A082111 -id:A082111 - cp's OEIS frontend

A176271 The odd numbers as a triangle read by rows.

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131

Offset: 1

Reinhard Zumkeller, Apr 13 2010

A108309(n) = number of primes in n-th row.

			From _Philippe Deléham_, Oct 03 2011: (Start)
Triangle begins:
   1;
   3,  5;
   7,  9, 11;
  13, 15, 17, 19;
  21, 23, 25, 27, 29;
  31, 33, 35, 37, 39, 41;
  43, 45, 47, 49, 51, 53, 55;
  57, 59, 61, 63, 65, 67, 69, 71;
  73, 75, 77, 79, 81, 83, 85, 87, 89; (End)

G. C. Greubel, Rows n = 1..100 of the triangle, flattened
Eric Weisstein's World of Mathematics, Nicomachus's Theorem
Wikipedia, Nikomachos von Gerasa

Cf. A000466, A000537, A000578, A001105, A002061, A005408, A014209.
Cf. A016754, A027688, A027690, A027692, A027694, A028387, A048058.
Cf. A053755, A065599, A082111, A108195, A108309, A214604, A214661.

a176271 n k = a176271_tabl !! (n-1) !! (k-1)
a176271_row n = a176271_tabl !! (n-1)
a176271_tabl = f 1 a005408_list where
   f x ws = us : f (x + 1) vs where (us, vs) = splitAt x ws
-- Reinhard Zumkeller, May 24 2012

[n^2-n+2*k-1: k in [1..n], n in [1..15]]; // G. C. Greubel, Mar 10 2024

A176271 := proc(n,k)
    n^2-n+2*k-1 ;
end proc: # R. J. Mathar, Jun 28 2013

Table[n^2-n+2*k-1, {n,15}, {k,n}]//Flatten (* G. C. Greubel, Mar 10 2024 *)

flatten([[n^2-n+2*k-1 for k in range(1,n+1)] for n in range(1,16)]) # G. C. Greubel, Mar 10 2024

T(n, k) = n^2 - n + 2*k - 1 for 1 <= k <= n.
T(n, k) = A005408(n*(n-1)/2 + k - 1).
T(2*n-1, n) = A016754(n-1) (main diagonal).
T(2*n, n) = A000466(n).
T(2*n, n+1) = A053755(n).
T(n, k) + T(n, n-k+1) = A001105(n), 1 <= k <= n.
T(n, 1)   = A002061(n), central polygonal numbers.
T(n, 2)   = A027688(n-1) for n > 1.
T(n, 3)   = A027690(n-1) for n > 2.
T(n, 4)   = A027692(n-1) for n > 3.
T(n, 5)   = A027694(n-1) for n > 4.
T(n, 6)   = A048058(n-1) for n > 5.
T(n, n-3) = A108195(n-2) for n > 3.
T(n, n-2) = A082111(n-2) for n > 2.
T(n, n-1) = A014209(n-1) for n > 1.
T(n, n) = A028387(n-1).
Sum_{k=1..n} T(n, k) = A000578(n) (Nicomachus's theorem).
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = (-1)^(n-1)*A065599(n) (alternating sign row sums).
Sum_{j=1..n} (Sum_{k=1..n} T(j, k)) = A000537(n) (sum of first n rows).

A028884 a(n) = (n + 3)^2 - 8.

Original entry on oeis.org

1, 8, 17, 28, 41, 56, 73, 92, 113, 136, 161, 188, 217, 248, 281, 316, 353, 392, 433, 476, 521, 568, 617, 668, 721, 776, 833, 892, 953, 1016, 1081, 1148, 1217, 1288, 1361, 1436, 1513, 1592, 1673, 1756, 1841, 1928, 2017, 2108, 2201, 2296, 2393

