A246558 -id:A246558 - cp's OEIS frontend

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

N. J. A. Sloane

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

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

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

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

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

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

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

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

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

A076564 List of numbers k such that the k-th Fibonacci number (A000045) does not contain the digit zero.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 22, 23, 24, 26, 27, 28, 29, 31, 33, 35, 37, 39, 42, 43, 53, 54, 55, 56, 57, 58, 78, 80, 85, 87, 97, 125, 184

Offset: 1

Robert G. Wilson v, Oct 19 2002

Probably finite, and probably 184 is the last term. No further terms below 10000. - N. J. A. Sloane, Dec 13 2019

A. Altassan and F. Luca, On a curious property of F_{184}, Fib. Q., 57:4 (2019), 363-365.

Shyam Sunder Gupta, Can You Find no. 9

Cf. A000045, A246558.

Select[ Range[50000], Union[ IntegerDigits[ Fibonacci[ # ]]] [[1]] != 0 & ]
Select[Range[200],DigitCount[Fibonacci[#],10,0]==0&] (* Harvey P. Dale, May 15 2019 *)

isok(k) = #select(x->(x==0), digits(fibonacci(k))) == 0;
select(isok, [1..1000]) \\ Michel Marcus, Feb 11 2025

A261598 Product of sexagesimal digits of Fibonacci numbers in base-60 representation.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 29, 48, 159, 102, 100, 432, 962, 172, 369, 2340, 156, 2420, 14763, 29952, 25000, 18447, 67716, 22848, 63800, 21420, 217854, 2250, 35264, 34944, 99330, 14364, 1300500, 0, 8726016, 2303910, 544272, 9728000, 5615610, 8419950

Offset: 0

Reinhard Zumkeller, Sep 09 2015

a(n) is the product of the terms in the n-th row of table A261575.

Conjecture: a(n) = 0 for n > 3329 (empirically checked up to 36000).

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Sexagesimal
Wikipedia, Sexagesimal

Cf. A261575, A000045, A246558, A261587.

Haskell
```
a261598 = product . a261575_row
```

a:= n-> mul(i, i=convert((<<0|1>, <1|1>>^n)[1, 2], base, 60)):
seq(a(n), n=0..44);  # Alois P. Heinz, Jan 22 2022

Apply[Times, IntegerDigits[Fibonacci[Range[0, 50]], 60], {1}] (* Paolo Xausa, Feb 19 2024 *)

a(n) = if (n, vecprod(digits(fibonacci(n), 60)), 0); \\ Michel Marcus, Jan 22 2022

A259773 Product of the digits of the n-th Lucas number.

Original entry on oeis.org

2, 1, 3, 4, 7, 1, 8, 18, 28, 42, 6, 81, 12, 10, 96, 72, 0, 105, 1960, 972, 70, 1344, 0, 0, 0, 1764, 672, 0, 0, 1440, 0, 0, 0, 24192, 0, 0, 34560, 0, 0, 1536, 43008, 0, 0, 0, 0, 0, 0, 41803776, 0, 0, 120960, 3024000, 0, 120960, 0, 0, 0, 6531840, 0, 440899200

Offset: 0

Vincenzo Librandi, Jul 05 2015

Probably, the last nonzero term is a(401) = 2^71*3^45*5^9*7^4. - Giovanni Resta, Jul 14 2015

			9349 is the 19th Lucas number; its digit product is 972, therefore a(19) = 972.
15127 is the 20th Lucas number; its digit product is 70, therefore a(20) = 70.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A000032, A007954, A139374, A246558.

Magma
```
[&*Intseq(Lucas(n)): n in [0..80]];
```

Table[Times@@IntegerDigits[LucasL[n]], {n, 0, 100}]

a(n) = A007954(A000032(n)). - Michel Marcus, Jul 05 2015

A373049 Integers k such that the product of the nonzero digits of the k-th Fibonacci number (A000045) is a perfect power.

