

A276553


Numbers n such that n^2 and (n + 1)^2 have the same number of divisors.


3



2, 14, 15, 21, 33, 34, 38, 44, 57, 75, 81, 85, 86, 93, 94, 98, 116, 118, 122, 133, 135, 141, 142, 145, 147, 158, 171, 177, 201, 202, 205, 213, 214, 217, 218, 230, 244, 253, 272, 285, 296, 298, 301, 302, 326, 332, 334, 375, 381, 387, 393, 394, 405, 429, 434, 445
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OFFSET

1,1


COMMENTS

Except for a(1), all the terms are composite.


LINKS



EXAMPLE

We see that 14^2 = 196, the divisors of which are 1, 2, 4, 7, 14, 28, 49, 98, 196, and there are nine of them. And we see that 15^2 = 225, the divisors of which are 1, 3, 5, 9, 15, 25, 45, 75, 225, and there are nine of them. Both 14^2 and 15^2 have the same number of divisors, hence 14 is in the sequence.
And we see that 16^2 = 256, the divisors of which are the powers of 2 from 2^0 to 2^8, that's nine divisors. Both 15^2 and 16^2 have the same number of divisors, hence 15 is also in the sequence.
But 16 is not in the sequence, since 17 is prime and 17^2 consequently only has three divisors.


MAPLE

N:= 1000: # to get all terms <= N
T:= map(t > numtheory:tau(t^2), [$1..N+1]):


MATHEMATICA

Select[Range[1000], DivisorSigma[0, #^2] == DivisorSigma[0, (# + 1)^2] &]


PROG

(PARI) k=[]; for(n=1, 1000, a=numdiv(n^2); b=numdiv((n+1)^2); if(a==b, k=concat(k, n))); k
(Python)
from sympy.ntheory import divisor_count
print([n for n in range(1, 501) if divisor_count(n**2) == divisor_count((n + 1)**2)]) # Indranil Ghosh, Apr 10 2017


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



