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The number of zeros in the fundamental Pisano period of the 3-Fibonacci numbers A006190 modulo n.
+20
24
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
COMMENTS
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.
FORMULA
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.
PROG
(PARI) 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))
CROSSREFS
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:
Primes p with 1 zero in a fundamental period of A006190 mod p.
+20
19
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
COMMENTS
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]
PROG
(PARI) forprime(p=2, 900, if( A322906(p)==1, print1(p, ", ")))
CROSSREFS
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
-----------------------------+----------+---------+----------
Primes p with 2 zeros in a fundamental period of A006190 mod p.
+20
19
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
COMMENTS
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]
PROG
(PARI) forprime(p=2, 1000, if( A322906(p)==2, print1(p, ", ")))
CROSSREFS
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
-----------------------------+----------+---------+----------
Primes p with 4 zeros in a fundamental period of A006190 mod p.
+20
19
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
COMMENTS
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]
PROG
(PARI) forprime(p=2, 900, if( A322906(p)==4, print1(p, ", ")))
CROSSREFS
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
-----------------------------+----------+---------+----------
Numbers k with 1 zero in a fundamental period of A006190 mod k.
+20
19
1, 2, 3, 4, 6, 9, 12, 18, 23, 27, 36, 43, 46, 53, 54, 61, 69, 79, 81, 86, 92, 101, 103, 106, 107, 108, 122, 127, 129, 131, 138, 139, 158, 159, 162, 172, 173, 179, 183, 191, 199, 202, 206, 207, 211, 212, 214, 237, 243, 244, 251, 254, 258, 262, 263, 276
COMMENTS
Numbers k such that A322906(k) = 1. The odd numbers k satisfy A175182(k) == 2 (mod 4).
PROG
(PARI) for(k=1, 300, if( A322906(k)==1, print1(k, ", ")))
CROSSREFS
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
-----------------------------+----------+---------+----------
Numbers k with 2 zeros in a fundamental period of A006190 mod k.
+20
19
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
COMMENTS
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).
PROG
(PARI) for(k=1, 100, if( A322906(k)==2, print1(k, ", ")))
CROSSREFS
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
-----------------------------+----------+---------+----------
Numbers k with 4 zeros in a fundamental period of A006190 mod k.
+20
19
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
COMMENTS
Numbers k such that A322906(k) = 4. Also numbers k such that A214027(k) is odd.
PROG
(PARI) for(k=1, 400, if( A322906(k)==4, print1(k, ", ")))
CROSSREFS
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
-----------------------------+----------+---------+----------
Pisano period of the 3-Fibonacci numbers A006190.
+20
18
1, 3, 2, 6, 12, 6, 16, 12, 6, 12, 8, 6, 52, 48, 12, 24, 16, 6, 40, 12, 16, 24, 22, 12, 60, 156, 18, 48, 28, 12, 64, 48, 8, 48, 48, 6, 76, 120, 52, 12, 28, 48, 42, 24, 12, 66, 96, 24, 112, 60, 16, 156, 26, 18, 24, 48, 40, 84, 24, 12, 30, 192, 48, 96, 156, 24, 136, 48, 22, 48, 144
COMMENTS
Period of the sequence defined by reading A006190 modulo n.
MAPLE
F := proc(k, n) option remember; if n <= 1 then n; else k*procname(k, n-1)+procname(k, n-2) ; end if; end proc:
Pper := proc(k, m) local cha, zer, n, fmodm ; cha := [] ; zer := [] ; for n from 0 do fmodm := F(k, n) mod m ; cha := [op(cha), fmodm] ; if fmodm = 0 then zer := [op(zer), n] ; end if; if nops(zer) = 5 then break; end if; end do ; if [op(1..zer[2], cha) ] = [ op(zer[2]+1..zer[3], cha) ] and [op(1..zer[2], cha)] = [ op(zer[3]+1..zer[4], cha) ] and [op(1..zer[2], cha)] = [ op(zer[4]+1..zer[5], cha) ] then return zer[2] ; elif [op(1..zer[3], cha) ] = [ op(zer[3]+1..zer[5], cha) ] then return zer[3] ; else return zer[5] ; end if; end proc:
k := 3 ; seq( Pper(k, m), m=1..80) ;
MATHEMATICA
Table[s = t = Mod[{0, 1}, n]; cnt = 1; While[tmp = Mod[3*t[[2]] + t[[1]], n]; t[[1]] = t[[2]]; t[[2]] = tmp; s!= t, cnt++]; cnt, {n, 100}] (* Vincenzo Librandi, Dec 20 2012, T. D. Noe *)
Composite numbers k coprime to 13 such that k divides A006190(k-Kronecker(13,k)).
+20
13
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
COMMENTS
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).
EXAMPLE
A006190(9) = 12970 which is divisible by 10, so 10 is a term.
PROG
(PARI) 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
CROSSREFS
m m=1 m=2 m=3
* k is composite and coprime to m^2 + 4.
Entry points for the 3-Fibonacci numbers A006190.
+20
10
1, 3, 2, 6, 3, 6, 8, 6, 6, 3, 4, 6, 13, 24, 6, 12, 8, 6, 20, 6, 8, 12, 22, 6, 15, 39, 18, 24, 7, 6, 32, 24, 4, 24, 24, 6, 19, 60, 26, 6, 7, 24, 42, 12, 6, 66, 48, 12, 56, 15, 8, 78, 26, 18, 12, 24, 20, 21, 12, 6, 30, 96, 24, 48, 39, 12, 68, 24, 22, 24, 72, 6
COMMENTS
a(n) is the smallest k > 0 such that n divides A006190(k). a(n) is also called the rank of A006190(n) modulo n. For primes p == 1, 9, 17, 25, 29, 49 (mod 52), a(p) divides (p - 1)/2.
For primes p == 3, 23, 27, 35, 43, 51 (mod 52), a(p) divides p - 1, but a(p) does not divide (p - 1)/2.
For primes p == 5, 21, 33, 37, 41, 45 (mod 52), a(p) divides (p + 1)/2.
For primes p == 7, 11, 15, 19, 31, 47 (mod 52), a(p) divides p + 1, but a(p) does not divide (p + 1)/2.
a(n) <= (12/7)*n for all n, where the equality holds if and only if n = 2*7^e, e >= 1.
FORMULA
a(m*n) = a(m)*a(n) if gcd(m, n) = 1.
For odd primes p, a(p^e) = p^(e-1)*a(p) if p^2 does not divide a(p). Any counterexample would be a 3-Wall-Sun-Sun prime.
a(2^e) = 3 if e = 1, 6 if e = 2 and 3*2^(e-2) if e >= 3. a(13^e) = 13^e, e >= 1.
PROG
(PARI) A006190(m) = ([3, 1; 1, 0]^m)[2, 1] a(n) = my(i=1); while( A006190(i)%n!=0, i++); i
CROSSREFS
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:
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