A089270 -id:A089270 - cp's OEIS frontend

A324598 Irregular triangle with the representative solutions of the Diophantine equation x^2 + x - 1 congruent to 0 modulo N(n), with N(n) = A089270(n), for n >= 1.

0, 2, 3, 7, 4, 14, 5, 23, 12, 18, 6, 34, 7, 47, 25, 33, 17, 43, 8, 62, 29, 49, 9, 79, 42, 52, 22, 78, 10, 98, 36, 84, 11, 119, 63, 75, 52, 93, 40, 108, 27, 123, 12, 142, 74, 104, 13, 167, 88, 102, 61, 137, 47, 157, 14, 194, 80, 128, 32, 178

Offset: 1

Wolfdieter Lang, Jul 08 2019

The length of row n is 1 for n = 1 and n = 2, and for n >= 3 it is 2^{r1 + r4} with the number r1 and r4 of distinct primes congruent to 1 and 4 modulo 5, respectively, in the prime number factorization of N(n). E.g., n = 29, N = 209 = 11*19, has r1 = 1 and r4 = 1, with four solutions. The next rows with four solutions are n = 41, 43, 59,..., with N = 319, 341, 451, ... ;  for n = 643, 688, 896, ..., with N  = 6061, 6479, 8569, ..., there are eight solutions.
For N(1) = 1 every integer solves this Diophantine equation, and the representative solution is 0.
For N(2) = 5 there is only one representative solution, namely 2.
For n >= 3 the representative solutions come in nonnegtive power of 2 pairs (x1, x2) with x2 = N - 1 - x1.
See the link in A089270 to the W. Lang paper, section 3, and Table 6.

			The irregular triangle T(n, k) begins (pairs (x, N - 1 - x) in brackets):
n,    N \ k   1   2     3   4  ...
----------------------------------
1,    1:      0
2,    5:      2
3,   11:     (3   7)
4,   19:     (4  14)
5,   29:     (5  23)
6,   31:    (12  18)
7,   41:     (6  34)
8,   55:     (7  47)
9,   59:    (25  33)
10,  61:    (17  43)
11,  71:     (8  62)
12,  79:    (29  49)
13,  89:     (9  79)
14,  95:    (42  52)
15, 101:    (22  78)
16, 109:    (10  98)
17, 121:    (36  84)
18, 131:    (11 119)
19, 139:    (63  75)
20, 145:    (52  93)
....
29, 209:    (14 194)  (80 128)
...
41, 319:   (139 179) (150 168)
...
43, 341:    (18 322)  (80 260)
...
59, 451:    (47 403) (157 293)
...

Cf. A089270, A324599 (x^2 - 5 == 0 (mod N)).

A324599 Irregular triangle with the representative solutions of the Diophantine equation x^2 - 5 congruent to 0 modulo N(n), with N(n) = A089270(n), for n >= 1.

Original entry on oeis.org

0, 0, 4, 7, 9, 10, 11, 18, 6, 25, 13, 28, 15, 40, 8, 51, 26, 35, 17, 54, 20, 59, 19, 70, 10, 85, 45, 56, 21, 88, 48, 73, 23, 108, 12, 127, 40, 105, 68, 81, 55, 96, 25, 130, 30, 149, 27, 154, 14, 177, 76, 123, 95, 110, 29, 180, 48, 161, 65, 146

Offset: 1

Wolfdieter Lang, Jul 08 2019

The length of row n is 1 for n = 1 and n = 2, and for n >= 3 it is 2^{r1 + r4} with the number r1 and r4 of distinct primes congruent to 1 and 4 modulo 5, respectively, in the prime number factorization of N(n). E.g., n = 29, N = 209 = 11*19, has r1 = 1 and r4 = 1, with four solutions.
For N(1) = 1 every integer solves this Diophantine equation, and the representative solution is 0.
For N(2) = 5 there is only one representative solution, namely 0.
For n >= 3 the solutions come in a nonnegative power of 2 pairs, each of the type (x1, x2) with x2 = N - x1.
See the link in A089270 to the W. Lang paper, section 3, and Table 7.

			The irregular triangle T(n, k) begins (pairs (x, N - x) in brackets):
n,    N \ k   1   2     3   4  ...
----------------------------------
1,    1:      0
2,    5:      0
3,   11:     (4   7)
4,   19:     (9  10)
5,   29:    (11  18)
6,   31:     (6  25)
7,   41:    (13  28)
8,   55:    (15  40)
9,   59:     (8  51)
10,  61:    (26  35)
11,  71:    (17  54)
12,  79:    (20  59)
13,  89:    (19  70)
14,  95:    (10  85)
15, 101:    (45  56)
16, 109:    (21  88)
17, 121:    (48  73)
18, 131:    (23 108)
19, 139:    (12 127)
20, 145:    (40 105)
....
29, 209:    (29 180)  (48 161)
...
41, 319:    (18 301)  (40 279)
...
43, 341:    (37 304) (161 180)
...
59, 451:    (95 356) (136 315)

Cf. A089270, A324598 (x^2 + x - 1 == 0 (mod N)).

A308686 Irregular triangle with the nonnegative proper fundamental solutions of the binary quadratic form x^2 + x*y - y^2 representing N = N(n) = A089270(n), for n >= 1.

