[seqfan] A187824, A220890

[Don Reble]

Seqfans:

> %N A187824 a(n) is the largest m such that n is congruent to
>            -1, 0 or 1 mod k for all k from 1 to m.
> %C A187824 First occurrence of k = 3, 4, 5,...: 2, 3, 4, 5, 29, 41,
>            55, 71, 881, 791, 9360, 10009, 1079, 30239, (17 unknown),
>            246960, (19 unknown), 636481, 1360800, 3160079, (23 unknown)...

    It's about about a trio {M-1, M, M+1} of numbers which has a
    multiple of each number 1 to N, but no multiple of N+1.

    The values 17, 19, 23, 27, 29, etc. don't occur. (See also A220890.)

    Suppose for two numbers A and B, D divides both A and B, D >= 3, and
    a trio has A*i and B*j. Then D divides A*i and B*j, and also A*i-B*j.
    That difference is within [-2,+2] and is therefore 0. So L = lcm(A,B)
    divides A*i, a member of the trio. And all factors F of L divide
    that member. If F > B > A, then F-1 can't occur in the sequence.

    For (A,B) = (6,9): L=18 eliminates 17
        (12,15)->60, no 59, 29, 19
        ( 8,12)->24, no 23
        (12,21)->84, no 83, 41, 27
        ( 6,15)->30, no 29
        (15,21)->105, no 104, 34
        (12,18)->36, no 35
        ( 8,20)->40, no 39
        (14,21)->42, no 41
        (12,33)->132, no 131, 65, 43

    L eliminates L-1 if
    - L is divisible by at least three primes p<q<r:
        gcd(L/p,L/q) >= r >= 5, so (L/p,L/q)->L
    - L is divisible by two primes 3<p<q: (3L/p,3L/q)->3L
    - L is divisible by 4 and prime p>=5: (3L/4,3L/p)->3L
    - L is divisible by 6 and prime p>=5: (L/2,L/3)->L
    - L is divisible by 8 and prime p/=2: (L/2,L/p)->L
    - L is divisible by 9 and prime p/=3: (L/3,L/p)->L

    That leaves p^k, 2*p (p>=5), 3*p (p>=5), and factors of 12.
    Let M = lcm(1,2,3,...,L-1).
    - if L = p^k, A187824(M)= L-1.
    - if L = 2*p, solve the bicongruence X = 0 mod M/p, X = 1 mod p.
        Then A187824(X)=L-1.
    - if L = 3*p, solve the bicongruence X = 1 mod M/p, X = -1 mod p.
        Then A187824(X)=L-1.
    - if L divides 12, then A187824(2,4,881) = 3,5,11.
        0, 1, 2 don't occur.

    BTW, those bicongruences rarely yield the least such X.

    ---

    I confirm Dr. Israel's b220890 table; here's more.

71 -1
72 10181233426044354312038401
73 1282872732472100546035459201
74 -1
75 -1
76 -1
77 -1
78 1710175123969695043870809601
79 -1
80 9109175486326305839498649599
81 137732553475447101916089945601
82 380861858126531933101078128001
83 -1
84 -1
85 7030952614125730395627416294399
86 10217107162503253807404669259199
87 -1
88 15621840662060409413009070297601
89 -1
90 -1
91 -1
92 204152243825236232321592891494399
93 313838190108888319510597097135999
94 -1
95 -1
96 416571285638527409262107210976001
97 -1
98 -1
99 -1
100 41425673724847598555537624113257599
101 -1
102 1548639393428933027443752413950267201
103 -1

-- 
Don Reble  djr at nk.ca