A244923 Numbers n such that the digit sum of Fibonacci(n) is equal to the digit sum of Lucas(n).

1, 13, 61, 73, 97, 217, 349, 649, 937, 1477, 1513, 1729, 2005, 2077, 2209, 3265, 3649, 3889, 4093, 4609, 4945, 5497, 5749, 5929, 6109, 7309, 7441, 8041, 8389, 8821, 9925, 10525, 10669, 11605, 13201, 13345, 16021, 18529, 18649, 20293, 21481, 22573, 22729, 24169

Offset: 1

Author

Michel Lagneau, Jul 08 2014

Keywords

• nonn
• base

Comments

Numbers n such that A004090(n) = A139374(n).

Subsequence of A017533.

It seems that n is odd. The primes of the sequence are: 13, 61, 73, 97, 349, 937, 3889, 4093, 5749, 7309, 8389, 8821, 21481, 22573, 24169, ...

Fibonacci(j) == Lucas(j) (mod 9) iff j == 1 (mod 12), so all a(n) == 1 (mod 12). - Robert Israel, Jul 10 2014

Examples

			13 is in the sequence because Fibonacci(13) = 233, Lucas(13) = 521 and 2+3+3 = 5+2+1 = 8.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..95

Crossrefs

Cf. A000032, A000045, A004090, A017533, A139374.

Programs

Mathematica

lst={}; Table[If[Total[IntegerDigits[LucasL[n]]] == Total[IntegerDigits[Fibonacci[n]]], AppendTo[lst, n]], {n, 0, 25000}]; lst
Select[Range[25000],Total[IntegerDigits[Fibonacci[#]]]==Total[IntegerDigits[LucasL[#]]]&] (* Harvey P. Dale, Mar 31 2024 *)