Original entry on oeis.org

1, 0, 2, 1, 3, 1, 3, 2, 4, 1, 4, 3, 5, 1, 5, 4, 5, 2, 5, 3, 6, 1, 6, 5, 7, 1, 7, 6, 7, 2, 7, 5, 7, 3, 7, 4, 8, 1, 8, 7, 8, 3, 8, 5, 9, 1, 9, 8, 9, 2, 9, 7, 9, 4, 9, 5, 10, 1, 10, 9, 10, 3, 10, 7, 11, 1, 11, 10, 11, 2, 11, 9, 11, 3, 11, 8, 11, 4, 11, 7, 11, 5, 11, 6, 12, 1, 12, 11, 12, 5, 12, 7, 13, 1, 13, 12, 13, 2, 13, 11, 13, 3, 13, 10, 13, 4, 13, 9, 13, 5, 13, 8, 14, 1, 14, 13, 13, 6, 13, 7

Offset: 1

Wolfdieter Lang, Jul 05 2019

The length of row n is 2 for n = 1, 2; 4 for n = 3..28, 30..40, 42, 44..58, 60...; 8 for 29, 41, 43, 59,...; 16 for 643, 688, 896, ...; ... .
The numbers N with row length 8 are 209, 319, 341, 451, 551, 589, 649, 671, 779, 781, 869, 899, 979, 1045, 1111, ...; with row length 16 they are 6061, 6479, 8569, 9889, ...; .... .
The fundamental solution (x, y) with gcd(x, y) = 1 (proper solutions) are listed pairwise for n >= 3 (N >= 11) and enclosed in square brackets in the example, Within a square bracket the numbers y always sum to x.
For the numbers N with a solution (x, 1) see A028387(n-1), for n >= 1. There N = 1 is included by taking the solution (1, 1) instead of (1, 0).
The general solutions are then obtained by applying integer powers of the automorphic matrix Auto(50) = Matrix([1, 1],[1, 2]) on these fundamental solutions. The matrix Auto(5) is related to the 2-cycle of the principal reduced form F_p = [1, 1, -1] and the reduced form F' = [-1, 1, 1].
See the W. Lang link in A089270 for proofs and Tables. Here Table 4.

			The irregular triangle T(n, k) begins (the solutions are (x, y)):
n,    N \ k  1  2    3   4       5  6    7   8    ...
1,    1:    (1  0) [sometimes (1, 1)]
2,    5:    (2  1)
3,   11:   [(3  1)  (3   2)]
4,   19:   [(4  1)  (4   3)]
5,   29:   [(5  1)  (5   4)]
6,   31:   [(5  2)  (5   3)]
7,   41:   [(6  1)  (6   5)]
8,   55:   [(7  1)  (7   6)]
9,   59:   [(7  2)  (7   5)]
10,  61:   [(7  3)  (7   4)]
11,  71:   [(8  1)  (8   7)]
12,  79:   [(8  3)  (8   5)]
13,  89:   [(9  1)  (9   8)]
14,  95:   [(9  2)  (9   7)]
15, 101:   [(9  4)  (9   5)]
16, 109:  [(10  1) (10   9)]
17, 121:  [(10  3) (10   7)]
18, 131:  [(11  1) (11  10)]
19, 139:  [(11  2) (11   9)]
20, 145:  [(11  3) (11   8)]
...
29, 209:  [(13  5) (13   8)]  [(14  1) (14  13)]
30, 211:  [(13  6) (13   7)]
...

Cf. A028387, A089270.

A028387 a(n) = n + (n+1)^2.

Original entry on oeis.org

1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255, 2351, 2449, 2549, 2651

