A322907 -id:A322907 - cp's OEIS frontend

A308948 a(n) = A006190(A322907(n)+1) mod n.

0, 1, 1, 1, 3, 1, 6, 5, 1, 3, 10, 1, 8, 13, 4, 9, 16, 1, 18, 9, 13, 21, 1, 13, 18, 5, 1, 13, 12, 19, 30, 17, 10, 33, 6, 1, 31, 37, 25, 29, 32, 13, 1, 21, 19, 1, 46, 25, 48, 43, 16, 25, 1, 1, 21, 41, 37, 17, 58, 49, 1, 61, 55, 33, 18, 43, 66, 33, 1, 41, 70, 37

Offset: 1

Jianing Song, Jul 02 2019

nonn

A322907(n) is the smallest k > 0 such that n divides A006190(k).
Let M = [{3, 1}, {1, 0}], I = [{1, 0}, {0, 1}] is the 2 X 2 identity matrix, then A322907(n) is the smallest k > 0 such that M^k == r*I (mod n) for some r such that 0 <= r < n, and a(n) gives the value r.
A322906(n) is the multiplicative order of a(n) modulo n, which can only take value 1, 2 or 4.

			For n = 7, {A006190(n) mod 7 : n > 0} = {1, 3, 3, 5, 4, 3, 6, 0, 6, ...}, so a(7) = 6. Also, A322907(7) = 8, and M^8 mod 7 = [{6, 0}, {0, 6}], so a(7) = 6.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A006190, A322906, A322907.

Similar sequences: A217036, A308947.

a[n_] := For[k = 1, True, k++, If[Divisible[Fibonacci[k, 3], n], Return[ Mod[Fibonacci[k + 1, 3], n]]]];
Array[a, 100] (* Jean-François Alcover, Jul 05 2019 *)

a(n) = my(M=[3, 1; 1, 0]); for(k=1, 12*n/7, if((Mod(M,n)^k)[2,1]==0, return(lift((Mod(M,n)^k)[1,1]))))

Also a(n) = A006190(A322907(n)-1) mod n.
a(2^e) = 1 if e = 1, 2, 2^(e-1) + 1 if e >= 3; a(p^e) = a(p)^(p^(e-1)) mod p^e for odd primes p.
For odd primes p, a(p^e) = 1 if and only if A322907(p) == 2 (mod 4); a(p^e) = p^e - 1 if and only if 4 divides A322907(p).

A172236 Array A(n,k) = n*A(n,k-1) + A(n,k-2) read by upward antidiagonals, starting A(n,0) = 0, A(n,1) = 1.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 5, 3, 0, 1, 4, 10, 12, 5, 0, 1, 5, 17, 33, 29, 8, 0, 1, 6, 26, 72, 109, 70, 13, 0, 1, 7, 37, 135, 305, 360, 169, 21, 0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 0, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 0, 1, 10, 82, 528, 2549, 8658, 18901, 23184, 12970, 2378, 89, 0, 1, 11, 101, 747, 4289, 18200, 53353, 98145, 98209, 42837, 5741, 144

