cp's OEIS Frontend

In range n=1..2215, each a(n) is a multiple of 41, with a(2215) = 177079 = 4319*41. In that range, all 2160 odd multiples 1, 3, 5, ..., 4319*41 occur, but only 55 even multiples, with a(34) = 66*41 = 2706 being the first one of them. The multipliers for even terms in that range seems to be given by 66*A047253(1..55) from 66*1 up to 66*A047253(55) = 66*65 = 4290. (Where A047253 gives the numbers not divisible by six).

From Antti Karttunen, Nov 20 2013 & Jan 26 2014: (Start)

Differs from A232098 for the first time at n=840, where a(840)=8, while A232098(840)=7. A232099 gives all the differing positions. See also the comments at A055926 and A232099.

The positions where a(n) is an odd prime is given by A017593 up to A017593(34)=414 (so far all 3's), after which comes the first 7 at a(420). (A017593 gives the positions of 3's.)

(Continued on Jan 26 2014):

Only terms of A181062 occur as values.

A235921 gives such n where a(n^2) (= A235918(n)) differs from A071222(n-1) (= A053669(n)-1). (End)

a(n) is the largest m such that A003418(m) divides n. - David W. Wilson, Nov 20 2014

a(n) is the largest number of consecutive integers dividing n. - David W. Wilson, Nov 20 2014

A051451 gives indices where record values occur. - Gionata Neri, Oct 17 2015

Yuri Matiyasevich calls this the maximum inheritable divisor of n. - N. J. A. Sloane, Dec 14 2023

From Antti Karttunen, Nov 20 - Dec 06 2013: (Start)

This sequence has several interpretations:

Numbers k such that A055874(k) differs from A055881(k). [Leroy Quet's original definition of the sequence. Note that A055874(k) >= A055881(k) for all k.]

Numbers k such that {largest m such that m! divides k^2} is different from {largest m such that m! divides k}, i.e., numbers k for which A232098(k) > A055881(k).

Numbers k which are either 12 times an odd number (A073762) or the largest m such that (m-1)! divides k is a composite number > 5 (A232743).

Please see my attached notes for the proof of the equivalence of these interpretations.

Additional implications based on that proof:

A232099 is a subset of this sequence.

A055881(a(n))+1 is always composite. In the range n = 1..17712, only values 4, 6, 8, 9 and 10 occur.

The new definition can be also rephrased by saying that the sequence contains all the positive integers k whose factorial base representation of (A007623(k)) either ends as '...200' (in which case k is an odd multiple of 12, 12 = '200', 36 = '1200', 60 = '2200', ...) or the number of trailing zeros + 2 in that representation is a composite number greater than or equal to 6, e.g. 120 = '10000' (in other words, A055881(k) is one of the terms of A072668 after the initial 3). Together these conditions also imply that all the terms are divisible by 12.

(End)

For all n, A055881(n) <= a(n), and probably also a(n) <= A055874(n).

Moreover, a(n) > A055881(n) if and only if A055874(n) > A055881(n), thus A055926 gives (also) all the positions where this sequence differs from A055881. Please see Comments section in A055926 for the proof.

Differs from A055874 for the first time at n=840, where a(840)=7, while A055874(840)=8. A232099 gives all the positions where such differences occur.

Equivalent definition is: numbers n such that {the largest m such that 1, 2, ..., m divide n^2 = A055874(n^2) = A235918(n)} is different from {the smallest k such that gcd(n-1,k) = gcd(n,k+1) = A071222(n-1)}.

All terms are multiples of 210 = 2*3*5*7, the fourth primorial, A002110(4).

The first term which is an even multiple of 210 (i.e., 210 times an even number), is 446185740 = 2124694 * 210 = 2*223092870 = 2*A002110(9) = 2*A034386(23). Note that 23 is the 9th prime, and 223092870 is its primorial. Thus this sequence differs from its subsequence, A236432, "the odd multiples of 210" = (2n-1)*210, for the first time at n = 1062348, where a(n) = 446185740, while A236432(n) = 446185950.

Note that a more comprehensive description for which terms are included is still lacking. Compare for example to the third definition of A055926.

At least we know the following:

If a number is not divisible by 210, then it cannot be a member, as then it is "missing" (i.e., not divisible by) one of those primes, 2, 3, 5 or 7, and thus its square is also "missing" the same prime. In more detail, this follows because:

If the least nondividing prime is 2, then A053669(n) = A236454(n) = 2. If the least nondividing prime is 3, then A053669(n) = A236454(n) = 3.

If the least nondividing prime is 5 (so 2 and 3 are present), then as 2|n and 4|(n^2), we have A053669(n) = A236454(n) = 5.

If the least nondividing prime is 7, but 2, 3 and 5 are present, then we have A053669(n) = A236454(n) = 7.

On the other hand, when n is an odd multiple of 210 (= 2*3*5*7), i.e., (2k+1)*210, so that its prime factorization is of the form 2*3*5*7*{zero or more additional odd prime factors}, then A053669(n) must be at least 11, the next prime after 7, which is certainly different from A236454(n) = A007978(n^2) which must be 8, as then 4 is the highest power of 2 dividing n^2.

In contrast to that, when n is an even multiple of 210, so that its prime factorization is of the form 2*2*3*5*7*{zero or more additional prime factors}, then also all the composites 8, 9, 10, 12, 14, 15, 16, 18 and 20 divide n^2, thus if A053669(n) is any prime from 11 to 19, A236454(n) will return the same result.

However, if n is of the form k*446185740, where k is not a multiple of 3, so that the prime factorization of n is 2*2*3*5*7*11*13*17*19*23*{zero or more additional prime factors, all different from 3}, then A053669(n) must be at least 29 (next prime after 23), but A236454(n) = 27, because then 9 is the highest power of 3 dividing n^2.

The pattern continues indefinitely: If n is of the form (2k+1)*2*3*200560490130, where 200560490130 = A002110(11), so that n has a prime factorization of the form 2*2*3*3*5*7*11*13*17*19*23*29*31*{zero or more additional odd prime factors}, then A053669(n) must be at least 37, while A236454(n) = 32 = 2^5, because then 16 is the highest power of 2 dividing n^2.