Offset: 0

Patrick De Geest

a(n+1) is the least k > a(n) + 1 such that A000217(a(n)) + A000217(k) is a square. - David Wasserman, Jun 30 2005
Values of Fibonacci polynomial n^2 - n - 1 for n = 2, 3, 4, 5, ... - Artur Jasinski, Nov 19 2006
A127701 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 24 2007
Row sums of triangle A135223. - Gary W. Adamson, Nov 23 2007
Equals row sums of triangle A143596. - Gary W. Adamson, Aug 26 2008
a(n-1) gives the number of n X k rectangles on an n X n chessboard (for k = 1, 2, 3, ..., n). - Aaron Dunigan AtLee, Feb 13 2009
sqrt(a(0) + sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + ...)))) = sqrt(1 + sqrt(5 + sqrt(11 + sqrt(19 + ...)))) = 2. - Miklos Kristof, Dec 24 2009
When n + 1 is prime, a(n) gives the number of irreducible representations of any nonabelian group of order (n+1)^3. - Andrew Rupinski, Mar 17 2010
a(n) = A176271(n+1, n+1). - Reinhard Zumkeller, Apr 13 2010
The product of any 4 consecutive integers plus 1 is a square (see A062938); the terms of this sequence are the square roots. - Harvey P. Dale, Oct 19 2011
Or numbers not expressed in the form m + floor(sqrt(m)) with integer m. - Vladimir Shevelev, Apr 09 2012
Left edge of the triangle in A214604: a(n) = A214604(n+1,1). - Reinhard Zumkeller, Jul 25 2012
Another expression involving phi = (1 + sqrt(5))/2 is a(n) = (n + phi)(n + 1 - phi). Therefore the numbers in this sequence, even if they are prime in Z, are not prime in Z[phi]. - Alonso del Arte, Aug 03 2013
a(n-1) = n*(n+1) - 1, n>=0, with a(-1) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 5 for b = 2*n+1. In general D = b^2 - 4ac > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
a(n) has prime factors given by A038872. - Richard R. Forberg, Dec 10 2014
A253607(a(n)) = -1. - Reinhard Zumkeller, Jan 05 2015
An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387. See Popular Computing link. - N. J. A. Sloane, May 03 2015
Left edge of the triangle in A260910: a(n) = A260910(n+2,1). - Reinhard Zumkeller, Aug 04 2015
Numbers m such that 4m+5 is a square. - Bruce J. Nicholson, Jul 19 2017
The numbers represented as 131 in base n: 131_4 = 29, 131_5 = 41, ... . If 'digits' larger than the base are allowed then 131_2 = 11 and 131_1 = 5 also. - Ron Knott, Nov 14 2017
From Klaus Purath, Mar 18 2019: (Start)
Let m be a(n) or a prime factor of a(n). Then, except for 1 and 5, there are, if m is a prime, exactly two squares y^2 such that the difference y^2 - m contains exactly one pair of factors {x,z} such that the following applies: x*z = y^2 - m, x + y = z with
x < y, where {x,y,z} are relatively prime numbers. {x,y,z} are the initial values of a sequence of the Fibonacci type. Thus each a(n) > 5, if it is a prime, and each prime factor p > 5 of an a(n) can be assigned to exactly two sequences of the Fibonacci type. a(0) = 1 belongs to the original Fibonacci sequence and a(1) = 5 to the Lucas sequence.
But also the reverse assignment applies. From any sequence (f(i)) of the Fibonacci type we get from its 3 initial values by f(i)^2 - f(i-1)*f(i+1) with f(i-1) < f(i) a term a(n) or a prime factor p of a(n). This relation is also valid for any i. In this case we get the absolute value |a(n)| or |p|. (End)
a(n-1) = 2*T(n) - 1, for n>=1, with T = A000217, is a proper subsequence of A089270, and the terms are 0,-1,+1 (mod 5). - Wolfdieter Lang, Jul 05 2019
a(n+1) is the number of wedged n-dimensional spheres in the homotopy of the neighborhood complex of Kneser graph KG_{2,n}. Here, KG_{2,n} is a graph whose vertex set is the collection of subsets of cardinality 2 of set {1,2,...,n+3,n+4} and two vertices are adjacent if and only if they are disjoint. - Anurag Singh, Mar 22 2021
Also the number of squares between (n+2)^2 and (n+2)^4. - Karl-Heinz Hofmann, Dec 07 2021
(x, y, z) = (A001105(n+1), -a(n-1), -a(n)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 8. - XU Pingya, Apr 25 2022
The least significant digit of terms of this sequence cycles through 1, 5, 1, 9, 9. - Torlach Rush, Jun 05 2024

			From _Ilya Gutkovskiy_, Apr 13 2016: (Start)
Illustration of initial terms:
                                        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 o o o o o o o
n=0  n=1       n=2           n=3               n=4
(End)
From _Klaus Purath_, Mar 18 2019: (Start)
Examples:
a(0) = 1: 1^1-0*1 = 1, 0+1 = 1 (Fibonacci A000045).
a(1) = 5: 3^2-1*4 = 5, 1+3 = 4 (Lucas A000032).
a(2) = 11: 4^2-1*5 = 11, 1+4 = 5 (A000285); 5^2-2*7 = 11, 2+5 = 7 (A001060).
a(3) = 19: 5^2-1*6 = 19, 1+5 = 6 (A022095); 7^2-3*10 = 19, 3+7 = 10 (A022120).
a(4) = 29: 6^2-1*7 = 29, 1+6 = 7 (A022096); 9^2-4*13 = 29, 4+9 = 13 (A022130).
a(11)/5 = 31: 7^2-2*9 = 31, 2+7 = 9 (A022113); 8^2-3*11 = 31, 3+8 = 11 (A022121).
a(24)/11 = 59: 9^2-2*11 = 59, 2+9 = 11 (A022114); 12^2-5*17 = 59, 5+12 = 17 (A022137).
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Patrick De Geest, World!Of Numbers, Palindromes of the form n+(n+1)^2.
Adalbert Kerber, A matrix of combinatorial numbers related to the symmetric groups, Discrete Math., 21 (1978), 319-321. [Annotated scanned copy]
Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
Nandini Nilakantan and Anurag Singh, Homotopy type of neighborhood complexes of Kneser graphs, KG_{2,k}, Proceeding-Mathematical Sciences, 128, Article number: 53(2018).
Yanni Pei and Jiang Zeng, Counting signed derangements with right-to-left minima and excedances, arXiv:2206.11236 [math.CO], 2022.
Popular Computing (Calabasas, CA), The CSR Function, Vol. 4 (No. 34, Jan 1976), pages PC34-10 to PC34-11. Annotated and scanned copy.
Zdzislaw Skupień and Andrzej Żak, Pair-sums packing and rainbow cliques, in Topics In Graph Theory, A tribute to A. A. and T. E. Zykovs on the occasion of A. A. Zykov's 90th birthday, ed. R. Tyshkevich, Univ. Illinois, 2013, pages 131-144, (in English and Russian).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Complement of A028392. Third column of array A094954.
Cf. A000217, A002522, A062392, A062786, A127701, A135223, A143596, A052905, A162997, A062938 (squares of this sequence).
A110331 and A165900 are signed versions.
Cf. A002327 (primes), A094210.
Frobenius number for k successive numbers: this sequence (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).
Cf. also A135223, A176271, A214604, A105728, A005408, A002378, A084990, A253607, A260910, A089270.

