A053028 -id:A053028 - cp's OEIS frontend

A112269 Least index k such that the n-th prime properly divides the k-th tribonacci number.

5, 8, 15, 13, 9, 19, 29, 19, 30, 78, 15, 20, 36, 83, 30, 34, 65, 69, 101, 133, 32, 19, 271, 110, 20, 187, 14, 185, 106, 173, 587, 80, 12, 35, 46, 224, 72, 38, 42, 315, 101, 26, 73, 172, 383, 27, 84, 362, 35, 250, 37, 29, 507, 305, 55, 38, 178, 332, 62, 537, 778, 459, 31

Offset: 1

Jonathan Vos Post, Nov 29 2005

The tribonacci numbers are indexed so that trib(0) = trib(1) = 0, trib(2) = 1, for n>2: trib(n) = trib(n-1) + trib(n-2) + trib(n-3). "Properly divides" means that this sequence is "Least index k such that the n-th prime divides the k-th tribonacci number not itself the n-th prime".

Since the tribonacci sequence is periodic mod p for any prime p, the sequence is well-defined. - T. D. Noe, Dec 01 2005

			a(1) = 5 because prime(1) = 2 and, although tribonacci(4) = 2, the first tribonacci number properly divided by 2 is tribonacci(5) = 4.
a(2) = 8 because prime(2) = 3 and tribonacci(8) = 24 = 2^3 * 3.
a(3) = 15 because prime(3) = 5 and tribonacci(15) = 1705 = 5 * 11 * 31.
a(4) = 13 because prime(4) = 7 and tribonacci(13) = 504 = 2^3 * 3^2 * 7.
a(5) = 9 because prime(5) = 11 and tribonacci(9) = 44 = 2^2 * 11.
a(6) = 19 because prime(6) = 13 and tribonacci(19) = 19513 = 13 * 19 * 79.
a(7) = 29 because prime(7) = 17 and tribonacci(29) = 646064 = 2^4 * 7 * 17 * 19 * 239.

Cf. A000040, A000045, A000073, A000204, A001644, A053028, A106299, A112312.

a[0] = a[1] = 0; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; f[n_] := Block[{k = 1, p = Prime[n]}, While[ Mod[a[k], p] != 0 || p >= a[k], k++ ]; k]; Array[f, 63] (* Robert G. Wilson v *)

a(n) = minimum k such that prime(n) | A000073(k) and A000073(k) > prime(n). a(n) = minimum k such that A000040(n) | A000073(k) and A000073(k) > A000040(n).

Corrected and extended by Robert G. Wilson v, Nov 30 2005

A116515 a(n) = the period of the Fibonacci numbers modulo p divided by the smallest m such that p divides Fibonacci(m), where p is the n-th prime.

Original entry on oeis.org

1, 2, 4, 2, 1, 4, 4, 1, 2, 1, 1, 4, 2, 2, 2, 4, 1, 4, 2, 1, 4, 1, 2, 4, 4, 1, 2, 2, 4, 4, 2, 1, 4, 1, 4, 1, 4, 2, 2, 4, 1, 1, 1, 4, 4, 1, 1, 2, 2, 1, 4, 1, 2, 1, 4, 2, 4, 1, 4, 2, 2, 4, 2, 1, 4, 4, 1, 4, 2, 1, 4, 1, 2, 4, 1, 2, 4, 4, 2, 2, 1, 4, 1, 4, 1, 2, 2, 4, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 4, 2, 2, 1

Offset: 1

Nick Krempel, Mar 24 2006

Conditions on p_n mod 4 and mod 5 restrict possible values of a(n). The unknown (?) case is p = 1 mod 4 and (5|p) = 1, equivalently, p = 1 or 9 mod 20, where {1, 2, 4} all occur.

Number of zeros in fundamental period of Fibonacci numbers mod prime(n). [From T. D. Noe, Jan 14 2009]

			a(4) = 2, as 7 is the 4th prime, the Fibonacci numbers mod 7 have period 16, the first Fibonacci number divisible by 7 is F(8) = 21 = 3*7 and 16 / 8 = 2.
One period of the Fibonacci numbers mod 7 is 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, which has two zeros. Hence a(4)=2. [From _T. D. Noe_, Jan 14 2009]

Cf. A060305, A001602.

Cf. A112860, A053027, A053028 (primes producing 1, 2 and 4 zeros) [From T. D. Noe, Jan 14 2009]

a(n) = A060305(n) / A001602(n). a(n) is always one of {1, 2, 4}.

a(n) = A001176(prime(n)) [From T. D. Noe, Jan 14 2009]

cp's OEIS Frontend

A112269 Least index k such that the n-th prime properly divides the k-th tribonacci number.

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