A248705 The cubes related to the strictly increasing subsequence of A053668(n), n >= 1.
1, 8, 27, 64, 343, 729, 2744, 3375, 6859, 35937, 46656, 148877, 287496, 438976, 778688, 2985984, 3869893, 8489664, 34645976, 43986977, 58863869, 75686967, 398688256, 426957777, 485587656, 596947688, 835896888, 1693669888, 2548895896, 2954987875, 4758586568
Offset: 1
Examples
a(4) = 64 = 4*4*4, which is a cube. Product of its digits = 6*4 = 24. a(5) = 343 = 7*7*7, which is a cube. Product of its digits = 3*4*3 = 36. Since 36 > 24, 64 and 343 appear in the sequence. As suggested by _Wolfdieter Lang_, examples further clarified: (Start) A053668 is sieved (from left to right): 1, 2, 3, 4, 5, 6, 7, 8, 9, ....(numbers: k) 1, 8, 27, 64, 125, 216, 343, 512, 729, ....(cubes: k^3) 1, 8, 14, 24, 10, 12, 36, 10, 126, ....(prod of digits of k^3) 1, 8, 14, 24, X, X, 36, X, 126, ....(sieved products) and related leftover cubes are: 1, 8, 27, 64, 343, 729, ....(leftover cubes) (End)
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..139 (first 116 terms from K. D. Bajpai)
Programs
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Mathematica
A248705 = {}; t = 0; Do[s = Apply[Times, IntegerDigits[n^3]]; If[s > t, t = s; AppendTo[A248705, n^3]], {n, 1, 10^4}]; A248705
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PARI
\\ For b-file 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))
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Python
from operator import mul from functools import reduce A248705_list, x, m = [], 0, [6, -6, 1, 0] for _ in range(10**9): for i in range(3): m[i+1]+= m[i] xn = reduce(mul,[int(d) for d in str(m[-1])],1) if xn > x: x = xn A248705_list.append(m[-1]) # Chai Wah Wu, Nov 19 2014
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