A139374 -id:A139374 - cp's OEIS frontend

A337449 The numbers k for which Lucas(k) are Niven numbers.

0, 1, 2, 3, 4, 6, 12, 18, 56, 81, 130, 225, 396, 637, 854, 2034, 4059, 4095, 5985, 7650, 21105, 31059, 41998, 46860, 83106, 114129, 120555, 150705, 201285, 287937, 338265, 359757, 475839, 512194, 583825, 606594, 627102, 717025, 877305, 922095, 991590, 1076355

Offset: 1

Marius A. Burtea, Sep 14 2020

For a(6) = 6, Lucas(6) = 18 and 18/digsum(18) = 2 is a prime number, so Lucas(6) is a Moran number (A001101).

For a(9) = 56, Lucas(56) = 505019158607 and 505019158607/digsum(505019158607) = 10745088481 is a prime number, so Lucas(56) is a Moran number.

			Lucas(0) = 2 = A005349(2), so 0 is a term.
Lucas(1) = 1 = A005349(1), so 1 is a term.
Lucas(6) = 12 = A005349(11), so 6 is a term.
Lucas(12) = 322 = A005349(90), so 12 is a term.
Lucas(18) = 5778 = A005349(1013), so 18 is a term.

Chai Wah Wu, Table of n, a(n) for n = 1..57

Cf. A000032, A139374, A001101, A005349, A337448.

niven:=func; [k:k in [0..70000]|niven(Lucas(k))];

nivenQ[n_] := Divisible[n, Plus @@ IntegerDigits[n]]; Select[Range[6000], nivenQ[LucasL[#]] &] (* Amiram Eldar, Sep 15 2020 *)

isok(k) = my(l=real((2+quadgen(5))*quadgen(5)^k)); (l % sumdigits(l)) == 0; \\ Michel Marcus, Sep 15 2020

A337449_list, k, p, q = [], 0, 2, 1
while k < 10**6:
    if p % sum(int(d) for d in str(p)) == 0:
        A337449_list.append(k)
    k += 1
    p, q = q, p+q # Chai Wah Wu, Sep 17 2020

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

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

Original entry on oeis.org

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

Michel Lagneau, Jul 08 2014

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

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

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

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

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 *)

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A337449 The numbers k for which Lucas(k) are Niven numbers.

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

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

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