Superset of A037074.
Cf. A059100 (curvatures for a Pappus chain).

[4*n^2-1: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

A000466:=n->4*n^2-1; seq(A000466(n), n=0..100); # Wesley Ivan Hurt, Nov 19 2013

4 Range[0,50]^2-1 (* Harvey P. Dale, Jan 23 2011 *)

makelist(4*n^2-1, n, 0, 50); /* Martin Ettl, Nov 12 2012 */

a(n)=4*n^2-1 \\ Charles R Greathouse IV, Oct 27 2011

[4*n^2-1 for n in (0..50)] # Bruno Berselli, Sep 24 2015

O.g.f.: ( 1-6*x-3*x^2 ) / (x-1)^3 . - R. J. Mathar, Mar 24 2011
E.g.f.: (-1 + 4*x + 4*x^2)*exp(x). - Ilya Gutkovskiy, May 26 2016
Sum_{n>=1} 1/a(n) = 1/2 [Jolley eq. 233]. - Benoit Cloitre, Apr 05 2002
Sum_{n>=1} 2/a(n) = 1 = 2/3 + 2/15 + 2/35 + 2/63 + 2/99 + 2/143, ..., with partial sums: 2/3, 4/5, 6/7, 8/9, 10/11, 12/13, 14/15, ... - Gary W. Adamson, Jun 16 2003
1/3 + Sum_{n>=2} 4/a(n) = 1 = 1/3 + 4/15 + 4/35 + 4/63, ..., with partial sums: 1/3, 3/5, 5/7, 7/9, 9/11, ..., (2n+1)/(2n+3). - Gary W. Adamson, Jun 18 2003
Sum_{n>=0} 2/a(2*n+1) = Pi/4 = 2/3 + 2/35 + 2/99, ... = (1 - 1/3) + (1/5 - 2/7) + (1/9 - 1/11) + ... = Sum_{n>=0} (-1)^n/(2*n+1). - Gary W. Adamson, Jun 22 2003
Product(n>=1, (a(n)+1)/a(n)) = Pi/2 (Wallis formula). - Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Mar 03 2004
a(n)+2 = A053755(n). - Zak Seidov, Jan 16 2007
a(n)^2 + A008586(n)^2 = A053755(n)^2 (Pythagorean triple). - Zak Seidov, Jan 16 2007
a(n) = a(n-1) + 8*n - 4 for n > 0, a(0)=-1. - Vincenzo Librandi, Dec 17 2010
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/4 - 1/2 = (A019669-1)/2. [Jolley eq (366)]. - R. J. Mathar, Mar 24 2011
For n>0, a(n) = 2/(Integral_{x=0..Pi/2} (sin(x))^3*(cos(x))^(2*n-2)). - Francesco Daddi, Aug 02 2011
Nonlinear recurrence for c(n) = a(n+1) (see the arbelos comment above) from Descartes' three circle theorem (see the links under A259555): c(n) = 4 + c(n-1) + 4*sqrt(c(n-1) + 1), with input c(0) = 3  =  A059100(1), for n >= 0. The appropriate solution of this recurrence is c(n-1) + 1 = 4*n^2. - Wolfdieter Lang, Jul 03 2015
a(n) = 3*Pochhammer(5/2,n-1)/Pochhammer(1/2,n-1). Hence, the e.g.f. for a(n+1), i.e., dropping the first term, is 3* 1F1(5/2;1/2;x), with 1F1 being the confluent hypergeometric function (also known as Kummer's). - Stanislav Sykora, May 26 2016
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/sqrt(2))/sqrt(2). - Amiram Eldar, Feb 04 2021

A014209 a(n) = n^2 + 3*n - 1.

Original entry on oeis.org

-1, 3, 9, 17, 27, 39, 53, 69, 87, 107, 129, 153, 179, 207, 237, 269, 303, 339, 377, 417, 459, 503, 549, 597, 647, 699, 753, 809, 867, 927, 989, 1053, 1119, 1187, 1257, 1329, 1403, 1479, 1557, 1637, 1719, 1803, 1889, 1977, 2067, 2159, 2253, 2349, 2447, 2547, 2649

Offset: 0

N. J. A. Sloane

Difference between n-th centered hexagonal number and (2*n)^2. - Alonso del Arte, Jul 06 2004
Given the roots to n^2 + 3*n - 1, a = -3.302775..., b = 0.302775...; then a(n) = (n + 3 + a)*(n + 3 + b). Example: a(3) = 17 = (6 - 3.302...)*(6 + 0.302775). - Gary W. Adamson, Jul 29 2009
a(n-1) = n*(n+1) - 3, with a(-1) = -3, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 13 for b = 2*n + 1. 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
Numbers m >= -1 such that 4*m + 13 is a square. - Bruce J. Nicholson, Jul 17 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Centered Hexagonal Numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002522, A003215, A176271.

