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
Keywords
Examples
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.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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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
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Mathematica
Select[Range[1000], DivisorSigma[0, #^2] == DivisorSigma[0, (# + 1)^2] &]
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PARI
k=[]; for(n=1, 1000, a=numdiv(n^2); b=numdiv((n+1)^2); if(a==b, k=concat(k, n))); k
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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
Comments