A317248 Semiprimes which when truncated arbitrarily on either side in base 10 yield semiprimes.

4, 6, 9, 46, 49, 69, 94, 469, 694, 949, 4694

Offset: 1

Author

Keith J. Bauer, Jul 24 2018

Keywords

• base
• fini
• full
• nonn

Comments

There are exactly 3 1-digit terms, 4 2-digit terms, 3 3-digit terms, and 1 4-digit term.

After the 4-digit term, there are no more terms in this sequence. This is provable by induction: There are no 5-digit terms. If there are no k-digit terms, there are no (k+1)-digit terms. (If there were, then said term, when truncated on either side, would produce a k-digit number that is in the sequence.) Therefore, there are no terms that have at least 5 digits.

The sequence 3, 4, 3, 1, 0, 0, ... does not appear to be significant.

This sequence depends on base 10 and is nonnegative.

Any truncation of a number in this sequence yields another number in this sequence. If one did not, then truncating the number more would yield a non-semiprime, which is impossible.

Base 10 is the first base in which this sequence contains 3-digit terms.

Examples

			4694 is a semiprime (2 * 2347), and its truncations are, too: 469 (7 * 67), 694 (2 * 347), 46 (2 * 23), etc.
		

Links

• Keith J. Bauer, "A317248 (Python v2.7.13)"

Crossrefs

Cf. A001358.

Subset of A107342 and A086698.

Programs

Mathematica

ok[w_, n_] := AllTrue[Flatten@ Table[ FromDigits@ Take[w, {i, j}], {i, n}, {j, i, n}], PrimeOmega[#] == 2 &]; Union @@ Reap[ Do[Sow[ FromDigits /@ Select[Tuples[{4, 6, 9}, n], ok[#, n] &]], {n, 5}]][[2, 1]] (* Giovanni Resta, Jul 26 2018 *)

Python

#v2.7.13, see LINKS to run it online.
#semitest(number, 0) returns True iff number is a semiprime
def semitest(number, factors):
    if number != 2:
        for p in [2] + range(3, int(number ** 0.5) + 1, 2):
            if number % p == 0:
                if factors < 2:
                    return semitest(number / p, factors + 1)
                else:
                    return False
    if factors == 1:
        return True
    else:
        return False
#main function
def doIt(base):
    #initialization
    numbers = [[]]
    indices_list = [[]]
    i = 0
    for number in range(1, base):
        if semitest(number, 0):
            numbers[0].append(number)
            indices_list[0].append([i])
            i += 1
    #numbers[0] is the digit pool
    #numbers[-1] is to be appended to
    #numbers[-2] is for reference to past numbers
    #indices_list records the indices of numbers
    numbers.append([])
    indices_list.append([])
    #main while loop, go until there are no numbers left in the sequence
    indices = [0, 0]
    while len(numbers[-2]) > 0:
        #test number
        if indices[:-1] in indices_list[-2]:
            if indices[1:] in indices_list[-2]:
                #little-endian
                number = 0
                power = 0
                for index in indices:
                    number += numbers[0][index] * base ** power
                    power += 1
                if semitest(number, 0):
                    numbers[-1].append(number)
                    indices_list[-1].append(indices[:])
        #increment indices
        for i in range(len(indices)):
            indices[i] += 1
            if indices[i] == len(numbers[0]):
                indices[i] = 0
                if i == len(indices) - 1:
                    indices = [0] * len(indices) + [0]
                    numbers.append([])
                    indices_list.append([])
            else:
                break
    #print results after while loop has run
    print base, sum(numbers, [])
    print numbers
#call main function
doIt(10)

Python

from sympy import factorint
A317248_list = xlist = [4,6,9]
for n in range(1,10):
    ylist = []
    for i in (4,6,9):
        for x in xlist:
            if sum(factorint(10*x+i).values()) == 2 and (10*x+i) % 10**n in xlist:
                ylist.append(10*x+i)
            elif sum(factorint(x+i*10**n).values()) == 2 and (x//10+i*10**(n-1)) in xlist:
                ylist.append(x+i*10**n)
    xlist = set(ylist)
    if not len(xlist):
        break
    A317248_list.extend(xlist)
A317248_list.sort() # Chai Wah Wu, Aug 23 2018