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A210529 a(n) is 1 or the smallest prime that makes |a(n)^2 - prime(n)^2| divisible by all primes smaller than sqrt(prime(n)). 3

%I #39 Oct 22 2023 02:24:47

%S 1,1,1,1,1,1,1,1,1,1,1,7,1,7,7,17,11,19,17,1,17,19,13,19,13,11,23,23,

%T 11,13,83,89,17,29,61,179,283,233,13,1213,1999,2029,719,1523,2927,

%U 2089,3221,5657,6857,541,1223,421,1319,3709,653,1277,3371,821,563,1721

%N a(n) is 1 or the smallest prime that makes |a(n)^2 - prime(n)^2| divisible by all primes smaller than sqrt(prime(n)).

%C Suppose a = a(n) + prime(n), b = |a(n) - prime(n)|, when a(n) > prime(n), prime(n) = (a - b)/2, and gcd(a,b) = 2. When a*b = |a(n)^2 - prime(n)^2|, (a - b)/2 is a primality proof of prime(n) since the list of prime factors of a and b contains all prime numbers smaller than sqrt(prime(n)) and gcd(a,b) = 2. - corrected by _Eric M. Schmidt_, Feb 02 2013

%C Conjecture: a(n) is defined for all positive integers n.

%H Lei Zhou and Charles R Greathouse IV, <a href="/A210529/b210529.txt">Table of n, a(n) for n = 1..270</a> (first 165 terms from Zhou)

%H T. Agoh, A. Granville, and P. Erdős, <a href="http://www.jstor.org/stable/2974476">Primes at a (somewhat lengthy) glance</a>, American Mathematical Monthly, 104(10):943-945, December 1997.

%H R. K. Guy, C. B. Lacampagne and J. L. Selfridge, <a href="http://dx.doi.org/10.1090/S0025-5718-1987-0866108-3">Primes at a glance</a>, Math. Comp. 48 (1987), 183-202.

%e n = 1, prime(1) = 2, the set of prime numbers smaller than sqrt(2) is {}, so a(1) = 1.

%e n = 11, prime(11) = 31, the set of prime numbers smaller than sqrt(31) is {2, 3, 5}, 961 - 1 = 960 is divisible by 2*3*5, so a(11) = 1.

%e n = 12, prime(12) = 37, the set of prime numbers smaller than sqrt(37) is {2, 3, 5}, 37^2 - 7^2 = 1320 is divisible by 2*3*5, so a(12) = 7.

%t Table[p = Prime[n]; t = Product[Prime[k], {k, 1, PrimePi[NextPrime[Floor[Sqrt[p]] + 1, -1]]}]; p1 = 1; While[r = Abs[p^2 - p1^2]; (r == 0) || (Mod[r, t] != 0), p1 = NextPrime[p1]]; p1, {n, 1, 60}]

%o (PARI) primorial(n)=vecprod(primes(n));

%o a(n)=if (n<=3, 1, my(p=prime(n),P=primorial(sqrtint(p)),p2=p^2);if(p2%P==1, return(1)); forprime(q=2,,if((q^2-p^2)%P==0&&p!=q,return(q)))) \\ _Charles R Greathouse IV_, Mar 01 2014; edited by _Michel Marcus_, Oct 22 2023

%K nonn,hard,nice

%O 1,12

%A _Lei Zhou_, Jan 27 2013

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