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A046758
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Equidigital numbers.
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11
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1, 2, 3, 5, 7, 10, 11, 13, 14, 15, 16, 17, 19, 21, 23, 25, 27, 29, 31, 32, 35, 37, 41, 43, 47, 49, 53, 59, 61, 64, 67, 71, 73, 79, 81, 83, 89, 97, 101, 103, 105, 106, 107, 109, 111, 112, 113, 115, 118, 119, 121, 122, 123, 127, 129, 131, 133, 134, 135, 137, 139
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OFFSET
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1,2
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COMMENTS
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Write n as product of primes raised to powers, let D(n) = A050252 = total number of digits in product representation (number of digits in all the primes plus number of digits in all the exponents that are greater than 1) and l(n) = number of digits in n; sequence gives n such that D(n)=l(n).
The term "equidigital number" was coined by Recamán (1995). - Amiram Eldar, Mar 10 2024
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REFERENCES
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Bernardo Recamán Santos, Equidigital representation: problem 2204, J. Rec. Math., Vol. 27, No. 1 (1995), pp. 58-59.
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LINKS
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FORMULA
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EXAMPLE
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For n = 125 = 5^3, l(n) = 3 but D(n) = 2. So 125 is not a member of this sequence.
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MATHEMATICA
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edQ[n_] := Total[IntegerLength[DeleteCases[Flatten[FactorInteger[n]], 1]]] == IntegerLength[n]; Join[{1}, Select[Range[140], edQ]] (* Jayanta Basu, Jun 28 2013 *)
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PROG
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(Haskell)
a046758 n = a046758_list !! (n-1)
a046758_list = filter (\n -> a050252 n == a055642 n) [1..]
(PARI) for(n=1, 100, s=""; F=factor(n); for(i=1, #F[, 1], s=concat(s, Str(F[i, 1])); s=concat(s, Str(F[i, 2]))); c=0; for(j=1, #F[, 2], if(F[j, 2]==1, c++)); if(#digits(n)==#s-c, print1(n, ", "))) \\ Derek Orr, Jan 30 2015
(Python)
from itertools import count, islice
from sympy import factorint
def A046758_gen(): # generator of terms
return (n for n in count(1) if n == 1 or len(str(n)) == sum(len(str(p))+(len(str(e)) if e > 1 else 0) for p, e in factorint(n).items()))
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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