Offset: 0

Patrick De Geest

From Klaus Purath, Jan 04 2023: (Start)
The product of two consecutive terms belongs to the sequence: a(n)*a(n+1) = a(a(n)+n) = (a(n)+n)*(a(n+1)-n-1) + 1.
a(n) is never divisible by primes given in A003629.
Each odd 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 == -6 (mod p).
The prime factors are listed in A038873 and the primes in A028886.
For n > 0, this is a proper subsequence of A079896.
Conjecture: a(n) = A079896(A265284(n-1)). -
(End)

			From _Stefano Spezia_, Nov 08 2022: (Start)
Illustrations for n = 0..4:
          *       * * *     * * * * *
      a(0) = 1    *   *     *       *
                  * * *     *   *   *
                a(1) = 8    *       *
                            * * * * *
                            a(2) = 17
.
   * * * * * * *    * * * * * * * * *
   *           *    *               *
   *   *   *   *    *   *   *   *   *
   *           *    *               *
   *   *   *   *    *   *   *   *   *
   *           *    *               *
   * * * * * * *    *   *   *   *   *
     a(3) = 28      *               *
                    * * * * * * * * *
                        a(4) = 41
(End)

Altug Alkan, Table of n, a(n) for n = 0..10000
Patrick De Geest, Palindromic Quasipronics of the form n(n+x).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005563, A014616, A028560, A082111, A190576.

a014616 n = (n * (n + 6) + 1) `div` 4 -- Reinhard Zumkeller, Apr 07 2013

Range[3, 50]^2 - 8 (* Alonso del Arte, Aug 15 2016 *)

a(n)=(n+3)^2-8 \\ Charles R Greathouse IV, Oct 07 2015

(3 to 49).map(n => n * n - 8) // Alonso del Arte, May 07 2020

a(n) = a(n-1) + 2*n + 5 (with a(0) = 1). - Vincenzo Librandi, Aug 05 2010
a(n) = A028560(n) + 1; A014616(n) = floor(a(n+1)/4). - Reinhard Zumkeller, Apr 07 2013
G.f.: (-1 - 5*x + 4*x^2)/(x - 1)^3. - R. J. Mathar, Mar 24 2013
Sum_{n >= 0} 1/a(n) = 51/112 - Pi*cot(2*Pi*sqrt(2))/(4*sqrt(2)) = 1.3839174974448... . - Vaclav Kotesovec, Apr 10 2016
E.g.f.: (1 + 7*x + x^2)*exp(x). - G. C. Greubel, Aug 19 2017
Sum_{n >= 0} (-1)^n/a(n) = (-19 + 14*sqrt(2)*Pi*cosec(2*sqrt(2)*Pi))/112. - Amiram Eldar, Nov 04 2020
From Klaus Purath, Jan 04 2023: (Start)
a(n) = 2*a(n-1) - a(n-2) + 2, n >= 2.
a(n) = A082111(n) + n.
a(n) = A190576(n+1) - n. (End)
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=1} (1 - 1/a(n)) = 7*Pi/(45*sqrt(2)*sin(2*sqrt(2)*Pi)).
Product_{n>=0} (1 + 1/a(n)) = (4*sqrt(14)/9)*sin(sqrt(7)*Pi)/sin(2*sqrt(2)*Pi). (End)

Definition corrected by Omar E. Pol, Jul 27 2009

A199898 T(n,k)=Number of -k..k arrays x(0..n-1) of n elements with zero sum, and adjacent elements not both strictly positive and not both strictly negative.

Original entry on oeis.org

1, 1, 3, 1, 5, 7, 1, 7, 15, 15, 1, 9, 25, 49, 33, 1, 11, 37, 111, 159, 75, 1, 13, 51, 209, 461, 533, 171, 1, 15, 67, 351, 1043, 2035, 1783, 391, 1, 17, 85, 545, 2031, 5725, 8823, 6027, 899, 1, 19, 105, 799, 3573, 13363, 30199, 39053, 20437, 2077, 1, 21, 127, 1121, 5839