Offset: 1

Roger L. Bagula, Jan 29 2010

Equals A073133 with an additional column A(.,0).
If the first column and top row are deleted, antidiagonal reading yields A118243.
Adding a top row of 1's and antidiagonal reading downwards yields A157103.
Antidiagonal sums are 0, 1, 2, 5, 12, 32, 93, 297, 1035, 3911, 15917, 69350, ....
From Jianing Song, Jul 14 2018: (Start)
All rows have strong divisibility, that is, gcd(A(n,k_1), A(n,k_2)) = A(n,gcd(k_1,k_2)) holds for all k_1, k_2 >= 0.
Let E(n,m) be the smallest number l such that m divides A(n,l), we have: for odd primes p that are not divisible by n^2 + 4, E(n,p) divides p - ((n^2+4)/p) if p == 3 (mod 4) and (p - ((n^2+4)/p))/2 if p == 1 (mod 4). E(n,p) = p for odd primes p that are divisible by n^2 + 4. E(n,2) = 2 for even n and 3 for odd n. Here ((n^2+4)/p) is the Legendre symbol. A prime p such that p^2 divides T(n,E(n,p)) is called an n-Wall-Sun-Sun prime.
E(n,p^e) <= p^(e-1)*E(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then E(n,p^e) = p^(e-1)*E(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n,p), so E(n,p^e) = p^e as described above. E(n,m_1*m_2) = lcm(E(n,m_1),E(n,m_2)) if gcd(m_1,m_2) = 1.
Let pi(n,m) be the Pisano period of A(n, k) modulo m, i.e, the smallest number l such that A(n, k+1) == T(n,k) (mod m) holds for all k >= 0, we have: for odd primes p that are not divisible by n^2 - 4, pi(n,p) divides p - 1 if ((n^2+4)/p) = 1 and 2(p+1) if ((n^2+4)/p) = -1. pi(n,p) = 4p for odd primes p that are divisible by n^2 + 4. pi(n,2) = 2 even n and 3 for odd n.
pi(n,p^e) <= p^(e-1)*pi(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then pi(n,p^e) = p^(e-1)*pi(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n, p), so pi(n,p^e) = 4p^e as described above. pi(n,m_1*m_2) = lcm(pi(n,m_1), pi(n,m_2)) if gcd(m_1,m_2) = 1.
If n != 2, the largest possible value of pi(n,m)/m is 4 for even n and 6 for odd n. For even n, pi(n,p^e) = 4p^e; for odd n, pi(n,2p^e) = 12p^e, where p is any odd prime factor of n^2 + 4. For n = 2 it is 8/3, obtained by m = 3^e.
Let z(n,m) be the number of zeros in a period of A(n, k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: z(n,p) = 4 for odd primes p that are divisible by n^2 + 4. For other odd primes p, z(n,p) = 4 if E(n,p) is odd; 1 if E(n,p) is even but not divisible by 4; 2 if E(n,p) is divisible by 4; see the table below. z(n,2) = z(n,4) = 1.
Among all values of z(n,p) when p runs through all odd primes that are not divisible by n^2 + 4, we have:
((n^2+4)/p)...p mod 8....proportion of 1.....proportion of 2.....proportion of 4
......1..........1......1/6 (conjectured)...2/3 (conjectured)...1/6 (conjectured)*
......1..........5......1/2 (conjectured)...........0...........1/2 (conjectured)*
......1.........3,7.............1...................0...................0
.....-1.........1,5.............0...................0...................1
.....-1.........3,7.............0...................1...................0
* The result is that among all odd primes that are not divisible by n^2 + 4, 7/24 of them are with z(n,p) = 1, 5/12 are with z(n,p) = 2 and 7/24 are with z(n,p) = 4 if n^2 + 4 is a twice a square; 1/3 of them are with z(n,p) = 1, 1/3 are with z(n,p) = 2 and 1/3 are with z(n,p) = 4 otherwise. [Corrected by Jianing Song, Jul 06 2019]
z(n,p^e) = z(n,p) for all odd primes p; z(n,2^e) = 1 for even n and 2 for odd n, e >= 3.
(End)
From Michael A. Allen, Mar 06 2023: (Start)
Removing the first (n=0) row of A352361 gives this sequence.
Row n is the n-metallonacci sequence.
A(n,k) is (for k>0) the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

			The array, A(n, k), starts in row n = 1 with columns k >= 0 as
  0      1      1      2      3      5      8
  0      1      2      5     12     29     70
  0      1      3     10     33    109    360
  0      1      4     17     72    305   1292
  0      1      5     26    135    701   3640
  0      1      6     37    228   1405   8658
  0      1      7     50    357   2549  18200
  0      1      8     65    528   4289  34840
  0      1      9     82    747   6805  61992
  0      1     10    101   1020  10301 104030
  0      1     11    122   1353  15005 166408
Antidiagonal triangle, T(n, k), begins as:
  0;
  0, 1;
  0, 1, 1;
  0, 1, 2,  2;
  0, 1, 3,  5,   3;
  0, 1, 4, 10,  12,   5;
  0, 1, 5, 17,  33,  29,    8;
  0, 1, 6, 26,  72, 109,   70,   13;
  0, 1, 7, 37, 135, 305,  360,  169,  21;
  0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34;

