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

Author

Don Reble, Sep 21 2001

Keywords

• easy
• nonn

Examples

			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)).
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..112
• Sean A. Irvine, Java program (github)

Programs

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