Offset: 1

R. H. Hardin Nov 11 2011

Table starts
....1.....1......1.......1........1........1.........1.........1..........1
....3.....5......7.......9.......11.......13........15........17.........19
....7....15.....25......37.......51.......67........85.......105........127
...15....49....111.....209......351......545.......799......1121.......1519
...33...159....461....1043.....2031.....3573......5839......9021......13333
...75...533...2035....5725....13363....27457.....51395.....89577.....147547
..171..1783...8823...30199....82555...193689....406575....783989....1413739
..391..6027..39053..164993...536967..1462859...3500269...7584081...15191479
..899.20437.172355..890299..3409609.10651367..28684325..68971571..151640029
.2077.69665.767425.4877477.22163661.80142549.245319361.661158741.1611184533

			Some solutions for n=7 k=6
..1....3....3....4...-3....4....4....0....3....3...-3...-6....4...-5....0....4
.-3...-4...-4...-6....5...-2...-6....6...-3...-5....0....3...-4....5...-3...-5
..0....1....2....2...-3....4....4...-5....0....4....3...-1....0...-4....5....1
.-1....0...-3....0....0...-1...-3....3....4....0....0....6....5....6....0...-1
..4....0....2....0....1....0....0...-6...-1....2...-2...-2...-6...-3...-3....5
.-6...-4...-2....3...-6...-5...-4....5....3....0....2....4....3....4....6....0
..5....4....2...-3....6....0....5...-3...-6...-4....0...-4...-2...-3...-5...-4

R. H. Hardin, Table of n, a(n) for n = 1..1096

Column 1 is A136029(n+1)

Row 3 is A082111

Empirical for rows:
T(1,k) = 1
T(2,k) = 2*k + 1
T(3,k) = k^2 + 5*k + 1
T(4,k) = (4/3)*k^3 + 6*k^2 + (20/3)*k + 1
T(5,k) = (11/12)*k^4 + (49/6)*k^3 + (193/12)*k^2 + (41/6)*k + 1
T(6,k) = (11/10)*k^5 + (55/6)*k^4 + (55/2)*k^3 + (173/6)*k^2 + (37/5)*k + 1
T(7,k) = (151/180)*k^6 + (163/15)*k^5 + (377/9)*k^4 + (395/6)*k^3 + (7429/180)*k^2 + (93/10)*k + 1

A181510 Number of permutations of the multiset {1,1,2,2,3,3,...,n+1,n+1} avoiding the permutation patterns {132, 231, 2134}.

Original entry on oeis.org

6, 18, 34, 54, 78, 106, 138, 174, 214, 258, 306, 358, 414, 474, 538, 606, 678, 754, 834, 918, 1006, 1098, 1194, 1294, 1398, 1506, 1618, 1734, 1854, 1978, 2106, 2238, 2374, 2514, 2658, 2806, 2958, 3114, 3274, 3438, 3606, 3778, 3954, 4134, 4318, 4506, 4698, 4894

Offset: 1

Lara Pudwell, Oct 25 2010

a(n) is also the surface ares of the n-th solid in the following recursive construction:
The first solid is a unit cube (hence a(1)=6).
To form the n-th solid from the (n-1)st solid, construct a row of 2n-1 cubes, then center the (n-1)st solid on top of this row. (For example, the second solid is a row of 3 unit cubes, with a single unit cube centered on top of the middle cube. This construction has surface area a(2)=18.)
The sequence provides all nonnegative integers m such that 2*m + 13 is a square. - Bruno Berselli, Mar 01 2013

			For n=1, the permutations of {1,1,2,2} avoiding the patterns {132, 231, 2134} are {1122, 1212, 1221, 2112, 2121, 2211}.
For n=2, the permutations of {1,1,2,2,3,3} avoiding the patterns {132, 231, 2134} are {112233, 121233, 122133, 211233, 212133, 221133, 311223, 312123, 312213, 321123, 321213, 322113, 331122, 331212, 331221, 332112, 332121, 332211}.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Lara K. Pudwell, Stacking Blocks and Counting Permutations, Mathematics Magazine, Vol. 83 (2010), pp. 297-302.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A014209, A051936, A082111.

