A069705 -id:A069705 - cp's OEIS frontend

A269266 a(n) = 2^n mod 31.

1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1

Offset: 0

Vincenzo Librandi, Mar 31 2016

Continued fraction expansion of (1651+sqrt(3236405))/2386. - Bruno Berselli, Mar 31 2016

Muniru A Asiru, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1).

Cf. A201912 (11th row of the triangle).

Cf. similar sequences of the type 2^n mod p, where p is a prime: A000034 (p=3), A070402 (p=5), A069705 (p=7), A036117 (p=11), A036118 (p=13), A062116 (p=17), A036120 (p=19), A070335 (p=23), A036122 (p=29), this sequence (p=31), A036124 (p=37), A070348 (p=41), A070349 (p=43), A070351 (p=47), A036128 (p=53), A036129 (p=59), A036130 (p=61), A036131 (p=67).

List([0..70],n->PowerMod(2,n,31)); # Muniru A Asiru, Jan 30 2019

Magma
```
[Modexp(2, n, 31): n in [0..100]];
```

&cat [[1,2,4,8,16]^^20] // Bruno Berselli, Mar 31 2016

Mathematica
```
PowerMod[2, Range[0, 100], 31]
```

a(n)=2^(n%5) \\ Charles R Greathouse IV, Mar 31 2016

x='x+O('x^99); Vec((1+2*x+4*x^2+8*x^3+16*x^4)/(1-x^5)) \\ Altug Alkan, Mar 31 2016

for n in range(0,100):print(2**n%31) # Soumil Mandal, Apr 03 2016

def A269266(n): return pow(2,n,31) # Chai Wah Wu, Jan 03 2022

[2^mod(n,5) for n in (0..100)] # Bruno Berselli, Mar 31 2016

G.f.: (1 + 2*x + 4*x^2 + 8*x^3 + 16*x^4)/(1 - x^5).
a(n) = a(n-5).
a(n) = 2^(n mod 5). - Bruno Berselli, Mar 31 2016

A178233 Decimal expansion of (11+sqrt(229))/18.

Original entry on oeis.org

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

Offset: 1

Klaus Brockhaus, May 23 2010

Continued fraction expansion of (11+sqrt(229))/18 is A069705.

			(11+sqrt(229))/18 = 1.45181921946786421817...

Cf. A166125 (decimal expansion of sqrt(229)), A069705 (repeat 1, 2, 4).

A202349 Lexicographically earliest sequence such that the sequence and its first and second differences share no terms, and the 3rd differences are equal to the original sequence.

Original entry on oeis.org

1, 3, 9, 20, 39, 75, 148, 297, 597, 1196, 2391, 4779, 9556, 19113, 38229, 76460, 152919, 305835, 611668, 1223337, 2446677, 4893356, 9786711, 19573419, 39146836, 78293673, 156587349, 313174700, 626349399, 1252698795, 2505397588, 5010795177, 10021590357

Offset: 1

Eric Angelini, Jun 21 2016

The sequence is completely determined by its first 3 terms. If the first terms are x, y, z, then the following terms are 2*x-3*y+3*z, 6*x-7*y+6*z, 12*x-12*y+11*z, 22*x-21*y+21*z, 42*x-41*y+42*z, 84*x-84*y+85*z, 170*x-171*y+171*z, 342*x-343*y+342*z. - Giovanni Resta, Jun 21 2016
Is it a theorem that, if x,y,z = 1,3,9, the sequence has the desired properties, or is it just a conjecture? - N. J. A. Sloane, Jun 21 2016
From Charlie Neder, Jan 10 2019: (Start)
No two terms among this sequence and its first and second differences are equal.
Proof: Representing the first and second differences by b(n) and c(n), we have that a-b is [-1, -3, -2, 1, 3, 2] with period 6, a-c is [-3, -2, -1, 3, 2, 1] with period 6, and b-c is [-2, 1, 3, 2, -1, -3] with period 6. Therefore, no two terms at the same index are equal. Since the sequence is forced to grow exponentially, only the first few terms need to be checked to confirm that no two terms at different indices are equal, proving the criterion always holds. (End)

			  1 3 9  20  39  75  148   297   597   1196
   2 6 11  19  36  73   149   300   599
    4 5   8  17  37  76   151    299
     1  3   9  20  39   75    148   <-- the starting sequence

David A. Corneth, Table of n, a(n) for n = 1..3318
Mathematics Stack Exchange, Conjectured recurrence/generating function for a binomial sum.
Sequence and first differences include all numbers, etc.
Index entries for linear recurrences with constant coefficients, signature (3,-3,2).

