A168607 -id:A168607 - cp's OEIS frontend

A253208 a(n) = 4^n + 3.

4, 7, 19, 67, 259, 1027, 4099, 16387, 65539, 262147, 1048579, 4194307, 16777219, 67108867, 268435459, 1073741827, 4294967299, 17179869187, 68719476739, 274877906947, 1099511627779, 4398046511107, 17592186044419, 70368744177667, 281474976710659

Offset: 0

Vincenzo Librandi, Dec 29 2014

Subsequence of A226807.

Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A141725, A226807.

Cf. Numbers of the form k^n+k-1: A000057 (k=2), A168607 (k=3), this sequence (k=4), A242329 (k=5), A253209 (k=6), A253210 (k=7), A253211 (k=8), A253212 (k=9), A253213 (k=10).

Magma
```
[4^n+3: n in [0..30]];
```

Table[4^n + 3, {n, 0, 30}] (* or *) CoefficientList[Series[(4 - 13 x) / ((1 - x) (1 - 4 x)), {x, 0, 40}], x]

a(n)=4^n+3 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (4 - 13*x)/((1 - x)*(1 - 4*x)).
a(n) = 5*a(n-1) - 4*a(n-2) for n > 1.
From Elmo R. Oliveira, Nov 14 2023: (Start)
a(n) = 4*a(n-1) - 9 with a(0) = 4.
E.g.f.: exp(4*x) + 3*exp(x). (End)

A178674 a(n) = 3^n + 3.

Original entry on oeis.org

4, 6, 12, 30, 84, 246, 732, 2190, 6564, 19686, 59052, 177150, 531444, 1594326, 4782972, 14348910, 43046724, 129140166, 387420492, 1162261470, 3486784404, 10460353206, 31381059612, 94143178830, 282429536484, 847288609446, 2541865828332, 7625597484990, 22876792454964

Offset: 0

Vincenzo Librandi, Dec 25 2010

a(n) is the deficiency of 3^n * 5. - Patrick J. McNab, May 27 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-3).

Cf. A034472, A115098, A168607, A168609, A168610.

List([0..40], n -> 3^n+3); # G. C. Greubel, Jan 27 2019

Magma
```
[3^n+3: n in [0..35]];
```

Table[3^n+3, {n, 0, 40}] (* or *) CoefficientList[Series[(4-10x)/((1-x) (1-3x)), {x, 0, 30}], x] (* Vincenzo Librandi, May 13 2014 *)

a(n)=3^n+3 \\ Charles R Greathouse IV, Oct 07 2015

[3^n+3 for n in range(40)] # G. C. Greubel, Jan 27 2019

a(n) = 3*(a(n-1) - 2), a(0)=4.
From R. J. Mathar, Jan 05 2011: (Start)
G.f.: (4-10*x)/((1-3*x)*(1-x)).
a(n) = 2*A115098(n). (End)
a(n) = 4*a(n-1) - 3*a(n-2) for n > 1. - Vincenzo Librandi, May 13 2014
E.g.f.: exp(x)*(exp(2*x) + 3). - Elmo R. Oliveira, Apr 02 2025

A168609 a(n) = 3^n + 4.

Original entry on oeis.org

5, 7, 13, 31, 85, 247, 733, 2191, 6565, 19687, 59053, 177151, 531445, 1594327, 4782973, 14348911, 43046725, 129140167, 387420493, 1162261471, 3486784405, 10460353207, 31381059613, 94143178831, 282429536485, 847288609447

Offset: 0

Vincenzo Librandi, Dec 01 2009

			a(1)=3*5-8=7; a(2)=3*7-8=13; a(3)=3*13-8=31.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-3).

Cf. A034472, A168607, A168610, A178674.

[3^n+4: n in [0..30]]; // Vincenzo Librandi, May 13 2014

Table[3^n + 4, {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, May 19 2011 *)
CoefficientList[Series[(5 - 13 x)/((1 - x) (1 - 3 x)), {x, 0, 30}], x] (* Vincenzo Librandi, May 13 2014 *)
LinearRecurrence[{4,-3},{5,7},30] (* Harvey P. Dale, Mar 11 2023 *)

a(n) = 3*a(n-1) - 8, with n>0, a(0)=5.
G.f.: (5 - 13*x)/((1-x)*(1-3 x)). - Vincenzo Librandi, May 13 2014
a(n) = 4*a(n-1) - 3*a(n-2) for n>1. - Vincenzo Librandi, May 13 2014
E.g.f.: exp(3*x) + 4*exp(x). - G. C. Greubel, Jul 27 2016

Formula and examples edited to use correct offset by Jon E. Schoenfield, Jun 19 2010

A301913 Primes which divide numbers of the form 3^k + 2 for k >= 1.

