A031348 2-multiplicative persistence: number of iterations of "multiply 2nd powers of digits" needed to reach 0 or 1.
0, 7, 6, 6, 3, 5, 5, 4, 5, 1, 1, 7, 6, 6, 3, 5, 5, 4, 5, 1, 7, 6, 5, 4, 2, 4, 5, 3, 4, 1, 6, 5, 5, 4, 3, 4, 4, 3, 4, 1, 6, 4, 4, 3, 2, 3, 3, 2, 4, 1, 3, 2, 3, 2, 3, 2, 3, 2, 2, 1, 5, 4, 4, 3, 2, 4, 5, 2, 4, 1, 5, 5, 4, 3, 3, 5, 2, 5, 4, 1, 4, 3, 3, 2, 2, 2, 5, 2, 3, 1, 5, 4, 4, 4, 2, 4, 4, 3, 3
Offset: 1
Examples
a(14) = 6 because 14 -> 1^2 * 4^2 = 16; 16 -> 1^2 * 6^2 = 36; 36 -> 3^2 * 6^2 = 324; 324 -> 3^2 * 2^2 * 4^2 = 576; 576 -> 5^2 * 7^2 * 6^2 = 44100; 44100 -> 0 => the trajectory is 14 -> 16 -> 36 -> 324 -> 576 -> 44100 -> 0 with 6 iterations. - _Michel Lagneau_, May 22 2013
References
- M. Gardner, Fractal Music, Hypercards and More Mathematical Recreations from Scientific American, Persistence of Numbers, pp. 120-1; 186-7, W. H. Freeman, NY, 1992.
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
- M. R. Diamond, Multiplicative persistence base 10: some new null results.
- N. J. A. Sloane, The persistence of a number, J. Recreational Math., 6 (1973), 97-98.
- Eric Weisstein's World of Mathematics, Multiplicative Persistence
Crossrefs
Cf. A031346.
Programs
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Mathematica
m2pd[n_]:=Length[NestWhileList[Times@@(IntegerDigits[#]^2)&,n,#>1&]]-1; Array[m2pd,100] (* Harvey P. Dale, Apr 19 2020 *)
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PARI
f(n) = my(d=digits(n)); prod(k=1, #d, d[k]^2); a(n) = if (n==1, 0, my(nb=1); while(((new = f(n)) > 1), n = new; nb++); nb); \\ Michel Marcus, Jun 13 2018
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Python
from math import prod from itertools import count, islice def f(n): return prod(map(lambda x: x*x, map(int, str(n)))) def a(n): c = 0 while n not in {0, 1}: n, c = f(n), c+1 return c print([a(n) for n in range(1, 100)]) # Michael S. Branicky, Oct 13 2022
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