A306454 a(n) = A261327(n)/A013946(n).

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 25, 1, 1, 25, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 169, 1, 1, 1, 1, 1, 1, 25, 1, 1, 25, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 25, 1, 1, 25, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 289, 1, 1, 1, 1, 1, 841, 1, 1, 1, 25, 1, 1, 25, 1, 1, 1

Offset: 1

Author

Paul Curtz, Feb 16 2019

Keywords

• nonn

Comments

Are all terms odd squares?

b(n) = A013946(n)*A261327(n) = 25, 4, 169, 25, 841, 100, 2809, 289, 7225, 676, 625, ... . Are all terms squares?

a(n) = A008833(n^2+4) if n is odd and A008833((n^2+4)/4) if n is even, so a(n) is always an odd square. - Jianing Song, Feb 27 2019

Are the square roots only primes?

The sequence of period 4: repeat [25, 1, 1, 25] appears apparently every 25 terms.

From Robert Israel, Mar 20 2019: (Start)

The first term whose square root is not 1 or a prime is a(261) = 25^2.

a(11+25*k) is divisible by 25. The first term where a(11+25*k) > 25 is a(261)=a(11+25*10)=625.

The first term where a(12+25*k) > 1 is a(1212)=a(12+25*48)=169.

The first term where a(13+25*k) > 1 is a(213)=a(13+25*8)=289.

a(14+25*k) is divisible by 25. The first term where a(14+25*k) > 25 is a(364)=a(14+25*14)=625.

All prime factors of members of the sequence are in A002144. For any p in A002144, there is k with 1 <= k < p^2/2 such that p^2 | a(n) if and only if n == k or -k (mod p^2). (End)

Examples

			A261327(n) = 5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 125, 37, 173, 50, ... .
A013946(n) = 5, 2, 13, 5, 29, 10, 53, 17, 85, 26,   5, 37, 173,  2, ... .
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A013946, A261327.

Cf. A002144, A008833.

Programs

Maple

core:= proc(n) local t; mul(t[1],t=select(s -> s[2]::odd, ifactors(n)[2])) end proc:
map(n -> numer((4+n^2)/4)/core(n^2+4), [$1..100]); # Robert Israel, Mar 20 2019

Mathematica

core[n_] := Times @@ Select[FactorInteger[n], OddQ[#[[2]]]&][[All, 1]];
a[n_] := Numerator[(n^2+4)/4]/core[n^2+4];
Array[a, 100] (* Jean-François Alcover, Jan 04 2022 *)

PARI

A013946(n) = core(n^2+4); \\ From A013946
A261327(n) = if(n%2, n^2+4, (n/2)^2+1); \\ From A261327
A306454(n) = (A261327(n)/A013946(n)); \\ Antti Karttunen, Feb 28 2019

PARI

A306454(n) = { my(k=((n^2)+4)/if(n%2,1,4)); k/core(k); }; \\ Antti Karttunen, Feb 28 2019, after Jianing Song's formula