Table[n^2+3*n-1, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Oct 08 2009 *)
CoefficientList[Series[(- 1 + 6 x - 3 x^2)/(1 - x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, Oct 15 2013 *)

a(n)=n^2+3*n-1 \\ Charles R Greathouse IV, Sep 24 2015

For n > 0: a(n) = A176271(n+1,n). - Reinhard Zumkeller, Apr 13 2010
a(n) = a(n-1) + 2*n + 2, with n > 0, a(0)=-1. - Vincenzo Librandi, Nov 20 2010
From Colin Barker, Feb 12 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (-1+6*x-3*x^2)/(1-x)^3. (End)
Sum_{n>=0} 1/a(n) = 1/3 + tan(sqrt(13)*Pi/2)*Pi/sqrt(13). - Amiram Eldar, Jan 08 2023
E.g.f.: exp(x)*(-1 + 4*x + x^2). - Elmo R. Oliveira, Oct 31 2024

A063494 a(n) = (2n - 1)(7n^2 - 7n + 3)/3.

Original entry on oeis.org

1, 17, 75, 203, 429, 781, 1287, 1975, 2873, 4009, 5411, 7107, 9125, 11493, 14239, 17391, 20977, 25025, 29563, 34619, 40221, 46397, 53175, 60583, 68649, 77401, 86867, 97075, 108053, 119829, 132431, 145887, 160225, 175473, 191659, 208811, 226957, 246125, 266343, 287639

Offset: 1

N. J. A. Sloane, Aug 01 2001

Interpret A176271 as an infinite square array read by antidiagonals, with rows 1,5,11,19,...; 3,9,17,27,... and so on. The sum of the terms in the n X n upper submatrix are s(n) = 1, 18, 93, 296, ... = n^2*(7*n^2-1)/6, and a(n) = s(n) - s(n-1) are the first differences. - J. M. Bergot, Jun 27 2013

Harry J. Smith, Table of n, a(n) for n = 1..1000
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

[(2*n - 1)*(7*n^2 - 7*n + 3)/3: n in [1..30]]; // G. C. Greubel, Dec 01 2017

Table[(2*n - 1)*(7*n^2 - 7*n + 3)/3, {n,1,30}] (* or *) LinearRecurrence[{4,-6,4,-1}, {1,17,75,203}, 30] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(7*n^2 - 7*n + 3)/3 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-3+6*x+21*x^2+14*x^3)*exp(x)/3 + 1)) \\ G. C. Greubel, Dec 01 2017

G.f.: x*(1+x)*(1+12*x+x^2)/(1-x)^4. - Colin Barker, Mar 02 2012
E.g.f.: (-3 + 6*x + 21*x^2 + 14*x^3)*exp(x)/3 + 1. - G. C. Greubel, Dec 01 2017
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Wesley Ivan Hurt, May 11 2023

A027688 a(n) = n^2 + n + 3.

Original entry on oeis.org

3, 5, 9, 15, 23, 33, 45, 59, 75, 93, 113, 135, 159, 185, 213, 243, 275, 309, 345, 383, 423, 465, 509, 555, 603, 653, 705, 759, 815, 873, 933, 995, 1059, 1125, 1193, 1263, 1335, 1409, 1485, 1563, 1643, 1725, 1809, 1895, 1983, 2073, 2165, 2259, 2355, 2453, 2553, 2655

Offset: 0

Patrick De Geest

Colin Barker, Table of n, a(n) for n = 0..1000
Patrick De Geest, Palindromic Quasi_Over_Squares of the form n^2+(n+X).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002522, A176271.

with (combinat):seq(fibonacci(3, n)+n+2, n=0..47); # Zerinvary Lajos, Jun 07 2008

Table[n^2 + n + 3, {n, 0, 50}] (* Bruno Berselli, Sep 03 2018 *)

Vec((3*x^2-4*x+3)/(1-x)^3 + O(x^100)) \\ Colin Barker, Dec 29 2014

For n > 0: a(n) = A176271(n+1,2). - Reinhard Zumkeller, Apr 13 2010
a(n) = 2*n + a(n-1) (with a(0)=3). - Vincenzo Librandi, Aug 05 2010
From Colin Barker, Dec 29 2014: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (3*x^2 - 4*x + 3)/(1 - x)^3. (End)
Sum_{n>=0} 1/a(n) = Pi*tanh(Pi*sqrt(11)/2)/sqrt(11). - Amiram Eldar, Jan 17 2021
E.g.f.: exp(x)*(3 + 2*x + x^2). - Elmo R. Oliveira, Oct 31 2024

Definition and offset corrected by Franklin T. Adams-Watters, Jul 06 2009

A048058 a(n) = n^2 + n + 11.