a028387 n = n + (n + 1) ^ 2  -- Reinhard Zumkeller, Jul 17 2014

[n + (n+1)^2: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

FoldList[## + 2 &, 1, 2 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
Table[n + (n + 1)^2, {n, 0, 100}] (* Vincenzo Librandi, Oct 17 2012 *)
Table[ FrobeniusNumber[{n, n + 1}], {n, 2, 30}] (* Zak Seidov, Jan 14 2015 *)

a(n)=n^2+3*n+1 \\ Charles R Greathouse IV, Jun 10 2011

def a(n): return (n**2+3*n+1) # Torlach Rush, May 07 2024

[n+(n+1)^2 for n in range(0,48)] # Zerinvary Lajos, Jul 03 2008

a(n) = sqrt(A062938(n)). - Floor van Lamoen, Oct 08 2001
a(0) = 1, a(1) = 5, a(n) = (n+1)*a(n-1) - (n+2)*a(n-2) for n > 1. - Gerald McGarvey, Sep 24 2004
a(n) = A105728(n+2, n+1). - Reinhard Zumkeller, Apr 18 2005
a(n) = A109128(n+2, 2). - Reinhard Zumkeller, Jun 20 2005
a(n) = 2*T(n+1) - 1, where T(n) = A000217(n). - Gary W. Adamson, Aug 15 2007
a(n) = A005408(n) + A002378(n); A084990(n+1) = Sum_{k=0..n} a(k). - Reinhard Zumkeller, Aug 20 2007
Binomial transform of [1, 4, 2, 0, 0, 0, ...] = (1, 5, 11, 19, ...). - Gary W. Adamson, Sep 20 2007
G.f.: (1+2*x-x^2)/(1-x)^3. a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - R. J. Mathar, Jul 11 2009
a(n) = (n + 2 + 1/phi) * (n + 2 - phi); where phi = 1.618033989... Example: a(3) = 19 = (5 + .6180339...) * (3.381966...). Cf. next to leftmost column in A162997 array. - Gary W. Adamson, Jul 23 2009
a(n) = a(n-1) + 2*(n+1), with n > 0, a(0) = 1. - Vincenzo Librandi, Nov 18 2010
For k < n, a(n) = (k+1)*a(n-k) - k*a(n-k-1) + k*(k+1); e.g., a(5) = 41 = 4*11 - 3*5 + 3*4. - Charlie Marion, Jan 13 2011
a(n) = lower right term in M^2, M = the 2 X 2 matrix [1, n; 1, (n+1)]. - Gary W. Adamson, Jun 29 2011
G.f.: (x^2-2*x-1)/(x-1)^3 = G(0) where G(k) = 1 + x*(k+1)*(k+4)/(1 - 1/(1 + (k+1)*(k+4)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 16 2012
Sum_{n>0} 1/a(n) = 1 + Pi*tan(sqrt(5)*Pi/2)/sqrt(5). - Enrique Pérez Herrero, Oct 11 2013
E.g.f.: exp(x) (1+4*x+x^2). - Tom Copeland, Dec 02 2013
a(n) = A005408(A000217(n)). - Tony Foster III, May 31 2016
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2)/6. (End)
a(5*n+1)/5 = A062786(n+1). - Torlach Rush, Jun 05 2024

Minor edits by N. J. A. Sloane, Jul 04 2010, following suggestions from the Sequence Fans Mailing List

A038872 Primes congruent to {0, 1, 4} mod 5.

Original entry on oeis.org

5, 11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, 131, 139, 149, 151, 179, 181, 191, 199, 211, 229, 239, 241, 251, 269, 271, 281, 311, 331, 349, 359, 379, 389, 401, 409, 419, 421, 431, 439, 449, 461, 479, 491, 499, 509, 521, 541, 569, 571, 599, 601, 619

