A236386 Numbers m such that phi(m) is an oblong number.

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

Offset: 1

Author

Joseph L. Pe, Jan 24 2014

Keywords

• nonn
• easy

Comments

An oblong number (A002378) is of the form k*(k+1) where k is a natural number.

From Bernard Schott, Feb 27 2023: (Start)

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)

Examples

			phi(13) = 12 = 3*4, an oblong number; so 13 is a term of the sequence.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000010, A002378, A002383, A359847.

Similar, but where phi(m) is: A039770 (square), A039771 (cube), A078164 (biquadrate), A096503 (repdigit), A117296 (palindrome), A360944 (triangular).

Programs

Maple

filter := m -> issqr(1 + 4*phi(m)) : select(filter, [$(1 .. 700)]); # Bernard Schott, Feb 26 2023

Mathematica

Select[Range[500], IntegerQ@Sqrt[1 + 4*EulerPhi[#]] &] (* Giovanni Resta, Jan 24 2014 *)

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)))
A236386_list = list(islice(A236386_gen(),20)) # Chai Wah Wu, Feb 28 2023

Extensions

a(16)-a(58) from Giovanni Resta, Jan 24 2014