A001175 -id:A001175 - cp's OEIS frontend

A308784 Primes p such that A001175(p) = 2*(p+1)/3.

47, 107, 113, 263, 347, 353, 563, 677, 743, 977, 1097, 1217, 1223, 1277, 1307, 1523, 1553, 1733, 1823, 1877, 1913, 1973, 2027, 2237, 2243, 2267, 2333, 2447, 2663, 2687, 2753, 2777, 3323, 3347, 3407, 3467, 3533, 3557, 3617, 3623, 3767, 3947, 4133, 4493, 4547, 4583

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+sqrt(5))/2,p) = 2*(p+1)/3, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
Also, primes p such that the least integer k > 0 such that M^k == I (mod p) is 2*(p+1)/3, where M = [{1, 1}, {1, 0}] and I is the identity matrix.
Also, primes p such that A001177(p) = (p+1)/3 or (p+1)/6. If p == 1 (mod 4), then A001177(p) = (p+1)/6, otherwise (p+1)/3.
Also, primes p such that ord(-(3+sqrt(5))/2,p) = (p+1)/3 or (p+1)/6. If p == 1 (mod 4), then ord(-(3+sqrt(5))/2,p) = (p+1)/6, otherwise (p+1)/3.
In general, let {T(n)} be a sequence defined by T(0) = 0, T(1) = 1, T(n) = k*T(n-1) + T(n-2), K be the quadratic field Q[sqrt(k^2+4)], O_K be the ring of integer of K, u = (k+sqrt(k^2+4))/2. For a prime p not dividing k^2 + 4, the Pisano period of {T(n)} modulo p (that is, the smallest m > 0 such that T(n+m) == T(n) (mod p) for all n) is ord(u,p); the entry point of {T(n)} modulo p (that is, the smallest m > 0 such that T(m) == 0 (mod p)) is ord(-u^2,p).
For an odd prime p:
(a) if p decomposes in K, then (O_K/pO_K)* (the multiplicative group of O_K modulo p) is congruent to C_(p-1) X C_(p-1), so the Pisano period of {T(n)} modulo p is equal to (p-1)/s, s = 1, 2, 3, 4, ...;
(b) if p is inert in K, then u^(p+1) == -1 (mod p) (see the Wikipedia link below), so the Pisano period of {T(n)} modulo p is equal to 2*(p+1)/r, r = 1, 3, 5, 7, ...
If (b) holds, then the entry point of {T(n)} modulo p is (p+1)/r if p == 3 (mod 4) and (p+1)/(2r) if p == 1 (mod 4). Proof: let d = ord(u,p) = 2*(p+1)/r, d' = ord(-u^2,p), then (-u^2)^d' == (u^(-p-1)*u^2)^d == u^(d'*(-p+1)) (mod p), so d divides d'*(p-1), d' = d/gcd(d, p-1). It is easy to see that gcd(d, p-1) = 4 if p == 1 (mod 4) and 2 if p == 3 (mod 4).
Here k = 1, and this sequence gives primes such that (b) holds and r = 3. For k = 1, r cannot be a multiple of 5 because if 5 divides p+1 then p decomposes in K = Q[sqrt(5)], which contradicts with (b).
Number of terms below 10^N:
  N | 1 mod 4 | 3 mod 4 |  Total | Inert primes*
  3 |       4 |       6 |     10 |         88
  4 |      41 |      43 |     84 |        618
  5 |     330 |     353 |    683 |       4813
  6 |    2745 |    2736 |   5481 |      39286
  7 |   23219 |   23250 |  46469 |     332441
  8 |  201805 |  201547 | 403352 |    2880969
  * Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5).

Michael De Vlieger, Table of n, a(n) for n = 1..714
Bob Bastasz, Lyndon words of a second-order recurrence, Fibonacci Quarterly (2020) Vol. 58, No. 5, 25-29.
Wikipedia, Pisano period

Similar sequences that give primes such that (b) holds: A071774 (r=1), this sequence (r=3), A308785 (r=7), A308786 (r=9).

