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%I #30 Jan 05 2025 19:51:40
%S 55,110,155,165,205,220,305,310,330,355,385,410,440,465,495,505,605,
%T 610,615,620,655,660,710,715,755,770,820,880,905,915,930,935,955,990,
%U 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
%N Multiples of 5 where Pisano periods of Fibonacci numbers A001175 and Lucas numbers A106291 agree.
%C Multiples of 5 where A001175 and A106291 agree. See 1st comment of A106291.
%H Patrick Flanagan, Marc S. Renault, and Josh Updike, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/53-1/FlanaganRenaultUpdike06032014.pdf">Symmetries of Fibonacci Points, Mod m</a>, Fibonacci Quart. 53 (2015), no. 1, 34-41. See p. 7. (Is this the same sequence?)
%e 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).
%o (JavaScript)
%o let bases = [],
%o basesd = [],
%o baselimit = 2000;
%o for (let base = 2; base <= baselimit; base++) {
%o let fibs = [1 % base,1 % base],
%o lucas = [2 % base,1 % base],
%o repeatingf = false,
%o repeatingl = false;
%o while (!repeatingf) {
%o fibs.push((fibs[fibs.length - 2] + fibs[fibs.length - 1]) % base);
%o if (1 == fibs[fibs.length - 2] &&
%o 0 == fibs[fibs.length - 1])
%o repeatingf = true;
%o }
%o while (!repeatingl) {
%o lucas.push((lucas[lucas.length - 2] + lucas[lucas.length - 1]) % base);
%o if ((lucas[0] == (lucas[lucas.length - 2] + lucas[lucas.length - 1]) % base) &&
%o (lucas[1] == (lucas[lucas.length - 2] + 2 *lucas[lucas.length - 1]) % base))
%o repeatingl = true;
%o }
%o if (fibs.length != lucas.length)
%o bases.push(base);
%o }
%o for (let i = 1; i <= baselimit/5; i++) {
%o if (!bases.includes(i * 5)) basesd.push(i * 5);
%o }
%o console.log(basesd.join(','));
%Y Cf. A001175, A106291.
%K nonn
%O 1,1
%A _Dan Dart_, Jul 18 2016