cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A112312 Least index k such that the n-th prime divides the k-th tribonacci number.

Original entry on oeis.org

4, 8, 15, 6, 9, 7, 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, 11, 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, 124
Offset: 1

Views

Author

Jonathan Vos Post, Nov 29 2005

Keywords

Comments

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). See A112618 for another version.
Brenner proves that every prime divides some tribonacci number T(n). For the similar 3-step Lucas sequence A001644, there are primes (A106299) that do not divide any term.

Examples

			a(1) = 4 because prime(1) = 2 and tribonacci( 4) = 2.
a(2) = 8 because prime(2) = 3 and tribonacci( 8) = 24 = 3 * 2^3.
a(3) = 15 because prime(3) = 5 and tribonacci(15) = 1705 = 5 *(11 * 31).
a(4) = 6 because prime(4) = 7 and tribonacci( 6) = 7.
a(5) = 9 because prime(5) = 11 and tribonacci( 9) = 44 = 11 * 4.
a(6) = 7 because prime(6) = 13 and tribonacci( 7) = 13.
a(7) = 29 because prime(7) = 17 and tribonacci(29) = 8646064 = 17 *(2^4 * 7 * 19 * 239).

Links

Crossrefs

Cf. A000040, A000045, A000073, A000204, A001644, A053028, A106299, A112312.
Cf. also A112618 = this sequence minus 1.

Programs

  • Mathematica
    a[0] = a[1] = 0; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; f[n_] := Module[{k = 2, p = Prime[n]}, While[Mod[a[k], p] != 0, k++ ]; k]; Array[f, 64] (* Robert G. Wilson v *)

Formula

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).

Extensions

Corrected and extended by Robert G. Wilson v, Dec 01 2005

A106300 Primes that do not divide any term of the Lucas 4-step sequence A073817.

Original entry on oeis.org

2789, 3847, 4451, 4751, 5431, 6203, 8317, 9533, 9629, 9907, 10093, 11839, 13903, 13907, 14207, 15823, 16319, 16759, 19543, 20939, 21379, 21859, 25303, 26683, 29483, 30871, 31267, 31699, 32003, 32771, 33967, 34963, 36229, 37061, 39983
Offset: 1

Views

Author

T. D. Noe, May 02 2005

Keywords

Comments

If a prime p divides a term a(k) of this sequence, then k must be less than the period of the sequence mod p. Hence these primes are found by computing A073817(k) mod p for increasing k and stopping when either A073817(k) mod p = 0 or the end of the period is reached. Interestingly, for all of these primes, the period of the sequence A073817(k) mod p appears to be (p-1)/d, where d is a small integer.

Links

Crossrefs

Cf. A053028 (primes not dividing any Lucas number), A106299 (primes not dividing any Lucas 3-step number), A106301 (primes not dividing any Lucas 5-step number).

Programs

  • Mathematica
    n=4; lst={}; Table[p=Prime[i]; a=Join[Table[ -1, {n-1}], {n}]; a=Mod[a, p]; a0=a; While[s=Mod[Plus@@a, p]; a=RotateLeft[a]; a[[n]]=s; !(a==a0 || s==0)]; If[s>0, AppendTo[lst, p]], {i, 10000}]; lst

A106301 Primes that do not divide any term of the Lucas 5-step sequence A074048.

Original entry on oeis.org

2, 691, 3163, 4259, 5419, 6637, 6733, 14923, 25111, 27947, 29339, 34123, 34421, 34757, 42859, 55207, 57529, 59693, 61643, 68897, 70249, 75991, 82763, 83177, 85607, 86441, 87103, 93169, 93283, 98573, 106121, 106433, 114847, 129589, 132313
Offset: 1

Views

Author

T. D. Noe, May 02 2005

Keywords

Comments

If a prime p divides a term a(k) of this sequence, then k must be less than the period of the sequence mod p. Hence these primes are found by computing A074048(k) mod p for increasing k and stopping when either A074048(k) mod p = 0 or the end of the period is reached. Interestingly, for all of these primes, the period of the sequence A074048(k) mod p appears to be (p-1)/d, where d is a small integer.

Links

Crossrefs

Cf. A053028 (primes not dividing any Lucas number), A106299 (primes not dividing any Lucas 3-step number), A106300 (primes not dividing any Lucas 4-step number).

Programs

  • Mathematica
    n=5; lst={}; Table[p=Prime[i]; a=Join[Table[ -1, {n-1}], {n}]; a=Mod[a, p]; a0=a; While[s=Mod[Plus@@a, p]; a=RotateLeft[a]; a[[n]]=s; !(a==a0 || s==0)]; If[s>0, AppendTo[lst, p]], {i, 10000}]; lst

A112618 Let T(n) = A000073(n+1), n >= 1; a(n) = smallest k such that prime(n) divides T(k).

Original entry on oeis.org

3, 7, 14, 5, 8, 6, 28, 18, 29, 77, 14, 19, 35, 82, 29, 33, 64, 68, 100, 132, 31, 18, 270, 109, 19, 186, 13, 184, 105, 172, 586, 79, 11, 34, 10, 223, 71, 37, 41, 314, 100, 25, 72, 171, 382, 26, 83, 361, 34, 249, 36, 28, 506, 304, 54, 37, 177, 331, 61, 536, 777, 458, 30, 123
Offset: 1

Views

Author

T. D. Noe, Dec 05 2005

Keywords

Comments

Brenner proves that every prime divides some tribonacci number T(n). For the similar 3-step Lucas sequence A001644, there are primes (A106299) that do not divide any term.

Examples

			Sequence T(n) starts 1,1,2,4,7,13,24,44. For the primes 2,3,7,11,13, it is easy to see that a(1)=3, a(2)=7, a(4)=5, a(5)=8, a(6)=6.

Links

Crossrefs

Equals A112312(n)-1.

Programs

  • Mathematica
    a[0]=0; a[1]=a[2]=1; a[n_]:=a[n]=a[n-1]+a[n-2]+a[n-3]; f[n_]:= Module[{k=2, p=Prime[n]}, While[Mod[a[k], p] != 0, k++ ]; k]; Array[f, 64] (* Robert G. Wilson v *)

Formula

a(n) = A112305(prime(n)).

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

Original entry on oeis.org

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

Views

Author

Jonathan Vos Post, Nov 29 2005

Keywords

Comments

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

Examples

			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.

Crossrefs

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

Programs

  • Mathematica
    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 *)

Formula

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).

Extensions

Corrected and extended by Robert G. Wilson v, Nov 30 2005
Showing 1-5 of 5 results.

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