A168610 -id:A168610 - cp's OEIS frontend

A168607 a(n) = 3^n + 2.

3, 5, 11, 29, 83, 245, 731, 2189, 6563, 19685, 59051, 177149, 531443, 1594325, 4782971, 14348909, 43046723, 129140165, 387420491, 1162261469, 3486784403, 10460353205, 31381059611, 94143178829, 282429536483, 847288609445

Offset: 0

Vincenzo Librandi, Dec 01 2009

Second bisection is A134752.
It appears that if s(n) is a first order rational sequence of the form s(1)=5, s(n)= (2*s(n-1)+1)/(s(n-1)+2),n>1, then s(n)= a(n)/(a(n)-4), n>1. - Gary Detlefs, Nov 16 2010
Mahler exhibits this sequence with n>=1 as a proof that there exists an infinite number of x coprime to 3, such that x belongs to A125293 and x^2 belongs to A005836. - Michel Marcus, Nov 12 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Gennady Eremin, Arithmetization of well-formed parenthesis strings. Motzkin Numbers of the Second Kind, arXiv:2012.12675 [math.CO], 2020.
Kurt Mahler, The representation of squares to the base 3, Acta Arith. Vol. 53, Issue 1 (1989), p. 99-106.
Index entries for linear recurrences with constant coefficients, signature (4,-3).

Cf. A008776 (2*3^n), A005051 (8*3^n), A034472 (3^n+1), A000244 (powers of 3), A024023 (3^n-1), A168609 (3^n+4), A168610 (3^n+5), A134752 (3^(2*n-1)+2).

Magma
```
[3^n+2: n in [0..30]];
```

A168607:=n->3^n + 2; seq(A168607(n), n=0..30); # Wesley Ivan Hurt, Mar 21 2014

CoefficientList[Series[(3 - 7 x)/((1-x) (1-3 x)), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 06 2013 *)
NestList[3 # - 4 & , 3, 25] (* Bruno Berselli, Feb 06 2013 *)

a(n)=3^n+2 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 3*a(n-1) - 4, a(0) = 3.
a(n+1) - a(n) = A008776(n).
a(n+2) - a(n) = A005051(n).
a(n) = A034472(n)+1 = A000244(n)+2 = A024023(n)+3 = A168609(n)-2 = A168610(n)-3.
G.f.: (3 - 7*x)/((1 - x)*(1 - 3*x)).
a(n) = 4*a(n-1) - 3*a(n-2), a(0) = 3, a(1) = 5. - Vincenzo Librandi, Feb 06 2013
E.g.f.: exp(3*x) + 2*exp(x). - Elmo R. Oliveira, Nov 09 2023

Edited by Klaus Brockhaus, Apr 13 2010

Further edited by N. J. A. Sloane, Aug 10 2010

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

A168613 a(n) = 3^n - 5.

Original entry on oeis.org

-4, -2, 4, 22, 76, 238, 724, 2182, 6556, 19678, 59044, 177142, 531436, 1594318, 4782964, 14348902, 43046716, 129140158, 387420484, 1162261462, 3486784396, 10460353198, 31381059604, 94143178822, 282429536476, 847288609438

Offset: 0

Vincenzo Librandi, Dec 01 2009

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

Cf. A168610.

I:=[-4, -2]; [n le 2 select I[n] else 4*Self(n-1)-3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 06 2012

CoefficientList[Series[2*(7*x-2)/((1-x)*(1-3*x)),{x,0,40}],x] (* Vincenzo Librandi, Jul 06 2012 *)
LinearRecurrence[{4,-3}, {-4, -2}, 25] (* G. C. Greubel, Jul 27 2016 *)
3^Range[0,30]-5 (* Harvey P. Dale, Sep 12 2022 *)

a(n) = 3*a(n-1) + 10 with a(0)=-4.
G.f.: 2*(7*x - 2)/((1-x)*(1-3*x)). - Vincenzo Librandi, Jul 06 2012
a(n) = 4*a(n-1) - 3*a(n-2). - Vincenzo Librandi, Jul 06 2012
a(n) = 2*A116970(n) + 2 with A116970(0)=-3, A116970(1)=-2. - Bruno Berselli, Jul 06 2012
E.g.f.: exp(3*x) - 5*exp(x). - G. C. Greubel, Jul 27 2016

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

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

A368350 a(n) is the least nonnegative integer k such that the 2-valuation of 3^k+5 is n, or -1 if no such number exists.

Original entry on oeis.org

0, -1, 1, 7, 3, 27, 43, 75, 139, 11, 779, 267, 1291, 3339, 7435, 32011, 48395, 81163, 146699, 277771, 15627, 1588491, 2637067, 539915, 4734219, 13122827, 63454475, 29900043, 231226635, 97008907, 902315275, 365444363, 1439186187, 3586669835, 7881637131

Offset: 1

Yifan Xie, Dec 22 2023

sign

a(n) is the least nonnegative integer k such that 3^k == 2^n-5 (mod 2^(n+1)), or -1 if no such number exists.
Theorem: For n >= 3, a(n+1) is the remainder of a(n) + 2^(n-1) +- 2^(n-2) modulo 2^n.
Proof: (Start)
We first prove that the 2-valuation of 3^(2^(k - 2)) - 1 is k for k >= 3. Using mathematical induction we can know that 3^(2^i) = 1 + 2^(i + 2)*u(i), where u(i)'s are odd positive integers, thus the theorem is the case where i = k - 2.
Then we prove the original theorem. Suppose that 3^a(n) = x*2^(n + 1) + 2^n - 5. Using the theorem proved before, we have 3^(2^(n-2)) = y*2^(n + 1) + 2^n + 1, where x and y are no negative integers. Therefore, 3^(a(n) + 2^(n - 2)) == (x - y)*2^(n + 1) - 5 (mod 2^(n + 2)), 3^(a(n) + 3*2^(n - 2)) == (x - y + 1)*2^(n + 1) - 5 (mod 2^(n + 2)). Since there must be an odd number between x - y and x - y + 1, suppose that 3^(a(n) + 2^(n-1) +- 2^(n-2)) == 2^(n + 1) - 5 (mod 2^(n + 2)). Since the multiplicative order of 3 mod 2^(n + 2) is 2^n, a(n + 1) is the remainder of a(n) + 2^(n-1) +- 2^(n-2) modulo 2^n. (End)

			a(2) = -1 because if 3^n+5 is divisible by 2^2, n must be odd, so 3^n+5 is divisible by 2^3.
a(10) = 11 because the 2-valuation of 3^11+5 is 10, and it's easy to verify that it is the least one.
Since a(13) = 1291 < 2^11, a(14) = 1291 + 2^12 +- 2^11. Then we can verify that the former is correct, thus a(14) = 3339.

Yifan Xie, Table of n, a(n) for n = 1..1000

Cf. A006519 (2-valuation of n), A168610.

a(n) = my(t=znlog(2^n-5, Mod(3, 2^(n+1)))); if(type(t)=="t_INT", t, -1);

a(n) < 2^(n-1).
a(n+1) = a(n) + 3*2^(n-2) or a(n) - 2^(n-2) if a(n) > 2^(n-2);
a(n+1) = a(n) + 3*2^(n-2) or a(n) + 2^(n-2) if a(n) < 2^(n-2). (See COMMENTS for proof)

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A168607 a(n) = 3^n + 2.

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

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A168613 a(n) = 3^n - 5.

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

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A368350 a(n) is the least nonnegative integer k such that the 2-valuation of 3^k+5 is n, or -1 if no such number exists.

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