A004171 -id:A004171 - cp's OEIS frontend

A083420 a(n) = 2*4^n - 1.

1, 7, 31, 127, 511, 2047, 8191, 32767, 131071, 524287, 2097151, 8388607, 33554431, 134217727, 536870911, 2147483647, 8589934591, 34359738367, 137438953471, 549755813887, 2199023255551, 8796093022207, 35184372088831, 140737488355327, 562949953421311

Offset: 0

Paul Barry, Apr 29 2003

Sum of divisors of 4^n. - Paul Barry, Oct 13 2005
Subsequence of A000069; A132680(a(n)) = A005408(n). - Reinhard Zumkeller, Aug 26 2007
If x = a(n), y = A000079(n+1) and z = A087289(n), then x^2 + 2*y^2 = z^2. - Vincenzo Librandi, Jun 09 2014
It seems that a(n) divides A001676(3+4n). Several other entries apparently have this sequence embedded in them, e.g., A014551, A168604, A213243, A213246-8, and A279872. - Tom Copeland, Dec 27 2016
To elaborate on Librandi's comment from 2014: all these numbers, even if prime in Z, are sure not to be prime in Z[sqrt(2)], since a(n) can at least be factored as ((2^(2n + 1) - 1) - (2^(2n) - 1)*sqrt(2))((2^(2n + 1) - 1) + (2^(2n) - 1)*sqrt(2)). For example, 7 = (3 - sqrt(2))(3 + sqrt(2)), 31 = (7 - 3*sqrt(2))(7 + 3*sqrt(2)), 127 = (15 - 7*sqrt(2))(15 + 7*sqrt(2)). - Alonso del Arte, Oct 17 2017
Largest odd factors of A147590. - César Aguilera, Jan 07 2020

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Roudy El Haddad, Recurrent Sums and Partition Identities, arXiv:2101.09089 [math.NT], 2021.
Roudy El Haddad, A generalization of multiple zeta value. Part 1: Recurrent sums. Notes on Number Theory and Discrete Mathematics, 28(2), 2022, 167-199, DOI: 10.7546/nntdm.2022.28.2.167-199.
A. J. Macfarlane, Generating functions for integer sequences defined by the evolution of cellular automata with even rule numbers, Fig 11.
Robert Schneider, Partition zeta functions, Research in Number Theory, 2(1):9, 2016.
Eric Weisstein's World of Mathematics, Rule 220
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A083421, A000668 (primes in this sequence), A004171, A000244.
Cf. A001676, A014551, A168604, A213243, A213246, A213247, A213248, A279872.
Cf. A000302.

a083420 = subtract 1 . (* 2) . (4 ^)  -- Reinhard Zumkeller, Dec 22 2015

[2*4^n-1 : n in [0..30]]; // Wesley Ivan Hurt, Mar 14 2015

seq(2*4^n-1, n = 0..22); # Peter Luschny, Aug 17 2011

2 * 4^Range[0, 31] - 1 (* Alonso del Arte, Oct 17 2017 *)

