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A080365
Composite numbers k whose smallest and largest prime factors are unitary prime factors.
2
6, 10, 14, 15, 21, 22, 26, 30, 33, 34, 35, 38, 39, 42, 46, 51, 55, 57, 58, 62, 65, 66, 69, 70, 74, 77, 78, 82, 85, 86, 87, 90, 91, 93, 94, 95, 102, 105, 106, 110, 111, 114, 115, 118, 119, 122, 123, 126, 129, 130, 133, 134, 138, 141, 142, 143, 145, 146, 154, 155, 158
OFFSET
1,1
LINKS
EXAMPLE
k=90 is not a prime; 90 = 2*3*3*5; extremal prime factors are 2 and 5; gcd(2, 90/2) = gcd(5, 90/5) = 1, so 2 and 5 are unitary prime divisors of 90, thus 90 is in the sequence.
MATHEMATICA
ma[x_] := Part[Reverse[Flatten[FactorInteger[x]]], 2] mi[x_] := Part[Flatten[FactorInteger[x]], 1] k=0; Do[s=mi[n]; s1=ma[n]; If[Equal[GCD[s, n/s], 1]&&Equal[GCD[s1, n/s1], 1]&&!PrimeQ[n], Print[n]], {n, 2, 256}]
PROG
(PARI) lista(nn) = {forcomposite(n=1, nn, my(f=factor(n)[, 1], p = vecmin(f), q = vecmax(f)); if ((gcd(p, n/p) == 1) && (gcd(q, n/q) == 1), print1(n, ", ")); ); } \\ Michel Marcus, Jul 09 2018
(GAP) D:=List(Filtered([2..160], i->not IsPrime(i)), Factors);;
a:=[];; for i in D do if Gcd(i[1], Product(i)/i[1])=1 and Gcd(i[Length(i)], Product(i)/i[Length(i)])=1 then Add(a, Product(i)); fi; od; a; # Muniru A Asiru, Jul 10 2018~
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Feb 21 2003
STATUS
approved