A181510:= func< n | ((2*n+3)^2 -13)/2 >;
[A181510(n): n in [1..60]]; // G. C. Greubel, Jan 21 2025

a[n_] := 2*n^2 + 6*n - 2; Array[a, 60] (* Amiram Eldar, Dec 23 2022 *)
LinearRecurrence[{3,-3,1},{6,18,34},50] (* Harvey P. Dale, May 10 2023 *)
((2*Range[1,60]+3)^2 -13)/2 (* G. C. Greubel, Jan 21 2025 *)

a(n)=2*n^2+6*n-2 \\ Charles R Greathouse IV, Jun 17 2017

def A181510(n): return (pow(2*n+3,2) -13)//2
print([A181510(n) for n in range(1,61)]) # G. C. Greubel, Jan 21 2025

a(n) = 2*n^2 + 6*n - 2.
From Bruno Berselli, Oct 29 2010: (Start)
G.f.: 2*x*(3-x^2)/(1-x)^3.
a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 0 for n > 3.
a(n) = 2*A014209(n) = 2*A082111(n-1) + 4 = A051936(2*n+2) + n + 4. (End)
Sum_{n>=1} 1/a(n) = 2/3 + Pi*tan(sqrt(13)*Pi/2)/(2*sqrt(13)). - Amiram Eldar, Dec 23 2022
E.g.f.: 2*(exp(x)*(x^2 + 4*x - 1) + 1). - Elmo R. Oliveira, Nov 17 2024

A108306 Expansion of (3x+1)/(1-3x-3*x^2).

Original entry on oeis.org

1, 6, 21, 81, 306, 1161, 4401, 16686, 63261, 239841, 909306, 3447441, 13070241, 49553046, 187869861, 712268721, 2700415746, 10238053401, 38815407441, 147160382526, 557927369901, 2115263257281, 8019571881546, 30404505416481, 115272231894081

Offset: 0

Creighton Dement, Jul 24 2005

Binomial transform is A055271. May be seen as a ibasefor-transform of the zero-sequence A000004 with respect to the floretion given in the program code.
The sequence is the INVERT transform of (1, 5, 10, 20, 40, 80, 160, ...) and can be obtained by extracting the upper left terms of matrix powers of [(1,5); (1,2)]. These results are a case (a=5, b=2) of the conjecture:  The INVERT transform of a sequence starting (1, a, a*b, a*b^2, a*b^3, ...) is equivalent to extracting the upper left terms of powers of the  2x2 matrix [(1,a); (1,b)]. - Gary W. Adamson, Jul 31 2016
From Klaus Purath, Mar 09 2023: (Start)
For any terms (a(n), a(n+1)) = (x, y), -3*x^2 - 3*x*y + y^2 = 15*(-3)^n = A082111(2)*(-3)^n. This is valid in general for all recursive sequences (t) with constant coefficients (3,3) and t(0) = 1: -3*x^2 - 3*x*y + y^2 = A082111(t(1)-4)*(-3)^n.
By analogy to this, for three consecutive terms (x, y, z) of any sequence (t) of the form (3,3) with t(0) = 1: y^2 - x*z = A082111(t(1)-4)*(-3)^n. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Martin Burtscher, Igor Szczyrba and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,3).

Cf. A055271.

Cf. A084057.

I:=[1,6]; [n le 2 select I[n] else 3*Self(n-1)+3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 01 2016

seriestolist(series((3*x+1)/(1-3*x-3*x^2), x=0,25));

CoefficientList[Series[(3 x + 1) / (1 - 3 x - 3 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 01 2016 *)

Recurrence: a(0)=1; a(1)=6; a(n) = 3a(n-1) + 3a(n-2) - N-E. Fahssi, Apr 20 2008

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)

A038893 Odd primes p such that 21 is a square mod p.

