A082039 -id:A082039 - cp's OEIS frontend

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

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

Enoch Haga, G. L. Honaker, Jr., Apr 10 2000

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

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Robert G. Wilson v, Cover of the March 1964 issue of Scientific American
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000384, A033951, A054552, A054554, A054556, A054567, A054568, A068225.
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.

List([1..50], n-> 4*n^2-6*n+3); # G. C. Greubel, Jul 04 2019

[4*n^2-6*n+3: n in [1..50]]; // G. C. Greubel, Jul 04 2019

f[n_]:= 4*n^2-6*n+3; Array[f, 50] (* Vladimir Joseph Stephan Orlovsky, Sep 02 2008 *)
LinearRecurrence[{3,-3,1},{1,7,21},50] (* Harvey P. Dale, Nov 17 2012 *)

a(n)=4*n^2-6*n+3 \\ Charles R Greathouse IV, Sep 24 2015

[4*n^2-6*n+3 for n in (1..50)] # G. C. Greubel, Jul 04 2019

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

Edited by Frank Ellermann, Feb 24 2002

A059826 a(n) = (n^2 - n + 1)*(n^2 + n + 1).

Original entry on oeis.org

1, 3, 21, 91, 273, 651, 1333, 2451, 4161, 6643, 10101, 14763, 20881, 28731, 38613, 50851, 65793, 83811, 105301, 130683, 160401, 194923, 234741, 280371, 332353, 391251, 457653, 532171, 615441, 708123, 810901, 924483, 1049601, 1187011, 1337493, 1501851, 1680913

Offset: 0

N. J. A. Sloane, Feb 24 2001

Main diagonal of A082039. - Paul Barry, Apr 02 2003

The base of the natural logarithms e = 2*Sum_{n>=0} 1/(a(n)*n!) and zeta(2) = Pi^2/6 = 1 + 2*Sum_{n>=1} (-1)^(n+1)/(a(n)*n^2). - Peter Bala, Jan 20 2008

Harry J. Smith, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A002061, A062159.

Main diagonal of A082039.

[n^4+n^2+1 : n in [0..50]]; // Wesley Ivan Hurt, Jun 09 2014

with(combinat): seq(fibonacci(3,n)+n^4, n=0..40); # Zerinvary Lajos, May 25 2008

Table[n^4 + n^2 + 1, {n, 0, 50}] (* Wesley Ivan Hurt, Jun 09 2014 *)

a(n) = { my(f=n^2 + 1); (f - n)*(f + n) } \\ Harry J. Smith, Jun 29 2009

a(n) = n^4+n^2+1. - Paul Barry, Apr 02 2003
a(n) = (n^2-n+1) * (n^2+n+1) = A002061(n) * A002061(n+1), products of two consecutive central polygonal numbers. a(n) = (n^6-1)/(n^2-1), n>1. a(n) = (n^5-n^4+n^3-n^2+n-1)/(n-1) = A062159(n)/(n-1), n>1. - Alexander Adamchuk, Apr 12 2006
O.g.f.: (-1+2*x-16*x^2-6*x^3-3*x^4) / (x-1)^5. - R. J. Mathar, Feb 26 2008
a(n) = A219069(n,1), for n > 0. - Reinhard Zumkeller, Nov 11 2012
a(n+2) = (n^2+3n+3) * (n^2+5n+7) = (t(n)+t(n+2)) * (t(n+1)+t(n+3)), where t=A000217 are triangular numbers. For n>=1, a(n+2) = t(2*t(n+2)+t(n)) -t(t(n)-1). - J. M. Bergot, Nov 29 2012
4*a(n) = (n^2+n+1)^2+(n^2-n+1)^2+(n^2+n-1)^2+(n^2-n-1)^2. - Bruno Berselli, Jul 03 2014
a(n) = A002061(n^2). - Franklin T. Adams-Watters, Aug 01 2014
Sum_{n>=0} 1/a(n) = 1/2 + sqrt(3)*Pi*tanh(sqrt(3)*Pi/2)/6. - Amiram Eldar, Feb 14 2021

A082040 a(n) = 9n^2 + 3n + 1.

