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 1-3 of 3 results.

A112305 Let T(n) = A000073(n+1), n >= 1; a(n) = smallest k such that n divides T(k).

Original entry on oeis.org

1, 3, 7, 4, 14, 7, 5, 7, 9, 19, 8, 7, 6, 12, 52, 15, 28, 12, 18, 31, 12, 8, 29, 7, 30, 39, 9, 12, 77, 52, 14, 15, 35, 28, 21, 12, 19, 28, 39, 31, 35, 12, 82, 8, 52, 55, 29, 64, 15, 52, 124, 39, 33, 35, 14, 12, 103, 123, 64, 52, 68, 60, 12, 15, 52, 35, 100, 28, 117
Offset: 1

Views

Author

N. J. A. Sloane, Nov 30 2005

Keywords

Comments

Brenner proves that every prime divides some tribonacci number T(n). The Mathematica program computes similar sequences for any n-step Fibonacci sequence.

Examples

			T(1), T(2), T(3), T(4), ... are 1,1,2,4,7,13,24,...; a(3) = 7 because 3 first divides T(7) = A000073(8) = 24.

References

  • Ed Pegg, Jr., Posting to Sequence Fan mailing list, Nov 30, 2005

Links

Crossrefs

Cf. A000073.
Cf. A112312 (least k such that prime(n) divides T(k)).

Programs

  • Mathematica
    n=3; Table[a=Join[{1}, Table[0, {n-1}]]; k=0; While[k++; s=Mod[Plus@@a, i]; a=RotateLeft[a]; a[[n]]=s; s!=0]; k, {i, 100}] (* T. D. Noe *)

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 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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