Robert Israel, <a href="/A362499/b362499_1.txt">Table of n, a(n) for n = 0..1000</a>

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(Python)

from sympy import factorint

from sympy.utilities.iterables import multiset_permutations as mp

from itertools import count, islice, combinations_with_replacement as mc

def ndgen():

yield from ((f, )+r for d in count(1) for f in "123456789" for r in mc("0123456789", d-1))

def c(n): # is_semiprime

return sum(factorint(n).values()) == 2

def f(digs):

return sum(1 for p in mp(digs) if p[0]!="0" and c(t:=int("".join(p))))

def agen(): # generator of terms

adict, n = dict(), 0

for t in ndgen():

v = f(t)

if v not in adict: adict[v] = int("".join(t))

while n in adict: yield adict[n]; n += 1

print(list(islice(agen(), 46))) # Michael S. Branicky, Jun 12 2023

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Robert Israel, <a href="/A362499/b362499_1.txt">Table of n, a(n) for n = 0..1000</a>

nonn,base,changed,look

allocated for Robert Israela(n) is the least positive integer that has exactly n anagrams that are semiprimes, or -1 if there is no such integer.

1, 4, 15, 123, 129, 134, 178, 1025, 1148, 1147, 1137, 1145, 1349, 1348, 1357, 10145, 3589, 10258, 10137, 10123, 11269, 10289, 10268, 10247, 10235, 10267, 10234, 10789, 10279, 11378, 10378, 12369, 10349, 10358, 12368, 10357, 12689, 12358, 10459, 12379, 12679, 13489, 12346, 12349, 16789, 12479

0,2

a(n) is the least k such that A131371(k) = n.

Leading zeros are not allowed.

a(3) = 123 because 123 has 3 anagrams that are semiprimes, namely 123 = 3 * 41, 213 = 3 * 71, and 321 = 3 * 107, and no smaller number works.

g:= proc(s, m) local t;

if s[1..m-1] = [0$(m-1)] then op(map(t -> [t, op(s)], [0, $(max(s) ..9)]))

else op(map(t -> [t, op(s)], [$(max(s) .. 9)]))

fi

end proc:

f:= proc(L, m) local P, t, i;

P:= select(t -> t[-1] <> 0 and numtheory:-bigomega(add(t[i]*10^(i-1), i=1..m))=2, combinat:-permute(L));

nops(P)

end proc:

V:= Array(0..100):

count:= 2: V[0]:= 1: V[1]:= 4:

L:= [seq(seq([b, a], b=[0, $a..9]), a=1..9)]:

for m from 2 while count < 101 do

for s in L while count < 101 do

v:= f(s, m);

if v <= 100 and V[v] = 0 then

V[v]:= add(s[i]*10^(i-1), i=1..m); count:= count+1;

fi

od;

L:= map(g, L, m)

od:

convert(V, list);

Cf. A131371, A001358.

allocated

nonn,base

Robert Israel, Jun 11 2023

approved

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allocated for Robert Israel

recycled

allocated

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a(n) is the maximum level of nesting of exponentiation required to write n in its prime exponent product form, with the exponents being written the same way, recursively.

1, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 3, 2, 2, 2, 4, 2, 3, 2, 3, 2, 2, 2, 3, 3, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 4, 3, 3, 2, 3, 2, 3, 2, 3, 2, 2, 2, 3, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 3, 3, 2, 2, 2, 4, 4, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 3, 2, 2, 2, 3, 2, 3, 3, 3

1,2

a(n) = 1 + max(a(e_i)), where n = 2^e_2 * 3^e_3 * 5^e_5 * 7^e_7 * ... * i^e_i, with i being a prime.

a(A005117(n)) = 2. - Jodi Spitz, Apr 22 2023

a(n) = A185102(n) + 1. - Pontus von Brömssen, Apr 22 2023

Take the height of the tallest power tower excluding the topmost 0:

For n = 1, 1 = 2^0 * 3^0 * 5^0 * 7^0 * ... * p_i^0, tallest power tower is p^0, so a(1) = 1.

For n = 2, 2 = 2^p^0, which is the tallest power tower for n = 2, so a(2) = 2.

For n = 12, 12 = 2^2 * 3^1 = 2^2^p^0 * 3^p^0, the tallest power tower is 2^2^p^0, so a(12) = 3.

For n = 16, 16 = 2^4 = 2^2^2 = 2^2^2^p^0, the tallest power tower is 2^2^2^p^0, so a(16) = 4.

p is an arbitrary prime in all of the examples.

Cf. A185102.

nonn

recycled

Daniyar Sarbasov, Apr 22 2023

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