

A161493


Positive integers, k, for which k mod d(k) and k have opposite (odd/even) parity, where d(k) is the number of divisors of k.


2



1, 4, 9, 16, 64, 100, 144, 196, 225, 324, 441, 484, 576, 625, 676, 900, 1024, 1089, 1296, 1521, 1764, 1936, 2025, 2116, 2304, 2601, 3136, 3249, 3364, 3844, 4096, 4225, 4356, 4761, 4900, 5625, 5776, 6084, 6400, 6561, 6724, 7396, 7569, 8649, 8836, 9216, 9801
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OFFSET

1,2


COMMENTS

It appears that the sequence {a(n)} consists entirely of squares. (This has been verified to a(431) = 998001 = 999^2.)
A number k appears in the sequence if and only if k is a square and floor(k/d(k)) is odd. This is because k mod d(k) = k  d(k) * floor(k/d(k)) and d(k) is odd if and only if k is square. [Hagen von Eitzen, Jun 12 2009]


LINKS

Ivan Neretin, Table of n, a(n) for n = 1..10000


EXAMPLE

k=4 has three divisors, so 4 mod d(4) = 1, which is odd. But 4 is even. Therefore 4 is a term of the sequence.
k=25 has three divisors, so 25 mod d(25) = 1, which is odd. 25 is also odd. Therefore 25 is not a term of the sequence.


MATHEMATICA

Select[Range[100]^2, OddQ@Quotient[#, DivisorSigma[0, #]] &] (* Ivan Neretin, Mar 23 2017 *)


PROG

(PARI) for(i=1, 999, k=i^2; if(floor(k/numdiv(k))%2, print1(k, ", "))) \\ Hagen von Eitzen, Jun 12 2009
(Python)
from sympy import divisor_count
print [n**2 for n in range(1, 10001) if int(math.floor(n**2/divisor_count(n**2)))%2 == 1] # Indranil Ghosh, Mar 23 2017


CROSSREFS

Cf. A000005, A161494 (gives the square roots).
Sequence in context: A023110 A277699 A073723 * A030075 A296111 A038784
Adjacent sequences: A161490 A161491 A161492 * A161494 A161495 A161496


KEYWORD

nonn


AUTHOR

John W. Layman, Jun 11 2009


STATUS

approved