Original entry on oeis.org

1, 13, 43, 91, 157, 241, 343, 463, 601, 757, 931, 1123, 1333, 1561, 1807, 2071, 2353, 2653, 2971, 3307, 3661, 4033, 4423, 4831, 5257, 5701, 6163, 6643, 7141, 7657, 8191, 8743, 9313, 9901, 10507, 11131, 11773, 12433, 13111, 13807, 14521, 15253, 16003, 16771, 17557

Offset: 0

Paul Barry, Apr 02 2003

4th row of A082039, case k = 3 of family T(n,k) = k^2*n^2 + k*n + 1.
a(n)^2 = 81*n^4 + 54*n^3 + 27*n^2 + 6*n + 1 = (24*((3*((3*n^2 + n)/2)^2 + ((3*n^2 + n)/2))/2) + 1). Therefore, (a(n)^2 - 1)/24 is a second pentagonal number (A005449) of index number equal to the n-th second pentagonal number. For example, a(30) = 8191 and (8191^2 - 1)/24 = (67092481 - 1)/24 = 2795520, the 1365th second pentagonal number. 1365 is the 30th second pentagonal number. - Raphie Frank, Sep 19 2012
For n >= 1, a(n) is the number of vertices in the hex derived network HDN1(n+1) from the Manuel et al. reference (see HFN1(4) in Fig. 8). - Emeric Deutsch, May 21 2018
4*a(n) - 3 is a square. - Muniru A Asiru, May 24 2018

Muniru A Asiru, Table of n, a(n) for n = 0..5000
Nicolay Avilov, Graphic illustration of sequence members
P. Manuel, R. Bharati, I. Rajasingh, and Chris Monica M, On minimum metric dimension of honeycomb networks, Journal of Discrete Algorithms, Vol. 6 (2008), pp. 20-27.
Leo Tavares, Snowflake illustration.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002061, A005449, A045945, A054569, A082039.
Partial sums of A019557.
Cf. A000290, A003215.

List([0..50], n->9*n^2+3*n+1); # Muniru A Asiru, May 21 2018

seq(9*n^2+3*n+1,n=0..50); # Muniru A Asiru, May 21 2018

a(n)=9*n^2+3*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 18*n + a(n-1) - 6 with n > 0, a(0)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = A045945(n) + 1: subsequence of A002061. - Muniru A Asiru, May 26 2018
a(n) = A003215(n) + 6*A000290(n). - Leo Tavares, Jul 14 2023
From Elmo R. Oliveira, Oct 23 2024: (Start)
G.f.: (1 + 10*x + 7*x^2)/(1 - x)^3.
E.g.f.: (1 + 12*x + 9*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A082043 Square array, A(n, k) = (kn)^2 + 2k*n + 1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 9, 9, 1, 1, 16, 25, 16, 1, 1, 25, 49, 49, 25, 1, 1, 36, 81, 100, 81, 36, 1, 1, 49, 121, 169, 169, 121, 49, 1, 1, 64, 169, 256, 289, 256, 169, 64, 1, 1, 81, 225, 361, 441, 441, 361, 225, 81, 1, 1, 100, 289, 484, 625, 676, 625, 484, 289, 100, 1