Original entry on oeis.org

3, 5, 7, 17, 37, 41, 43, 47, 59, 67, 79, 83, 89, 101, 109, 127, 131, 151, 163, 167, 173, 193, 211, 227, 251, 257, 269, 277, 293, 311, 331, 337, 353, 373, 379, 383, 419, 421, 457, 461, 463, 467, 479, 487, 499, 503

Offset: 1

N. J. A. Sloane

nonn

These primes correspond to the representation of the two classes of discriminant 21 of binary quadratic forms with principal reduced forms [1, 3, -3] and [3, 3, -1]. The first class represents the primes given in A141159 (or A139492). The second class gives the prime 3 (which divides 21), and primes congruent to 2 (mod 3) and also to 3, 5, 6 (mod 7). The solution of x^2 - 21 == 0 (mod p) leads to the representative primitive parallel forms for discriminant 21 and representation of primes p. - Wolfdieter Lang, Jun 19 2019

Prime factors of A082111 and excluding the 3, prime factors of A004538. - Klaus Purath, Jan 04 2023

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A139492, A141159.

Select[Prime[Range[100]], JacobiSymbol[21, #] != -1 &] (* Vincenzo Librandi, Sep 07 2012 *)

isok(p) = (p>2) && isprime(p) && issquare(Mod(21, p)); \\ Michel Marcus, Jun 19 2019

Name clarified by Michel Marcus, Jun 22 2019

A185877 Array T given by T(n,k) = k^2 +(2n-3)k -2*n +3, by antidiagonals.

Original entry on oeis.org

1, 3, 1, 7, 5, 1, 13, 11, 7, 1, 21, 19, 15, 9, 1, 31, 29, 25, 19, 11, 1, 43, 41, 37, 31, 23, 13, 1, 57, 55, 51, 45, 37, 27, 15, 1, 73, 71, 67, 61, 53, 43, 31, 17, 1, 91, 89, 85, 79, 71, 61, 49, 35, 19, 1, 111, 109, 105, 99, 91, 81, 69, 55, 39, 21, 1, 133, 131, 127, 121, 113, 103, 91, 77, 61, 43, 23, 1, 157, 155, 151, 145, 137, 127, 115, 101, 85, 67, 47, 25, 1, 183, 181, 177, 171, 163, 153, 141, 127, 111, 93, 73, 51, 27, 1

Offset: 1

Clark Kimberling, Feb 05 2011

A member of the accumulation chain ... < A185879 < A185877 < A185878 < A185880 < ... (See A144112 for the definition of accumulation array).

			Northwest corner:
  1, 3,  7, 13, 21
  1, 5, 11, 19, 29
  1, 7, 15, 25, 45
  1, 9, 19, 31, 45

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A144112, A185878, A185879, A185880.
Row 1 to 3: A002061, A028387, A082111.
diag (1,5,...): A056108;
diag (3,11,...): A056106;
diag (7,19,...): A003215;
diag (13,29,...): A144391;
diag (1,7,...): A003215;
diag (1,9,...): A144390.

(* This program generates A185877, its accumulation array A185878, and its weight array A185879. *)
f[n_,0]:=0;f[0,k_]:=0;
f[n_,k_]:=k^2+(2n-3)k-2n+3;
TableForm[Table[f[n,k],{n,1,10},{k,1,15}]] (* A185877 *)
Table[f[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten
s[n_,k_]:=Sum[f[i,j],{i,1,n},{j,1,k}]; (* accumulation array of {f(n,k)} *)
FullSimplify[s[n,k]]  (* formula for A185878 *)
TableForm[Table[s[n,k],{n,1,10},{k,1,15}]]
Table[s[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten
w[m_,n_]:=f[m,n]+f[m-1,n-1]-f[m,n-1]-f[m-1,n]/;Or[m>0,n>0];
TableForm[Table[w[n,k],{n,1,10},{k,1,15}]] (* A185879 *)
Table[w[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten

T(n,k) = k^2 + (2*n-3)*k - 2*n + 3, k>=1, n>=1.