Offset: 1

N. J. A. Sloane

Also odd primes p such that 5 is a square mod p: (5/p) = +1 for p > 5.
Primes of the form x^2 + x*y - y^2 (as well as of the form x^2 + 3*x*y + y^2), both with discriminant = 5 and class number = 1. Binary quadratic forms a*x^2 + b*x*y + c*y^2 have discriminant d = b^2 - 4ac and gcd(a, b, c) = 1. [This was originally a separate entry, submitted by Laura Caballero Fernandez, Lourdes Calvo Moguer, Maria Josefa Cano Marquez, Oscar Jesus Falcon Ganfornina and Sergio Garrido Morales, Jun 06 2008. R. J. Mathar proved on Jul 22 2008 that this coincides with the present sequence.]
Also primes of the form 5x^2 - y^2 (cf. A031363). - N. J. A. Sloane, May 30 2014
Also primes of the form x^2 + 4*x*y - y^2. Every binary quadratic primitive form of discriminant 20 or 5 has proper solutions for positive integers N given in A089270, including the present primes. Proof from computing the corresponding representative parallel primitive forms, which leads to x^2 - 5 == 0 (mod N) or x^2 + x - 1 == 0 (mod N) which have solutions precisely for these positive N values, including these primes. - Wolfdieter Lang, Jun 19 2019
For a Pythagorean triple a, b, c, these primes (and 2) are the possible prime factors of 2a + b, |2a - b|, 2b + a, and 2b - a. - J. Lowell, Nov 05 2011
The prime factors of A028387(n^2+3n+1). - Richard R. Forberg, Dec 12 2014
Except for p = 5, these are primes p that divide Fibonacci(p-1). - Dmitry Kamenetsky, Jul 27 2015
Apart from the first term, these are rational primes that decompose in the field Q[sqrt(5)]. For example, 11 = ((7 + sqrt(5))/2)*((7 - sqrt(5))/2), 19 = ((9 + sqrt(5))/2)*((9 - sqrt(5))/2). - Jianing Song, Nov 23 2018
The possible prime factors of x^2 - x - 1. - Charles R Greathouse IV, Mar 18 2022

Z. I. Borevich and I. R. Shafarevich, Number Theory.

T. D. Noe, Table of n, a(n) for n = 1..1000
C. Banderier, Calcul de (5/p)
Henri Darmon, Andrew Wiles's Marvelous Proof, Notices of the AMS (2017), Volume 64, Number 3 pp. 209-216. See p. 211.
Tamara M. Lavshuk, Regular polygons and polyhedra over finite field, Mathematical Notes of NEFU, Vol 22 No 4 (2015). Mentions this sequence.
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)
D. B. Zagier, Zetafunktionen und quadratische Körper, Springer, 1981.

Cf. A045468, A089270.
Cf. A038872 (d=5); A038873 (d=8); A068228, A141123 (d=12); A038883 (d=13). A038889 (d=17); A141111, A141112 (d=65).
Cf. A003631 (complement with respect to A000040).

Filtered(Concatenation([5],Flat(List([1..140],k->[5*k-1,5*k+1]))),IsPrime); # Muniru A Asiru, Nov 24 2018

[ p: p in PrimesUpTo(700) | p mod 5 in {0,1,4}]; // Vincenzo Librandi, Aug 21 2012

select(isprime, [5, seq(op([5*k-1,5*k+1]),k=1..1000)]); # Robert Israel, Dec 22 2014

Join[{5}, Select[Prime[Range[4, 100]], Mod[#, 5] == 1 || Mod[#, 5] == 4 &]] (* Alonso del Arte, Nov 27 2011 *)

forprime(p=2,1e3,if(kronecker(5,p)>=0,print1(p", "))) \\ Charles R Greathouse IV, Jun 16 2011

a(n) = A045468(n-1) for n > 1. - Robert Israel, Dec 22 2014

a(n) ~ 2n*log(n). - Charles R Greathouse IV, Nov 29 2016

Corrected and extended by Peter K. Pearson, May 29 2005

Edited by N. J. A. Sloane, Jul 28 2008 at the suggestion of R. J. Mathar

A062786 Centered 10-gonal numbers.

Original entry on oeis.org

1, 11, 31, 61, 101, 151, 211, 281, 361, 451, 551, 661, 781, 911, 1051, 1201, 1361, 1531, 1711, 1901, 2101, 2311, 2531, 2761, 3001, 3251, 3511, 3781, 4061, 4351, 4651, 4961, 5281, 5611, 5951, 6301, 6661, 7031, 7411, 7801, 8201, 8611, 9031, 9461, 9901, 10351, 10811

Offset: 1

Jason Earls, Jul 19 2001

Deleting the least significant digit yields the (n-1)-st triangular number: a(n) = 10*A000217(n-1) + 1. - Amarnath Murthy, Dec 11 2003
All divisors of a(n) are congruent to 1 or -1, modulo 10; that is, they end in the decimal digit 1 or 9. Proof: If p is an odd prime different from 5 then 5n^2 - 5n + 1 == 0 (mod p) implies 25(2n - 1)^2 == 5 (mod p), whence p == 1 or -1 (mod 10). - Nick Hobson, Nov 13 2006
Centered decagonal numbers. - Omar E. Pol, Oct 03 2011
The partial sums of this sequence give A004466. - Leo Tavares, Oct 04 2021
The continued fraction expansion of sqrt(5*a(n)) is [5n-3; {2, 2n-2, 2, 10n-6}]. For n=1, this collapses to [2; {4}]. - Magus K. Chu, Sep 12 2022
Numbers m such that 20*m + 5 is a square. Also values of the Fibonacci polynomial y^2 - x*y - x^2 for x = n and y = 3*n - 1. This is a subsequence of A089270. - Klaus Purath, Oct 30 2022
All terms can be written as a difference of two consecutive squares a(n) = A005891(n-1)^2 - A028895(n-1)^2, and they can be represented by the forms (x^2 + 2mxy + (m^2-1)y^2) and (3x^2 + (6m-2)xy + (3m^2-2m)y^2), both of discriminant 4. - Klaus Purath, Oct 17 2023