Offset: 0

Paul Barry, Apr 03 2003

			Array, A(n, k), begins as:
  1,   1,   1,    1,    1,    1,    1,    1,     1, ... A000012;
  1,   4,   9,   16,   25,   36,   49,   64,    81, ... A000290;
  1,   9,  25,   49,   81,  121,  169,  225,   289, ... A016754;
  1,  16,  49,  100,  169,  256,  361,  484,   625, ... A016778;
  1,  25,  81,  169,  289,  441,  625,  841,  1089, ... A016814;
  1,  36, 121,  256,  441,  676,  961, 1296,  1681, ... A016862;
  1,  49, 169,  361,  625,  961, 1369, 1849,  2401, ... A016922;
  1,  64, 225,  484,  841, 1296, 1849, 2500,  3249, ... A016994;
  1,  81, 289,  625, 1089, 1681, 2401, 3249,  4225, ... A017078;
  1, 100, 361,  784, 1369, 2116, 3025, 4096,  5329, ... A017174;
  1, 121, 441,  961, 1681, 2601, 3721, 5041,  6561, ... A017282;
  1, 144, 529, 1156, 2025, 3136, 4489, 6084,  7921, ... A017402;
  1, 169, 625, 1369, 2401, 3721, 5329, 7225,  9409, ... A017534;
  1, 196, 729, 1600, 2809, 4356, 6241, 8464, 11025, ... ;
Antidiagonals, T(n, k), begin as:
  1;
  1,   1;
  1,   4,   1;
  1,   9,   9,   1;
  1,  16,  25,  16,   1;
  1,  25,  49,  49,  25,   1;
  1,  36,  81, 100,  81,  36,   1;
  1,  49, 121, 169, 169, 121,  49,   1;
  1,  64, 169, 256, 289, 256, 169,  64,   1;
  1,  81, 225, 361, 441, 441, 361, 225,  81,   1;
  1, 100, 289, 484, 625, 676, 625, 484, 289, 100,  1;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Rows include A000290, A016754, A016778, A016814, A016862, A016922, A016994, A017078, A017174, A017282, A017402, A017534.
Diagonals include A000583, A058031, A062938, A082044 (main diagonal).
Diagonal sums (row sums if viewed as number triangle) are A082045.
Cf. A082039, A082045, A082046, A082105.

A082043:= func< n,k | (k*(n-k))^2 +2*k*(n-k) +1 >;
[A082043(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Dec 24 2022

T[n_, k_]:= (k*(n-k))^2 +2*k*(n-k) +1;
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 24 2022 *)

def A082043(n,k): return (k*(n-k))^2 +2*k*(n-k) +1
flatten([[A082043(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Dec 24 2022

A(n, k) = (k*n)^2 + 2*k*n + 1 (square array).
T(n, k) = (k*(n-k))^2 + 2*k*(n-k) + 1 (number triangle).
A(k, n) = A(n, k).
T(n, n-k) = T(n, k).
A(n, n) = T(2*n, n) = A082044(n).
A(n, n-1) = T(2*n+1, n-1) = A058031(n), n >= 1.
A(n, n-2) = T(2*(n-1), n) = A000583(n-1), n >= 2.
A(n, n-3) = T(2*n-3, n) = A062938(n-3), n >= 3.
Sum_{k=0..n} T(n, k) = A082045(n) (diagonal sums).
Sum_{k=0..n} (-1)^k * T(n, k) = (1/4)*(1+(-1)^n)*(2 - 3*n). - G. C. Greubel, Dec 24 2022

A082046 Square array, A(n, k) = (kn)^2 + 3k*n + 1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 11, 11, 1, 1, 19, 29, 19, 1, 1, 29, 55, 55, 29, 1, 1, 41, 89, 109, 89, 41, 1, 1, 55, 131, 181, 181, 131, 55, 1, 1, 71, 181, 271, 305, 271, 181, 71, 1, 1, 89, 239, 379, 461, 461, 379, 239, 89, 1, 1, 109, 305, 505, 649, 701, 649, 505, 305, 109, 1

