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(PARI) isok(k, vp) = {for (i=1, #vp, if ((k % vp[i]) >= vp[i]/2, return (0)); ); return (1); }
a(n) = {my(k=1, vp = primes(n)); while (!isok(k, vp), k++); k; } \\ Michel Marcus, Sep 08 2019
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Bert Dobbelaere, <a href="/A327408/b327408.txt">Table of n, a(n) for n = 1..53</a>
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allocated for Bert DobbelaereSmallest integer > 0 so that its remainders modulo the first n primes are less than half their respective moduli.
2, 4, 6, 10, 16, 16, 70, 136, 210, 210, 442, 786, 786, 786, 6450, 53110, 53110, 247690, 303810, 303810, 813450, 3443146, 5889382, 9327220, 10068256, 63916062, 63916062, 63916062, 285847290, 285847290, 285847290, 285847290, 370793956, 370793956, 370793956, 370793956
1,1
a(6) = 16.
16 mod 2 = 0 < 2/2
16 mod 3 = 1 < 3/2
16 mod 5 = 1 < 5/2
16 mod 7 = 2 < 7/2
16 mod 11 = 5 < 11/2
16 mod 13 = 3 < 13/2
16 is the smallest integer > 0 satisfying these inequalities for the first 6 primes.
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Bert Dobbelaere, Sep 07 2019
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