Brilliant numbers: Difference between revisions

blocksprimes = [primes(10**n, 10**(n + 1)) for n in range(max_order)]

def smallest_brilliant(thresh=0lb):

forroot blk= in blocks:isqrt(lb)

for blk in if best < threshprimes:

hii = nnp.searchsorted(blk, -root, 1'left')

for i, p+= in enumerate(blk):[i]*blk[i] < lb

if not best = min(prod, best)i:

return blk[0]*blk[0], breakpos

sel = idx < lo = in

p, idx = p[sel], while hi > lo:idx[sel]

q = if p*blk[(j := (lo + hi)//2)idx] >= thresh:

sel = q pos +>= hi - ip

p, q, idx if (prod := p*blk[hisel]), >=q[sel], thresh:idx[sel]

return bestnp.min(p*q), pos

p, first100res = 0, []

p, pos_ = smallest_brilliant(p + 1)

first100res.append(p)

print(f'Firstfirst 100: are {first100res}')

for ni in range(2*max_order*2):

p, posthresh = smallest_brilliant(10**n)i

print(f'first aboveAbove 10**^{ni:2d}: {p:18d20d} at position #{pos:14d}')</lang>

<pre>Firstfirst 100: are [4, 6, 9, 10, 14, 15, 21, 25, 35, 49, 121, 143, 169, 187, 209, 221, 247, 253, 289, 299, 319, 323, 341, 361, 377, 391, 403, 407, 437, 451, 473, 481, 493, 517, 527, 529, 533, 551, 559, 583, 589, 611, 629, 649, 667, 671, 689, 697, 703, 713, 731, 737, 767, 779, 781, 793, 799, 803, 817, 841, 851, 869, 871, 893, 899, 901, 913, 923, 943, 949, 961, 979, 989, 1003, 1007, 1027, 1037, 1067, 1073, 1079, 1081, 1121, 1139, 1147, 1157, 1159, 1189, 1207, 1219, 1241, 1247, 1261, 1271, 1273, 1333, 1343, 1349, 1357, 1363, 1369]