A064219 a(1) = 1; a(n) > 0; for each k from 1 to n, k divides a(n) or a(n)+1 and a(n) is the least such integer.

1, 1, 2, 3, 15, 24, 35, 119, 504, 720, 2519, 2519, 41040, 83160, 83160, 196559, 524160, 524160, 3160079, 3160079, 3160079, 3160079, 68468400, 68468400, 68468400, 68468400, 4724319600, 4724319600, 26702675999, 26702675999

Offset

1,3

Links

Chai Wah Wu, Table of n, a(n) for n = 1..112

Sean A. Irvine, Java program (github)

Example

a(5)=15 because (2 divides a(5)+1) and (3 divides a(5)) and (4 divides a(5)+1) and (5 divides a(5)).

Prog

(PARI) { a=1; for (n=1, 100, if (a%n && (a+1)%n, until (b, b=1; a++; for (k=1, n, if (a%k && (a+1)%k, b=0; break)))); write("b064219.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 10 2009
(Python)
from math import lcm
from itertools import product
from sympy.ntheory.modular import solve_congruence
def A064219(n):
    if n == 1: return 1
    alist, blist, c, klist = [], [], 1, list(range(n, 1, -1))
    while klist:
        k = klist.pop(0)
        if not c%k:
            blist.append(k)
        else:
            c = lcm(c, k)
            alist.append(k)
            for m in klist.copy():
                if not k%m:
                   klist.remove(m)
    for d in product([0, 1], repeat=len(alist)):
        x = solve_congruence(*list(zip(d, alist)))
        if x is not None:
            y = x[0]
            if y > 1:
                for b in blist:
                    if y%b > 1:
                        break
                else:
                    if y < c:
                        c = y
    return int(c-1) # Chai Wah Wu, Jun 19 2023

Crossrefs

Sequence in context: A048613 A162890 A030072 * A244377 A244330 A342867

Adjacent sequences: A064216 A064217 A064218 * A064220 A064221 A064222

Keyword

easy,nonn

Author

Don Reble, Sep 21 2001

Status

approved