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