a(n)=2*4^n-1 \\ Charles R Greathouse IV, Sep 24 2015

def A083420(n): return (1<<(n<<1|1))-1 # Chai Wah Wu, Mar 10 2025

G.f.: (1+2*x)/((1-x)*(1-4*x)).
E.g.f.: 2*exp(4*x)-exp(x).
With a leading zero, this is a(n) = (4^n - 2 + 0^n)/2, the binomial transform of A080925. - Paul Barry, May 19 2003
From Benoit Cloitre, Jun 18 2004: (Start)
a(n) = (-16^n/2)*B(2n, 1/4)/B(2n) where B(n, x) is the n-th Bernoulli polynomial and B(k) = B(k, 0) is the k-th Bernoulli number.
a(n) = 5*a(n-1) - 4*a(n-2).
a(n) = (-4^n/2)*B(2*n, 1/2)/B(2*n). (End)
a(n) = A099393(n) + A020522(n) = A000302(n) + A024036(n). - Reinhard Zumkeller, Feb 07 2006
a(n) = Stirling2(2*(n+1), 2). - Zerinvary Lajos, Dec 06 2006
a(n) = 4*a(n-1) + 3 with n > 0, a(0) = 1. - Vincenzo Librandi, Dec 30 2010
a(n) = A001576(n+1) - 2*A001576(n). - Brad Clardy, Mar 26 2011
a(n) = 6*A002450(n) + 1. - Roderick MacPhee, Jul 06 2012
a(n) = A000203(A000302(n)). - Michel Marcus, Jan 20 2014
a(n) = Sum_{i = 0..n} binomial(2n+2, 2i). - Wesley Ivan Hurt, Mar 14 2015
a(n) = (1/4^n) * Sum_{k = 0..n} binomial(2*n+1,2*k)*9^k. - Peter Bala, Feb 06 2019
a(n) = A147590(n)/A000079(n). - César Aguilera, Jan 07 2020
a(n) = numerator(zeta_star({2}(n + 1))/zeta(2*n + 2)) where zeta_star is the multiple zeta star values and ({2}_n) represents (2, ..., 2) where the multiplicity of 2 is n. - _Roudy El Haddad, Feb 22 2022

A039770 Numbers k such that phi(k) is a perfect square.

Original entry on oeis.org

1, 2, 5, 8, 10, 12, 17, 32, 34, 37, 40, 48, 57, 60, 63, 74, 76, 85, 101, 108, 114, 125, 126, 128, 136, 160, 170, 185, 192, 197, 202, 204, 219, 240, 250, 257, 273, 285, 292, 296, 304, 315, 364, 370, 380, 394, 401, 432, 438, 444, 451, 456, 468, 489, 504, 505

Offset: 1

Olivier Gérard

A004171 is a subsequence because phi(2^(2k+1)) = (2^k)^2. - Enrique Pérez Herrero, Aug 25 2011
Subsequence of primes is A002496 since in this case phi(k^2+1) = k^2. - Bernard Schott, Mar 06 2023
Products of distinct terms of A002496 form a subsequence. - Chai Wah Wu, Aug 22 2025

			phi(34) = 16 = 4*4.

D. M. Burton, Elementary Number Theory, Allyn and Bacon Inc., Boston MA, 1976, p. 141.

T. D. Noe, Table of n, a(n) for n = 1..10000
W. D. Banks, J. B. Friedlander, C. Pomerance and I. E. Shparlinski, Multiplicative structure of values of the Euler function, in High Primes and Misdemeanours: Lectures in Honour of the Sixtieth Birthday of Hugh Cowie Williams (A. Van der Poorten, ed.), Fields Inst. Comm. 41 (2004), pp. 29-47.
P. Pollack and C. Pomerance, Square values of Euler's function, submitted for publication.
Bernard Schott, Subfamilies and subsequences

Cf. A000010, A007614. A062732 gives the squares. A306882 (squares not totient).
Cf. A068560, A114063, A078164.
Subsequences: A054755, A002496, A004171, A262406, A324745, A324746, A247129, A324747, A306908.

with(numtheory); isA039770 := proc (n) return issqr(phi(n)) end proc; seq(`if`(isA039770(n), n, NULL), n = 1 .. 505); # Nathaniel Johnston, Oct 09 2013

Select[ Range[ 600 ], IntegerQ[ Sqrt[ EulerPhi[ # ] ] ]& ]

for(n=1, 120, if (issquare(eulerphi(n)), print1(n, ", ")))

from math import isqrt
from sympy import totient as phi
def ok(n): return isqrt(p:=phi(n))**2 == p
print([k for k in range(1, 506) if ok(k)]) # Michael S. Branicky, Aug 17 2025

a(n) seems to be asymptotic to c*n^(3/2) with 1 < c < 1.3. - Benoit Cloitre, Sep 08 2002

Banks, Friedlander, Pomerance, and Shparlinski show that a(n) = O(n^1.421). - Charles R Greathouse IV, Aug 24 2009

A000980 Number of ways of writing 0 as Sum_{k=-n..n} e(k)*k, where e(k) is 0 or 1.

