A085645 -id:A085645 - cp's OEIS frontend

A331029 Least integer of each composite prime signature where primes ending in 1 or 9 are treated distinctly from those ending in 3 or 7.

1, 3, 9, 11, 21, 27, 33, 63, 81, 99, 121, 189, 209, 231, 243, 273, 297, 363, 441, 567, 627, 693, 729, 819, 891, 1089, 1323, 1331, 1701, 1881, 2079, 2187, 2299, 2457, 2541, 2673, 3003, 3267, 3969, 3993, 4389, 4641, 4851, 5103, 5643, 5733, 6061, 6237, 6561, 6897

Offset: 1

Andrew Howroyd, Jan 07 2020

This sequence is analogous to A025487. The two primes 2 and 5 are ignored and the remainder are divided into two distinct classes depending on the final digit of the prime. A combined prime signature is then created from the prime signatures of the two classes of prime.
Consider the problem of finding the smallest number having n divisors ending with 1 or 9 (sequence A085645). Solutions must lie in this sequence since numbers with the same composite prime signature as defined here will have the same number of divisors ending with 1 or 9.
Primes ending in either 1 or 9 are 11, 19, 29, 31, 41, 59, ... (A045468).
Primes ending in either 3 or 7 are 3, 7, 13, 17, 23, 37, ... (A097957).
The partial products of these two sequences form two sequences analogous to the primorial numbers (11, 11*19, 11*19*29, ... and 3, 3*7, 3*7*13, ...). In the same manner that A025487 can be defined as products of primorial numbers, an alternative description of this sequence is that it is the set of all products of the two primorial analogs.

			Primes in this sequence are 3 and 11 because these are the smallest primes in the two classes.
Semiprimes in this sequence are 9 = 3^2, 21 = 3*7, 33 = 3*11, 121 = 11^2,  209 = 11*19 because 3, 7 are the smallest primes ending with either 3 or 7 and 11, 19 are the smallest primes ending with either 1 or 9.

Andrew Howroyd, Table of n, a(n) for n = 1..1000

Cf. A025487, A045468 (primes ending in 1 or 9), A085645, A097957 (primes ending in 3 or 7), A331082.

GenS(lim, pred)={my(L=List(), S=[1]); forprime(p=2, oo, if(pred(p), listput(L,S); my(pp=vector(logint(lim, p), i, p^i)); S=concat([k*pp[1..min(if(k>1, my(f=factor(k)[,2]); f[#f], oo), logint(lim\k, p))] | k<-S]); if(!#S, return(Set(concat(L)))) ))}
Merge(s1, s2, lim)={Set(concat(vector(#s1, i, [t | t<-s1[i]*s2, t<=lim])))}
lista331029(lim)={Merge(GenS(lim, k->abs(k%10-5)==2), GenS(lim, k->abs(k%10-5)==4), lim)}
{ lista331029(10^4) }

A331082 Smallest number having exactly n divisors ending with 3 or 7.

Original entry on oeis.org

3, 21, 63, 189, 567, 693, 3969, 2079, 5733, 6237, 413343, 9009, 74529, 43659, 51597, 27027, 1750329, 63063, 26040609, 81081, 464373, 670761, 219667417263, 153153, 2528253, 819819, 693693, 729729, 160137547184727, 567567, 6036849, 459459, 37614213, 19253619

Offset: 1

Andrew Howroyd, Jan 08 2020

A term cannot be a multiple of either 2 or 5 since removing these factors does not alter the number of divisors ending with 3 or 7.

All terms must be in A331029.

			a(1) = 3: divisors of 3 are 1, 3*. (* marks divisors ending in 3 or 7.)
a(2) = 21: divisors of 21 are 1, 3*, 7*, 21.
a(3) = 63: divisors of 63 are 1, 3*, 7*, 9, 21, 63*.
a(4) = 189: divisors of 189 are 1, 3*, 7*, 9, 21, 27*, 63*, 189.

Cf. A085645, A331029.

a(n)={forstep(k=1, oo, 2, if(sumdiv(k, d, abs(d%10-5)==2) == n, return(k)))}

\\ faster program: needs lista331029 defined in A331029.
a(n)={my(lim=1); while(1, lim*=10; my(S=lista331029(lim)); for(i=1, #S, my(k=S[i]); if(sumdiv(k, d, abs(d%10-5)==2)==n, return(k))))}

from sympy import divisors
def count37(iterable): return sum(i%10 in [3, 7] for i in iterable)
def a(n):
  m = 2
  while count37(divisors(m)) != n: m += 1
  return m
print([a(n) for n in range(1, 17)]) # Michael S. Branicky, Mar 21 2021

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A331029 Least integer of each composite prime signature where primes ending in 1 or 9 are treated distinctly from those ending in 3 or 7.

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A331082 Smallest number having exactly n divisors ending with 3 or 7.

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