Jianing Song, Antidiagonals n = 1..100, flattened (the first 30 antidiagonals from Vincenzo Librandi)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Rows n include: A000045 (n=1), A000129 (n=2), A006190 (n=3), A001076 (n=4), A052918 (n=5), A005668 (n=6), A054413 (n=7), A041025 (n=8), A099371 (n=9), A041041 (n=10), A049666 (n=11), A041061 (n=12), A140455 (n=13), A041085 (n=14), A154597 (n=15), A041113 (n=16), A178765 (n=17), A041145 (n=18), A243399 (n=19), A041181 (n=20). (Note that there are offset shifts for rows n = 5, 7, 8, 10, 12, 14, 16..20.)
Columns k include: A000004 (k=0), A000012 (k=1), A000027 (k=2), A002522 (k=3), A054602 (k=4), A057721 (k=5), A124152 (k=6).
Entry points for A(n,k) modulo m: A001177 (n=1), A214028 (n=2), A322907 (n=3).
Pisano period for A(n,k) modulo m: A001175 (n=1), A175181 (n=2), A175182 (n=3), A175183 (n=4), A175184 (n=5), A175185 (n=6).
Number of zeros in a period for A(n,k) modulo m: A001176 (n=1), A214027 (n=2), A322906 (n=3).
Cf. A084844, A084845, A118243, A157103, A271782, A316269.
Sums include: A304357, A304359.
Similar to: A073133.

A172236:= func< n,k | k le 1 select k else Evaluate(DicksonSecond(k-1,-1), n-k) >;
[A172236(n,k): k in [0..n-1], n in [1..13]]; // G. C. Greubel, Sep 29 2024

A172236[n_,k_]:=Fibonacci[k, n-k];
Table[A172236[n, k], {n,15}, {k,0,n-1}]//Flatten

A(n, k) = if (k==0, 0, if (k==1, 1, n*A(n, k-1) + A(n, k-2)));
tabl(nn) = for(n=1, nn, for (k=0, nn, print1(A(n, k), ", ")); print); \\ Jianing Song, Jul 14 2018 (program from Michel Marcus; see also A316269)

A(n, k) = ([n, 1; 1, 0]^k)[2, 1] \\ Jianing Song, Nov 10 2018

def A172236(n,k): return sum(binomial(k-j-1,j)*(n-k)^(k-2*j-1) for j in range(1+(k-1)//2))
flatten([[A172236(n,k) for k in range(n)] for n in range(1,14)]) # G. C. Greubel, Sep 29 2024

A(n,k) = (((n + sqrt(n^2 + 4))/2)^k - ((n-sqrt(n^2 + 4))/2)^k)/sqrt(n^2 + 4), n >= 1, k >= 0. - Jianing Song, Jun 27 2018
For n >= 1, Sum_{i=1..k} 1/A(n,2^i) = ((u^(2^k-1) + v^(2^k-1))/(u + v)) * (1/A(n,2^k)), where u = (n + sqrt(n^2 + 4))/2, v = (n - sqrt(n^2 + 4))/2 are the two roots of the polynomial x^2 - n*x - 1. As a result, Sum_{i>=1} 1/A(n,2^i) = (n^2 + 4 - n*sqrt(n^2 + 4))/(2*n). - Jianing Song, Apr 21 2019
From G. C. Greubel, Sep 29 2024: (Start)
A(n, k) = F_{k}(n) (Fibonacci polynomials F_{n}(x)) (array).
T(n, k) = F_{k}(n-k) (antidiagonal triangle).
Sum_{k=0..n-1} T(n, k) = A304357(n) - (1-(-1)^n)/2.
Sum_{k=0..n-1} (-1)^k*T(n, k) = (-1)*A304359(n) + (1-(-1)^n)/2.
T(2*n, n) = A084844(n).
T(2*n+1, n+1) = A084845(n). (End)

