A144837 -id:A144837 - cp's OEIS frontend

A268922 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 1 mod 5 numbers, except for n = 0.

0, 1, 11, 11, 261, 2136, 2136, 64636, 220886, 1392761, 7252136, 46314636, 241627136, 974049011, 2194752136, 8298267761, 99851002136, 710202564636, 710202564636, 12154294361511, 31227780689636

Offset: 0

Wolfdieter Lang, Mar 02 2016

The other approximation for the 5-adic integer sqrt(-4) with numbers 4 (mod 5) is given in A269590.
For the two approximations of the 5-adic integer sqrt(-1) see A048899 and A048898. For comments and some proofs see A210848.
For the digits of this 5-adic integer sqrt(-4), that is the scaled first differences, see A269591. This 5-adic number has the digits read from the right to the left ...32234111132111101130213043113443324032021 = u.
  The companion 5-adic number -u has digits ....12210333312333343314231401331001120412424. See A269592.
The recurrence given below (for n >= 1) has been derived from the following facts (i) x^2 + 4 == 0 (mod 5) has the two distinct solutions x(1) = 1 and x(2) = 5 - x(1) = 4. This guarantees the existence of a unique solution x = x1(n) of  x^2 + 4 == 0 (mod 5^n) , for n >= 2 , which satisfies also x1(n) == 1 (mod 5). See e.g., Theorem 50, p. 87 of the Nagell reference. The same is true for the solution x = x2(n) with x2(n) == 4 (mod 5^n). (ii) As a consequence of Hensel's lemma (see e.g., the Wikipedia reference under Hensel lifting) one knows that  x1(n) (which is treated here) satisfies the congruence x1(n) == x1(n-1) (mod 5^(n-1)) with x1(1) = x1 = 1. (A similar statement holds for x2(n) with input x2(1) = x(2) = 4. This is used in A269590). These two facts allow one to derive a recurrence for x1(n) (and for x2(n)).

			n=2:  11^2 + 4 = 125  == 0 (mod 5^2), and 125 is the only solution from {0, 1, ..., 24} which is congruent to 1 modulo 5.
n=3:  the only solution of x1^2 + 4 == 0 (mod 5^3) with x1 from {0, ..., 124} and x1 ==  1 (mod 4) is also 11. The number 114 satisfies also the first congruence but not the second one: 114 == 2 (mod 4).

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, p. 87.

Seiichi Manyama, Table of n, a(n) for n = 0..1430
Wolfdieter Lang, Note on a Recurrence for Approximation Sequences of p-adic Square Roots
Wikipedia, Hensel's lemma.

Cf. A000032, A048899, A048898, A144837, A210848, A269590, A269591, A269592.

with(padic):D1:=op(3,op([evalp(RootOf(x^2+4),5,20)][1])): 0,seq(sum('D1[k]*5^(k-1)','k'=1..n), n=1..20);
# alternative program - see A144837
a := proc (n) option remember; if n = 1 then 1 else irem( a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1), 5^n) end if; end: seq(a(n), n = 1..20); # Peter Bala, Nov 14 2022

a(n) = truncate(sqrt(-4+O(5^(n)))); \\ Michel Marcus, Mar 04 2016

Recurrence for n >= 1: a(n) = modp( a(n-1) + 2*(a(n-1)^2 + 4), 5^n), n >= 2, with a(1) = 1. Here modp(a, m) is used to pick the representative of the residue class a modulo m from the smallest nonnegative complete residue system {0, 1, ... , m-1}.
a(n) = 5^n - A269590(n), n >= 1.
a(n) == Lucas(5^n) (mod 5^n). - Peter Bala, Nov 10 2022

A144836 a(n) = round(phi^(4^n)) where phi is the golden ratio (A001622).

Original entry on oeis.org

2, 7, 2207, 23725150497407, 316837008400094222150776738483768236006420971486980607

Offset: 0

Artur Jasinski, Sep 22 2008

Amiram Eldar, Table of n, a(n) for n = 0..6

Cf. A000032, A001566, A001622, A006267, A144837, A144838, A144839.