Original entry on oeis.org

5, 7, 11, 17, 19, 29, 31, 43, 53, 59, 79, 83, 89, 97, 101, 107, 113, 127, 131, 137, 139, 149, 163, 173, 179, 197, 199, 211, 223, 227, 233, 241, 251, 257, 269, 281, 283, 293, 317, 331, 337, 347, 353, 379, 389, 401, 409, 419, 439, 443, 449, 457, 461, 463, 467

Offset: 1

Luke W. Richards, Mar 28 2018

nonn

The first odd prime not to appear in the sequence is 3 because 3^k + 2 == 2 mod 3 for k >= 1.

Primes p such that the order of -2 (mod p) divides the order of 3 (mod p). - Joerg Arndt, Mar 31 2018, corrected by Robert Israel, May 04 2018

			5 divides 245 which is 3^5+2 so 5 is in the sequence.
7 divides 245 which is 3^5+2 so 7 is in the sequence.
The values of x = (3^k+2) mod 13 for k = 0, 1, 2, 3, ... are 3, 5, 11, 3, 5, 11, ...; 13 never divides any 3^k + 2, so 13 is not in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A168607.

select(t -> numtheory:-mlog(-2,3,t)<>FAIL, [seq(ithprime(i),i=3..100)]);

fQ[p_] := IntegerQ@ MultiplicativeOrder[3, p, -2]; Select[ Prime@ Range@ 100, fQ] (* Robert G. Wilson v, Apr 07 2018 *)

is(n)=n>4 && isprime(n) && znorder(Mod(-2,n))%znorder(Mod(3,n))==0 \\ Charles R Greathouse IV, May 04 2018

A340131 Numbers whose ternary expansions have the same number of 1's and 2's and, in each prefix (initial fragment), at least as many 1's as 2's.

Original entry on oeis.org

0, 5, 11, 15, 29, 33, 44, 45, 50, 83, 87, 98, 99, 104, 116, 128, 132, 135, 140, 146, 150, 245, 249, 260, 261, 266, 278, 290, 294, 297, 302, 308, 312, 332, 344, 348, 377, 380, 384, 395, 396, 401, 405, 410, 416, 420, 434, 438, 449, 450, 455, 731, 735, 746, 747

Offset: 1

Gennady Eremin, Dec 29 2020

For a nonzero term, the ternary code starts with 1, otherwise the balance of 1's and 2's is broken already in the one-digit prefix. Therefore 7, 19, 21, etc. (see A039001) are not terms.
As another example, for the integer 52 the balance is broken in the three-digit prefix 122 (the entire ternary code is 1221).
Each term with a ternary code of length k corresponds one-to-one to the Motzkin path of length k that starts with an up step. Therefore, the terms can be called digitized Motzkin paths.
The number of terms with a ternary code of length k is equal to A244884(k). Example: five terms 29, 33, 44, 45 and 50 have a ternary length of 4, respectively A244884(4)=5.

			The first terms 0 and 5 are obvious, because the four intermediate ternary codes 1, 2, 10[3], and 11[4] are rejected due to a violation of the balance of 1's and 2's. Next, the successor function S works: for any term x, the next term is S(x).
Iterating over numbers is inefficient; code suffixes (final digits) can be processed faster. The transition from 0 to 12[5] is generalized for terms that are multiples of 9. For example,
S(10200[99]) = 10212[104], S(1122000[1188]) = 1122012[1193], etc.
In this case, the calculation of the subsequent term is reduced to simply replacing the suffix s = 00 with the subsequent suffix s'= 12.
Another common suffix is s = 02..2 = 02^k (twos are repeated at the end of the ternary code). Then the subsequent suffix is s'= 202..2 = 202^(k-1), i.e., within such a suffix, the first two digits are reversed. Here are some examples:
k = 1, S(1002[29]) = 1020[33], the increment is 4*3^0 = 4;
k = 2, S(110022[332]) = 110202[344], the increment is 4*3^1 = 12;
k = 3, S(10110222[2537]) = 10112022[2573], the increment is 4*3^2 = 36;
k = 4, S(111102222[9800]) = 111120222[9908], the increment is 4*3^3 = 108.
There are 5 such group suffixes.

