A248648
The squares related to the strictly increasing subsequence of A053667(n), n >= 1.
Original entry on oeis.org
1, 4, 9, 25, 36, 49, 169, 256, 289, 576, 676, 1849, 3844, 3969, 5776, 6889, 26896, 27889, 55696, 69696, 97969, 339889, 376996, 499849, 678976, 698896, 779689, 2679769, 2768896, 2778889, 4695889, 4999696, 9696996, 26697889, 28879876, 36759969, 37994896
Offset: 1
5 * 5 = 25 is a square and the product of its digits = 2 * 5 = 10. Because a(3) = 9, and 4 * 4 = 16 has product of digits 6 < 9, a(4) = 25 because 10 > 9. The next entry a(5) comes from 6 * 6 = 36 with product of digits 18 > 10.
From _Wolfdieter Lang_, Oct 31 2014: (Start)
A053667 is sieved (from the left to the right):
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...
1, 4, 9, 6, 10, 18, 36, 24, 8, 0, 2, 16, 54, 54, ...
1, 4, 9, x, 10, 18, 36, x, x, x, x, x, 54, x, ...
and the related leftover squares are
1, 4, 9, 25 36, 49, 169, ...
(End)
-------------------------------------------------------
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A248648 = {}; k = 0; Do[s = Apply[Times, IntegerDigits[n^2]];If[s > k, k = s; AppendTo[A248648, n^2]], {n, 1, 10^4}]; A248648
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product=0;for(n=1,10^5,d=digits(n^2);p=prod(i=1,#d,d[i]);while(p>product,print1(n^2,", ");product=p)) \\ Derek Orr, Oct 11 2014
A248705
The cubes related to the strictly increasing subsequence of A053668(n), n >= 1.
Original entry on oeis.org
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
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)
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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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\\ 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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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
A257652
The semiprimes which set new records for the sum of their decimal digits.
Original entry on oeis.org
4, 6, 9, 38, 39, 49, 69, 169, 278, 289, 299, 489, 589, 689, 699, 799, 899, 2899, 3899, 4989, 5899, 5999, 6999, 7999, 9899, 19999, 29999, 48999, 58999, 68999, 69999, 88999, 99899, 299899, 398999, 589989, 589999, 689999, 798999, 889999, 899999, 2899999, 3899999
Offset: 1
a(4) = 38 = 2 * 19, which is a semiprime with sum of digits = 3 + 8 = 11.
a(5) = 39 = 3 * 13, which is a semiprime with sum of digits = 3 + 9 = 12. Since 12 > 11, 38 and 39 are in list.
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t = {}; s = 0; Do[If[(x = Total[IntegerDigits[n]]) > s && PrimeOmega[n] == 2, AppendTo[t, n]; s = x], {n, 1000000}];t
DeleteDuplicates[{#,Total[IntegerDigits[#]]}&/@Select[Range[4*10^6],PrimeOmega[#] == 2&],GreaterEqual[ #1[[2]],#2[[2]]]&][[;;,1]] (* Harvey P. Dale, Apr 12 2024 *)
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