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0 && Mod[ Fibonacci[k + 1], n] == 1, Return[k]]];
Reap[For[p = 2, p <= 4583, p = NextPrime[p], If[pn[p] == 2(p+1)/3, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Jul 02 2019 *)

Pisano_for_inert_prime(p) = my(k=1, M=[k, 1; 1, 0], Id=[1, 0; 0, 1]); if(isprime(p)&&kronecker(k^2+4,p)==-1, my(v=divisors(2*(p+1))); for(d=1, #v, if(Mod(M,p)^v[d]==Id, return(v[d]))))
forprime(p=2, 4000, if(Pisano_for_inert_prime(p)==2*(p+1)/3, print1(p, ", ")))

A235249 Fixed point when A001175 (Pisano periods) is iterated with starting value n.

Original entry on oeis.org

1, 24, 24, 24, 120, 24, 24, 24, 24, 120, 120, 24, 24, 24, 120, 24, 24, 24, 24, 120, 24, 120, 24, 24, 600, 24, 24, 24, 24, 120, 120, 24, 120, 24, 120, 24, 24, 24, 24, 120, 120, 24, 120, 120, 120, 24, 24, 24, 24, 600, 24, 24, 24, 24, 120, 24, 24, 24, 24, 120

Offset: 1

Reinhard Zumkeller, Jan 15 2014

nonn

Are the powers of 5 (together with 2) the indices of records in this sequence? - Charles R Greathouse IV, Aug 11 2022

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
J. D. Fulton and W. L. Morris, On arithmetical functions related to the Fibonacci numbers, Acta Arithmetica, 16 (1969), 105-110.
Wikipedia, Pisano period

Cf. A001175, A001179.

a235249 n = if y == n then n else a235249 y  where y = a001175 n

a(n) = 24*5^(A001179(n)-1) for n > 1.

A001175(a(n)) = a(n).

A275124 Multiples of 5 where Pisano periods of Fibonacci numbers A001175 and Lucas numbers A106291 agree.

Original entry on oeis.org

55, 110, 155, 165, 205, 220, 305, 310, 330, 355, 385, 410, 440, 465, 495, 505, 605, 610, 615, 620, 655, 660, 710, 715, 755, 770, 820, 880, 905, 915, 930, 935, 955, 990, 1010, 1045, 1065, 1085, 1155, 1205, 1210, 1220, 1230, 1240, 1255, 1265, 1310, 1320, 1355, 1395, 1420, 1430, 1435, 1485, 1510, 1515, 1540, 1555, 1595, 1640, 1655, 1705, 1760, 1810, 1815, 1830

Offset: 1

Dan Dart, Jul 18 2016

nonn

Multiples of 5 where A001175 and A106291 agree. See 1st comment of A106291.

			55 is the first multiple of 5 where the Pisano period (Fibonacci) of n = 55 and the Pisano period (Lucas) of n = 55 agree (this is in this case 20).

Patrick Flanagan, Marc S. Renault, and Josh Updike, Symmetries of Fibonacci Points, Mod m, Fibonacci Quart. 53 (2015), no. 1, 34-41. See p. 7. (Is this the same sequence?)

Cf. A001175, A106291.

let bases = [],
    basesd = [],
    baselimit = 2000;
for (let base = 2; base <= baselimit; base++) {
    let fibs = [1 % base,1 % base],
        lucas = [2 % base,1 % base],
        repeatingf = false,
        repeatingl = false;
    while (!repeatingf) {
        fibs.push((fibs[fibs.length - 2] + fibs[fibs.length - 1]) % base);
        if (1 == fibs[fibs.length - 2] &&
            0 == fibs[fibs.length - 1])
            repeatingf = true;
    }
    while (!repeatingl) {
        lucas.push((lucas[lucas.length - 2] + lucas[lucas.length - 1]) % base);
        if ((lucas[0] == (lucas[lucas.length - 2] + lucas[lucas.length - 1]) % base) &&
            (lucas[1] == (lucas[lucas.length - 2] + 2 *lucas[lucas.length - 1]) % base))
            repeatingl = true;
    }
    if (fibs.length != lucas.length)
        bases.push(base);
}
for (let i = 1; i <= baselimit/5; i++) {
    if (!bases.includes(i * 5)) basesd.push(i * 5);
}
console.log(basesd.join(','));

A308785 Primes p such that A001175(p) = 2*(p+1)/7.

