A218072 -id:A218072 - cp's OEIS frontend

A218013 Numbers that divide the product of the nonzero digits (in base 10) of their square.

1, 2, 3, 5, 6, 8, 28, 36, 42, 54, 75, 192, 216, 288, 486, 525, 648, 768, 864, 882, 1728, 2160, 3024, 6048, 6075, 7056, 7680, 17280, 18144, 20736, 30240, 40824, 56448, 60480, 61236, 62208, 64512, 84672, 122472, 138915, 150528, 387072, 408240, 497664, 622080

Offset: 1

Nels Olson, Oct 18 2012

			For n=5, n^2 is 25; the product of the digits of 25 is 2 * 5 = 10, which is divisible by n=5.

David A. Corneth, Table of n, a(n) for n = 1..10919 (first 473 terms from Chai Wah Wu; terms <= 10^100)

Cf. A002473.

Related to A218072. Subsets of this sequence include A218029 and A218030.

isok(n) = digs = digits(n^2); (prod(i=1, #digs, if (digs[i], digs[i], 1)) % n) == 0; \\ Michel Marcus, Aug 12 2013

from operator import mul
from functools import reduce
from gmpy2 import t_mod, mpz
A218013 = [n for n in range(1,10**6) if not t_mod(reduce(mul,(mpz(d) for d in str(n**2) if d != '0')),n)] # Chai Wah Wu, Aug 23 2014

A218030 Numbers k equal to half of the product of the nonzero (base-10) digits of k^2.

Original entry on oeis.org

2, 5, 54, 648, 2160, 337169025526136832000, 685506275314921762068267522458966662115416623590907309075726336000000, 46641846972427276691124922228108091690332947069125333309512419901440000000000

Offset: 1

Nels Olson, Oct 18 2012

The first 5 terms of the sequence were found by the author around 1980 using his Commodore PET computer. He found the subsequent terms in 1991 by means of an improved program. The author has always referred to these as the "Faithy numbers" after his mother, Faith, who posed the problem.

			For n=5, n^2 is 25; the product of the digits of 25 is 2*5 = 10, which is equal to 2*n.

Michael S. Branicky, Table of n, a(n) for n = 1..12 (all terms < 10^300)
Michael S. Branicky, Python program.
Giovanni Resta, C program for this and related sequences.

Special case of A218013 where the ratio of the digit-product to the original number is 2. Related to A218072.

mx = 2^255; L = {};
p2 = 1; While[p2 < mx, Print["--> 2^", Log[2, p2]];
p3 = p2; While [p3 < mx,
  p5 = p3; While[p5 < mx,
   n = p5; While[n < mx,
    If[2 n == Times @@ Select[IntegerDigits[n^2], # > 0 &],
     AppendTo[L, n]; Print[n]]; n *= 7]; p5 *= 5]; p3 *= 3];
p2 *= 2]; Sort[L] (* Giovanni Resta, Oct 19 2012 *)

is_A218030(n)={my(d=digits(n^2));n*=2;for(i=1,#d,d[i]||next;n%d[i]&return;n\=d[i]);n==1} \\ M. F. Hasler, Oct 19 2012

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 *)

A218029 Numbers that are equal to the product of the nonzero digits (in base 10) of their square.

Original entry on oeis.org

1, 30240, 60480, 51597803520, 687009790646587541893939200, 2639216811747930700939756830720000000, 399953837788563897626396208106026886244605021117749733129069000654848000000000

Offset: 1

Nels Olson, Oct 18 2012

			For n=30240, n^2 = 914,457,600; the product of the nonzero digits of this square, 9*1*4*4*5*7*6 = 30240 = n.

Special case of A218013 where the ratio of the digit-product to the original number is 1. Related to A218072.

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

A218013 Numbers that divide the product of the nonzero digits (in base 10) of their square.

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A218030 Numbers k equal to half of the product of the nonzero (base-10) digits of k^2.

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

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A218029 Numbers that are equal to the product of the nonzero digits (in base 10) of their square.

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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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