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A248705 The cubes related to the strictly increasing subsequence of A053668(n), n >= 1.

%I A248705 #65 Jan 13 2024 14:05:23
%S A248705 1,8,27,64,343,729,2744,3375,6859,35937,46656,148877,287496,438976,
%T A248705 778688,2985984,3869893,8489664,34645976,43986977,58863869,75686967,
%U A248705 398688256,426957777,485587656,596947688,835896888,1693669888,2548895896,2954987875,4758586568
%N A248705 The cubes related to the strictly increasing subsequence of A053668(n), n >= 1.
%C A248705 The triangular numbers of this form are at A246753.
%C A248705 The squares of this form are at A248648.
%H A248705 Chai Wah Wu, <a href="/A248705/b248705.txt">Table of n, a(n) for n = 1..139</a> (first 116 terms from K. D. Bajpai)
%e A248705 a(4) = 64 = 4*4*4, which is a cube. Product of its digits = 6*4 = 24.
%e A248705 a(5) = 343 = 7*7*7, which is a cube. Product of its digits = 3*4*3 = 36.
%e A248705 Since 36 > 24, 64 and 343 appear in the sequence.
%e A248705 As suggested by _Wolfdieter Lang_, examples further clarified:
%e A248705 (Start)
%e A248705 A053668 is sieved (from left to right):
%e A248705 1, 2,  3,  4,   5,   6,   7,   8,   9, ....(numbers: k)
%e A248705 1, 8, 27, 64, 125, 216, 343, 512, 729, ....(cubes: k^3)
%e A248705 1, 8, 14, 24,  10,  12,  36,  10, 126, ....(prod of digits of k^3)
%e A248705 1, 8, 14, 24,   X,   X,  36,   X, 126, ....(sieved products)
%e A248705 and related leftover cubes are:
%e A248705 1, 8, 27, 64,           343,      729, ....(leftover cubes)
%e A248705 (End)
%t A248705 A248705 = {}; t = 0; Do[s = Apply[Times, IntegerDigits[n^3]]; If[s > t, t = s; AppendTo[A248705, n^3]], {n, 1, 10^4}]; A248705
%o A248705 (PARI) \\ For b-file
%o A248705 c = 0; k = 0; for(n=1, 5*10^8, d = digits(n^3); p = prod(i = 1, #d, d[i]); while(p > k, c++; print(c, "  ", n^3); k = p))
%o A248705 (Python)
%o A248705 from operator import mul
%o A248705 from functools import reduce
%o A248705 A248705_list, x, m  = [], 0, [6, -6, 1, 0]
%o A248705 for _ in range(10**9):
%o A248705     for i in range(3):
%o A248705         m[i+1]+= m[i]
%o A248705     xn = reduce(mul,[int(d) for d in str(m[-1])],1)
%o A248705     if xn > x:
%o A248705         x = xn
%o A248705         A248705_list.append(m[-1]) # _Chai Wah Wu_, Nov 19 2014
%Y A248705 Cf. A000578, A053668, A230041, A246569, A246753, A248648.
%K A248705 nonn,base
%O A248705 1,2
%A A248705 _K. D. Bajpai_, Oct 13 2014