Cf. A374883 (phi^4).

a := proc(n) option remember; if n = 0 then 2 elif n = 1 then 7 else a(n-1)^4 - 4*a(n-1)^2 + 2 end if; end proc: seq(a(n), n = 0..4); # Peter Bala, Nov 28 2022

Table[Round[GoldenRatio^(4^n)], {n, 0, 5}]
c = (1 + Sqrt[5])/2; Join[{2}, Table[Expand[c^(4^n) + (1 - c)^(4^n)], {n, 1, 5}]] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[2*Cosh[4^n*ArcCosh[Sqrt[5]/2]]], {n, 0, 5}] (* Artur Jasinski, Oct 09 2008 *)
a[n_] := LucasL[4^n]; a[0] = 2; Array[a, 5, 0] (* Amiram Eldar, Jul 12 2025 *)

a(n)=round(((1+sqrt(5))/2)^4^n) \\ Charles R Greathouse IV, Jul 29 2011

a(n) = Lucas(4^n) = A000032(4^n), n>0.
a(n) = phi^(4^n) + (1 - phi)^(4^n) = phi^(4^n) + (-phi)^(-4^n), where phi is golden ratio = (1 + sqrt(5))/2 = 1.6180339887..., n>0. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(4^n*arccosh(sqrt(5)/2)), n>0. - Artur Jasinski, Oct 09 2008
a(n+1) = a(n)^4 - 4*a(n-1)^2 + 2 with a(1) = 7. - Peter Bala, Nov 28 2022

Offset corrected by Charles R Greathouse IV, May 15 2013
Offset changed to 0 by Georg Fischer, Sep 02 2022
New name from Peter Bala, Nov 18 2022
Revised by editors, Jul 12 2025

A144838 a(n) = Lucas(6^n).

Original entry on oeis.org

18, 33385282, 1384619022984618483717737087933569992335566082

Offset: 1

Artur Jasinski, Sep 22 2008

nonn

Previous name was: a(n) = round(phi^(6^n)) where phi = 1.6180339887498948482... = (sqrt(5)+1)/2.

General (hyperbolic) trigonometric formula for a(n) = round(phi^((2*k)^n)) = 2*cosh((2*k)^n*arccosh(sqrt(5)/2)) where phi = 1.6180339887498948482... = (sqrt(5)+1)/2. - Artur Jasinski, Oct 09 2008

Amiram Eldar, Table of n, a(n) for n = 1..4

Cf. A001566, A006267, A144836, A144837, A144838, A144839.

a := proc(n) option remember; if n = 1 then 18 else a(n-1)^6 - 6*a(n-1)^4 + 9*a(n-1)^2 - 2 end if; end;
seq(a(n), n = 1..5); # Peter Bala, Nov 28 2022

Table[Round[GoldenRatio^(6^n)], {n, 1, 5}]
c = (1 + Sqrt[5])/2; Table[Expand[c^(6^n) + (1 - c)^(6^n)], {n, 1, 5}] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[2*Cosh[6^n*ArcCosh[Sqrt[5]/2]]], {n, 1, 4}] (* Artur Jasinski, Oct 09 2008 *)
Table[LucasL[6^n], {n, 1, 4}] (* Amiram Eldar, Jul 13 2025 *)

a(n) = G^(6^n) + (1 - G)^(6^n) = G^(6^n) + (-G)^(-6^n) where G is the golden ratio A001622. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(6^n*arccosh(sqrt(5)/2)). - Artur Jasinski, Oct 09 2008
From Peter Bala, Nov 28 2022: (Start)
a(n) = Lucas(6^n).
a(n+1) = a(n)^6 - 6*a(n)^4 + 9*a(n)^2 - 2 with a(1) = 18. (End)

New name from Peter Bala, Nov 28 2022

A144839 a(n) = Lucas(7^n).

Original entry on oeis.org

29, 17393796001, 481682208844384447843365760878364816732549453120338354329505085763436029

Offset: 1

Artur Jasinski, Sep 22 2008

nonn

Previous name was: a(n) = round(phi^(7^n)) where phi = 1.6180339887498948482... = (sqrt(5)+1)/2.

Amiram Eldar, Table of n, a(n) for n = 1..4

Cf. A000032, A001622, A001566, A006267, A144836, A144837, A144838, A144839.

a := proc(n) option remember; if n = 0 then 1 else a(n-1)^7 + 7*a(n-1)^5 + 14*a(n-1)^3 +7*a(n-1) end if; end;
seq(a(n), n = 1..5); # Peter Bala, Nov 28 2022

