A215199 Smallest number k such that k and k+1 are both of the form p*q^n where p and q are distinct primes.

14, 44, 135, 2511, 8991, 29888, 916352, 12393728, 155161088, 2200933376, 6856828928, 689278976, 481758175232, 3684603215871, 35419114668032, 2035980763136, 174123685117952, 9399153082499072, 19047348965998592, 203368956137832447, 24217192574746623, 2503092614937444351

Offset: 1

Author

Michel Lagneau, Aug 05 2012

Keywords

• nonn

Comments

a(15) <= 35419114668032. - Donovan Johnson, Aug 22 2012

If k is a term such that k = p*q^n and k+1 = r*s^n, where p,q,r,s are primes, then clearly q != s. Conjecture: q and s are either 2 or 3 for all terms. - Chai Wah Wu, Mar 10 2019

Since q^n and s^n are coprime, the Chinese Remainder Theorem can be used to find candidate terms to test, i.e., numbers k such that k+1 == 0 (mod s^n) and k+1 == 1 (mod q^n) (see Python code). - Chai Wah Wu, Mar 12 2019

From David A. Corneth, Mar 13 2019: (Start)

Conjecture: Let 1 <= D < 2^n be the denominator of N/D of (3/2)^n. Without loss of generality, if the conjecture above holds that (q, s) = (2, 3) then r = D + k*2^n for some n.

Example: for n = 100, we have the continued fraction of (3/2)^100 to be 406561177535215237, 2, 1, 1, 14, 9, 1, 1, 2, 2, 1, 4, 1, 2, 6, 5, 1, 195, 3, 26, 39, 6, 1, 1, 1, 2, 7, 1, 4, 2, 1, 11, 1, 25, 6, 1, 4, 3, 2, 112, 1, 2, 1, 3, 1, 3, 4, 8, 1, 1, 12, 2, 1, 3, 2, 2 from which we compute D = 519502503658624787456021964081. We find r = 1100840223501761745286594404230449 = D + 868 * 2^100 giving a(100) + 1 = r*3^100. (End)

Examples

			a(3) = 135 because 135 = 5*3^3 and 136 = 17*2^3;
a(4) = 2511 because 2511 = 31*3^4 and 2512 = 157*2^4.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..1279 (terms 25..32 from David A. Corneth)

Crossrefs

Cf. A074172, A215173, A215197, A215198.

Programs

Maple

psig := proc(n)
    local s,p ;
    s := [] ;
    for p in ifactors(n)[2] do
        s := [op(s),op(2,p)] ;
    end do:
    sort(s) ;
end proc:
A215199 := proc(n)
    local slim,smi,sma,ca,qi,q,p,k ;
    for slim from 0 do
        smi := slim*1000 ;
        sma := (slim+1)*1000 ;
        ca := sma ;
        q := 2 ;
        for qi from 1 do
            p := nextprime(floor(smi/q^n)-1) ;
            while p*q^n < sma do
                if p <> q then
                    k := p*q^n ;
                    if psig(k+1) = [1,n] then
                        ca := min(ca,k) ;
                    end if;
                end if;
                p := nextprime(p) ;
            end do:
            if q^n >= sma then
                break;
            end if;
            q := nextprime(q) ;
        end do:
        if ca < sma then
            return ca ;
        end if;
    end do:
end proc:
for n from 1 do
    print(A215199(n)) ;
end do; # R. J. Mathar, Aug 07 2012

Python

from sympy import isprime, nextprime
from sympy.ntheory.modular import crt
def A215199(n):
    l = len(str(3**n))-1
    l10, result = 10**l, 2*10**l
    while result >= 2*l10:
        l += 1
        l102, result = l10, 20*l10
        l10 *= 10
        q, qn = 2, 2**n
        while qn <= l10:
            s, sn = 2, 2**n
            while sn <= l10:
                if s != q:
                    a, b = crt([qn,sn],[0,1])
                    if a <= l102:
                        a = b*(l102//b) + a
                    while a < l10:
                        p, t = a//qn, (a-1)//sn
                        if p != q and t != s and isprime(p) and isprime(t):
                            result = min(result,a-1)
                        a += b
                s = nextprime(s)
                sn = s**n
            q = nextprime(q)
            qn = q**n
    return result # Chai Wah Wu, Mar 12 2019

Extensions

a(10)-a(14) from Donovan Johnson, Aug 22 2012

a(15)-a(17) from Chai Wah Wu, Mar 09 2019

a(18)-a(22) from Chai Wah Wu, Mar 10 2019