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%I #51 Mar 18 2023 08:49:14
%S 1,5,10,11,15,25,50,51,52,100,101,105,110,111,115,125,150,151,152,205,
%T 215,250,251,255,357,375,455,500,501,502,510,511,512,520,521,525,537,
%U 545,552,554,573,735,753,1000,1001,1005,1010,1011,1015,1025,1050,1051
%N Numbers whose trajectory under iteration of the product of squares of nonzero digits map includes 1.
%C To determine whether a given number k is a term of this sequence, start with k, take the square of the product of its nonzero digits, apply the same process to the result, and continue until 1 is reached or a loop is entered. If 1 is reached, k is a term of this sequence.
%C Every power 10^k is a term of this sequence.
%C If k is a term, the numbers obtained by inserting zeros anywhere in k are terms.
%C If k is a term, the numbers obtained by inserting ones anywhere in k are terms.
%C If k is a term, each distinct permutation of the digits of k gives another term.
%C If k is a term, the number of iterations required to converge to 1 is less than or equal to 3 (conjectured).
%C From _Michael S. Branicky_, Feb 07 2022: (Start)
%C The product of squares of nonzero digits map, f, has fixed points given in A115385.
%C The map f has (at least) the following cycles:
%C - 324, 576, 44100, 256, 3600;
%C - 11664, 20736, 63504, 129600;
%C - 15876, 2822400, 65536, 7290000;
%C - 5308416, 8294400;
%C - 49787136000000, 64524128256, 849346560000, 386983526400, 55725627801600.
%C (End)
%H Luca Onnis, <a href="https://arxiv.org/abs/2203.03381">On a variant of the happy numbers and their generalizations</a>, arXiv:2203.03381 [math.GM], 2022.
%e 255 is a term of the sequence: the square of the product of its nonzero digits is (2*5*5)^2=2500, the square of the product of its nonzero digits is (2*5)^2=100, and the square of the product of its nonzero digits is 1^2=1.
%e 2 is not a term of the sequence because its trajectory under the map is 2 -> 4 -> 16 -> 36 -> 324 -> 576 -> 44100 -> 256 -> 3600 -> 324 (reached earlier), so it enters a loop and never reaches 1.
%p b:= proc() false end:
%p q:= proc(n) local m, s; m, s:= n, {};
%p do if m=1 then return true
%p elif m in s or b(m) then b(n):= true; return false
%p else s, m:= {s[], m}, mul(max(1, i)^2, i=convert(m, base, 10))
%p fi
%p od
%p end:
%p select(q, [$1..2000])[]; # _Alois P. Heinz_, Feb 11 2022
%t Select[Range[1000],
%t FixedPoint[
%t Product[ReplaceAll[0 -> 1][IntegerDigits[#]][[i]]^2, {i, 1,
%t Length[ReplaceAll[0 -> 1][IntegerDigits[#]]]}] &, #, 10] == 1 &]
%o (Python)
%o from math import prod
%o def psd(n): return prod(int(d)**2 for d in str(n) if d != "0")
%o def ok(n):
%o seen = set()
%o while n not in seen: # iterate until fixed point or in cycle
%o seen.add(n)
%o n = psd(n)
%o return n == 1
%o def aupto(n): return [k for k in range(1, n+1) if ok(k)]
%o print(aupto(1205)) # _Michael S. Branicky_, Feb 07 2022
%o (PARI) f(n) = vecprod(apply(d -> if (d, d^2, 1), digits(n)))
%o is(n) = { my (m=f(n)); while (1, if (n==1, return (1), n==m, return (0), n=f(n); m=f(f(m)))) } \\ _Rémy Sigrist_, Feb 11 2022
%Y Cf. A007770, A051801, A115385.
%K nonn,base
%O 1,2
%A _Luca Onnis_, Feb 07 2022
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