Offset: 0

Paul Barry, Apr 03 2003

			Array, A(n, k), begins as:
  1,  1,   1,   1,   1,    1,    1,    1, ... A000012;
  1,  5,  11,  19,  29,   41,   55,   71, ... A028387;
  1, 11,  29,  55,  89,  131,  181,  239, ... A082108;
  1, 19,  55, 109, 181,  271,  379,  505, ... A069131;
  1, 29,  89, 181, 305,  461,  649,  869, ... ;
  1, 41, 131, 271, 461,  701,  991, 1331, ... ;
  1, 55, 181, 379, 649,  991, 1405, 1891, ... ;
  1, 71, 239, 505, 869, 1331, 1891, 2549, ... ;
Antidiagonals, T(n, k), begin as:
  1;
  1,  1;
  1,  5,   1;
  1, 11,  11,   1;
  1, 19,  29,  19,   1;
  1, 29,  55,  55,  29,   1;
  1, 41,  89, 109,  89,  41,   1;
  1, 55, 131, 181, 181, 131,  55,  1;
  1, 71, 181, 271, 305, 271, 181, 71,  1;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A028387, A057721, A069131, A072025, A082039, A082043, A082047, A082105, A082108.

[(k*(n-k))^2 + 3*(k*(n-k)) + 1: k in [0..n], n in [0..13]]; // G. C. Greubel, Dec 22 2022

T[n_, k_]:= (k*(n-k))^2 + 3*(k*(n-k)) + 1;
Table[T[n,k], {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 22 2022 *)

def A082046(n,k): return (k*(n-k))^2 + 3*(k*(n-k)) + 1
flatten([[A082046(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Dec 22 2022

A(n, k) = (k*n)^2 + 3*k*n + 1 (square array).
A(k, n) = A(n, k).
A(n, n) = T(2*n, n) = A057721(n).
A(n, n+1) = A072025(n).
T(n, k) = (k*(n-k))^2 + 3*k*(n-k) + 1 (antidiagonals).
Sum_{k=0..n} T(n, k) = A082047(n) (antidiagonal sums).
From G. C. Greubel, Dec 22 2022: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = (1/2)*(1 + (-1)^n)*(1 - 2*n).
T(2*n+1, n-1) = T(2*n-1, n-1) = A072025(n-1). (End)

A082110 Array A(n,k) = (kn)^2 + 5(k*n) + 1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 15, 15, 1, 1, 25, 37, 25, 1, 1, 37, 67, 67, 37, 1, 1, 51, 105, 127, 105, 51, 1, 1, 67, 151, 205, 205, 151, 67, 1, 1, 85, 205, 301, 337, 301, 205, 85, 1, 1, 105, 267, 415, 501, 501, 415, 267, 105, 1, 1, 127, 337, 547, 697, 751, 697, 547, 337, 127, 1

Offset: 0

Paul Barry, Apr 04 2003

			Square array, A(n, k), begins as:
  1,   1,   1,   1,    1,    1,    1,    1,    1, ... A000012;
  1,   7,  15,  25,   37,   51,   67,   85,  105, ... A082111;
  1,  15,  37,  67,  105,  151,  205,  267,  337, ... A082112;
  1,  25,  67, 127,  205,  301,  415,  547,  697, ...
  1,  37, 105, 205,  337,  501,  697,  925, 1185, ...
  1,  51, 151, 301,  501,  751, 1051, 1401, 1801, ...
  1,  67, 205, 415,  697, 1051, 1477, 1975, 2545, ...
  1,  85, 267, 547,  925, 1401, 1975, 2647, 3417, ...
  1, 105, 337, 697, 1185, 1801, 2545, 3417, 4417, ...
Antidiagonals, T(n, k), begins as:
  1;
  1,  1;
  1,  7,   1;
  1, 15,  15,   1;
  1, 25,  37,  25,   1;
  1, 37,  67,  67,  37,   1;
  1, 51, 105, 127, 105,  51,   1;
  1, 67, 151, 205, 205, 151,  67,  1;
  1, 85, 205, 301, 337, 301, 205, 85,  1;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A082039, A082043, A082046, A082105, A082111, A082112, A082113, A082114.