T. D. Noe, Table of n, a(n) for n = 1..1000
Leo Tavares, Illustration: Pentagonal Stars.
Leo Tavares, Illustration: Mid-section Stars.
Leo Tavares, Illustration: Mid-section Star Pillars.
Leo Tavares, Illustration: Trapezoidal Rays.
R. Yin, J. Mu, and T. Komatsu, The p-Frobenius Number for the Triple of the Generalized Star Numbers, Preprints 2024, 2024072280. See p. 2.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001263, A124080, A101321, A028387, A016861, A003154, A005891, A000217, A004466, A144390, A000326, A060544.

Cf. also A016754, A002378, A069099, A045943, A003215, A046092, A001844, A028896, A005448, A024966, A082970.

List([1..50], n-> 1+5*n*(n-1)); # G. C. Greubel, Mar 30 2019

[1+5*n*(n-1): n in [1..50]]; // G. C. Greubel, Mar 30 2019

FoldList[#1+#2 &, 1, 10Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
1+5*Pochhammer[Range[50]-1, 2] (* G. C. Greubel, Mar 30 2019 *)

j=[]; for(n=1,75,j=concat(j,(5*n*(n-1)+1))); j

for (n=1, 1000, write("b062786.txt", n, " ", 5*n*(n - 1) + 1) ) \\ Harry J. Smith, Aug 11 2009

def a(n): return(5*n**2-5*n+1) # Torlach Rush, May 10 2024

[1+5*rising_factorial(n-1, 2) for n in (1..50)] # G. C. Greubel, Mar 30 2019

a(n) = 5*n*(n-1) + 1.
From  Gary W. Adamson, Dec 29 2007: (Start)
Binomial transform of [1, 10, 10, 0, 0, 0, ...];
Narayana transform (A001263) of [1, 10, 0, 0, 0, ...]. (End)
G.f.: x*(1+8*x+x^2) / (1-x)^3. - R. J. Mathar, Feb 04 2011
a(n) = A124080(n-1) + 1. - Omar E. Pol, Oct 03 2011
a(n) = A101321(10,n-1). - R. J. Mathar, Jul 28 2016
a(n) = A028387(A016861(n-1))/5 for n > 0. - Art Baker, Mar 28 2019
E.g.f.: (1+5*x^2)*exp(x) - 1. - G. C. Greubel, Mar 30 2019
Sum_{n>=1} 1/a(n) = Pi * tan(Pi/(2*sqrt(5))) / sqrt(5). - Vaclav Kotesovec, Jul 23 2019
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=1} a(n)/n! = 6*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 6/e - 1. (End)
a(n) = A005891(n-1) + 5*A000217(n-1). - Leo Tavares, Jul 14 2021
a(n) = A003154(n) - 2*A000217(n-1). See Mid-section Stars illustration. - Leo Tavares, Sep 06 2021
From Leo Tavares, Oct 06 2021: (Start)
a(n) = A144390(n-1) + 2*A028387(n-1). See Mid-section Star Pillars illustration.
a(n) = A000326(n) + A000217(n) + 3*A000217(n-1). See Trapezoidal Rays illustration.
a(n) = A060544(n) + A000217(n-1). (End)
From Leo Tavares, Oct 31 2021: (Start)
a(n) = A016754(n-1) + 2*A000217(n-1).
a(n) = A016754(n-1) + A002378(n-1).
a(n) = A069099(n) + 3*A000217(n-1).
a(n) = A069099(n) + A045943(n-1).
a(n) = A003215(n-1) + 4*A000217(n-1).
a(n) = A003215(n-1) + A046092(n-1).
a(n) = A001844(n-1) + 6*A000217(n-1).
a(n) = A001844(n-1) + A028896(n-1).
a(n) = A005448(n) + 7*A000217(n).
a(n) = A005448(n) + A024966(n). (End)
From Klaus Purath, Oct 30 2022: (Start)
a(n) = a(n-2) + 10*(2*n-3).
a(n) = 2*a(n-1) - a(n-2) + 10.
a(n) = A135705(n-1) + n.
a(n) = A190816(n) - n.
a(n) = 2*A005891(n-1) - 1. (End)

Better description from Terrel Trotter, Jr., Apr 06 2002

A013655 a(n) = F(n+1) + L(n), where F(n) and L(n) are Fibonacci and Lucas numbers, respectively.

