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 A331544 Primes that begin a run of consecutive primes whose first differences are strictly increasing. 1
 2, 7, 13, 19, 31, 41, 59, 71, 83, 101, 109, 131, 139, 151, 163, 179, 193, 199, 223, 229, 241, 257, 269, 281, 311, 337, 349, 373, 383, 401, 421, 433, 443, 461, 487, 503, 523, 547, 563, 571, 593, 601, 613, 619, 641, 659, 677, 709, 727, 739, 757, 773, 797, 811, 823 (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 must keep increasing. Conjecture: Average length of each run = e, converging from below. Comment from Peter Woodward, Apr 22 2023: (Start) Heuristic justification: Consider that e = 2 + 1/2! + 1/3! + 1/4! + ... We always have at least two values in a run. The probability that there is a third value is 1/2!. The probability that there is a fourth value is 1/3! (because there are exactly 3! ways to sort three differences, and only one of these ways is in increasing order). And so on. The process is distorted by the possibility of equal increases. This sequence disallows equal increases in the runs, causing the convergence to e to approach from below. 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 nondecreasing", includes equal increases in its rows, thus it approaches e from above. Averaging the runs of the two sequences should negate the aberration, giving immediate convergence to e. However, while using a set of random numbers arranged in this fashion does immediately converge to e, with primes the convergence is slowed for unknown reasons, and the average run length remains slightly below e with extensive calculation (after 10^7 runs it reaches 2.712 vs. 2.718...). The probability that a randomly chosen integer n is prime is approximately 1/log(n) (by the prime number theorem). Thus, the expected gap between consecutive primes of size k is approximately (1/log(n)) * (1 - 1/log(n))^k - 1. Using the above formula, we can calculate the expected length of the k-th set as follows: E(length of k-th set) = Sum_{n>=m+1} ((1/log(n)) * (1 - 1/log(n))^k - 1) where m is the largest prime less than or equal to the starting prime of the k-th set. (End) LINKS Table of n, a(n) for n=1..55. Reddit blog, Found "e" in the primes (maybe), 2019. EXAMPLE The first run is 2, 3, 5, thus the first value is 2; the second run is 7, 11, thus the second value is 7; the third run is 13, 17, thus the third value is 13; the fourth run is 19, 23, 29, thus the fourth value is 19. PROG (PARI) lista(nn) = {my(m=2, d=0); forprime(p=2, nn, if(p-m>d, d=p-m, d=0; print1(p, ", ")); m=p); } \\ Jinyuan Wang, Jan 21 2020 CROSSREFS Cf. A309663. Sequence in context: A106011 A045375 A249419 * A237499 A231476 A138042 Adjacent sequences: A331541 A331542 A331543 * A331545 A331546 A331547 KEYWORD nonn AUTHOR Peter Woodward, Jan 19 2020 EXTENSIONS More terms from Jinyuan Wang, Jan 21 2020 STATUS approved

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Last modified August 11 19:17 EDT 2024. Contains 375073 sequences. (Running on oeis4.)