More terms from Jianing Song, Jul 14 2018

A322906 The number of zeros in the fundamental Pisano period of the 3-Fibonacci numbers A006190 modulo n.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 2, 2, 1, 4, 2, 1, 4, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 4, 4, 1, 2, 4, 2, 2, 2, 2, 2, 2, 1, 4, 2, 2, 2, 4, 2, 1, 2, 2, 1, 2, 2, 2, 4, 2, 2, 1, 1, 2, 2, 2, 4, 2, 2, 1, 2, 2, 2, 4, 2, 2, 2, 1, 2, 2, 2, 4, 4, 2, 2, 2, 2, 1, 2, 1, 4, 2, 2, 2, 1, 2

Offset: 1

Jianing Song, Jan 05 2019

nonn

a(n) is the multiplicative order of A006190(A322907(n)+1) modulo n.
a(n) has value 1, 2 or 4. This is because A006190(k,m+1)^4 == 1 (mod A006190(k,m)).
Conjecture: For primes p == 1, 9, 17, 25, 49, 81 (mod 104), the probability of a(p^e) taking on the value 1, 2, 4 is 1/6, 2/3, 1/6, respectively; for primes p == 29, 53, 61, 69, 77, 101 (mod 104), the probability of a(p^e) taking on the value 1, 4 is 1/2, 1/2, respectively.

Jianing Song, Table of n, a(n) for n = 1..1000

Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = k*x(n+1) + x(n). Then the periods, ranks and the ratios of the periods to the ranks modulo a given integer n are given by:
k = 1: A001175 (periods), A001177 (ranks), A001176 (ratios).
k = 2: A175181 (periods), A214028 (ranks), A214027 (ratios).
k = 3: A175182 (periods), A322907 (ranks), this sequence (ratios).

A006190(m) = ([3, 1; 1, 0]^m)[2, 1]
a(n) = my(i=1); while(A006190(i)%n!=0, i++); znorder(Mod(A006190(i+1), n))

For n > 2, T(n,k) = 4 iff A322907(n) is odd; 1 iff A322907(n) is even but not divisible by 4; 2 iff A322907(n) is divisible by 4.
For primes p == 3, 23, 27, 35, 43, 51 (mod 52), a(p^e) = 1.
For primes p == 5, 21, 33, 37, 41, 45 (mod 52), a(p^e) = 4.
For primes p == 7, 11, 15, 19, 31, 47 (mod 52), a(p^e) = 2.
a(13^e) = 4. a(2^e) = 1 if e = 1, 2 and 2 if e >= 3.
a(n) = A175182(n)/A322907(n).

A309586 Primes p with 1 zero in a fundamental period of A006190 mod p.

Original entry on oeis.org

2, 3, 23, 43, 53, 61, 79, 101, 103, 107, 127, 131, 139, 173, 179, 191, 199, 211, 251, 263, 277, 283, 311, 347, 367, 419, 433, 439, 443, 467, 491, 503, 523, 547, 563, 569, 571, 599, 607, 647, 659, 677, 719, 727, 751, 757, 823, 829, 859, 881, 883, 887, 907

Offset: 1

Jianing Song, Aug 10 2019

nonn

Primes p such that A322906(p) = 1.
For p > 2, p is in this sequence if and only if A175182(p) == 2 (mod 4), and if and only if A322907(p) == 2 (mod 4). For a proof of the equivalence between A322906(p) = 1 and A322907(p) == 2 (mod 4), see Section 2 of my link below.
This sequence contains all primes congruent to 3, 23, 27, 35, 43, 51 modulo 52. This corresponds to case (3) for k = 11 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

Jianing Song, Table of n, a(n) for n = 1..1200
Jianing Song, Lucas sequences and entry point modulo p

Cf. A175182, A322907.
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+----------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | this seq
Primes p such that w(p) = 2  | A053027  | A309581 | A309587
Primes p such that w(p) = 4  | A053028  | A261580 | A309588
Numbers k such that w(k) = 1 | A053031  | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | A309592
Numbers k such that w(k) = 4 | A053029  | A309585 | A309593
* and also A053032 U {2}

forprime(p=2, 900, if(A322906(p)==1, print1(p, ", ")))

A309587 Primes p with 2 zeros in a fundamental period of A006190 mod p.

