cp's OEIS Frontend

From Bob Selcoe, Aug 25 2016: (Start)

Once a term with a given prime signature S (i.e., multiset of prime exponents) appears, then all numbers with the same prime signature follow. So either all or no terms with the same prime signature appear (first conjectured by Charles R Greathouse IV).

Proof:

i. Let S = {i,j,k..,y}, i>=j>=k.. be prime signatures, i.e., numbers of the form (p^i*q^j*r^k..*z^y) where {p,q,r,..,z} are distinct primes; denote S = {} as the signature for p^0 = 1.

ii. By definition, 1 and all primes p appear in the sequence; so terms where S = {k} (i.e., one-digit signatures denoting prime powers p^k) are k=0 and k = A026474(n) = {1,2,4,8,15,22,29..}, because k cannot be the sum of any combination of 3 smaller k. So the only prime power terms are p^0 = 1 and p^A026474(n), or S = {}, {1}, {2}, {4}, {8}, {15}, {22}, {29}...

iii. By induction, once S = {k} is determined, all other terms with the same prime signature appear (or do not) depending on the various combinations of signature exponents of previous terms with smaller signature sums. So for example, terms with the same two-digit signature S = {k,k} (i.e. (pq)^k) are constrained by the following: since p = {1} and q = {1} appear, then (pq) = {1,1} does not because in this case {1}*{1}*{} = {1,1}; since p^2 = {2} and q^2 = {2} appear, then (pq)^2 = {2,2} does not; since {4} appears but {3} does not, then {3,3} appears but {4,4} and {5,5} do not (the latter because {3,3}*{2}*{2} = {5,5} when p^2,q^2 = {2}). Note that {3,3} would not appear if {3,2} appeared because {3,2}*{1}*{} = {3,3} when q = {1}; but because p = {1} and p^2,q^2 = {2} appear, then {2}*{2}*{1} = {3,2} does not. {6,6} does not appear because {6,5} appears (by virtue of other constraints) and {6,5}*{1}*{} = {6,6} when q = {1}. Determining which signatures appear and which do not becomes increasingly complicated as the sequence increases.

Don Reble offered a proof on the Sequence Fans Mailing List which seems to be different (and certainly more formal) than mine. Perhaps mine is more of an "explanation" than a "proof"? (End)

Number of repeated prime divisors of a(n) is a square for a(n) < 128. - Jason Earls, Jul 01 2001

Numbers a(n) such that A001222(a(n)) is not a square are 128, 192, 288, 320, 432, 448, 480, 648, 672, 704, 720, 800, 832, 972, ... - Altug Alkan, Sep 26 2016

From David A. Corneth, Sep 24 2016 (Start):

Let t be a term of this sequence. Let v = (i, j, m) be a vector of three elements where i is A007814(t), j = A007949(t) and m = A001222(t) - i - j. Then possible vectors v for t up to 100000 are {[0, 0, 0], [0, 0, 1], [0, 0, 4], [0, 0, 7], [0, 1, 1], [0, 1, 4], [0, 2, 0], [0, 3, 0], [0, 4, 3], [0, 5, 3], [0, 6, 2], [0, 7, 2], [0, 8, 1], [0, 9, 1], [0, 10, 0], [1, 0, 0], [1, 0, 3], [1, 0, 6], [1, 1, 0], [1, 1, 3], [1, 1, 6], [1, 4, 2], [1, 5, 2], [1, 6, 1], [1, 7, 1], [1, 8, 0], [1, 9, 0], [2, 0, 0], [2, 3, 2], [2, 5, 1], [2, 7, 0], [3, 0, 2], [3, 0, 5], [3, 1, 2], [3, 1, 5], [3, 3, 1], [3, 4, 1], [3, 5, 0], [3, 6, 0], [4, 2, 1], [4, 2, 4], [4, 4, 0], [5, 0, 1], [5, 0, 4], [5, 1, 1], [5, 1, 4], [5, 2, 0], [5, 3, 0], [6, 0, 3], [6, 1, 0], [6, 1, 3], [7, 0, 0], [7, 3, 2], [8, 0, 2], [8, 1, 2], [8, 3, 1], [8, 5, 0], [9, 2, 1], [9, 4, 0], [10, 0, 1], [10, 1, 1], [10, 2, 0], [10, 3, 0], [11, 1, 0], [12, 0, 0]}.

(End)