A053668 -id:A053668 - cp's OEIS frontend

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

K. D. Bajpai, Oct 13 2014

The triangular numbers of this form are at A246753.

The squares of this form are at A248648.

			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)

Chai Wah Wu, Table of n, a(n) for n = 1..139 (first 116 terms from K. D. Bajpai)

Cf. A000578, A053668, A230041, A246569, A246753, A248648.

A248705 = {}; t = 0; Do[s = Apply[Times, IntegerDigits[n^3]]; If[s > t, t = s; AppendTo[A248705, n^3]], {n, 1, 10^4}]; A248705

\\ 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))

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

A218145 Product of the nonzero digits (in base 10) of n^3.

Original entry on oeis.org

1, 8, 14, 24, 10, 12, 36, 10, 126, 1, 9, 112, 126, 224, 315, 216, 108, 240, 2160, 8, 108, 192, 84, 192, 300, 1470, 1296, 180, 1728, 14, 1134, 2016, 2835, 324, 2240, 4320, 450, 2240, 1215, 24, 864, 1792, 2205, 1280, 90, 3402, 144, 90, 1512, 10, 180, 192

Offset: 1

Jonathan Vos Post, Oct 21 2012

Similar to A053668, which does not exclude zero digits from the product. This is to the cubes A000578 as A218072 is to the squares A000290.

			a(4) = A007954(4^3) = product of digits of 64 = 6 * 4 = 24, since none of the digits is 0.
a(10) = product of nonzero digits of 1000 = 1.

Cf. A000578, A007954, A053668, A218072.

Table[Times @@ Select[IntegerDigits[n^3], # > 0 &], {n, 60}] (* T. D. Noe, Oct 22 2012 *)

A117690 Cubes for which both the sum of the digits and the product of the digits are squares.

Original entry on oeis.org

0, 1, 1000, 27000, 216000, 970299, 1000000, 1860867, 2146689, 4019679, 5000211, 10077696, 14348907, 21717639, 26198073, 27000000, 27818127, 29503629, 36926037, 37933056, 57960603, 62099136, 67917312, 80621568, 84027672, 106496424, 116930169, 123505992

Offset: 0

Luc Stevens (lms022(AT)yahoo.com), Apr 12 2006

			2146689 is in the sequence because it is a cube and (1)the sum of its digits 2+1+4+6+6+8+9=36 is a square and (2)the product of its digits 2*1*4*6*6*8*9=20736 is also a square.

Cf. A000290, A000578, A004164, A053668.

okQ[n_]:=Module[{idn=IntegerDigits[n]},IntegerQ[Sqrt[Total[idn]]] && IntegerQ[Sqrt[Times@@idn]]]; Select[Range[0,500]^3,okQ] (* Harvey P. Dale, Aug 24 2011 *)

Corrected (by inserting a(15)=26198073) by Harvey P. Dale, Aug 24 2011

More terms from Jason Yuen, Aug 30 2025

A218215 Product of the nonzero digits (in base 10) of n^4.

Original entry on oeis.org

1, 6, 8, 60, 60, 108, 8, 216, 180, 1, 96, 252, 480, 576, 300, 2700, 240, 1512, 18, 6, 1152, 1440, 4032, 2646, 1620, 45360, 240, 4320, 784, 8, 540, 6720, 720, 2916, 300, 13608, 1344, 1440, 288, 60, 6720, 972, 768, 36288, 240, 94080, 96768, 2880, 6720, 60, 2520, 756, 16128, 3600, 2700, 186624, 150, 3888, 252, 108

Offset: 1

Jonathan Vos Post, Oct 23 2012

This is to 4 as A218145 Product of the nonzero digits (in base 10) of n^3, is to 3. Similar to A053668, which does not exclude zero digits from the product. This is to the 4th powers A000583 as cubes A000578 are to A218145, and as A218072 is to the squares A000290.

Cf. A000583, A007954, A053668, A218072, A218145.

Table[Times @@ Select[IntegerDigits[n^4], # > 0 &], {n, 60}]

A218311 Product of the nonzero digits (in base 10) of n^5.

Original entry on oeis.org

1, 36, 576, 64, 900, 4235364, 112896, 4064256, 2624400, 1, 900, 9437184, 1285956, 45158400, 1093955625, 45158400, 101606400, 19110297600, 740710656, 36, 16384, 7290000, 241864704, 1316818944, 12859560000, 65028096, 585252864, 4064256, 129600, 576, 74649600

Offset: 1

Jonathan Vos Post, Oct 25 2012

This is to 5 as A218145, product of the nonzero digits (in base 10) of
n^3, is to 3. Similar to A053668, which does not exclude zero digits
from the product.  This is to the 5th powers A000584 as cubes A000578
are to A218145, and as A218072 is to the squares A000290.

			a(3) = 36 because 3^5 = 243, and 2*4*3 = 36.

Cf. A000584, A007954, A053668, A218072, A218145.

Table[Times @@ Select[IntegerDigits[n^5], # > 0 &], {n, 60}]

cp's OEIS Frontend

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

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A218145 Product of the nonzero digits (in base 10) of n^3.

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A117690 Cubes for which both the sum of the digits and the product of the digits are squares.

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A218215 Product of the nonzero digits (in base 10) of n^4.

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A218311 Product of the nonzero digits (in base 10) of n^5.

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