A276095 -id:A276095 - cp's OEIS frontend

A072882 A nonlinear recurrence of order 3: a(1)=a(2)=a(3)=1; a(n)=(a(n-1)+a(n-2))^2/a(n-3).

1, 1, 1, 4, 25, 841, 187489, 1418727556, 2393959458891025, 30567386265691995561839449, 658593751358960570203157512237008273218521, 181183406309644143341701434158730639946454023369335051404405528107396

Offset: 1

Benoit Cloitre, Jul 28 2002

All terms are perfect squares.

Seiichi Manyama, Table of n, a(n) for n = 1..17

Cf. A064098, A276095, A276097.

RecurrenceTable[{a[1]==1,a[2]==1,a[3]==1, a[n]==(a[n-1]+a[n-2])^2/a[n-3]},a,{n,1,10}] (* Vaclav Kotesovec, May 06 2015 *)

a(n) ~ 1/9 * c^(((1+sqrt(5))/2)^n), where c = 1.6403763522562240514693138664331346215549... . - Vaclav Kotesovec, May 06 2015
a(n) = A064098(n)^2. - Seiichi Manyama, Aug 18 2016
From Seiichi Manyama, Aug 26 2016: (Start)
a(n) = 9*a(n-1)*a(n-2) - 2*a(n-1) - 2*a(n-2) - a(n-3).
a(n)*a(n-1)*a(n-2) = ((a(n) + a(n-1) + a(n-2))/3)^2. (End)

A276045 Primes p such that d(p*(2p+1)) = 8 where d(n) is the number of divisors of n (A000005).

Original entry on oeis.org

7, 13, 17, 19, 43, 47, 59, 61, 71, 79, 101, 107, 109, 149, 151, 163, 167, 197, 223, 257, 263, 271, 311, 317, 347, 349, 353, 383, 389, 401, 421, 439, 449, 461, 479, 503, 521, 523, 557, 569, 599, 601, 613, 631, 673, 677, 691, 701, 811, 821, 827, 839, 853, 863, 881, 919

Offset: 1

Anthony Hernandez, Aug 17 2016

Primes p such that 2p+1 is in A030513. - Robert Israel, Aug 17 2016
From Anthony Hernandez, Aug 29 2016: (Start)
Conjecture: this sequence is infinite.
It appears that the prime numbers in this sequence which have 7 for as final digit form the sequence A104164.
Conjecture: this sequence contains infinitely many twin primes. The first few twin primes in this sequence are 17,19,59,61,107,109,521,523,599,601,... (End)
From Bernard Schott, Apr 28 2020: (Start)
This sequence equals the union of {13} and A234095; proof by double inclusion:
-> 1st inclusion: {13} Union A234095 is included in A276045.
1) if p = 13, then 13*27 = 351 = 3^3 * 13, hence d(351) = 8 and 13 belongs to A276045.
2) if p is in A234095, then p*(2*p+1) = p*r*s (p,r,s primes) and d(p*r*s) = 8, hence p is in 276045.
-> 2nd inclusion: A276045 is included in {13} Union A234095.
If p is in A276095, then m=p*(2*p+1) has 8 divisors and there are only three possibilities: m = u*v*w, or m = u^3*v or m = u^7 with u, v, w are distinct primes.
1st case: if p*(2*p+1) = u*v*w then u=p, and 2p+1=v*w is semiprime; hence, p is in A234095 Union {13}.
2nd case: if p*(2p+1) = u^3*v then  p=v and 2*p+1=u^3 ==> 2*p = u^3-1 = (u-1)*(u^2+u+1) with 2 and p are primes; then (u-1=2, u^2+u+1=p) so u=3, and p=3^2+3+1=13; hence p = 13 belongs to {13} Union A234095.
3rd case: p*(2p+1) = u^7 is impossible.
Conclusion: this sequence = {13} Union A234095. (End)

			d(7*(2*7+1))=d(105)=8 so 7 is a term.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000005, A030513, A030626.

Equals {13} Union A234095.

select(n -> isprime(n) and numtheory:-tau(n*(2*n+1))=8,
[seq(i, i=3..1000, 2)]); # Robert Israel, Aug 17 2016

Select[Prime@ Range@ 160, DivisorSigma[0, # (2 # + 1)] == 8 &] (* Michael De Vlieger, Aug 28 2016 *)

lista(nn) = forprime(p=2, nn, if (numdiv(p*(2*p+1))==8, print1(p, ", "))); \\ Michel Marcus, Aug 17 2016

A276097 A nonlinear recurrence of order 5: a(1)=a(2)=a(3)=a(4)=a(5)=1; a(n)=(a(n-1)+a(n-2)+a(n-3)+a(n-4))^2/a(n-5).

Original entry on oeis.org

1, 1, 1, 1, 1, 16, 361, 143641, 20741472361, 430214650013601071641, 11567790319010747187536221088708755344001, 370675271093071368960746074163948008803845834307486807769098691609909105887376

Offset: 1

Seiichi Manyama, Aug 18 2016

nonn

All terms are perfect squares.

Seiichi Manyama, Table of n, a(n) for n = 1..15

Cf. A072879, A072882, A276095.

RecurrenceTable[{a[1] == a[2] == a[3] == a[4] == a[5] == 1, a[n] == (a[n-1] + a[n-2] + a[n-3] + a[n-4])^2 / a[n-5]}, a, {n, 15}] (* Vincenzo Librandi, Aug 21 2016 *)

def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n
    i = a[1..-1].inject(:+)
    j = i * i
    break if j % a[0] > 0
    a = *a[1..-1], j / a[0]
    ary << a[0]
  end
  ary
end
def A276097(n)
  A(5, n)
end

a(n) = A072879(n)^2.
a(n) = 25*a(n-1)*a(n-2)*a(n-3)*a(n-4) - 2a(n-1) - 2a(n-2) - 2a(n-3) - 2a(n-4) - a(n-5).
a(n)*a(n-1)*a(n-2)*a(n-3)*a(n-4) = ((a(n) + a(n-1) + a(n-2) + a(n-3) + a(n-4))/5)^2.

A276256 A nonlinear recurrence of order 3: a(1)=a(2)=a(3)=1; a(n)=(a(n-1)+a(n-2)+1)^2/a(n-3).

Original entry on oeis.org

1, 1, 1, 9, 121, 17161, 33189121, 9112869675009, 4839170130591864288001, 705579627764469400107962287271157001, 54630717750911142085059359537536676834981970103054482555001

Offset: 1

Seiichi Manyama, Aug 25 2016

nonn

All terms are perfect squares.

Seiichi Manyama, Table of n, a(n) for n = 1..16

Cf. A276095.

a(n) = A276258(n)^2.
a(n) = 16*a(n-1)*a(n-2) - 2a(n-1) - 2a(n-2) - 2 - a(n-3).
a(n)*a(n-1)*a(n-2) = ((a(n) + a(n-1) + a(n-2) + 1)/4)^2.

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A072882 A nonlinear recurrence of order 3: a(1)=a(2)=a(3)=1; a(n)=(a(n-1)+a(n-2))^2/a(n-3).

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A276045 Primes p such that d(p*(2p+1)) = 8 where d(n) is the number of divisors of n (A000005).

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A276097 A nonlinear recurrence of order 5: a(1)=a(2)=a(3)=a(4)=a(5)=1; a(n)=(a(n-1)+a(n-2)+a(n-3)+a(n-4))^2/a(n-5).

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A276256 A nonlinear recurrence of order 3: a(1)=a(2)=a(3)=1; a(n)=(a(n-1)+a(n-2)+1)^2/a(n-3).

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