Cf. A024493, A130781, A069705 (inverse binomial transform assuming offset 0).

For many similar sequences, see the Index link.

d = Differences; i = Intersection; sol = Solve[d@ d@ d@ Array[x, 50] == Array[x, 47], Array[x, 47, 4]][[1]]; a = (Array[x, 50] /. sol) /. {x[1] -> 1, x[2] -> 3, x[3] -> 9}; Print["Check = ", {i[a, d@ a], i[a, d@ d@ a], i[d@ a, d@ d@ a]}]; a (* Giovanni Resta, Jun 21 2016 *)

first(n) = {n = max(n, 4); my(res = vector(n)); for(i = 1, 3, res[i] = 3^(i - 1)); for(i = 4, n, res[i] = 3 * res[i - 1] - 3 * res[i - 2] + 2 * res[i - 3]); res } \\ David A. Corneth, Jan 11 2019

Vec(x*(1 + 3*x^2) / ((1 - 2*x)*(1 - x + x^2)) + O(x^40)) \\ Colin Barker, Jan 12 2019

From Colin Barker, Jan 11 2019: (Start)
G.f.: x*(1 + 3*x^2) / ((1 - 2*x)*(1 - x + x^2)).
a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) for n>2. (End)
a(n) = Sum_{k=0..n-1} binomial(n-1,k)*(2^k mod 7). - Fabio Visonà, Sep 05 2023

a(18)-a(33) from Giovanni Resta, Jun 21 2016

A145644 Cubefree part of 10^n.

Original entry on oeis.org

1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100, 1, 10, 100

Offset: 0

Artur Jasinski, Oct 15 2008

Period 3: repeat [1, 10, 100].

Index entries for linear recurrences with constant coefficients, signature (0,0,1).

Cf. A004709, A010872, A011557, A050985, A069705, A145642, A145643.

&cat [[1, 10,100]^^30]; // Wesley Ivan Hurt, Apr 18 2021

CubefreePart[n_Integer?Positive] := Times @@ Power @@@ ({#[[1]], Mod[ #[[2]], 3]} & /@ FactorInteger[n]); Table[CubefreePart[10^n], {n, 1, 40}]
PadRight[{}, 50, {1, 10, 100}] (* Wesley Ivan Hurt, Apr 18 2021 *)

a(n) = 10^(n mod 3) = A050985(A011557(n)) = A011557(A010872(n)). - Amiram Eldar, Feb 14 2021
From Wesley Ivan Hurt, Apr 18 2021: (Start)
G.f.: (1+10*x+100*x^2)/(1-x^3).
a(n) = a(n-3). (End)

Offset corrected by Amiram Eldar, Feb 14 2021

A302588 a(n) = a(n-3) + 7*(n-2), a(0)=1, a(1)=2, a(2)=4.

Original entry on oeis.org

1, 2, 4, 8, 16, 25, 36, 51, 67, 85, 107, 130, 155, 184, 214, 246, 282, 319, 358, 401, 445, 491, 541, 592, 645, 702, 760, 820, 884, 949, 1016, 1087, 1159, 1233, 1311, 1390, 1471, 1556, 1642, 1730, 1822, 1915, 2010

Offset: 0

Paul Curtz, Jul 17 2018

nonn

Third of a family after A000124 and A084684. Built from the second differences. The fourth sequence is 1, 2, 4, 8, 16, 32, 49, 91, ..., from periodic [1, 2, 4, 8].

Index entries for linear recurrences with constant coefficients, signature (2, -1, 1, -2, 1).

Cf. A069705, A047350 (first differences), A130518.

CoefficientList[ Series[-(4x^4 +x^3 +x^2 +1)/((x -1)^3 (x^2 +x +1)), {x, 0, 50}], x] (* or *)
LinearRecurrence[{2, -1, 1, -2, 1}, {1, 2, 4, 8, 16}, 50] (* Robert G. Wilson v, Jul 18 2018 *)

Repeat the second differences of 1, 2, 4, 8, 16, i.e., repeat [1, 2, 4].
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).
a(n) = A069705(n) + 7*A130518(n).

cp's OEIS Frontend

A269266 a(n) = 2^n mod 31.

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A178233 Decimal expansion of (11+sqrt(229))/18.

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A202349 Lexicographically earliest sequence such that the sequence and its first and second differences share no terms, and the 3rd differences are equal to the original sequence.

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A145644 Cubefree part of 10^n.

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A302588 a(n) = a(n-3) + 7*(n-2), a(0)=1, a(1)=2, a(2)=4.

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