Original entry on oeis.org

307, 797, 1483, 3023, 4157, 4283, 6397, 6733, 7027, 7433, 7867, 9337, 9743, 9883, 10177, 10303, 10597, 11423, 12823, 14293, 18493, 19963, 20593, 20873, 24247, 24793, 25703, 28433, 29917, 30113, 31387, 31723, 31793, 32353, 33347, 34537, 34747, 37057, 38653, 38723

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+sqrt(5))/2,p) = 2*(p+1)/7, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
Also, primes p such that the least integer k > 0 such that M^k == I (mod p) is 2*(p+1)/7, where M = [{1, 1}, {1, 0}] and I is the identity matrix.
Also, primes p such that A001177(p) = (p+1)/7 or (p+1)/14. If p == 1 (mod 4), then A001177(p) = (p+1)/14, otherwise (p+1)/7.
Also, primes p such that ord(-(3+sqrt(5))/2,p) = (p+1)/7 or (p+1)/14. If p == 1 (mod 4), then ord(-(3+sqrt(5))/2,p) = (p+1)/14, otherwise (p+1)/7.
In general, let {T(n)} be a sequence defined by T(0) = 0, T(1) = 1, T(n) = k*T(n-1) + T(n-2), K be the quadratic field Q[sqrt(k^2+4)], O_K be the ring of integer of K, u = (k+sqrt(k^2+4))/2. For a prime p not dividing k^2 + 4, the Pisano period of {T(n)} modulo p (that is, the smallest m > 0 such that T(n+m) == T(n) (mod p) for all n) is ord(u,p); the entry point of {T(n)} modulo p (that is, the smallest m > 0 such that T(m) == 0 (mod p)) is ord(-u^2,p).
For an odd prime p:
(a) if p decomposes in K, then (O_K/pO_K)* (the multiplicative group of O_K modulo p) is congruent to C_(p-1) X C_(p-1), so the Pisano period of {T(n)} modulo p is equal to (p-1)/s, s = 1, 2, 3, 4, ...;
(b) if p is inert in K, then u^(p+1) == -1 (mod p) (see the Wikipedia link below), so the Pisano period of {T(n)} modulo p is equal to 2*(p+1)/r, r = 1, 3, 5, 7, ...
If (b) holds, then the entry point of {T(n)} modulo p is (p+1)/r if p == 3 (mod 4) and (p+1)/(2r) if p == 1 (mod 4). Proof: let d = ord(u,p) = 2*(p+1)/r, d' = ord(-u^2,p), then (-u^2)^d' == (u^(-p-1)*u^2)^d == u^(d'*(-p+1)) (mod p), so d divides d'*(p-1), d' = d/gcd(d, p-1). It is easy to see that gcd(d, p-1) = 4 if p == 1 (mod 4) and 2 if p == 3 (mod 4).
Here k = 1, and this sequence gives primes such that (b) holds and r = 7. For k = 1, r cannot be a multiple of 5 because if 5 divides p+1 then p decomposes in K = Q[sqrt(5)], which contradicts with (b).
Number of terms below 10^N:
  N | 1 mod 4 | 3 mod 4 | Total | Inert primes*
  3 |       1 |       1 |     2 |         88
  4 |       6 |       8 |    14 |        618
  5 |      48 |      42 |    90 |       4813
  6 |     371 |     350 |   721 |      39286
  7 |    3098 |    3086 |  6184 |     332441
  8 |   27035 |   26989 | 54024 |    2880969
  * Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5).

Bob Bastasz, Lyndon words of a second-order recurrence, Fibonacci Quarterly (2020) Vol. 58, No. 5, 25-29.
Wikipedia, Pisano period

Similar sequences that give primes such that (b) holds: A071774 (r=1), A308784 (r=3), this sequence (r=7), A308786 (r=9).