Original entry on oeis.org

7, 11, 17, 19, 31, 47, 59, 67, 71, 83, 113, 151, 163, 167, 223, 227, 239, 257, 271, 307, 313, 331, 337, 359, 379, 383, 431, 463, 479, 487, 499, 521, 587, 601, 619, 631, 641, 643, 673, 683, 691, 739, 743, 787, 809, 811, 827, 839, 863, 947, 967, 983

Offset: 1

Jianing Song, Aug 10 2019

nonn

Primes p such that A322906(p) = 2.
For p > 2, p is in this sequence if and only if 8 divides A175182(p), and if and only if 4 divides A322907(p). For a proof of the equivalence between A322906(p) = 2 and 4 dividing A322907(p), see Section 2 of my link below.
This sequence contains all primes congruent to 7, 11, 15, 19, 31, 47 modulo 52. This corresponds to case (2) for k = 11 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

Jianing Song, Table of n, a(n) for n = 1..1200
Jianing Song, Lucas sequences and entry point modulo p

Cf. A006190, A175182, A322907.
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+----------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | A309586
Primes p such that w(p) = 2  | A053027  | A309581 | this seq
Primes p such that w(p) = 4  | A053028  | A261580 | A309588
Numbers k such that w(k) = 1 | A053031  | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | A309592
Numbers k such that w(k) = 4 | A053029  | A309585 | A309593
* and also A053032 U {2}

forprime(p=2, 1000, if(A322906(p)==2, print1(p, ", ")))

A309588 Primes p with 4 zeros in a fundamental period of A006190 mod p.

Original entry on oeis.org

5, 13, 29, 37, 41, 73, 89, 97, 109, 137, 149, 157, 181, 193, 197, 229, 233, 241, 269, 281, 293, 317, 349, 353, 373, 389, 397, 401, 409, 421, 449, 457, 461, 509, 541, 557, 577, 593, 613, 617, 653, 661, 701, 709, 733, 761, 769, 773, 797, 821, 853, 857, 877

Offset: 1

Jianing Song, Aug 10 2019

nonn

Primes p such that A322906(p) = 4.
For p > 2, p is in this sequence if and only if A175182(p) == 4 (mod 8), and if and only if A322907(p) is odd. For a proof of the equivalence between A322906(p) = 4 and A322907(p) being odd, see Section 2 of my link below.
This sequence contains all primes congruent to 5, 21, 33, 37, 41, 45 modulo 52. This corresponds to case (1) for k = 11 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

Jianing Song, Table of n, a(n) for n = 1..1200
Jianing Song, Lucas sequences and entry point modulo p

Cf. A006190, A175182, A322907.
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+----------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | A309586
Primes p such that w(p) = 2  | A053027  | A309581 | A309587
Primes p such that w(p) = 4  | A053028  | A261580 | this seq
Numbers k such that w(k) = 1 | A053031  | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | A309592
Numbers k such that w(k) = 4 | A053029  | A309585 | A309593
* and also A053032 U {2}

forprime(p=2, 900, if(A322906(p)==4, print1(p, ", ")))

A309592 Numbers k with 2 zeros in a fundamental period of A006190 mod k.

Original entry on oeis.org

7, 8, 11, 14, 15, 16, 17, 19, 20, 21, 22, 24, 28, 30, 31, 32, 33, 34, 35, 38, 39, 40, 42, 44, 45, 47, 48, 49, 51, 52, 55, 56, 57, 59, 60, 62, 63, 64, 66, 67, 68, 70, 71, 72, 75, 76, 77, 78, 80, 83, 84, 85, 87, 88, 90, 91, 93, 94, 95, 96, 98, 99, 100, 102, 104

Offset: 1

Jianing Song, Aug 10 2019

nonn

Numbers k such that A322906(k) = 2.
This sequence contains all numbers k such that 4 divides A322907(k). As a consequence, this sequence contains all numbers congruent to 7, 11, 15, 19, 31, 47 modulo 52.
This sequence contains all odd numbers k such that 8 divides A175182(k).

Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A175182, A322907.
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+----------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | A309586
Primes p such that w(p) = 2  | A053027  | A309581 | A309587
Primes p such that w(p) = 4  | A053028  | A261580 | A309588
Numbers k such that w(k) = 1 | A053031  | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | this seq
Numbers k such that w(k) = 4 | A053029  | A309585 | A309593
* and also A053032 U {2}

for(k=1, 100, if(A322906(k)==2, print1(k, ", ")))

A309593 Numbers k with 4 zeros in a fundamental period of A006190 mod k.