Original entry on oeis.org

2, 4, 8, 20, 52, 152, 472, 1520, 5044, 17112, 59008, 206260, 729096, 2601640, 9358944, 33904324, 123580884, 452902072, 1667837680, 6168510256, 22903260088, 85338450344, 318995297200, 1195901750512, 4495448217544, 16940411201280, 63983233268592

Offset: 0

N. J. A. Sloane

The 4-term sequence 2,4,8,20 is the answer to the "Solitaire Army" problem, or checker-jumping puzzle. It is too short to have its own entry. See Conway et a;., Winning Ways, Vol. 2, pp. 715-717. - N. J. A. Sloane, Mar 01 2018

Number of subsets of {-n..n} with sum 0. Also the number of subsets of {0..2n} that are empty or have mean n. For median instead of mean we have twice A024718. - Gus Wiseman, Apr 23 2023

			From _Gus Wiseman_, Apr 23 2023: (Start)
The a(0) = 2 through a(2) = 8 subsets of {-n..n} with sum 0 are:
  {}   {}        {}
  {0}  {0}       {0}
       {-1,1}    {-1,1}
       {-1,0,1}  {-2,2}
                 {-1,0,1}
                 {-2,0,2}
                 {-2,-1,1,2}
                 {-2,-1,0,1,2}
(End)

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 294.
E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see pp. 715-717.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Ray Chandler, Table of n, a(n) for n = 0..1668 (terms < 10^1000; terms 0..200 from T. D. Noe, terms 201..400 from Alois P. Heinz)
Eunice Y. S. Chan and R. M. Corless, Narayana, Mandelbrot, and A New Kind of Companion Matrix, arXiv preprint arXiv:1606.09132 [math.CO], 2016.
R. C. Entringer, Representation of m as Sum_{k=-n..n} epsilon_k k, Canad. Math. Bull., 11 (1968), 289-293.
Steven R. Finch, Signum equations and extremal coefficients, February 7, 2009. [Cached copy, with permission of the author]
J. H. van Lint, Representations of 0 as Sum_{k = -N..N} epsilon_k*k, Proc. Amer. Math. Soc., 18 (1967), 182-184.

A047653(n) = a(n)/2.
Bisection of A084239. Cf. A063865, A141000.
A007318 counts subsets by length, A327481 by integer mean.
A327475 counts subsets with integer mean, A000975 integer median.
Cf. A024718, A047997, A070925, A079309, A133406, A212352, A359893, A362046.

a000980 n = length $ filter ((== 0) . sum) $ subsequences [-n..n]

b:= proc(n, i) option remember; `if`(n>i*(i+1)/2, 0,
      `if`(i=0, 1, 2*b(n, i-1)+b(n+i, i-1)+b(abs(n-i), i-1)))
    end:
a:=n-> 2*b(0, n):
seq(a(n), n=0..40); # Alois P. Heinz, Mar 10 2014

a[n_] := SeriesCoefficient[ Product[1+x^k, {k, -n, n}], {x, 0, 0}]; a[0] = 2; Table[a[n], {n, 0, 24}](* Jean-François Alcover, Nov 28 2011 *)
nmax = 26; d = {2}; a1 = {};
Do[
  i = Ceiling[Length[d]/2];
  AppendTo[a1, If[i > Length[d], 0, d[[i]]]];
  d = PadLeft[d, Length[d] + 2 n] + PadRight[d, Length[d] + 2 n] +
    2 PadLeft[PadRight[d, Length[d] + n], Length[d] + 2 n];
  , {n, nmax}];
a1 (* Ray Chandler, Mar 15 2014 *)
Table[Length[Select[Subsets[Range[-n,n]],Total[#]==0&]],{n,0,5}] (* Gus Wiseman, Apr 23 2023 *)

PARI
```
a(n)=polcoeff(prod(k=-n,n,1+x^k),0)
```

Constant term of Product_{k=-n..n} (1+x^k).
a(n) = Sum_i A067059(2n+1-i, i) = 2+2*Sum_j A047997(n, j); i.e., sum of alternate antidiagonals of A067059 and two more than twice row sums of A047997. - Henry Bottomley, Aug 11 2002
a(n) = A004171(n) - 2*A181765(n).
Coefficient of x^(n*(n+1)/2) in 2*Product_{k=1..n} (1+x^k)^2. - Sean A. Irvine, Oct 03 2011
From Gus Wiseman, Apr 23 2023: (Start)
a(n) = 2*A047653(n).
a(n) = A070925(2n+1) + 1.
a(n) = 2*A133406(2n+1).
a(n) = 2*(A212352(n) + 1).
a(n) = A222955(2n+1).
a(n) = 2*(A362046(2n) + 1).
(End)

More terms from Michael Somos, Jun 10 2000

A002063 a(n) = 9*4^n.