Select[Prime@ Range[1000], Function[n, Mod[Last@ NestWhile[{Mod[#2, n], Mod[#1 + #2, n], #3 + 1} & @@ # &, {1, 1, 1}, #[[1 ;; 2]] != {0, 1} &], n] == Mod[2 (n + 1)/7, n] ]] (* Michael De Vlieger, Mar 31 2021, after Leo C. Stein at A001175 *)

Pisano_for_inert_prime(p) = my(k=1, M=[k, 1; 1, 0], Id=[1, 0; 0, 1]); if(isprime(p)&&kronecker(k^2+4,p)==-1, my(v=divisors(2*(p+1))); for(d=1, #v, if(Mod(M,p)^v[d]==Id, return(v[d]))))
forprime(p=2, 40000, if(Pisano_for_inert_prime(p)==2*(p+1)/7, print1(p, ", ")))

A308786 Primes p such that A001175(p) = 2*(p+1)/9.

Original entry on oeis.org

233, 557, 953, 4013, 4733, 5147, 6983, 7307, 7883, 9377, 10133, 12923, 14867, 15767, 17747, 19403, 20753, 22877, 23813, 26387, 26783, 27737, 29483, 32057, 33533, 35117, 39383, 40013, 40787, 41543, 41903, 42767, 43613, 45557, 46187, 48473, 48563, 50993, 51263, 53927

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+sqrt(5))/2,p) = 2*(p+1)/9, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
Also, primes p such that the least integer k > 0 such that M^k == I (mod p) is 2*(p+1)/9, where M = [{1, 1}, {1, 0}] and I is the identity matrix.
Also, primes p such that A001177(p) = (p+1)/9 or (p+1)/18. If p == 1 (mod 4), then A001177(p) = (p+1)/18, otherwise (p+1)/9.
Also, primes p such that ord(-(3+sqrt(5))/2,p) = (p+1)/9 or (p+1)/18. If p == 1 (mod 4), then ord(-(3+sqrt(5))/2,p) = (p+1)/18, otherwise (p+1)/9.
In general, let {T(n)} be a sequence defined by T(0) = 0, T(1) = 1, T(n) = k*T(n-1) + T(n-2), K be the quadratic field Q[sqrt(k^2+4)], O_K be the ring of integer of K, u = (k+sqrt(k^2+4))/2. For a prime p not dividing k^2 + 4, the Pisano period of {T(n)} modulo p (that is, the smallest m > 0 such that T(n+m) == T(n) (mod p) for all n) is ord(u,p); the entry point of {T(n)} modulo p (that is, the smallest m > 0 such that T(m) == 0 (mod p)) is ord(-u^2,p).
For an odd prime p:
(a) if p decomposes in K, then (O_K/pO_K)* (the multiplicative group of O_K modulo p) is congruent to C_(p-1) X C_(p-1), so the Pisano period of {T(n)} modulo p is equal to (p-1)/s, s = 1, 2, 3, 4, ...;
(b) if p is inert in K, then u^(p+1) == -1 (mod p) (see the Wikipedia link below), so the Pisano period of {T(n)} modulo p is equal to 2*(p+1)/r, r = 1, 3, 5, 7, ...
If (b) holds, then the entry point of {T(n)} modulo p is (p+1)/r if p == 3 (mod 4) and (p+1)/(2r) if p == 1 (mod 4). Proof: let d = ord(u,p) = 2*(p+1)/r, d' = ord(-u^2,p), then (-u^2)^d' == (u^(-p-1)*u^2)^d == u^(d'*(-p+1)) (mod p), so d divides d'*(p-1), d' = d/gcd(d, p-1). It is easy to see that gcd(d, p-1) = 4 if p == 1 (mod 4) and 2 if p == 3 (mod 4).
Here k = 1, and this sequence gives primes such that (b) holds and r = 9. For k = 1, r cannot be a multiple of 5 because if 5 divides p+1 then p decomposes in K = Q[sqrt(5)], which contradicts with (b).
Number of terms below 10^N:
  N | 1 mod 4 | 3 mod 4 | Total | Inert primes*
  3 |       3 |       0 |     3 |         88
  4 |       6 |       4 |    10 |        618
  5 |      36 |      28 |    64 |       4813
  6 |     313 |     300 |   613 |      39286
  7 |    2563 |    2597 |  5160 |     332441
  8 |   22377 |   22350 | 44727 |    2880969
  * Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5).