Gennady Eremin, Table of n, a(n) for n = 1..1000
Gennady Eremin, Arithmetization of well-formed parenthesis strings. Motzkin Numbers of the Second Kind, arXiv:2012.12675 [math.CO], 2020.

Subsequence of A039001.
Subsequences: A134752, A168607.
Cf. A244884.

is(n) = {my(d = digits(n, 3), v = [0, 0]); for(i = 1, #d, if(d[i] > 0, v[d[i]]++); if(v[1] < v[2], return(0))); v[1] == v[2] } \\ David A. Corneth, Dec 29 2020

def digits(n, b):
  out = []
  while n >= b:
    out.append(n % b)
    n //= b
  return [n] + out[::-1]
def ok(n):
  t = digits(n, 3)
  if t.count(1) != t.count(2): return False
  return all(t[:i].count(1) >= t[:i].count(2) for i in range(1, len(t)))
print([n for n in range(750) if ok(n)]) # Michael S. Branicky, Dec 29 2020

A176260 Periodic sequence: Repeat 5, 1.

Original entry on oeis.org

5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5

Offset: 0

Klaus Brockhaus, Apr 13 2010

Interleaving of A010716 and A000012.
Also continued fraction expansion of (5+3*sqrt(5))/2.
Also decimal expansion of 17/33.
Essentially first differences of A047264.
Binomial transform of 5 followed by -A122803 without initial terms 1, -2.
Inverse binomial transform of 5 followed by A007283 without initial term 3.
Second inverse binomial transform of A168607 without initial term 3.
Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 3*x^3 + 6*x^4 + 6*x^5 + ... is the o.g.f. for A008805. - Peter Bala, Mar 13 2015

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

Cf. A010716 (all 5's sequence), A000012 (all 1's sequence), A090550 (decimal expansion of (5+3*sqrt(5))/2), A010686 (repeat 1, 5), A047264 (congruent to 0 or 5 mod 6), A122803 (powers of -2), A007283 (3*2^n), A168607 (3^n+2), A008805.

&cat[ [5, 1]: n in [0..52] ];
[ 3+2*(-1)^n: n in [0..104] ];

a(n) = 3+2*(-1)^n.
a(n) = a(n-2) for n > 1; a(0) = 5, a(1) = 1.
a(n) = -a(n-1)+6 for n > 0; a(0) = 5.
a(n) = 5*((n+1) mod 2)+(n mod 2).
a(n) = A010686(n+1).
G.f.: (5+x)/(1-x^2).
From Amiram Eldar, Jan 01 2023: (Start)
Multiplicative with a(2^e) = 5, and a(p^e) = 1 for p >= 3.
Dirichlet g.f.: zeta(s)*(1+2^(2-s)). (End)
E.g.f.: 5*cosh(x) + sinh(x). - Stefano Spezia, Feb 09 2025

A298827 a(n) is the smallest positive integer k such that 3^n+2 divides 3^(n+k)+2.

Original entry on oeis.org

4, 5, 28, 41, 84, 336, 990, 193, 1260, 5905, 75918, 10065, 318860, 2391485, 14348908, 20390382, 5031420, 31624326, 5985168, 1743333144, 8569036, 668070480, 547062516, 141214768241, 167874004756, 1270932914165, 385131186110, 2837770056420, 784347169884, 475536631360, 149093578413164, 139370386996590

Offset: 1

Luke W. Richards, Jan 27 2018

nonn

3^n+2 divides 3^(n+a(n)*m)+2 for all nonnegative integers m.

Jon E. Schoenfield noted that a(n) coincides with the multiplicative order of 3 modulo 3^n+2. This is true because 3^(n+a(n)) == 3^n mod 3^n+2 and since 3^n and 3^n+2 are coprime, 3^a(n) == 1 mod 3^n+2 and the multiplicative order is the smallest positive such number. - Chai Wah Wu, Jan 29 2018

			For n = 1, f(1) = 3^1 + 2 = 5, where f(x) = 3^x + 2. Given the last digits of f(x) form a recurring sequence of 5, 1, 9, 3 [, 5, 1, 9, 3] then whenever x = 1 mod 4, f(x) will be a multiple of f(1).
For n = 2, f(2) = 3^2 + 2 = 11. a(2) = 5. So any x = 2 mod 5 will be a multiple of 11. For instance, 27 = 2 mod 5, and f(27) = 3^27 + 2 = 7625597474989 = 11 * 693236134999.