Original entry on oeis.org

3, 2, 5, 7, 12, 19, 31, 50, 81, 131, 212, 343, 555, 898, 1453, 2351, 3804, 6155, 9959, 16114, 26073, 42187, 68260, 110447, 178707, 289154, 467861, 757015, 1224876, 1981891, 3206767, 5188658, 8395425, 13584083, 21979508, 35563591, 57543099, 93106690, 150649789, 243756479

Offset: 0

Mohammad K. Azarian

Apart from initial term, same as A001060.
Pisano period lengths same as for A001060. - R. J. Mathar, Aug 10 2012
The beginning of this sequence is the only sequence of four consecutive primes in a Fibonacci-type sequence. - Franklin T. Adams-Watters, Mar 21 2015
(a(2*k), a(2*k+1)) give for k >= 0 the proper positive solutions of one of two families (or classes) of solutions (x, y) of the indefinite binary quadratic form x^2 + x*y - y^2 of discriminant 5 representing 11. The other family of such solutions is given by (x2, y2) = (b(2*k), b(2*k+1)) with b = A104449. See the formula in terms of Chebyshev S polynomials S(n, 3) = A001906(n+1) below, which follows from the fundamental solution (3, 2) by applying positive powers of the automorphic matrix A^k = Matrix([A(k), B(k)], [B(k), A(k+1)]), with A(k) = S(k-1, 3) - S(k-2, 3) and B(k) = S(k-1, 3). See also A089270 with the Alfred Brousseau link with D = 11. - Wolfdieter Lang, May 28 2019
For n>1, a(n) is the number of ways to tile a strip of length n+1 with black and white squares and white dominos, where there must be exactly one black square and it must appear amongst the first three cells. - Greg Dresden and Emma Li, Aug 24 2024

G. C. Greubel, Table of n, a(n) for n = 0..2500
Mark W. Coffey, James L. Hindmarsh, Matthew C. Lettington, and John Pryce, On Higher Dimensional Interlacing Fibonacci Sequences, Continued Fractions and Chebyshev Polynomials, arXiv:1502.03085 [math.NT], 2015 (see p. 31).
Rigoberto Flórez, Robinson A. Higuita, and Antara Mukherjee, The Geometry of some Fibonacci Identities in the Hosoya Triangle, arXiv:1804.02481 [math.NT], 2018.
Tanya Khovanova, Recursive Sequences
Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (1,1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A001060, A001906, A089270, A104449.

[2*Fibonacci(n-3)+Fibonacci(n): n in [2..41]]; // Vincenzo Librandi, Apr 16 2011

[GeneralizedFibonacciNumber(3, 2, n): n in [0..39]]; // Arkadiusz Wesolowski, Mar 16 2016

with(combinat): a:=n->2*fibonacci(n-1)+fibonacci(n+2): seq(a(n), n=0..40); # Zerinvary Lajos, Oct 05 2007

LinearRecurrence[{1, 1}, {3, 2}, 40] (* or *)
Table[Fibonacci[n + 1] + LucasL[n], {n, 0, 40}] (* or *)
Table[Fibonacci[n + 3] + Fibonacci[n - 3] - 3*Fibonacci[n], {n,2,40}] (* Bruno Berselli, Dec 30 2016 *)

a(n)=([0,1; 1,1]^n*[3;2])[1,1] \\ Charles R Greathouse IV, Sep 24 2015

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

a(n) = A000045(n+1) + A000032(n).
a(n) = a(n-1) + a(n-2).
a(n) = F(n+3) - F(n-2) for n>1, where F=A000045. - Gerald McGarvey, Jul 10 2004
a(n) = 2*F(n-3) + F(n) for n>1. - Zerinvary Lajos, Oct 05 2007
G.f.: (3-x)/(1-x-x^2). - Philippe Deléham, Nov 19 2008
a(n) = Sum_{k = n-3..n+1} F(k). - Gary Detlefs, Dec 30 2012
a(n) = ((3*sqrt(5)+1)*(((1+sqrt(5))/2)^n)+(3*sqrt(5)-1)*(((1-sqrt(5))/2)^n))/(2*sqrt(5)). - Bogart B. Strauss, Jul 19 2013
a(n) = F(n+3) + F(n-3) - 3*F(n) for n>1. - Bruno Berselli, Dec 29 2016
Bisection: a(2*k) = 3*S(k, 3) - 4*S(k-1, 3), a(2*k+1) = 2*S(k, 3) + S(k-1, 3), for k >= 0, with the Chebyshev S(n, 3) polynomials from A001906(n+1) for n >= -1. - Wolfdieter Lang, May 28 2019
a(3n + 2)/a(3n - 1) = continued fraction 4,4,4,...,4,-5 (that's n 4's followed by a single -5). - Greg Dresden and Shaoxiong Yuan, Jul 16 2019
E.g.f.: ((- 1 + 3*sqrt(5))*exp((1/2)*(1 - sqrt(5))*x) + (1 + 3*sqrt(5))*exp((1/2)*(1 + sqrt(5))*x))/(2*sqrt(5)). - Stefano Spezia, Jul 17 2019
a(n) = (F(3n+1) - F(n+1)^3)/(F(n)^2) for n>1, where F(n) = A000045(n). - Michael Tulskikh, Jul 22 2020
a(n) = 3 * Sum_{k=0..n-2} A168561(n-2,k) + 2 * Sum_{k=0..n-1} A168561(n-1,k), n>0. - R. J. Mathar, Feb 14 2024

Definition corrected by Gary Detlefs, Dec 30 2012

A104449 Fibonacci sequence with initial values a(0) = 3 and a(1) = 1.