Original entry on oeis.org

9, 36, 144, 576, 2304, 9216, 36864, 147456, 589824, 2359296, 9437184, 37748736, 150994944, 603979776, 2415919104, 9663676416, 38654705664, 154618822656, 618475290624, 2473901162496, 9895604649984, 39582418599936, 158329674399744, 633318697598976

Offset: 0

N. J. A. Sloane

a(n) is twice the area of the trapezoid created by the four points (2^n,2^(n+1)), (2^(n+1), 2^n), (2^(n+1), 2^(n+2)), (2^(n+2), 2^(n+1)). - J. M. Bergot, May 23 2014
These are squares that can be expressed as sum of exactly two distinct powers of two. For instance, a(4) = 9*4^4 = 2304 = 2^11 + 2^8 . It is conjectured that these are the only squares with this characteristic (tested on squares up to (10^7)^2). - Andres Cicuttin, Apr 23 2016
Conjecture is true. It is equivalent to prove that the Diophantine equation m^2 = 2^k*(1+2^h), where h>0, has solutions only when h=3. Dividing by 2^k we must obtain an odd square on the left, since 1+2^h is odd, so we can write (2*r+1)^2 = 1+2^h. Expanding, we have 4*r*(r+1) = 2^h, from which it follows that r must be equal to 1 and thus h=3, since r and r+1 must be powers of 2. - Giovanni Resta, Jul 27 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (4).

Essentially the same as A055841. First differences of A002001.

Cf. A000302.

[9*4^n: n in [0..30]]; // Vincenzo Librandi, May 19 2011

9*4^Range[0, 100] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
NestList[4#&,9,30] (* Harvey P. Dale, Jan 15 2019 *)

a(n)=9<Charles R Greathouse IV, Apr 17 2012

def A002063(n): return 9<<(n<<1) # Chai Wah Wu, Apr 09 2025

From Philippe Deléham, Nov 23 2008: (Start)
a(n) = 4*a(n-1), n > 0; a(0)=9.
G.f.: 9/(1-4*x). (End)
a(n) = 9*A000302(n). - Michel Marcus, Apr 23 2016
E.g.f.: 9*exp(4*x). - Ilya Gutkovskiy, Apr 23 2016
a(n) = 2^(2*n+3) + 2^(2*n). - Andres Cicuttin, Apr 26 2016
a(n) = A004171(n+1) + A000302(n). - Zhandos Mambetaliyev, Nov 19 2016

A087289 a(n) = 2^(2*n+1) + 1.

Original entry on oeis.org

3, 9, 33, 129, 513, 2049, 8193, 32769, 131073, 524289, 2097153, 8388609, 33554433, 134217729, 536870913, 2147483649, 8589934593, 34359738369, 137438953473, 549755813889, 2199023255553, 8796093022209, 35184372088833, 140737488355329, 562949953421313, 2251799813685249

