A350184 Numbers of multiplicative persistence 5 which are themselves the product of digits of a number.
2688, 18816, 26244, 98784, 222264, 262144, 331776, 333396, 666792, 688128, 1769472, 2939328, 3687936, 4214784, 4917248, 13226976, 19361664, 38118276, 71663616, 111476736, 133413966, 161414428, 169869312, 184473632, 267846264, 368947264, 476171136, 1783627776
Offset: 1
Examples
2688 is in this sequence because: - 2688 goes to a single digit in 5 steps: p(2688)=768, p(768)=336, p(336)=54, p(54)=20, p(20)=0. - p(27648) = p(47628) = 2688, etc. 331776 is in this sequence because: - 331776 goes to a single digit in 5 steps: p(331776)=2646, p(2646)=288, p(288)=128, p(128)=16, p(16)=6. - p(914838624) = p(888899) = 331776, etc.
Links
- Daniel Mondot, Table of n, a(n) for n = 1..41
- Daniel Mondot, Multiplicative Persistence Tree
- Eric Weisstein's World of Mathematics, Multiplicative Persistence
Crossrefs
Programs
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Mathematica
mx=10^10;lst=Sort@Flatten@Table[2^i*3^j*5^k*7^l,{i,0,Log[2,mx]},{j,0,Log[3,mx/2^i]},{k,0,Log[5,mx/(2^i*3^j)]},{l,0,Log[7,mx/(2^i*3^j*5^k)]}]; Select[lst,Length@Most@NestWhileList[Times@@IntegerDigits@#&,#,#>9&]==5&] (* code for 7-smooth numbers from A002473. - Giorgos Kalogeropoulos, Jan 16 2022 *)
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Python
from math import prod def hd(n): while (n&1) == 0: n >>= 1 while (n%3) == 0: n /= 3 while (n%5) == 0: n /= 5 while (n%7) == 0: n /= 7 return(n) def pd(n): return prod(map(int, str(n))) def ok(n): if hd(n) > 9: return False return (p := pd(n)) > 9 and (q := pd(p)) > 9 and (r := pd(q)) > 9 and (s := pd(r)) > 9 and pd(s) < 10 print([k for k in range(10,476200000) if ok(k)])
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