[(k*(n-k))^2 + 5*(k*(n-k)) + 1: k in [0..n], n in [0..13]]; // G. C. Greubel, Dec 22 2022

T[n_, k_]:= (k*(n-k))^2 + 5*(k*(n-k)) + 1;
Table[T[n,k], {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 22 2022 *)

def A082110(n,k): return (k*(n-k))^2 + 5*(k*(n-k)) + 1
flatten([[A082110(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Dec 22 2022

A(n, k) = (k*n)^2 + 5*(k*n) + 1 (Square array).
A(k, n) = A(n, k).
A(2, k) = A082111(k).
A(3, k) = A082112(k).
A(n, n) = T(2*n, n) = A082113(n) (main diagonal).
T(n, k) = (k*(n-k))^2 + 5*k*(n-k) + 1 (number triangle).
Sum_{k=0..n} T(n, k) = A082114(n) (diagonal sums of the array).
From G. C. Greubel, Dec 22 2022: (Start)
T(n, n-k) = T(n, k).
Sum_{k=0..n} (-1)^k*T(n, k) = (1 - 3*n)*(1 + (-1)^n)/2. (End)

A082038 A square array of quadratic-factorial numbers, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 7, 14, 6, 1, 13, 42, 78, 24, 1, 21, 86, 258, 504, 120, 1, 31, 146, 546, 1752, 3720, 720, 1, 43, 222, 942, 3768, 13320, 30960, 5040, 1, 57, 314, 1446, 6552, 28920, 113040, 287280, 40320, 1, 73, 422, 2058, 10104, 50520, 246960, 1063440

Offset: 0

Paul Barry, Apr 02 2003

Rows include A000142, A001564, A082035, A082036.

			Rows begin
1 1 2 6 24 ...
1 3 14 78 504 ...
1 7 42 258 1752 ...
1 13 86 546 3768 ...
1 21 146 942 6552 ...

Cf. A082037, A082039.

Square array defined by T(n, k)=((kn)^2+kn+1)n!

A082041 a(n) = 16n^2 + 4n + 1.

Original entry on oeis.org

1, 21, 73, 157, 273, 421, 601, 813, 1057, 1333, 1641, 1981, 2353, 2757, 3193, 3661, 4161, 4693, 5257, 5853, 6481, 7141, 7833, 8557, 9313, 10101, 10921, 11773, 12657, 13573, 14521, 15501, 16513, 17557, 18633, 19741, 20881, 22053, 23257, 24493

Offset: 0

Paul Barry, Apr 02 2003

Also sequence found by reading the segment (1,21) together with the line from 21, in the direction 21, 73, ..., in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Nov 02 2012

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Column k=4 of A082039.

Cf. A054569, A082040, A074377.

Table[16n^2+4n+1,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{1,21,73},50] (* Harvey P. Dale, Sep 28 2024 *)

a(n)=16*n^2+4*n+1 \\ Charles R Greathouse IV, Jun 17 2017

G.f.: (-1-18*x-13*x^2)/(x-1)^3 . - R. J. Mathar, Dec 03 2014
From Elmo R. Oliveira, Oct 28 2024: (Start)
E.g.f.: exp(x)*(1 + 20*x + 16*x^2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

cp's OEIS Frontend

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

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A059826 a(n) = (n^2 - n + 1)*(n^2 + n + 1).

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A082040 a(n) = 9*n^2 + 3*n + 1.

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A082043 Square array, A(n, k) = (k*n)^2 + 2*k*n + 1, read by antidiagonals.

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A082046 Square array, A(n, k) = (k*n)^2 + 3*k*n + 1, read by antidiagonals.

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A082110 Array A(n,k) = (k*n)^2 + 5*(k*n) + 1, read by antidiagonals.

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A082038 A square array of quadratic-factorial numbers, read by antidiagonals.

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

A082040 a(n) = 9n^2 + 3n + 1.

A082043 Square array, A(n, k) = (kn)^2 + 2k*n + 1, read by antidiagonals.

A082046 Square array, A(n, k) = (kn)^2 + 3k*n + 1, read by antidiagonals.

A082110 Array A(n,k) = (kn)^2 + 5(k*n) + 1, read by antidiagonals.

A082041 a(n) = 16n^2 + 4n + 1.