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%I #39 Jan 04 2022 05:53:09
%S 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,
%T 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,
%U 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
%N a(n) = A261327(n)/A013946(n).
%C Are all terms odd squares?
%C b(n) = A013946(n)*A261327(n) = 25, 4, 169, 25, 841, 100, 2809, 289, 7225, 676, 625, ... . Are all terms squares?
%C 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
%C Are the square roots only primes?
%C The sequence of period 4: repeat [25, 1, 1, 25] appears apparently every 25 terms.
%C From _Robert Israel_, Mar 20 2019: (Start)
%C The first term whose square root is not 1 or a prime is a(261) = 25^2.
%C 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.
%C The first term where a(12+25*k) > 1 is a(1212)=a(12+25*48)=169.
%C The first term where a(13+25*k) > 1 is a(213)=a(13+25*8)=289.
%C 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.
%C 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).
%C - _Robert Israel_, Mar 20 2019
%H Robert Israel, <a href="/A306454/b306454.txt">Table of n, a(n) for n = 1..10000</a>
%e A261327(n) = 5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 125, 37, 173, 50, ... .
%e A013946(n) = 5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 5, 37, 173, 2, ... .
%p core:= proc(n) local t; mul(t[1],t=select(s -> s[2]::odd, ifactors(n)[2])) end proc:
%p map(n -> numer((4+n^2)/4)/core(n^2+4), [$1..100]); # _Robert Israel_, Mar 20 2019
%t core[n_] := Times @@ Select[FactorInteger[n], OddQ[#[[2]]]&][[All, 1]];
%t a[n_] := Numerator[(n^2+4)/4]/core[n^2+4];
%t Array[a, 100] (* _Jean-François Alcover_, Jan 04 2022 *)
%o (PARI)
%o A013946(n) = core(n^2+4); \\ From A013946
%o A261327(n) = if(n%2, n^2+4, (n/2)^2+1); \\ From A261327
%o A306454(n) = (A261327(n)/A013946(n)); \\ _Antti Karttunen_, Feb 28 2019
%o (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
%Y Cf. A013946, A261327.
%Y Cf. A002144, A008833.
%K nonn
%O 1,11
%A _Paul Curtz_, Feb 16 2019
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