Robert Israel, Table of n, a(n) for n = 1..175

Cf. A168607.

[Modorder(3,3^n+2): n in [1..29]]; // Jon E. Schoenfield, Jan 28 2018

seq(numtheory:-order(3, 3^n+2), n=1..100); # Robert Israel, Feb 05 2018

Array[Block[{k = 1}, While[! Divisible[3^(# + k) + 2, 3^# + 2], k++]; k] &, 12] (* Michael De Vlieger, Feb 05 2018 *)
Table[MultiplicativeOrder[3, 3^n + 2], {n, 32}] (* Jean-François Alcover, Feb 06 2018 *)

a(n) = znorder(Mod(3, 3^n+2)); \\ Michel Marcus, Jan 29 2018

def fmod(n, mod):
    return (pow(3, n, mod) + 2) % mod
def f(n):
    return pow(3, n) + 2
#terms is the number of terms to generate
terms = 20
for x in range(1, terms + 1):
    div = f(x)
    y = x + 1
    while fmod(y, div) != 0:
        y += 1
    print(y - x)

from sympy import n_order
def A298827(n):
    return n_order(3,3**n+2) # Chai Wah Wu, Jan 29 2018

a(22)-a(32) from Robert Israel, Feb 05 2018

A323767 A(n,k) = Sum_{j=0..floor(n/2)} binomial(n-j,j)^k, square array A(n,k) read by antidiagonals, for n >= 0, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 3, 1, 1, 2, 5, 5, 3, 1, 1, 2, 9, 11, 8, 4, 1, 1, 2, 17, 29, 26, 13, 4, 1, 1, 2, 33, 83, 92, 63, 21, 5, 1, 1, 2, 65, 245, 338, 343, 153, 34, 5, 1, 1, 2, 129, 731, 1268, 1923, 1281, 376, 55, 6

Offset: 0

Seiichi Manyama, Jan 27 2019

			Square array begins:
   1,  1,   1,    1,     1,      1,       1, ...
   1,  1,   1,    1,     1,      1,       1, ...
   2,  2,   2,    2,     2,      2,       2, ...
   2,  3,   5,    9,    17,     33,      65, ...
   3,  5,  11,   29,    83,    245,     731, ...
   3,  8,  26,   92,   338,   1268,    4826, ...
   4, 13,  63,  343,  1923,  10903,   62283, ...
   4, 21, 153, 1281, 11553, 108801, 1050753, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns 0-5 give A004526(n+2), A000045(n+1), A051286, A181545, A181546, A181547.
Rows 0-5 give A000012, A000012, A007395, A000051, A168607, A074506.
Main diagonal gives A323769.
Cf. A011973,

f := Sum[Power[Binomial[#1 - i, i], #2], {i, 0, #1/2}] &;a = Flatten[Reverse[DeleteCases[Table[Table[f[m - n, n], {n, 0, 20}], {m, 0, 20}], 0, Infinity], 2]] (* Elijah Beregovsky, Nov 24 2020 *)

A259821 a(n) = floor( (3^n+1)^2/3^n ).

Original entry on oeis.org

4, 5, 11, 29, 83, 245, 731, 2189, 6563, 19685, 59051, 177149, 531443, 1594325, 4782971, 14348909, 43046723, 129140165, 387420491, 1162261469, 3486784403, 10460353205, 31381059611, 94143178829, 282429536483

Offset: 0

Kival Ngaokrajang, Jul 07 2015

a(n) is the curvature (rounded down) of circles inscribed between a unit circle and a circumscribed equilateral triangle. See illustration.

Apart from the first term the same as A168607. - R. J. Mathar, Jul 09 2015

Colin Barker, Table of n, a(n) for n = 0..1000
Kival Ngaokrajang, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (4,-3).

Cf. A034472, A000244, A168607.