Original entry on oeis.org

3, 1, 4, 5, 9, 14, 23, 37, 60, 97, 157, 254, 411, 665, 1076, 1741, 2817, 4558, 7375, 11933, 19308, 31241, 50549, 81790, 132339, 214129, 346468, 560597, 907065, 1467662, 2374727, 3842389, 6217116, 10059505, 16276621, 26336126, 42612747, 68948873, 111561620

Offset: 0

Casey Mongoven, Mar 08 2005

The old name was: The Pibonacci numbers (a Fibonacci-type sequence): each term is the sum of the two previous terms.
The 6th row in the Wythoff array begins with the 6th term of the sequence (14, 23, 37, 60, 97, 157, ...). a(n) = f(n-3) + f(n+2) for the Fibonacci numbers f(n) = f(n-1) + f(n-2); f(0) = 0, f(1) = 1.
(a(2*k), a(2*k+1)) give for k >= 0 the proper positive solutions of one of two families (or classes) of solutions (x, y) of the indefinite binary quadratic form x^2 + x*y - y^2 of discriminant 5 representing 11. The other family of such solutions is given by (x2, y2) = (b(2*k), b(2*k+1)) with b = A013655. See the formula in terms of Chebyshev S polynomials S(n, 3) = A001906(n+1) below, which follows from the fundamental solution (3, 1) by applying positive powers of the automorphic matrix given in a comment in A013655. See also A089270 with the Alfred Brousseau link with D = 11. - Wolfdieter Lang, May 28 2019

V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.

G. C. Greubel, Table of n, a(n) for n = 0..1000
John Conway, Alex Ryba, The extra Fibonacci series and the Empire State Building, Math. Intelligencer 38 (2016), no. 1, 41-48. (Uses the name Pibonacci.)
Tanya Khovanova, Recursive Sequences
Ron Knott, Fibonacci Numbers and the Golden Section .
Eric Weisstein's World of Mathematics, Fibonacci Number
Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (1,1).
Index entries for sequences related to Chebyshev polynomials.

Cf. Other Fibonacci-type sequences: A000045, A000032, A013655. Other related sequences: A001906, A013655, A089270, A103343, A103344.
Wythoff array: A035513.
Essentially the same as A000285.

a:=[3,1];; for n in [3..40] do a[n]:=a[n-1]+a[n-2]; od; a; # G. C. Greubel, May 29 2019

[Fibonacci(n-1) + Lucas(n): n in [0..40]]; // G. C. Greubel, May 29 2019

a:=n->3*fibonacci(n-1)+fibonacci(n): seq(a(n), n=0..40); # Zerinvary Lajos, Oct 05 2007

LinearRecurrence[{1,1},{3,1},40] (* Harvey P. Dale, May 23 2014 *)

a(n)=3*fibonacci(n-1)+fibonacci(n) \\ Charles R Greathouse IV, Jun 05 2011

((3-2*x)/(1-x-x^2)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, May 29 2019

a(n) = a(n-1) + a(n-2) with a(0) = 3, a(1) = 1.
a(n) = 3*Fibonacci(n-1) + Fibonacci(n). - Zerinvary Lajos, Oct 05 2007
G.f.: (3-2*x)/(1-x-x^2). - Philippe Deléham, Nov 19 2008
a(n) = ( (3*sqrt(5)-1)*((1+sqrt(5))/2)^n + (3*sqrt(5)+1)*((1-sqrt(5) )/2)^n )/(2*sqrt(5)). - Bogart B. Strauss, Jul 19 2013
Bisection: a(2*k) = 4*S(k-1, 3) - 3*S(k-2, 3), a(2*k+1) = 2*S(k-1, 3) + S(k, 3) for k >= 0, with the Chebyshev S(n, 3) polynomials from A001906(n+1) for n >= -1. - Wolfdieter Lang, May 28 2019
a(n) = Fibonacci(n-1) + Lucas(n). - G. C. Greubel, May 29 2019
a(3n + 4)/a(3n + 1) = continued fraction 4,4,4,...,4,9 (that's n 4's followed by a single 9). - Greg Dresden and Shaoxiong Yuan, Jul 16 2019
E.g.f.: (exp((1/2)*(1 - sqrt(5))*x)*(1 + 3*sqrt(5) + (- 1 + 3*sqrt(5))*exp(sqrt(5)*x)))/(2*sqrt(5)). - Stefano Spezia, Jul 18 2019

Name changed by Wolfdieter Lang, Jun 17 2019

cp's OEIS Frontend

A324598 Irregular triangle with the representative solutions of the Diophantine equation x^2 + x - 1 congruent to 0 modulo N(n), with N(n) = A089270(n), for n >= 1.

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A324599 Irregular triangle with the representative solutions of the Diophantine equation x^2 - 5 congruent to 0 modulo N(n), with N(n) = A089270(n), for n >= 1.

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A308686 Irregular triangle with the nonnegative proper fundamental solutions of the binary quadratic form x^2 + x*y - y^2 representing N = N(n) = A089270(n), for n >= 1.

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A028387 a(n) = n + (n+1)^2.

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A038872 Primes congruent to {0, 1, 4} mod 5.

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A062786 Centered 10-gonal numbers.

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A013655 a(n) = F(n+1) + L(n), where F(n) and L(n) are Fibonacci and Lucas numbers, respectively.

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A155006 Primes p such that (p-2)(p+2)-+2p are primes.