OFFSET

1,1

COMMENTS

a(n) = A001248(n+1) - A001248(n) = A000040(n+1)^2 - A000040(n)^2 = (A000040(n+1) - A000040(n))*(A000040(n+1) + A000040(n)) = A001223(n)*A001043(n); together with A069484(n) and A069486(n) a Pythagorean triangle is formed with area = A069487(n).

For n>2: A078701(a(n)) = 3.

Except for the first two terms, these numbers are divisible by 24. Let p, q be consecutive primes. Then p > 3 = 3k+-1 and q = 3m+-1 and (3k+-1)^2 - (3m+-1)^2 is divisible by 3. Similarly, p = 4k+-1 and q=4m+-1 and (4k+-1)^2 - (4m+-1)^2 is divisible by 8. So 8 and 3 divide q^2 - p^2 => 24 divides q^2 - p^2. - Cino Hilliard, May 28 2009

Repetition of a(n) values occurs with decreasing frequency but increasing tallies (i.e., number of repetitions of a given value).

Tally = 2, first a(n) value is 72, with first n=4, prime=7.

Tally = 3, first a(n) value is 1848, with first n=36, prime=151.

Tally = 4, first a(n) value is 4920, with first n=46, prime=199.

Tally = 5, first a(n) value is 187117320, with first n=224752, prime 3118607.

Three a(n) values have a tally = 5, and none with tally > 5 for n<10,000,000. Note: Tallies for a given a(n) value are "confirmed" (i.e., not to be greater) only after examining a(n) values for all p(n) <= r/4-1, where r is the a(n) value in question, because twin primes provide the last chance for adding to the tally of any a(n) value. Tallies for the four a(n) values above are "confirmed" and all of them rely on twin primes for their last repetition. Thus r/4 +-1 is prime for the above four cases. However this is not true for all a(n) values that repeat.

Conjecture: The sum of prime factors with repetition (sopfr) applied to a(n), A001414(a(n)), covers all integers covered by sopfr, except 2,3,4,6,7,10,13,15. See A001414 for the sopfr sequence, which does not cover 0 and 1. - Richard R. Forberg, Feb 07 2015

Conjecture: There is no upper bound on the number of repetitions (i.e., size of a tally) that will occur for some a(n) values, because the number of possible ways of producing a value of a(n) grows with increasing n, despite decreasing prime density. This happens because there is increasing range in the size of prime gaps which increases the range of primes that can produce the same a(n) value much faster than the decrease in prime density which is decelerating with larger n. - Richard R. Forberg, Feb 17 2015

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Eric Weisstein's World of Mathematics, Pythagorean Triple

EXAMPLE

MATHEMATICA

Table[Prime[n+1]^2 - Prime[n]^2, {n, 1, 40}] (* Vladimir Joseph Stephan Orlovsky, Mar 01 2009; modified by G. C. Greubel, May 19 2019 *)

#[[2]]-#[[1]]&/@Partition[Prime[Range[60]]^2, 2, 1] (* Harvey P. Dale, Jan 13 2011 *)

Differences[Prime[Range[100]]^2](* Waldemar Puszkarz, Feb 09 2015 *)

PROG

(Haskell)

a069482 n = a069482_list !! (n-1)

a069482_list = zipWith (-) (tail a001248_list) a001248_list

-- Reinhard Zumkeller, Jun 08 2015

(PARI) {a(n) = prime(n+1)^2 - prime(n)^2}; \\ G. C. Greubel, May 19 2019

(Magma) [NthPrime(n+1)^2 - NthPrime(n)^2: n in [1..40]]; // G. C. Greubel, May 19 2019

(Sage) [nth_prime(n+1)^2 - nth_prime(n)^2 for n in (1..40)] # G. C. Greubel, May 19 2019

(Python)

from sympy import prime, primerange

def aupton(terms):

p = list(primerange(1, prime(terms+1)+1))

return [p[n+1]**2-p[n]**2 for n in range(terms)]

print(aupton(50)) # Michael S. Branicky, May 16 2021

CROSSREFS

KEYWORD

nonn

AUTHOR

Reinhard Zumkeller, Mar 29 2002, Aug 05 2007

STATUS

approved