A190576 a(n) = n^2 + 5*n - 5.

Original entry on oeis.org

1, 9, 19, 31, 45, 61, 79, 99, 121, 145, 171, 199, 229, 261, 295, 331, 369, 409, 451, 495, 541, 589, 639, 691, 745, 801, 859, 919, 981, 1045, 1111, 1179, 1249, 1321, 1395, 1471, 1549, 1629, 1711, 1795, 1881, 1969, 2059, 2151, 2245, 2341

Offset: 1

Vladimir Joseph Stephan Orlovsky, May 12 2011

Also a(n) = n^2 + 9*n + 9 if the offset is changed to -1. - R. J. Mathar, May 18 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. sequences of the form n^2 + k*n - k : A000290 (k=0), A028387 (k=1), A028872 (k=2), A082111 (k=3), A028884 (k=4).

[n^2+5*n-5: n in [1..50]]; // Vincenzo Librandi, Sep 30 2011

k = 5; Table[n^2 + k*n - k, {n, 100}]
LinearRecurrence[{3,-3,1},{1,9,19},50] (* Harvey P. Dale, May 28 2015 *)

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

G.f.: x*(-1 - 6*x + 5*x^2) / (x-1)^3. - R. J. Mathar, May 18 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=1, a(2)=9, a(3)=19. - Harvey P. Dale, May 28 2015
Sum_{n>=1} 1/a(n) = 199/495 + Pi*tan(3*sqrt(5)*Pi/2)/(3*sqrt(5)). - Amiram Eldar, Jan 18 2021

A099867 a(n) = 5*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=9.

Original entry on oeis.org

1, 9, 44, 211, 1011, 4844, 23209, 111201, 532796, 2552779, 12231099, 58602716, 280782481, 1345309689, 6445765964, 30883520131, 147971834691, 708975653324, 3396906431929, 16275556506321, 77980876099676, 373628823992059, 1790163243860619, 8577187395311036

Offset: 0

Creighton Dement, Oct 28 2004

From Klaus Purath, Mar 07 2023: (Start)
For any two terms (a(n), a(n+1)) = (x, y), x^2 - 5*x*y + y^2 = 37 = A082111(4). This is valid in general for all recursive sequences (t) with constant coefficients (5,-1) and t(0) =  1: x^2 - 5*x*y + y^2 = A082111(t(1)-5). This includes and interprets the Feb 04 2014 comment in A004253 by Colin Barker.
By analogy to all this, for three consecutive terms (x, y, z) of any sequence (t) of the form (5,-1) with t(0) = 1: y^2 - x*z = A082111(t(1)-5). (End)

Colin Barker, Table of n, a(n) for n = 0..1000
A. F. Horadam, Pell Identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (5,-1).

Cf. A099868, A003501, A004254.

I:=[1,9]; [n le 2 select I[n] else 5*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 30 2015

a[0] = 1; a[1] = 9; a[n_] := a[n] = 5 a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 21}] (* Robert G. Wilson v, Dec 14 2004 *)
LinearRecurrence[{5, -1}, {1, 9}, 30] (* or *) CoefficientList[Series[(1 + 4 x)/(1 - 5 x + x^2), {x, 0, 30}], x] (* Harvey P. Dale, Jun 26 2011 *)

Vec((1+4*x) / (1-5*x+x^2) + O(x^30)) \\ Colin Barker, Mar 31 2017

|2*a(n) + A099868(n) - A003501(n+1)| = 20*A004254(n).
From R. J. Mathar, Sep 11 2008: (Start)
G.f.: (1+4*x) / (1-5*x+x^2).
a(n) = A004254(n+1) + 4*A004254(n).
(End)
a(n) = 2^(-1-n)*((5-sqrt(21))^n*(-13+sqrt(21)) + (5+sqrt(21))^n*(13+sqrt(21))) / sqrt(21). - Colin Barker, Mar 31 2017

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