A276553 Numbers n such that n^2 and (n + 1)^2 have the same number of divisors.

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

Offset

1,1

Comments

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

Links

Antti Karttunen, Table of n, a(n) for n = 1..10000

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]):
select(t -> T[t]=T[t+1], [$1..N]); # Robert Israel, Apr 10 2017

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
(Scheme, with Antti Karttunen's IntSeq-library) (define A276553 (ZERO-POS 1 1 A284570)) ;; Antti Karttunen, Apr 15 2017

Crossrefs

Cf. A000290, A048691, A062832, A276542, A284378.

Cf. A052213 (a subsequence).

Positions of zeros in A284570.

Sequence in context: A072163 A084674 A009774 * A032933 A194230 A109993

Adjacent sequences: A276550 A276551 A276552 * A276554 A276555 A276556

Keyword

nonn

Author

K. D. Bajpai, Apr 10 2017

Status

approved