A321170 Numbers k such that Fibonacci(k) in base 5 ends with k in base 5.

1, 5, 9, 21, 25, 45, 49, 65, 85, 101, 105, 125, 145, 165, 185, 205, 225, 245, 249, 265, 285, 305, 325, 345, 365, 385, 405, 425, 445, 465, 485, 501, 505, 525, 545, 565, 585, 605, 625, 645, 705, 725, 745, 805, 825, 845, 905, 925, 945, 1005, 1025, 1045, 1105

Offset

1,2

Comments

Conjecture: For m >= 4, a(39 + (m-4)*77) = 5^m and the 77 first differences to the next m are ((x,y,x)^6, x, 4, y-4, x, (x,y,x)^11, x, y-4, 4, x, (x,y,x)^6) with x = 5^(m-2)-5, y = 2*(5^(m-2)+5). Verified to m=30.

Links

Lars Blomberg, Table of n, a(n) for n = 1..2100

Example

21 is a term because 21 = 41_5 and Fibonacci(21) = 10946 = 322241_5 which ends in 41.

Mathematica

fk5Q[n_]:=Module[{idn5=IntegerDigits[n, 5]}, Take[IntegerDigits[ Fibonacci[ n], 5], -Length[idn5]]==idn5]; Select[Range[1200], fk5Q] (* Harvey P. Dale, May 10 2020 *)

Crossrefs

Cf. A193806 (similar for base 3), A000045 (Fibonacci), A007091 (numbers in base 5).

Sequence in context: A256138 A246335 A255264 * A328407 A147407 A186297

Adjacent sequences: A321167 A321168 A321169 * A321171 A321172 A321173

Keyword

nonn,base

Author

Lars Blomberg, Jan 10 2019

Status

approved