Original entry on oeis.org

5, 10, 13, 25, 26, 29, 37, 41, 50, 58, 65, 73, 74, 82, 89, 97, 109, 125, 130, 137, 145, 146, 149, 157, 169, 178, 181, 185, 193, 194, 197, 205, 218, 229, 233, 241, 250, 269, 274, 281, 290, 293, 298, 314, 317, 325, 338, 349, 353, 362, 365, 370, 373, 377

Offset: 1

Jianing Song, Aug 10 2019

nonn

Numbers k such that A322906(k) = 4.

Also numbers k such that A214027(k) is odd.

Jianing Song, Table of n, a(n) for n = 1..2000

Cf. A322907.
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+----------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | A309586
Primes p such that w(p) = 2  | A053027  | A309581 | A309587
Primes p such that w(p) = 4  | A053028  | A261580 | A309588
Numbers k such that w(k) = 1 | A053031  | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | A309592
Numbers k such that w(k) = 4 | A053029  | A309585 | this seq
* and also A053032 U {2}

for(k=1, 400, if(A322906(k)==4, print1(k, ", ")))

A327653 Composite numbers k coprime to 13 such that k divides A006190(k-Kronecker(13,k)).

Original entry on oeis.org

10, 119, 649, 1189, 1763, 3599, 4187, 5559, 6681, 12095, 12403, 12685, 12871, 12970, 14041, 14279, 15051, 16109, 19043, 22847, 23479, 24769, 26795, 28421, 30743, 30889, 31631, 31647, 33919, 34997, 37949, 38503, 39203, 41441, 46079, 48577, 49141, 50523, 50545, 53301, 56279, 58081, 58589

Offset: 1

Jianing Song, Sep 20 2019

nonn

Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n) = m*x(n-1) + x(n-2) for k >= 2. For primes p, we have (a) p divides x(p-((m^2+4)/p)); (b) x(p) == ((m^2+4)/p) (mod p), where (D/p) is the Kronecker symbol. This sequence gives composite numbers k such that gcd(k, m^2+4) = 1 and that a condition similar to (a) holds for k, where m = 3.
If k is not required to be coprime to m^2 + 4 (= 13), then there are 360 such k <= 10^5 and 1506 such k <= 10^6, while there are only 62 terms <= 10^5 and 197 terms <= 10^6 in this sequence.
Also composite numbers k coprime to 13 such that A322907(k) divides k - Kronecker(13,k).

			A006190(9) = 12970 which is divisible by 10, so 10 is a term.

Amiram Eldar, Table of n, a(n) for n = 1..10000

             m                      m=1            m=2      m=3
k | x(k-Kronecker(m^2+4,k))*  A081264 U A141137  A327651  this seq
k | x(k)-Kronecker(m^2+4,k)        A049062       A099011  A327654
            both                   A212424       A327652  A327655
* k is composite and coprime to m^2 + 4.
Cf. A006190, A322907, A011583 ({Kronecker(13,n)}).

seqmod(n, m)=((Mod([3, 1; 1, 0], m))^n)[1, 2]
isA327653(n)=!isprime(n) && !seqmod(n-Kronecker(13,n), n) && gcd(n,13)==1 && n>1

cp's OEIS Frontend

A308948 a(n) = A006190(A322907(n)+1) mod n.

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A172236 Array A(n,k) = n*A(n,k-1) + A(n,k-2) read by upward antidiagonals, starting A(n,0) = 0, A(n,1) = 1.

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A322906 The number of zeros in the fundamental Pisano period of the 3-Fibonacci numbers A006190 modulo n.

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A309586 Primes p with 1 zero in a fundamental period of A006190 mod p.

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A309587 Primes p with 2 zeros in a fundamental period of A006190 mod p.

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A309588 Primes p with 4 zeros in a fundamental period of A006190 mod p.

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A309592 Numbers k with 2 zeros in a fundamental period of A006190 mod k.

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A309593 Numbers k with 4 zeros in a fundamental period of A006190 mod k.

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