Offset: 0

W. Edwin Clark, Aug 29 2003

Number of pairs of polynomials (f,g) in GF(2)[x] satisfying deg(f) <= n, deg(g) <= n and gcd(f,g) = 1.
An unpublished result due to Stephen Suen, David desJardins, and W. Edwin Clark. This is the case k = 2, q = 2 of their formula q^((n+1)*k) * (1 - 1/q^(k-1) + (q-1)/q^((n+1)*k)) for the number of ordered k-tuples (f_1, ..., f_k) of polynomials in GF(q)[x] such that deg(f_i) <= n for all i and gcd(f_1, ..., f_k) = 1.
Apparently the same as A084508 shifted left.
Terms in binary are palindromes of the form 1x1 where x is a string of 2*n zeros (A152577). - Brad Clardy, Sep 01 2011
For n > 0, a(n) is the number k such that the number of iterations of the map k -> (3k +1)/8 == 4 (mod 8) until reaching (3k +1)/8 <> 4 (mod 8) equals n. (see the Collatz problem: the start of the parity trajectory of a(n) is n times {100} = 100100100100...100abcd...). - Michel Lagneau, Jan 23 2012
An Engel expansion of 2 to the base 4 as defined in A181565, with the associated series expansion 2 = 4/3 + 4^2/(3*9) + 4^3/(3*9*33) + 4^4/(3*9*33*129) + .... Cf. A199561 and A207262. - Peter Bala, Oct 29 2013
For x = A083420(n), y = A000079(n+1), z = a(n) then x^2 + 2*y^2 = z^2. - Vincenzo Librandi, Jun 09 2014
A254046(n+1) is the 3-adic valuation of a(n). - Fred Daniel Kline, Jan 11 2017

			a(0) = 3 since there are three pairs, (0,1), (1,0) and (1,1) of polynomials (f,g) in GF(2)[x] of degree at most 0 such that gcd(f,g) = 1.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
K. Morrison, Random polynomials over finite fields.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A087290, A087291, A087292, A099393.
Equals A004171 + 1.
Cf. also A007583, A181565, A199561, A207262, A254046, A283070.
Cf. A000079, A083420, A084508, A152577.

[2^(2*n+1) + 1: n in [0..30]]; // Vincenzo Librandi, May 16 2011

Table[2^(2 n + 1) + 1, {n, 0, 20}] (* or *) 3 NestList[4 # - 1 &, 1, 20]
(* or *) CoefficientList[Series[(3 - 6 x)/((1 - x) (1 - 4 x)), {x, 0, 20}], x] (* Michael De Vlieger, Mar 03 2017 *)

a(n)=2^(2*n+1)+1 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: (3-6*x)/((1-x)*(1-4*x)).
a(n) = 3*A007583(n).
a(n) = 4*a(n-1) - 3. - Lekraj Beedassy, Apr 29 2005
a(n) = A099393(n+1) - 2*A099393(n). - Brad Clardy, Sep 01 2011
a(n) = 2^(2*n + 1)*a(-1-n) for all n in Z. - Michael Somos, Jan 11 2017
a(n) = A283070(n) - 1. - Peter M. Chema, Mar 02 2017
From Elmo R. Oliveira, Feb 22 2025: (Start)
E.g.f.: exp(x)*(2*exp(3*x) + 1).
a(n) = 5*a(n-1) - 4*a(n-2). (End)

A152978 a(n) = A139251(n+2)/4 = A152968(n+1)/2.

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 1, 2, 3, 3, 4, 7, 8, 4, 1, 2, 3, 3, 4, 7, 8, 5, 4, 7, 9, 10, 15, 22, 20, 8, 1, 2, 3, 3, 4, 7, 8, 5, 4, 7, 9, 10, 15, 22, 20, 9, 4, 7, 9, 10, 15, 22, 21, 14, 15, 23, 28, 35, 52, 64, 48, 16, 1, 2, 3, 3, 4, 7, 8, 5, 4, 7, 9, 10, 15, 22, 20, 9, 4, 7, 9, 10, 15, 22, 21, 14, 15, 23

Offset: 1

Omar E. Pol, Dec 16 2008, Dec 20 2008

Also, first differences of toothpick numbers A153000.

			If written as a triangle, begins:
.1,1;
.1,2,3,2;
.1,2,3,3,4,7,8,4;
.1,2,3,3,4,7,8,5,4,7,9,10,15,22,20,8;
....
Rows converge to A152980.
It appears that row sums give A004171. [From _Omar E. Pol_, May 25 2010]

David Applegate, The movie version
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS

Cf. toothpick sequence A139250.
Cf. A139251, A139252, A139560, A152968, A152973.
Cf. A152980, A152998, A153000, A153001, A153003, A153004, A153006.
Cf. A004171. [From Omar E. Pol, May 25 2010]

G.f.: (1+x)*(Prod(1+x^(2^k-1)+2*x^(2^k),k=1..oo)-1)/(1+2*x). - N. J. A. Sloane, May 20 2009

More terms from Omar E. Pol, Jul 26 2009

A083233 a(n) = (3*8^n + 0^n)/4.

