

A024702


a(n) = (prime(n)^2  1)/24.


28



1, 2, 5, 7, 12, 15, 22, 35, 40, 57, 70, 77, 92, 117, 145, 155, 187, 210, 222, 260, 287, 330, 392, 425, 442, 477, 495, 532, 672, 715, 782, 805, 925, 950, 1027, 1107, 1162, 1247, 1335, 1365, 1520, 1552, 1617, 1650, 1855, 2072, 2147, 2185, 2262, 2380, 2420, 2625, 2752, 2882, 3015
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OFFSET

3,2


COMMENTS

Note that p^2  1 is always divisible by 24 since p == 1 or 2 (mod 3), so p^2 == 1 (mod 3) and p == 1, 3, 5, or 7 (mod 8) so p^2 == 1 (mod 8).  Michael B. Porter, Sep 02 2016
For n > 3 and m > 1, a(n) = A000330(m)/(2*m + 1), where 2*m + 1 = prime(n). For example, for m = 8, 2*m + 1 = 17 = prime(7), A000330(8) = 204, 204/17 = 12 = a(7).  Richard R. Forberg, Aug 20 2013
For primes => 5, a(n) == 0 or 2 (mod 5).  Richard R. Forberg, Aug 28 2013
The only primes in this sequence are 2, 5 and 7 (checked up to n = 10^7). The set of prime factors, however, appears to include all primes.  Richard R. Forberg, Feb 28 2015
Subsequence of generalized pentagonal numbers (cf. A001318): a(n) = k_n*(3*k_n  1)/2, for k_n in {1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 9, 10, 10, ...} = A024699(n2)*((A000040(n) mod 6)  3)/2, n >= 3.  Daniel Forgues, Aug 02 2016
The only primes in this sequence are indeed 2, 5 and 7. For a prime p >= 5, if both p + 1 and p  1 contains a prime factor > 3, then (p^2  1)/24 = (p + 1)*(p  1)/24 contains at least 2 prime factors, so at least one of p + 1 and p  1 is 3smooth. Let's call it s. Also, If (p^2  1)/24 is a prime, then A001222(p^21) = 5. Since that A001222(p+1) and A001222(p1) are both at least 2, A001222(s) <= 5  2 = 3. From these we can see the only possible cases are p = 7, 11 and 13.  Jianing Song, Dec 28 2018


LINKS

Charles R Greathouse IV, Table of n, a(n) for n = 3..10000
Brady Haran and Matt Parker, Squaring Primes, Numberphile video (2018)


FORMULA

a(n) = (A000040(n)^2  1)/24 = (A001248(n)  1)/24.  Omar E. Pol, Dec 07 2011
a(n) = A005097(n1)*A006254(n1)/6.  Bruno Berselli, Dec 08 2011
a(n) = A084920(n)/24.  R. J. Mathar, Aug 23 2013


EXAMPLE

For n = 6, the 6th prime is 13, so a(6) = (13^2  1)/24 = 168/24 = 7.


MAPLE

A024702:=n>(ithprime(n)^21)/24: seq(A024702(n), n=3..70); # Wesley Ivan Hurt, Mar 01 2015


MATHEMATICA

(Prime[Range[3, 100]]^21)/24 (* Vladimir Joseph Stephan Orlovsky, Mar 15 2011 *)


PROG

(PARI) a(n)=prime(n)^2\24 \\ Charles R Greathouse IV, May 30 2013
(PARI) is(n)=my(k); issquare(24*n+1, &k)&&isprime(k) \\ Charles R Greathouse IV, May 31 2013


CROSSREFS

Subsequence of generalized pentagonal numbers A001318.
Cf. A075888.
Sequence in context: A088822 A080182 A001318 * A226084 A294861 A161664
Adjacent sequences: A024699 A024700 A024701 * A024703 A024704 A024705


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Dec 11 1999


STATUS

approved