Original entry on oeis.org

11, 13, 17, 23, 31, 41, 53, 67, 83, 101, 121, 143, 167, 193, 221, 251, 283, 317, 353, 391, 431, 473, 517, 563, 611, 661, 713, 767, 823, 881, 941, 1003, 1067, 1133, 1201, 1271, 1343, 1417, 1493, 1571, 1651, 1733, 1817, 1903, 1991, 2081, 2173, 2267, 2363, 2461, 2561

Offset: 0

N. J. A. Sloane

Fontebasso lists this as a prime-generating polynomial due to Legendre, but doesn't give a reference. - Charles R Greathouse IV, Jun 30 2021

Seiichi Manyama, Table of n, a(n) for n = 0..10000
P. A. Fontebasso, Un'altra formula che dà una serie limitata di numeri primi, Supplemento al Periodico di matematica (1899), p. 130.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002522, A048059, A048097, A176271.

with(combinat):seq(fibonacci(3, n)+n+10, n=0..45); # Zerinvary Lajos, Jun 07 2008

f[n_]:=n^2+n+11;f[Range[0,100]] (* Vladimir Joseph Stephan Orlovsky, Mar 12 2011*)

a(n)=n^2+n+11 \\ Charles R Greathouse IV, Jun 29 2015

For n > 4: a(n) = A176271(n+1,6). - Reinhard Zumkeller, Apr 13 2010
a(n) = 2*n + a(n-1) (with a(0)=11). - Vincenzo Librandi, Aug 06 2010
Sum_{n>=0} 1/a(n) = Pi*tanh(Pi*sqrt(43)/2)/sqrt(43). - Amiram Eldar, Jan 17 2021
From Elmo R. Oliveira, Oct 28 2024: (Start)
G.f.: (11 - 20*x + 11*x^2)/(1 - x)^3.
E.g.f.: (11 + 2*x + x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A082111 a(n) = n^2 + 5*n + 1.

Original entry on oeis.org

1, 7, 15, 25, 37, 51, 67, 85, 105, 127, 151, 177, 205, 235, 267, 301, 337, 375, 415, 457, 501, 547, 595, 645, 697, 751, 807, 865, 925, 987, 1051, 1117, 1185, 1255, 1327, 1401, 1477, 1555, 1635, 1717, 1801, 1887, 1975, 2065, 2157, 2251, 2347, 2445, 2545, 2647

Offset: 0

Paul Barry, Apr 04 2003

From Gary W. Adamson, Jul 29 2009: (Start)
Let (a,b) = roots to x^2 - 5*x + 1 = 0 = 4.79128... and 0.208712...
Then a(n) = (n + a) * (n + b). Example: a(5) = 51 = (5 + 4.79128...) * (5 + 0.208712...) (End)
For n > 0: a(n) = A176271(n+2,n). - Reinhard Zumkeller, Apr 13 2010
a(n-2) = n*(n+1) - 5, n >= 0, with a(-2) = -5 and a(-1) = -3, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 21 for b = 2*n + 1. 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
Numbers m > 0 such that 4m+21 is a square. - Bruce J. Nicholson, Jul 19 2017
Numbers represented as 151 in number base B. If 'digits' from B upwards are allowed then 151(2)=15, 151(3)=25, 151(4)=37, 151(5)=51 also. - Ron Knott, Nov 14 2017
If A and B are sequences satisfying the recurrence t(n) = 5*t(n-1) - t(n-2) with initial values A(0) = 1, A(1) = n+5 and B(0) = -1, B(1) = n, then a(n) = A(i)^2 - A(i-1)*A(i+1) = B(j)^2 - B(j-1)*B(j+1) for i, j > 0. - Klaus Purath, Oct 18 2020
The prime terms in this sequence are listed in A089376. The prime factors are given in A038893. With the exception of 3 and 7, each prime factor p divides exactly 2 out of any p consecutive terms. If a(i) and a(k) form such a pair that are divisible by p, then i + k == -5 (mod p). - Klaus Purath, Nov 24 2020

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

First row of A082110.