Original entry on oeis.org

1, 6, 48, 384, 3072, 24576, 196608, 1572864, 12582912, 100663296, 805306368, 6442450944, 51539607552, 412316860416, 3298534883328, 26388279066624, 211106232532992, 1688849860263936, 13510798882111488, 108086391056891904, 864691128455135232

Offset: 0

Paul Barry, Apr 23 2003

Binomial transform of A083232. Inverse binomial transform of A066443.
Numbers k such that, except for some first term, k^2 = [A000302]^3 + [A004171]^3 + [A002001]^3; e.g., 3072^2 = 64^3 + 128^3 + 192^3; 51539607552^2 = 4194304^3 + 8388608^3 + 12582912^3. - Vincenzo Librandi, Aug 08 2010
With the exception of the first term, these numbers cannot be written as the sum of two integer cubes but can be written as the sum of two positive rational cubes (i.e., 6*8^n = (17*2^n/21)^3 + (37*2^n/21)^3). - Arkadiusz Wesolowski, Aug 15 2013
a(n+1) is the number of unit square faces on the convex hull of a level n Menger sponge. This follows since it has six exterior faces, each of which is a Sierpinski carpet with 8^n squares. - Allan Bickle, Nov 28 2022

			a(0) = (3*8^0 + 0^0)/4 = 4/4 = 1 (using 0^0 = 1).

Allan Bickle, Degrees of Menger and Sierpinski Graphs, Congr. Num. 227 (2016) 197-208.
Allan Bickle, MegaMenger Graphs, The College Mathematics Journal, 49 1 (2018) 20-26.
Eric Weisstein's World of Mathematics, Menger Sponge
Wikipedia, Menger sponge
Index entries for linear recurrences with constant coefficients, signature (8).

Cf. A083234. Subsequence of A159843.
Cf. A000302, A004171, A002001.
Cf. A291066, A083233, and A332705 on the surface area of the n-Menger sponge graph.

Join[{1},NestList[8#&,6,20]] (* Harvey P. Dale, Sep 25 2020 *)

a(n)=(3*8^n+0^n)/4 \\ Charles R Greathouse IV, Oct 07 2015

def A083233(n): return 3<<3*n-2 if n else 1 # Chai Wah Wu, Nov 27 2023

a(n) = (3*8^n + 0^n)/4.
G.f.: (1-2x)/(1-8x).
E.g.f.: (3*exp(8x) + exp(0))/4.
a(0) = 1, a(n+1) = 6*8^n. - Arkadiusz Wesolowski, Aug 15 2013

A002023 a(n) = 6*4^n.

Original entry on oeis.org

6, 24, 96, 384, 1536, 6144, 24576, 98304, 393216, 1572864, 6291456, 25165824, 100663296, 402653184, 1610612736, 6442450944, 25769803776, 103079215104, 412316860416, 1649267441664, 6597069766656, 26388279066624, 105553116266496, 422212465065984

Offset: 0

N. J. A. Sloane

From Peter M. Chema, Mar 02 2017: (Start)
Number of rods (line segments) required to make a Sierpinski tetrahedron of side length 2^n.
Also equals the number of balls (vertices) in a Sierpinski tetrahedron of side length 2^n+1 minus the number of balls in a Sierpinski tetrahedron of side length 2^n (the first difference in the tetrix numbers). See formula. (End)
Equivalently, the number of edges in the (n+1)-Sierpinski tetrahedron graph. - Eric W. Weisstein, Aug 17 2017
These numbers a(n) together with the 13 numbers from A337217 give the positive integers m represented uniquely by the ternary form x^2 + y^2 + 2*z^2, with integers 0 <= x <= y and 0 <= z. This is theorem 2.1 of Kaplansky, p. 87 with proof on p. 90. - Wolfdieter Lang, Aug 20 2020
a(n) is also the domination number of the (n+3)-Sierpinski tetrahedron graph. - Eric W. Weisstein, Sep 13 2021

