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A236386
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Numbers m such that phi(m) is an oblong number.
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3
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3, 4, 6, 7, 9, 13, 14, 18, 21, 25, 26, 28, 31, 33, 36, 42, 43, 44, 49, 50, 62, 66, 73, 86, 87, 91, 95, 98, 111, 116, 117, 121, 135, 146, 148, 152, 157, 161, 169, 174, 182, 190, 201, 207, 211, 216, 222, 228, 234, 237, 241, 242, 252, 268, 270, 287, 289, 305
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OFFSET
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1,1
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COMMENTS
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An oblong number (A002378) is of the form k*(k+1) where k is a natural number.
Subsequence of primes is A002383 because in this case phi(k^2+k+1) = k*(k+1).
Subsequence of oblong numbers is A359847 where k and phi(k) are both oblong numbers. (End)
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LINKS
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EXAMPLE
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phi(13) = 12 = 3*4, an oblong number; so 13 is a term of the sequence.
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MAPLE
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filter := m -> issqr(1 + 4*phi(m)) : select(filter, [$(1 .. 700)]); # Bernard Schott, Feb 26 2023
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MATHEMATICA
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Select[Range[500], IntegerQ@Sqrt[1 + 4*EulerPhi[#]] &] (* Giovanni Resta, Jan 24 2014 *)
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PROG
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(PARI) isok(m) = my(t=eulerphi(m)); !(t%2) && ispolygonal(t/2, 3); \\ Michel Marcus, Feb 27 2023
(Python)
from itertools import count, islice
from sympy.ntheory.primetest import is_square
from sympy import totient
def A236386_gen(startvalue=1): # generator of terms >= startvalue
return filter(lambda n:is_square((totient(n)<<2)+1), count(max(1, startvalue)))
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CROSSREFS
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KEYWORD
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nonn,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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