Table[Floor[(3^n + 1)^2/3^n], {n, 0, 30}] (* Michael De Vlieger, Jul 07 2015 *)
LinearRecurrence[{4,-3},{4,5,11},30] (* Harvey P. Dale, Sep 30 2024 *)

a(n)=floor((3^n+1)^2/3^n)
for (n=0, 100, print1(a(n),", "))

Vec((3*x^2-11*x+4)/((x-1)*(3*x-1)) + O(x^100)) \\ Colin Barker, Jul 07 2015

a(n) = floor( A034472(n)^2/A000244(n) ).
From Colin Barker, Jul 07 2015: (Start)
a(n) = 3^n + 2 for n>0.
a(n) = 4*a(n-1) - 3*a(n-2) for n>2.
G.f.: (3*x^2-11*x+4) / ((x-1)*(3*x-1)).
(End)

A298940 a(n) is the smallest positive integer k such that 3^n - 2 divides 3^(n + k) + 2, or 0 if there is no such k.

Original entry on oeis.org

1, 3, 10, 39, 60, 121, 0, 117, 4920, 0, 0, 0, 28322, 0, 1434890, 0, 0, 0, 116226146, 0, 0, 15690529803, 0, 108443565, 66891206007, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 22514195294549868, 0, 405255515301897626, 0, 1823649818858539320, 0, 0, 5861731560616733529, 0, 0, 0

Offset: 1

Luke W. Richards, Jan 29 2018

nonn

3^n - 2 divides 3^(n + (2m + 1) * a(n)) + 2 for all nonnegative integers m.

a(n) is the least positive integer k, if any, such that 3^k == -1 (mod 3^n-2). If the order of 3 mod p is odd for some prime p dividing 3^n-2, a(n)=0. - Robert Israel, Feb 05 2018

			a(2) = 3 because 3^2 - 2 divides 3^5 + 2 and 3^2 - 2 does not divide any 3^x - 2 for 2 < x < 5.
a(5) = 60 because 3^5 - 2 divides 3^65 + 2 and 3^5 - 2 does not divide any 3^x - 2 for 5 < x < 65.

Robert Israel, Table of n, a(n) for n = 1..166

Cf. A168607, A298827.

# This requires Maple 2016 or later
f:= proc(n) local m,ps,a,p,q,phiq,v,br,ar;
  m:= 3^n-2;
   ps:= ifactors(m)[2];
   a:= 0;
   for p in ps do
     q:= p[1]^p[2];
     phiq:= (p[1]-1)*p[1]^(p[2]-1);
     v:= NumberTheory:-MultiplicativeOrder(3,q);
     if v::odd then return 0 fi;
     if p[2]=1 then br:= v/2
     else br:= traperror(NumberTheory:-ModularLog(-1,3,q));
          if br = lasterror then return 0 fi;
     fi;
     if a = 0 then a:= v; ar:= br
     else
        ar:= NumberTheory:-ChineseRemainder([ar,br],[a,v]);
        if ar = FAIL then return 0 fi;
        a:= ilcm(a, v);
     fi
   od:
   ar;
end proc:
f(1):= 1:
map(f, [$1..50]); # Robert Israel, Feb 06 2018

a[1] = 1; a[n_] := If[IntegerQ[order = MultiplicativeOrder[3, 3^n - 2, {-1}]], order, 0]; Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 1, 20}] (* Jean-François Alcover, Feb 06 2018, after Robert Israel *)

a(n) = if(n==1, return(1)); my(l = znlog(-1, Mod(3, 3^n - 2))); if(l == [], return(0), return(l)) \\ Iain Fox, Feb 06 2018

from sympy import discrete_log
def A298940(n):
    if n == 1:
        return 1
    try:
        return discrete_log(3**n-2,-1,3)
    except ValueError:
        return 0 # Chai Wah Wu, Feb 05 2018

Corrected by Robert Israel, Feb 05 2018

cp's OEIS Frontend

A253208 a(n) = 4^n + 3.

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A178674 a(n) = 3^n + 3.

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A168609 a(n) = 3^n + 4.

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A301913 Primes which divide numbers of the form 3^k + 2 for k >= 1.

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A340131 Numbers whose ternary expansions have the same number of 1's and 2's and, in each prefix (initial fragment), at least as many 1's as 2's.

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A176260 Periodic sequence: Repeat 5, 1.

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A298827 a(n) is the smallest positive integer k such that 3^n+2 divides 3^(n+k)+2.

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