A004090 Sum of digits of Fibonacci numbers.

0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 10, 17, 9, 8, 17, 7, 24, 22, 19, 14, 24, 20, 17, 28, 27, 19, 19, 29, 21, 23, 17, 31, 30, 34, 37, 35, 27, 35, 44, 43, 24, 31, 46, 41, 33, 29, 35, 37, 54, 55, 46, 29, 48, 41, 53, 58, 48, 52, 73, 44, 54, 53, 62, 61, 51, 67, 73, 59

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• base
• easy

Comments

a(n) and Fibonacci(n) are congruent modulo 9 which implies that (a(n) mod 9) is equal to (Fibonacci(n) mod 9) A007887(n). Thus (a(n) mod 9) is periodic with Pisano period A001175(9) = 24. - Hieronymus Fischer, Jun 25 2007

It appears that a(n) - n stays negative for n > 5832, which explains why A020995 is finite. - T. D. Noe, Mar 19 2012

Links

• T. D. Noe, Table of n, a(n) for n = 0..10000
• T. D. Noe, Plot of a(n)-n for n = 0..100000

Crossrefs

Cf. A000045 (Fibonacci), A007953 (digit sum), A030132 (digital root of A45), A010888 (digital root), A246558, A261587, A068500.

Programs

Haskell

a004090 = a007953 . a000045 -- Reinhard Zumkeller, Nov 17 2014

Magma

[&+Intseq(Fibonacci(n)): n in [0..80] ]; // Vincenzo Librandi, Jun 18 2015

Mathematica

Table[Plus@@IntegerDigits@(Fibonacci[n]), {n, 0, 90}] (* Vincenzo Librandi, Jun 18 2015 *)

PARI

a(n)=sumdigits(fibonacci(n)) \\ Charles R Greathouse IV, Feb 03 2014

Python

A004090 = lambda n: sum(int(d)for d in str(A000045(n))) # M. F. Hasler, Jul 07 2025

Formula

a(n) = Fibonacci(n) - 9*Sum_{k>0} floor(Fibonacci(n)/10^k). - Hieronymus Fischer, Jun 25 2007

a(n) = A007953(A000045(n)). - Reinhard Zumkeller, Nov 17 2014

A010888(a(n)) = A030132(n) == a(n) (mod 9). - M. F. Hasler, Jul 07 2025