Bob Bastasz, Lyndon words of a second-order recurrence, Fibonacci Quarterly (2020) Vol. 58, No. 5, 25-29.
Wikipedia, Pisano period

Similar sequences that give primes such that (b) holds: A071774 (r=1), A308784 (r=3), A308785 (r=7), this sequence (r=9).

Pisano_for_inert_prime(p) = my(k=1, M=[k, 1; 1, 0], Id=[1, 0; 0, 1]); if(isprime(p)&&kronecker(k^2+4,p)==-1, my(v=divisors(2*(p+1))); for(d=1, #v, if(Mod(M,p)^v[d]==Id, return(v[d]))))
forprime(p=2, 55000, if(Pisano_for_inert_prime(p)==2*(p+1)/9, print1(p, ", ")))

A213278 Least common multiple of A001175(n) and n.

Original entry on oeis.org

1, 6, 24, 12, 20, 24, 112, 24, 72, 60, 110, 24, 364, 336, 120, 48, 612, 72, 342, 60, 336, 330, 1104, 24, 100, 1092, 216, 336, 406, 120, 930, 96, 1320, 612, 560, 72, 2812, 342, 2184, 120, 1640, 336, 3784, 660, 360, 1104, 1504, 48, 784, 300, 1224, 1092, 5724

Offset: 1

Lars Blomberg, Jun 09 2012

nonn

If n>1 then a(n) is even (see A001175). - Jon Maiga, Mar 25 2019

			Example with n=3:
Fib(k): 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368
Fib(k) mod 3: 0,1,1,2,0,2,2,1,0,1,1,2,0,2,2,1,0,1,1,2,0,2,2,1,0
k mod 3:      0,1,2,0,1,2,0,1,2,0,1,2,0,1,2,0,1,2,0,1,2,0,1,2,0
k increases by 24 before it realigns with Fib(k) mod 3 therefore a(3) = lcm(A001175(3), 3) = lcm(8, 3) = 24.

Jon Maiga, Table of n, a(n) for n = 1..23002 (2..23002 from Lars Blomberg)

Cf. A000045, A001175, A213277.

a(n) = lcm(A001175(n), n). - Jon Maiga, Mar 24 2019

Changed offset to 1, added a(1)=1 and simplified name by Jon Maiga, Mar 25 2019

A222414 a(n) = A001175(A222413(n)).

Original entry on oeis.org

14, 32, 44, 50, 72, 76, 46, 50, 90, 22, 42, 114, 52, 176, 56, 88, 110, 232, 174, 236, 200, 84, 46, 254, 26, 90, 124, 376, 206, 220, 452, 138, 118, 496, 380, 192, 228, 202, 270, 276, 78, 176, 70, 102, 470, 212, 176, 652, 198, 126, 510, 206, 262, 530, 356, 128, 732, 96, 554, 230, 812, 816, 614, 410, 624, 852

Offset: 1

N. J. A. Sloane, Feb 28 2013

nonn

See A222413.

D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.

Cf. A001175, A222413.

A317971 Numbers m such that the Pisano period A001175(m) divides m.

Original entry on oeis.org

1, 24, 48, 72, 96, 120, 144, 192, 216, 240, 288, 336, 360, 384, 432, 480, 576, 600, 648, 672, 720, 768, 864, 960, 1008, 1080, 1104, 1152, 1200, 1224, 1296, 1320, 1344, 1368, 1440, 1536, 1680, 1728, 1800, 1920, 1944, 2016, 2160, 2208, 2304, 2352, 2400, 2448, 2592, 2640, 2688, 2736, 2880, 3000, 3024, 3072

Offset: 1

N. J. A. Sloane, Sep 01 2018

nonn

More terms than usual are displayed because there are some very similar sequences in the OEIS.
The terms > 1 are divisible by 24, and the quotients give A072378.
In their paper, Shtefan and Dobrovolska (2018) prove that all terms m > 1 of this sequence are such that the sum of any m consecutive Fibonacci numbers is divisible by m. - Petros Hadjicostas, May 19 2019

N. J. A. Sloane, Table of n, a(n) for n = 1..10001
D. Shtefan and I. Dobrovolska, The sums of the consecutive Fibonacci numbers,, Fibonacci Quarterly, 56 (2018), 229-236.
Eric Weisstein's World of Mathematics, Pisano period.
Wikipedia, Pisano period.

