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

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

Author

Dan Dart, Jul 18 2016

Keywords

• nonn

Comments

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

Examples

			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).
		

Links

• 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?)

Crossrefs

Cf. A001175, A106291.

Programs

JavaScript

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(','));