Irving Kaplansky, Integers Uniquely Represented by Certain Ternary Forms, in "The Mathematics of Paul Erdős I", Ronald. L. Graham and Jaroslav Nešetřil (Eds.), Springer, 1997, pp. 86 - 94.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Shaoshi Chen, Hanqian Fang, Sergey Kitaev, and Candice X.T. Zhang, Patterns in Multi-dimensional Permutations, arXiv:2411.02897 [math.CO], 2024. See pp. 2, 17, 26.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Domination Number
Eric Weisstein's World of Mathematics, Sierpinski Tetrahedron Graph
Index entries for linear recurrences with constant coefficients, signature (4).

Cf. A283070 (vertex count).

Cf. A004171.

[6*4^n: n in [0..30]]; // Vincenzo Librandi, May 16 2011

6*4^Range[0, 100] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
Table[6 4^n, {n, 0, 20}] (* Eric W. Weisstein, Aug 17 2017 *)
LinearRecurrence[{4}, {6}, 20] (* Eric W. Weisstein, Aug 17 2017 *)
CoefficientList[Series[6/(1 - 4 x), {x, 0, 20}], x] (* Eric W. Weisstein, Aug 17 2017 *)
NestList[4#&,6,30] (* Harvey P. Dale, Mar 17 2024 *)

a(n)=6<<(2*n) \\ Charles R Greathouse IV, Apr 17 2012

From Philippe Deléham, Nov 23 2008: (Start)
a(n) = 4*a(n-1) for n > 0, a(0)=6.
G.f.: 6/(1-4*x). (End)
a(n) = 3*A004171(n). - R. J. Mathar, Mar 08 2011
From Peter M. Chema, Mar 03 2017: (Start)
a(n) = A283070(n+1) - A283070(n).
a(n) = A004171(n+1) - A004171(n). (End)
E.g.f.: 6*exp(4*x). - G. C. Greubel, Aug 17 2017

A360677 Sum of the right half (exclusive) of the prime indices of n.

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 1, 2, 3, 0, 2, 0, 4, 3, 2, 0, 2, 0, 3, 4, 5, 0, 3, 3, 6, 2, 4, 0, 3, 0, 2, 5, 7, 4, 4, 0, 8, 6, 4, 0, 4, 0, 5, 3, 9, 0, 3, 4, 3, 7, 6, 0, 4, 5, 5, 8, 10, 0, 5, 0, 11, 4, 3, 6, 5, 0, 7, 9, 4, 0, 4, 0, 12, 3, 8, 5, 6, 0, 4, 4, 13, 0, 6, 7

Offset: 1

Gus Wiseman, Mar 05 2023

nonn

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

			The prime indices of 810 are {1,2,2,2,2,3}, with right half (exclusive) {2,2,3}, so a(810) = 7.
The prime indices of 3675 are {2,3,3,4,4}, with right half (exclusive) {4,4}, so a(3675) = 8.

Positions of 0's are 1 and A000040.
Positions of last appearances are A004171.
Positions of first appearances are A100484.
These partitions are counted by A360672.
The value k > 0 appears A360673(k) times, inclusive A360671.
The left version is A360676.
The inclusive version is A360679.
A112798 lists prime indices, length A001222, sum A056239, median* A360005.
A360616 gives half of bigomega (exclusive), inclusive A360617.
First for prime indices, second for partitions, third for prime factors:
- A360676 gives left sum (exclusive), counted by A360672, product A361200.
- A360677 gives right sum (exclusive), counted by A360675, product A361201.
- A360678 gives left sum (inclusive), counted by A360675, product A347043.
- A360679 gives right sum (inclusive), counted by A360672, product A347044.
Cf. A001248, A026424, A359908, A359912, A360006, A360457.

prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
Table[Total[Take[prix[n],-Floor[Length[prix[n]]/2]]],{n,100}]

Last position of k is 2^(2k+1).
A360676(n) + A360679(n) = A001222(n).
A360677(n) + A360678(n) = A001222(n).

A096773 a(n) = 4*a(n-2) + 1 with a(1) = 0, a(2) = 3.

