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.

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A168556 Smallest primes p = p(k) with (p(k)+p(k+1)+p(k+2))/15 an integer.

Original entry on oeis.org

3, 4691, 6397, 6911, 8893, 10181, 11833, 12113, 13063, 13267, 14251, 16661, 17851, 20563, 21341, 21881, 22229, 22877, 23339, 23633, 24859, 26561, 26591, 27337, 27361, 28151, 29531, 30029, 30211, 30881, 30983, 31271, 33961, 34439, 35227
Offset: 1

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Author

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Nov 29 2009

Keywords

Comments

(1) Denominator 15 = p(2)+ p(3)+ p(4) is sum of the first 3 consecutive odd primes.
(2) First three consecutive odd primes 3, 5, 7 are an exceptional case, because of p(k+1)=p(3)=5, no other prime has end digit 5.
(3) Four out of 19 possibilities for end digit permutations (which all "happen") of three consecutive primes p(k), p(k+1) and p(k+2) are permutations of (1,1,3), (1,7,7), (3,3,9) and (7,9,9), as sum of these digits has to be necessarily a multiple of 5.
(4) Note the "gap" for a(1) and a(2).

Examples

			(1) 3 + 5 + 7 = 15 * 1 => a(1)=3 for k=2.
(2) 4691 + 4703 + 4721 = 14115 = 15 * 941 => a(2)=4691 for k=634.
(3) 6397 + 6421 + 6427 = 19245 = 15 * 1283 => a(3)=6397 for k=834.
(4) 8893 + 8923 + 8929 = 26745 = 15 * 1783 => a(5)=8893 for k=1108.
(5) 23339 + 23357 + 23369 = 70065 = 15 * 4671 => a(19)=23339 for k=2603.
		

References

  • Leonard E. Dickson: History of the Theory of numbers, vol. I, Dover Publications 2005
  • Theo Kempermann, Zahlentheoretische Kostproben, Harri Deutsch, 2. aktualisierte Auflage 2005

Crossrefs

Cf. A000040 The prime numbers
Cf. A168069 a(n) is the index k of the smallest prime p(k) with (p(k) + p(k+1))/n an integer

Programs

  • Mathematica
    Select[Partition[Prime[Range[4000]],3,1],IntegerQ[Total[#]/15]&][[;;,1]] (* Harvey P. Dale, Jul 20 2025 *)
  • PARI
    lista(nn) = {vp = primes(nn); for (k = 1, nn - 2, if ((vp[k] + vp[k+1] + vp[k+2]) % 15 == 0, print1(vp[k], ", ")););} \\ Michel Marcus, Aug 27 2013
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