Table[Round[GoldenRatio^(7^n)], {n, 1, 5}]
c = (1 + Sqrt[5])/2; Table[Expand[c^(7^n) + (1 - c)^(7^n)], {n, 1, 5}] (* Artur Jasinski, Oct 05 2008 *)
Table[LucasL[7^n], {n, 1, 4}] (* Amiram Eldar, Jul 13 2025 *)

a(n) = G^(7^n) + (1 - G)^(7^n) = G^(7^n) + (-G)^(-7^n) where G is the golden ratio A001622. [Artur Jasinski, Oct 05 2008]
From Peter Bala, Nov 28 2022: (Start)
a(n) = Lucas(7^n).
a(n+1) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n) with a(0) = 1.
a(n) == 1 (mod 7).
a(n+1) == a(n) (mod 7^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
In the ring of 7-adic integers, the limit_{n -> oo} a(n) exists and is a root of the quartic equation x^4 + 4*x^2 + 2 = 0. (End)

New name from Peter Bala, Nov 28 2022

A371649 Decimal expansion of Sum_{k>=0} 1/Lucas(5^k).

Original entry on oeis.org

1, 0, 9, 0, 9, 1, 5, 0, 5, 1, 7, 7, 0, 0, 7, 7, 6, 7, 0, 0, 1, 8, 6, 5, 7, 5, 0, 2, 4, 1, 4, 2, 2, 8, 2, 0, 5, 7, 1, 5, 1, 7, 5, 1, 0, 2, 3, 1, 9, 9, 0, 6, 6, 9, 8, 9, 0, 5, 0, 3, 2, 1, 7, 0, 9, 2, 2, 2, 4, 3, 0, 8, 1, 7, 5, 8, 2, 8, 8, 4, 4, 6, 4, 9, 0, 2, 6, 3, 1, 9, 6, 7, 3, 7, 2, 4, 8, 1, 8, 3, 1, 2, 4, 1, 7

Offset: 1

Amiram Eldar, Mar 31 2024

This constant is a transcendental number (Nyblom, 2001).

			1.09091505177007767001865750241422820571517510231990...

M. A. Nyblom, A Theorem on Transcendence of Infinite Series II, Journal of Number Theory, Vol. 91, No. 1 (2001), pp. 71-80.
Index entries for transcendental numbers.

Cf. A000032, A144837, A371650.

Similar constants: A093540, A338304, A338612, A371647.

RealDigits[Sum[1/LucasL[5^k], {k, 0, 10}], 10, 120][[1]]

suminf(k = 0, 1/(fibonacci(5^k-1) + fibonacci(5^k+1)))

Equals Sum_{k>=0} 1/A144837(k).

A371650 Decimal expansion of Product_{k>=0} (1 + 1/Lucas(5^k)).

Original entry on oeis.org

2, 1, 8, 1, 8, 3, 1, 1, 8, 7, 3, 3, 3, 0, 6, 2, 0, 2, 3, 8, 4, 2, 3, 6, 0, 1, 7, 0, 5, 5, 6, 7, 6, 1, 0, 6, 7, 8, 0, 2, 4, 3, 5, 6, 4, 6, 9, 5, 7, 9, 6, 9, 2, 4, 1, 3, 1, 7, 0, 3, 2, 2, 5, 2, 9, 1, 3, 8, 0, 9, 1, 2, 6, 4, 3, 6, 0, 1, 6, 1, 9, 9, 2, 4, 7, 4, 9, 2, 0, 7, 3, 4, 0, 6, 4, 4, 1, 6, 0, 3, 5, 5, 7, 5, 0, 0, 2

Offset: 1

Amiram Eldar, Mar 31 2024

This constant is a transcendental number (Nyblom, 2004).

			2.18183118733306202384236017055676106780243564695796...

M. A. Nyblom, On the Construction of a Family of Transcendental Valued Infinite Products, Fibonacci Quarterly, Vol. 42, No. 4 (2004), pp. 353-358.
Index entries for transcendental numbers.

Cf. A000032, A144837, A371649.

Similar constants: A371525, A371526, A371648.

RealDigits[Product[1 + 1/LucasL[5^k], {k, 0, 10}], 10, 120][[1]]

prodinf(k = 0, 1 + 1/(fibonacci(5^k-1) + fibonacci(5^k+1)))

Equals Product_{k>=0} (1 + 1/A144837(k)).

cp's OEIS Frontend

A268922 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 1 mod 5 numbers, except for n = 0.

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A144836 a(n) = round(phi^(4^n)) where phi is the golden ratio (A001622).

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A144838 a(n) = Lucas(6^n).

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A144839 a(n) = Lucas(7^n).

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A371649 Decimal expansion of Sum_{k>=0} 1/Lucas(5^k).

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A371650 Decimal expansion of Product_{k>=0} (1 + 1/Lucas(5^k)).

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Examples

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Programs

Mathematica

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Formula