A350183 Numbers of multiplicative persistence 4 which are themselves the product of digits of a number.
378, 384, 686, 768, 1575, 1764, 2646, 4374, 6144, 6174, 6272, 7168, 8232, 8748, 16128, 21168, 23328, 27216, 28672, 32928, 34992, 49392, 59535, 67228, 77175, 96768, 112896, 139968, 148176, 163296, 214326, 236196, 393216, 642978, 691488, 774144, 777924
Offset: 1
Examples
384 is in this sequence because: - 384 goes to a single digit in 4 steps: p(384)=96, p(96)=54, p(54)=20, p(20)=0. - p(886)=384, p(6248)=384, p(18816)=384, etc. 378 is in this sequence because: - 378 goes to a single digits in 4 steps: p(378)=168, p(168)=48, p(48)=32, p(32)=6. - p(679)=378, p(2397)=378, p(12379)=378, etc.
Links
- Daniel Mondot, Table of n, a(n) for n = 1..142
- Eric Weisstein's World of Mathematics, Multiplicative Persistence
- Daniel Mondot, Multiplicative Persistence Tree
Crossrefs
Programs
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Mathematica
mx=10^6;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)]}]; (* from A002473 *) Select[lst,Length@Most@NestWhileList[Times@@IntegerDigits@#&,#,#>9&]==4&] (* Giorgos Kalogeropoulos, Jan 16 2022 *)
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PARI
pd(n) = if (n, vecprod(digits(n)), 0); \\ A007954 mp(n) = my(k=n, i=0); while(#Str(k) > 1, k=pd(k); i++); i; \\ A031346 isok(k) = (mp(k)==4) && (vecmax(factor(k)[,1]) <= 7); \\ Michel Marcus, Jan 25 2022
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Python
from math import prod from sympy import factorint def pd(n): return prod(map(int, str(n))) def ok(n): if n <= 9 or max(factorint(n)) > 9: return False return (p := pd(n)) > 9 and (q := pd(p)) > 9 and (r := pd(q)) > 9 and pd(r) < 10 print([k for k in range(778000) if ok(k)])
Comments