A077217 Prime(k) such that the prime power with largest exponent that divides the product P(k) of composite numbers between prime(k) and prime(k+1) is an odd number, i.e., if p^r and 2^s divide P(k) then r >= s, p is an odd prime.

2, 5, 17, 29, 41, 101, 107, 137, 149, 179, 197, 269, 281, 457, 461, 499, 521, 569, 593, 617, 641, 673, 727, 809, 821, 827, 857, 881, 1049, 1061, 1229, 1277, 1289, 1301, 1321, 1451, 1453, 1481, 1483, 1619, 1697, 1721, 1753, 1777, 1861, 1873, 1877, 1949, 1997, 2027

Offset: 1

Author

Amarnath Murthy, Nov 02 2002

Keywords

• nonn

Comments

In most cases a power of 2 has a larger exponent than any odd prime power.

Primes p = prime(k) such that A051903(A000265(A061214(k))) >= A007814(A061214(k)). - Amiram Eldar, Apr 01 2021

Examples

			5 is a member as 6 is divisible by 3^1 as well as by 2^1.
17 is a member as 18 is divisible by 3^2 but not by 2^2.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000265, A007814, A051903, A061214.

Programs

Mathematica

q[p_] := Module[{prod = Product[k, {k, p + 1, NextPrime[p] - 1}], e2}, e2 = IntegerExponent[prod, 2]; Max[FactorInteger[prod/2^e2][[;; , 2]]] >= e2]; Select[Range[2000], PrimeQ[#] && q[#] &] (* Amiram Eldar, Apr 01 2021 *)

PARI

f(p) = prod(k=p+1, nextprime(p+1)-1, k);\\ A061214
isok(p) = {my(prd = f(p), e = valuation(prd, 2), ofprd = prd/2^e); if (prd > 1, (ofprd == 1) || (e <= vecmax(factor(ofprd)[,2])));} \\ Michel Marcus, Apr 01 2021

Extensions

Wrong term removed and more terms added by Amiram Eldar, Apr 01 2021