A131662 Numbers n where either n or n+1 is divisible by the numbers from 1 to 12.

2519, 11879, 13320, 14399, 15840, 25200, 27719, 27720, 30239, 39599, 41040, 42119, 43560, 52920, 55439, 55440, 57959, 67319, 68760, 69839, 71280, 80640, 83159, 83160, 85679, 95039, 96480, 97559, 99000, 108360, 110879, 110880, 113399, 122759

Offset

1,1

Comments

Equivalent to numbers n where either n or n+1 is divisible by the numbers from 7 to 12. n is a term if n or n+1 is a multiple of 27720. - Chai Wah Wu, Jun 15 2020

Links

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Formula

Conjectures from Colin Barker, Jun 15 2020: (Start)

G.f.: x*(2519 + 9360*x + 1441*x^2 + 1079*x^3 + 1441*x^4 + 9360*x^5 + 2519*x^6 + x^7) / ((1 - x)^2*(1 + x)*(1 + x^2)*(1 + x^4)).

a(n) = a(n-1) + a(n-8) - a(n-9) for n>9. (End)

Example

2519 = 11*229 and 2520 = 2^3*3^2*5*7; that is, 2520 is divisible by all number from 1 to 12 except 11, while 2519 is divisible by 11.

Mathematica

Select[Range[150000], Mod[ #, 8]*Mod[ # + 1, 8] == 0 && Mod[ #, 9]*Mod[ # + 1, 9] == 0 && Mod[ #, 5]*Mod[ # + 1, 5] == 0 && Mod[ #, 7]*Mod[ # + 1, 7] == 0 && Mod[ #, 8]*Mod[ # + 1, 8] == 0 && Mod[ #, 12]*Mod[ # + 1, 12] == 0 && Mod[ #, 10]*Mod[ # + 1, 10] == 0 && Mod[ #, 11]*Mod[ # + 1, 11] == 0 &]

Prog

(Python)
A131662_list, n = [], 12
while len(A131662_list) < 10000:
    for i in range(7, 12):
        if (n-1) % i and n % i:
            break
    else:
        A131662_list.append(n-1)
    for i in range(7, 12):
        if n % i and (n+1) % i:
            break
    else:
        A131662_list.append(n)
    n += 12 # Chai Wah Wu, Jun 15 2020

Crossrefs

Sequence in context: A250679 A145534 A166931 * A068352 A175754 A068547

Adjacent sequences: A131659 A131660 A131661 * A131663 A131664 A131665

Keyword

nonn,easy

Author

Tanya Khovanova, Sep 13 2007

Status

approved