Cf. A001175, A072378.

A326612 Indices where A001175 (Pisano period) sets a new record value.

Original entry on oeis.org

1, 2, 3, 5, 6, 10, 25, 30, 50, 98, 125, 150, 206, 243, 250, 490, 566, 590, 625, 750, 1030, 1046, 1094, 1154, 1214, 1226, 1250, 2450, 2738, 2830, 2846, 2894, 2906, 3086, 3125, 3750, 4802, 5534, 5594, 5606, 5666, 5714, 5770, 5834, 5906, 5990, 6070, 6130, 6250

Offset: 1

Richard N. Smith, Jul 14 2019

nonn

Record values: 1, 3, 8, 20, 24, 60, 100, 120, 300, 336, 500, 600, 624, 648, 1500, 1680, 1704, 1740, 2500, 3000, ...

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..1547 from Richard N. Smith)

Cf. A001175.

fibmod(n, m)=((Mod([1, 1; 1, 0], m))^n)[1, 2]
entryp(p)=my(k=p+[0, -1, 1, 1, -1][p%5+1], f=factor(k)); for(i=1, #f[, 1], for(j=1, f[i, 2], if((Mod([1, 1; 1, 0], p)^(k/f[i, 1]))[1, 2], break); k/=f[i, 1])); k
entry(n)=if(n==1, return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i, 1]>1e14, entryp(f[i, 1]^f[i, 2]), entryp(f[i, 1])*f[i, 1]^(f[i, 2] - 1))); if(f[1, 1]==2&&f[1, 2]>1, v[1]=3<r, r=a(n); print1(", "n))) \\ after Charles R Greathouse IV in A001175

A253806 One half of the maximal values of the length of the period for Fibonacci numbers modulo p (A001175(p)) for primes p > 5, according to Wall's Theorems 6 and 7.

Original entry on oeis.org

8, 5, 14, 18, 9, 24, 14, 15, 38, 20, 44, 48, 54, 29, 30, 68, 35, 74, 39, 84, 44, 98, 50, 104, 108, 54, 114, 128, 65, 138, 69, 74, 75, 158, 164, 168, 174

Offset: 1

Wolfdieter Lang, Jan 16 2015

			a(1) = 8 = 7 + 1 because prime(4) = 7 == 7 (mod 10). The length of the period for 7 is 2*8 = 16 = A001175(7).
a(2) = 5 = (11 - 1)/2 because prime(4) = 11 = 1 (mod 10). The length of the period for 11 is 10 = A001175(11).

D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.

Cf. A001175, A222413, A222414.

a(n) = (prime(n+3) - 1)/2 if prime(n+3) == 1 or 9 (mod 10) and a(n) = (prime(n+3) + 1) if

prime(n+3) == 3 or 7 (mod 10), n >= 1.

cp's OEIS Frontend

A308784 Primes p such that A001175(p) = 2*(p+1)/3.

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A235249 Fixed point when A001175 (Pisano periods) is iterated with starting value n.

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A275124 Multiples of 5 where Pisano periods of Fibonacci numbers A001175 and Lucas numbers A106291 agree.

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A308785 Primes p such that A001175(p) = 2*(p+1)/7.

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A308786 Primes p such that A001175(p) = 2*(p+1)/9.

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A213278 Least common multiple of A001175(n) and n.

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A222414 a(n) = A001175(A222413(n)).

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A317971 Numbers m such that the Pisano period A001175(m) divides m.

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A326612 Indices where A001175 (Pisano period) sets a new record value.

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