A120905 -id:A120905 - cp's OEIS frontend

A176594 a(n) = 5^(2^n).

5, 25, 625, 390625, 152587890625, 23283064365386962890625, 542101086242752217003726400434970855712890625, 293873587705571876992184134305561419454666389193021880377187926569604314863681793212890625

Offset: 0

Vincenzo Librandi, Apr 21 2010

nonn

Also the hypotenuse of primitive Pythagorean triangles obtained by repeated application of basic formula c(n)=p(n)^2+q(n)^2 starting p(0)=2, q(0)=1, see A100686, A098122. Example: a(2)=25 since starting (2,1) gives Pythagorean triple (3,4,5) using (3,4) as new generators gives triple (7,24,25) hypotenuse 25=a(2). - Carmine Suriano, Feb 04 2011

Paolo Xausa, Table of n, a(n) for n = 0..10

Cf. A001146, A011764, A078886, A098122, A100686, A165423.

Cf. A185457, A120905, A139011. - Carmine Suriano, Feb 04 2011

NestList[#^2 &, 5, 8] (* Paolo Xausa, Jul 13 2025 *)

a(n) = 5^(2^n); \\ Michel Marcus, Jan 26 2016

a(n) = A165423(n+3).
a(n+1) = a(n)^2 with a(0)=5.
a(n-1) = (Im((2+i)^(2^n))^2 + Re((2+i)^(2^n))^2)^(1/2). - Carmine Suriano, Feb 04 2011
Sum_{n>=0} 1/a(n) = A078886. - Amiram Eldar, Nov 09 2020
Product_{n>=0} (1 + 1/a(n)) = 5/4. - Amiram Eldar, Jan 29 2021

Offset corrected by R. J. Mathar, Jun 18 2010

A121705 Triangle read by rows: 5^n expressed as the sum of two squares.

Original entry on oeis.org

0, 1, 1, 2, 0, 5, 3, 4, 2, 11, 5, 10, 0, 25, 7, 24, 15, 20, 10, 55, 25, 50, 38, 41, 0, 125, 35, 120, 44, 117, 75, 100, 29, 278, 50, 275, 125, 250, 190, 205, 0, 625, 175, 600, 220, 585, 336, 527, 375, 500, 145, 1390, 250, 1375, 625, 1250, 718, 1199, 950, 1025, 0, 3125

Offset: 0

Zak Seidov, Sep 10 2006

			5^n expressed as the sum of two squares: 5^n = x^2 + y^2, 0 <= x < y.
Number of solutions for n=0,1,...: a(n)=1,1,2,2,3,3,4,4,5,5,6,6,...
Triangle of solutions for n=0,1,...:
  {x,y}
  {{0,1}},
  {{1,2}},
  {{0,5},{3,4}},
  {{2,11},{5,10}},
  {{0,25},{7,24},{15,20}},
  {{10,55},{25,50},{38,41}},
  {{0,125},{35,120},{44,117},{75,100}},
  {{29,278},{50,275},{125,250},{190,205}},
  {{0,625},{175,600},{220,585},{336,527},{375,500}},
  {{145,1390},{250,1375},{625,1250},{718,1199},{950,1025}},
  {{0,3125},{237,3116},{875,3000},{1100,2925},{1680,2635},{1875,2500}},
  {{725,6950},{1250,6875},{2642,6469},{3125,6250},{3590,5995},{4750,5125}},
  {{0,15625},{1185,15580},{4375,15000},{5500,14625},{8400,13175},{9375,12500},{10296,11753}}

Cf. A004431, A006496, A024507, A120905.

A185457 a(n) = abs( Im((2+i)^(2^n)) ).

Original entry on oeis.org

1, 4, 24, 336, 354144, 116749235904, 22940770664883067253376, 182503181432559739767250904458105698387204864

Offset: 0

Carmine Suriano, Feb 04 2011

The next term is too large to be displayed here.

Old name was: Leg of primitive Pythagorean triangle generated by repeated application of the basic formula y(n) = 2*x(n-1)*y(n-1), x(1)=2, y(1)=1.

			y(2)=24 since x(1)=3, y(1)=4 are the two legs of Pythagorean triangle obtained by p=2, q=1; second iteration p=3, q=4 gives 2*3*4=24.

Alois P. Heinz, Table of n, a(n) for n = 0..11

Cf. A176594, A120905.

Cf. A099456 ( imaginary part of (2+i)^n ).

a:= n-> abs(Im((2+I)^(2^n))):
seq(a(n), n=0..8);  # Alois P. Heinz, Apr 25 2013

Table[Abs[Im[(2 + I)^(2^n)]], {n, 0, 10}] (* G. C. Greubel, Jul 07 2017 *)

a(n) = abs(imag((2+I)^(2^n))); \\ Joerg Arndt, Apr 25 2013

from sympy import im, I
def a(n): return abs(im((2 + I)**(2**n)))
print([a(n) for n in range(11)]) # Indranil Ghosh, Jul 08 2017

a(n) = abs( Im((2+i)^(2^n)) ).

Better name from Joerg Arndt, Apr 25 2013

cp's OEIS Frontend

A176594 a(n) = 5^(2^n).

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A121705 Triangle read by rows: 5^n expressed as the sum of two squares.

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A185457 a(n) = abs( Im((2+i)^(2^n)) ).

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