Cf. A002522, A014209, A028387, A028872, A338041.

Table[n^2 + 5*n + 1,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 19 2011 *)
LinearRecurrence[{3,-3,1},{1,7,15},80] (* Harvey P. Dale, Apr 22 2012 *)

a(n)=n^2+5*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 2*n + a(n-1) + 4 (with a(0)=1). - Vincenzo Librandi, Aug 08 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=1, a(1)=7, a(2)=15. - Harvey P. Dale, Apr 22 2012
Sum_{n>=0} 1/a(n) = 8/15 + Pi*tan(sqrt(21)*Pi/2)/sqrt(21) = 1.424563592286456286... . - Vaclav Kotesovec, Apr 10 2016
From G. C. Greubel, Jul 19 2017: (Start)
G.f.: (1 + 4*x - 3*x^2)/(1 - x)^3.
E.g.f.: (x^2 + 6*x + 1)*exp(x). (End)
a(n) = A014209(n+1) - 2 = A338041(2*n+1). - Hugo Pfoertner, Oct 08 2020
a(n) = A249547(n+1) - A024206(n-4), n >= 5. - Klaus Purath, Nov 24 2020

New title (using given formula) from Hugo Pfoertner, Oct 08 2020

A027690 a(n) = n^2 + n + 5.

Original entry on oeis.org

5, 7, 11, 17, 25, 35, 47, 61, 77, 95, 115, 137, 161, 187, 215, 245, 277, 311, 347, 385, 425, 467, 511, 557, 605, 655, 707, 761, 817, 875, 935, 997, 1061, 1127, 1195, 1265, 1337, 1411, 1487, 1565, 1645, 1727, 1811, 1897, 1985, 2075, 2167, 2261, 2357, 2455, 2555

Offset: 0

Patrick De Geest

Patrick De Geest, Palindromic Quasi_Over_Squares of the form n^2+(n+X).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002061, A002522, A176271.

with(combinat): seq(fibonacci(3, n)+n+4, n=0..47); # Zerinvary Lajos, Jun 07 2008

Table[n^2 + n + 5, {n, 0, 100}] (* T. D. Noe, Oct 29 2009 *)

a(n)=n^2+n+5 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = A176271(n+1,3) for n > 1. - Reinhard Zumkeller, Apr 13 2010
a(n) = 2*n + a(n-1) for n > 0, a(0)=5. - Vincenzo Librandi, Aug 05 2010
From Ilya Gutkovskiy, Nov 25 2016: (Start)
G.f.: (5 - 8*x + 5*x^2)/(1 - x)^3.
Sum_{n>=0} 1/a(n) = Pi*tanh(sqrt(19)*Pi/2)/sqrt(19) = 0.720729156259... (End)
From Elmo R. Oliveira, Oct 28 2024: (Start)
E.g.f.: (5 + 2*x + x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

Corrected by T. D. Noe, Nov 09 2006

Definition and offset fixed by Franklin T. Adams-Watters, Jul 06 2009

A027692 a(n) = n^2 + n + 7.

Original entry on oeis.org

7, 9, 13, 19, 27, 37, 49, 63, 79, 97, 117, 139, 163, 189, 217, 247, 279, 313, 349, 387, 427, 469, 513, 559, 607, 657, 709, 763, 819, 877, 937, 999, 1063, 1129, 1197, 1267, 1339, 1413, 1489, 1567, 1647, 1729, 1813, 1899, 1987, 2077, 2169, 2263, 2359, 2457, 2557

Offset: 0

Patrick De Geest

Integers k for which the discriminant of x^3 - k*x - k is a square. - Jacob A. Siehler, Mar 14 2009
Integers k for which the Galois group of the polynomial x^3 - k*x - k over Q is a cyclic group of order 3. See Conrad, Corollary 2.5. - Peter Bala, Oct 17 2021
From Peter Bala, Nov 18 2021: (Start)
Integers k such that 4*k - 27 is a square.
Integers k for which the Galois group of the polynomial x^3 + k*(x + 1)^2 over Q is the cyclic group C_3 (apply Conrad, Corollary 2.5 and Uchida, Lemma 1).
For the primes in this list see A005471. (End)

Muniru A Asiru, Table of n, a(n) for n = 0..3000
Keith Conrad, Galois groups of cubics and quartics (not in characteristic 2).
Patrick De Geest, Palindromic Quasi_Over_Squares of the form n^2+(n+X).
Koji Uchida, Class numbers of cubic cyclic fields, Journal of the Mathematical Society of Japan, Vol. 26, No. 3, 1974, pp. 447-453.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002522, A005471 (subset of primes), A109007, A176271.