Original entry on oeis.org

0, 1, 2, 6, 10, 12, 19, 21, 22, 27, 31, 46, 49, 50, 73, 79, 85, 102, 108, 116, 117, 160, 161, 179, 181, 237, 247, 250, 257, 281, 285, 302, 309, 351, 354, 359, 373, 376, 377, 380, 415, 419, 434, 449, 470, 479, 497, 498, 515, 521, 543, 565, 569, 584, 590, 599, 602, 609, 615, 665, 696

Offset: 1

Gonzalo Martínez, May 20 2024

For most of the terms in this list, the product of their nonzero digits is a perfect square.

Conjecture: this sequence has infinitely many terms. Since the product of the nonzero digits of Fibonacci(k) is of the form 2^a * 3^b * 5^c * 7^d, a sufficient condition for Fibonacci(k) to belong to the sequence is that a, b, c and d are all even.

			21 is a term, because Fibonacci(21) = 10946 and the product of its nonzero digits is 1*9*4*6 = 6^3.
46 is a term, because Fibonacci(46) = 1836311903 and the product of its nonzero digits is 1*8*3*6*3*1*1*9*3 = 108^2.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000045, A246558, A076564, A370071, A373116.

filter:= proc(n) local L,t,q2,q3,q5,q7;
  L:=convert(combinat:-fibonacci(n),base,10);
  q2:= 0: q3:= 0: q5:= 0: q7:= 0:
  for t in L do
    if t = 2 then q2:= q2+1
    elif t = 3 then q3:= q3+1
    elif t = 4 then q2:= q2+2
    elif t = 5 then q5:= q5+1
    elif t = 6 then q2:= q2+1; q3:= q3+1
    elif t = 7 then q7:= q7+1
    elif t = 8 then q2:= q2+3
    elif t = 9 then q3:= q3+2
    fi
  od;
  igcd(q2,q3,q5,q7) > 1
end proc:
filter(0):= true: filter(1):= true: filter(2):= true:
select(filter, [$0..1000]); # Robert Israel, May 26 2025

powQ[n_] := n == 1 || GCD @@ FactorInteger[n][[;; , 2]] > 1; Select[Range[0, 700], powQ[Times @@ Select[IntegerDigits[Fibonacci[#]], #1 > 0 &]] &] (* Amiram Eldar, May 25 2024 *)

isok(k) = my(x=vecprod(select(x->(x>0), digits(fibonacci(k))))); (x==1) || ispower(x); \\ Michel Marcus, May 20 2024

More terms from Michel Marcus, May 20 2024

A373116 Fibonacci numbers whose digits product is a positive perfect power (A001597).

Original entry on oeis.org

1, 8, 55, 144, 4181, 17711, 196418, 1346269, 259695496911122585

Offset: 1

Gonzalo Martínez, May 25 2024

Since the product of the digits of Fibonacci(k) is required to be positive, Fibonacci(k) does not have zero as a digit. For this reason this list is probably finite, since it is conjectured that there are only finitely many Fibonacci numbers that do not contain the digit zero (see A076564). If the conjecture is true, the largest number possessing the property would be Fibonacci(85) = 259695496911122585 whose digit product is 194400^2.

Unlike A373049, here the product uses all the digits of Fibonacci(k).

			196418 is a term, because Fibonacci(27) = 196418 and the product of its digits is 1*9*6*4*1*8 = 12^3.

Cf. A000045, A001597, A076564, A246558, A370071, A373049.

powQ[n_] := n == 1 || GCD @@ FactorInteger[n][[;; , 2]] > 1; Select[Fibonacci[Range[2, 100]], powQ[Times @@ IntegerDigits[#]] &] (* Amiram Eldar, May 25 2024 *)

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A004090 Sum of digits of Fibonacci numbers.

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A076564 List of numbers k such that the k-th Fibonacci number (A000045) does not contain the digit zero.

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A261598 Product of sexagesimal digits of Fibonacci numbers in base-60 representation.

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A259773 Product of the digits of the n-th Lucas number.

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A373049 Integers k such that the product of the nonzero digits of the k-th Fibonacci number (A000045) is a perfect power.

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A373116 Fibonacci numbers whose digits product is a positive perfect power (A001597).

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