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 A309663 Primes that begin a run of consecutive primes whose first differences are nondecreasing. 1
 2, 13, 19, 31, 41, 61, 71, 83, 101, 109, 131, 139, 151, 167, 181, 193, 199, 227, 241, 257, 271, 281, 311, 337, 349, 373, 383, 401, 421, 433, 443, 461, 487, 503, 523, 547, 563, 571, 593, 601, 617, 641, 661, 677, 709, 727, 743, 757, 773, 797, 811, 823, 829, 857 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Arrange primes in rows where the value of the increase between consecutive primes cannot shrink. Conjecture: Average length of each run = e, converging from above. Heuristic justification: Consider that e = 2 + 1/2! + 1/3! + 1/4! + ... We always have at least two values in a run. The odds of there being a third value = 1/2!. The odds of there being a fourth value = 1/3! (because exactly 3! ways to sort three differences, and only one of these ways is in increasing order). etc... The process is aberrated by the possibility of equal increases. This sequence allows equal increases in the runs, causing the convergence to e to approach from above. However, as the scale increases, these equal increases occur less frequently and their effect approaches zero. The sister sequence, "Primes that begin a run of consecutive primes whose first differences are strictly increasing", disallows equal increases in its rows, thus it approaches e from below. Averaging the runs of the two sequences negates the aberration, giving immediate convergence to e. LINKS Reddit blog, Found "e" in the primes (maybe), 2019. EXAMPLE The first run is 2, 3, 5, 7, 11, thus the first value is 2; The second run is 13, 17, thus the second value is 13; The third run is 19, 23, 29, thus the third value is 19; The fourth run is 31, 37, thus the fourth value is 31. PROG (PARI) lista(nn) = {my(d=m=2); forprime(p=2, nn, if(p-m

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Last modified May 20 09:27 EDT 2022. Contains 353871 sequences. (Running on oeis4.)