List([0..50],n->n^2+n+7); # Muniru A Asiru, Jul 15 2018

A027692 := proc(n)
    n*(n+1)+7 ;
end proc: # R. J. Mathar, Jun 06 2019

f[n_]:=n^2+n+7;f[Range[0,60]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011 *)

a(n)=n^2+n+7 \\ Charles R Greathouse IV, Jun 11 2015

For n > 2: a(n) = A176271(n+1,4). - Reinhard Zumkeller, Apr 13 2010
a(n) = 2*n + a(n-1) (with a(0)=7). - Vincenzo Librandi, Aug 05 2010
G.f.: (-7 + 12*x - 7*x^2)/(x-1)^3. - R. J. Mathar, Feb 06 2011
a(n+1) = n^2 + 3*n + 9, see A005471. - R. J. Mathar, Jun 06 2019
a(n) mod 6 = A109007(n+2). - R. J. Mathar, Jun 06 2019
Sum_{n>=0} 1/a(n) = Pi*tanh(Pi*3*sqrt(3)/2)/(3*sqrt(3)). - Amiram Eldar, Jan 17 2021
From Elmo R. Oliveira, Oct 30 2024: (Start)
E.g.f.: exp(x)*(7 + 2*x + x^2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A108195 a(n) = n^2 + 5*n - 1.

Original entry on oeis.org

5, 13, 23, 35, 49, 65, 83, 103, 125, 149, 175, 203, 233, 265, 299, 335, 373, 413, 455, 499, 545, 593, 643, 695, 749, 805, 863, 923, 985, 1049, 1115, 1183, 1253, 1325, 1399, 1475, 1553, 1633, 1715, 1799, 1885, 1973, 2063, 2155, 2249, 2345, 2443, 2543, 2645, 2749

Offset: 1

Reinhard Zumkeller, Jun 15 2005

a(n-2) = n*(n + 1) - 7, n >= 0, with a(-2) = -7, a(-1) = -5 and a(0) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 29 for b = 2*n + 1. 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 16 2013

Numbers m such that 4*m + 29 is an odd square, starting with 7^2 = A016754(3). - Bruce J. Nicholson, Jul 11 2017

			The Cross of Lorraine having n=2 crossbeams consists of a(2)=13 squares
        +---+
        | + |
    +---+---+---+
    | + | + | + |
    +---+---+---+
        | + |
+---+---+---+---+---+
| + | + | + | + | + |
+---+---+---+---+---+
        | + |
        +---+
        | + |
        +---+
        | + |
        +---+

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Gaullist Cross.
Eric Weisstein's World of Mathematics, Greek Cross.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002522, A016754, A176271.

[n^2+5*n-1: n in [1..40]]; // Vincenzo Librandi, Jun 11 2014

with (combinat):seq(fibonacci(3, n)+n-8, n=3..51); # Zerinvary Lajos, Jun 07 2008

CoefficientList[Series[(5 + 3 x - x^2 - 5 x)/(1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Jun 11 2014 *)
Array[#^2 + 5 # - 1 &, 49] (* Michael De Vlieger, Jul 12 2017 *)

a(n)=n^2+5*n-1 \\ Charles R Greathouse IV, Oct 07 2015

For n > 1: a(n) = A176271(n+2,n-1). - Reinhard Zumkeller, Apr 13 2010
a(n) = 2*n + a(n-1) + 4, with n > 1, a(1)=5. - Vincenzo Librandi, Nov 13 2010
G.f.: x*(5 - 2*x - x^2)/(1 - x)^3. - Vincenzo Librandi, Jun 11 2014
From Elmo R. Oliveira, Nov 01 2024: (Start)
E.g.f.: exp(x)*(x^2 + 6*x - 1) + 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)
Sum_{n>=1} 1/a(n) = 47/35 + tan(sqrt(29)*Pi/2)*Pi/sqrt(29). - Amiram Eldar, May 12 2025

cp's OEIS Frontend

A005917 Rhombic dodecahedral numbers: a(n) = n^4 - (n - 1)^4.

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A000466 a(n) = 4*n^2 - 1.

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A014209 a(n) = n^2 + 3*n - 1.

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A063494 a(n) = (2*n - 1)*(7*n^2 - 7*n + 3)/3.

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A027688 a(n) = n^2 + n + 3.

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A048058 a(n) = n^2 + n + 11.

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A063494 a(n) = (2n - 1)(7n^2 - 7n + 3)/3.