A337081 Primitive complement of A337037: terms of A337080 that are not multiples of previous terms.

4, 90, 546, 675, 850, 918, 945, 1026, 1050, 1134, 1242, 1365, 1386, 1575, 1650, 1750, 1782, 1950, 2205, 2295, 2310, 2450, 2475, 2646, 2793, 2850, 3250, 3366, 3465, 3626, 3654, 3762, 3850, 3969, 3990, 4218, 4290, 4374, 4455, 4510, 4550, 4650, 4875, 4998, 5022, 5166, 5382, 5390, 5610

Offset: 1

Author

Matej Veselovac, Aug 14 2020

Keywords

• nonn
• easy

Comments

The only semiprime in the sequence is a(1) = 4, and there are no terms with exactly 3 prime factors.

Numbers of form p^k where p >= 5 is a prime number are terms of the sequence if and only if k = 4p+6. The only terms of the form 2^k or 3^k have k = 2, 12 respectively.

Examples

			Numbers of the form m = 2*p*q*((p-1)*q-(p-2)) where p, q and (p-1)*q-(p-2) are odd prime numbers are even terms of the sequence. First, notice that m is a term of A337080 because the factorizations m = (2*((p-1)*q-(p-2)))*(p)*(q) = (2)*(((p-1)*q-(p-2)))*(p*q) have equal sums of factors. Second, m is not a multiple of any of the previous terms of the sequence because m has exactly 4 prime factors and the only term with less than 4 prime factors is 4, but 4 does not divide m.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000
• Math StackExchange, Smallest power of a prime whose factorizations don't have distinct sums of factors, 2020.

Crossrefs

Cf. A337037, A337080, A337112 (smallest term with n factors).

Cf. A001055 (number of unordered factorizations of n), A074206 (number of ordered factorizations of n).

Cf. A056472 (all factorizations of n), A069016 (number of distinct sums).

Programs

PARI

factz(n, minn) = {my(v=[]); fordiv(n, d, if ((d>=minn) && (d<=sqrtint(n)), w = factz(n/d, d); for (i=1, #w, w[i] = concat([d], w[i]);); v = concat(v, w););); concat(v, [[n]]);}
factorz(n) = factz(n, 2);
isok(n) = my(vs = apply(x->vecsum(x), factorz(n))); #vs != #Set(vs);
isprimitive(n, va) = {for (k=1, #va, if ((n % va[k]) == 0, return (0));); return (1);}
lista(nn) = {my(va = []); for (n=1, nn, if (isok(n) && isprimitive(n, va), va = concat(va, n));); va;} \\ Michel Marcus, Aug 15 2020