A281447 Refactorable numbers n such that 3*n + 1 is also a refactorable number.

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

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

From Altug Alkan, Jan 25 2017: (Start)

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., (6239^16-1)/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

Table of n, a(n) for n=1..19.

S. Colton, Refactorable Numbers - A Machine Invention, J. Integer Sequences, Vol. 2, 1999.

Joshua Zelinsky, Tau Numbers: A Partial Proof of a Conjecture and Other Results , Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.8

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

Cf. A001318, A033950, A281294.

Sequence in context: A187438 A184374 A205932 * A246249 A183696 A104963

Adjacent sequences: A281444 A281445 A281446 * A281448 A281449 A281450

Keyword

nonn

Author

Altug Alkan, Jan 21 2017

Status

approved