

A281447


Refactorable numbers n such that 3*n + 1 is also a refactorable number.


0



3050208, 27150208, 712250208, 4198150208, 9887150208, 29407950208, 186613550208, 254756450208, 412941550208, 496967350208, 553174550208, 1710112750208, 8023681250208, 9908919150208, 20053008750208, 20931113950208, 22635692110208, 24734957450208, 39291663950208
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OFFSET

1,1


COMMENTS

Corresponding first four values of 3*n + 1 are 5^4 * 11^4, 5^4 * 19^4, 5^4 * 43^4, 5^4 * 67^4.
Primes p such that both (5*p)^4 and ((5*p)^4  1)/3 are refactorable numbers begin 11, 19, 43, 67, 83, 109, 173, 211, 227, 443, 467, 557, 563, 587, 659, 739, 787, 821, 829, 853, 1123, 1187, 1229, 1277, 1453, 1523, 1571, 1709, 1901, 1973, 2083, 2099, 2237, 2467, 2531, 2621, 2909, 3347, 3517, 3877, 3923, 4099, 4243, 4253, 4259, 4483, 4547, ...; for each, p == 3 or 5 (mod 8).  Jon E. Schoenfield, Jan 21 2017
Although numbers of the form ((5*p)^4  1)/3 appear in the beginning of sequence, note that not all terms are of the form ((5*p)^4  1)/3, i.e., (6239^161)/3.
However we can show that all terms are of the form 8 * A001318(m).
Proof: If an odd number n is in this sequence, then n must be a square and 3*n + 1 = 3 * (2*k + 1)^2 + 1 = 12 * k * (k + 1) + 4 = 24 * A000217(k) + 4 is a refactorable number. 4 = 2^2 is the highest power of 2 that divides 24 * A000217(k) + 4 because 6 * A000217(k) + 1 is an odd number. Since 24 * A000217(k) + 4 is not divisible by 3, 3*n + 1 cannot be a refactorable number when n is an odd refactorable number.
Since we proved that n is an even number, 3 * n + 1 is odd and it must be a square. If 3 * n + 1 = (2 * t + 1)^2, then n = ((2 * t + 1)^2  1) / 3 = 4 * t * (t + 1) / 3 = 8 * A001318(m). (End)


LINKS



EXAMPLE

3050208 is a term because d(3050208) = 144 divides 3050208 and 3050208*3 + 1 = 5^4 * 11^4 is divisible by d(55^4) = 25.


PROG

(PARI) isA033950(n) = n % numdiv(n) == 0;
is(n) = isA033950(n) && isA033950(3*n+1);


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