Original entry on oeis.org

0, 3, 1, 13, 5, 53, 21, 213, 85, 853, 341, 3413, 1365, 13653, 5461, 54613, 21845, 218453, 87381, 873813, 349525, 3495253, 1398101, 13981013, 5592405, 55924053, 22369621, 223696213, 89478485, 894784853, 357913941, 3579139413, 1431655765

Offset: 1

Gottfried Helms, Aug 15 2004

Remainders for classes m of integers n (mod 2^(m+1)). After applying one Collatz (3x+1)-transformation to the so-classified integers the result can be written in two classes (mod 6) only.
This classifying scheme covers all positive integers.
With one 3x+1-transformation T(x;p) := x' = (3x+1)/2^p, all numbers x, described in the form, with the free parameter i >= 0, x = i*2^N + a(N) result in x', describable by the two classes with the same parameter i:
x' = i*6 + 1 (for odd N>2), or x' = i*6 + 5 (for even N). Thus
x =  4*i +  3 -> x' = 6*i + 5, x =   8*i +  1 -> x' = 6*i + 1,
x = 16*i + 13 -> x' = 6*i + 5, x =  32*i +  5 -> x' = 6*i + 1,
x = 64*i + 53 -> x' = 6*i + 5, x = 128*i + 21 -> x' = 6*i + 1,
....
all with "i" as a free parameter >= 0 covering all positive integers.

			a(1) = (2^0-1)/3 =  0, a(2) = (5*2^1 - 1) / 3 =  3,
a(3) = (2^2-1)/3 =  1, a(4) = (5*2^3 - 1) / 3 = 13,
a(5) = (2^4-1)/3 =  5, a(6) = (5*2^5 - 1) / 3 = 53,
a(7) = (2^6-1)/3 = 21.
....

Michael De Vlieger, Table of n, a(n) for n = 1..3323
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for linear recurrences with constant coefficients, signature (1,4,-4).

Bisections are A002450 & A072197.
After the initial 0, column 1 of A257852.
Cf. A176965.

[(2^(n-1)*(3 + 2*(-1)^n) - 1)/3: n in [1..40]]; // Vincenzo Librandi, Jul 12 2015

a[1] = 0; a[2] = 3; a[n_] := a[n] = 4a[n - 2] + 1; Table[ a[n], {n, 35}] (* Robert G. Wilson v, Aug 20 2004 *)
Table[(2^(n - 1)*(3 + 2*(-1)^(n)) - 1)/3, {n, 10}] (* L. Edson Jeffery, Jul 12 2015 *)
nxt[{a_,b_}]:={b,4a+1}; NestList[nxt,{0,3},40][[;;,1]] (* or *) LinearRecurrence[{1,4,-4},{0,3,1},40] (* Harvey P. Dale, Mar 19 2025 *)

apply( {A096773(n) = if(n%2, 1, 5)<<(n-1)\3}, [1..55]) \\ M. F. Hasler, May 28 2024

# To map any (odd) v to its (r,c):
use bigint; $v=149; $r=$c=0; while(1){ $b=($v&1); $v>>=1; if ($b==($v&1)){ $c=($v>>1); last} $r++} $r&=1; # this splits the binary representation into two parts, at the first repeated digit from the right: the number of bits on the right is the row value, and the binary value on the left is the column value. Example: 149 => 1.00.10101 => (r,c)=(5,1). Ruud H.G. van Tol, Sep 23 2021

A096773=lambda n:((1 if n&1 else 5)<M. F. Hasler, May 28 2024

a(2m) = (5*2^(2m-1) - 1)/3, a(2m-1) = (2^(2m-2)-1)/3.
From Paul Curtz, Jul 01 2008; corrected by Bob Selcoe, Jul 28 2018: (Start)
a(2n) = 10*a(2n-1) + 3.
a(n+1) - 2*a(n) = A001045(n+2), signed. (End)
a(n) = (2^(n-1)*(3 + 2*(-1)^n) - 1)/3. - L. Edson Jeffery, Jul 12 2015
a(2n) = A086893(2n), a(2n+1) = A086893(2n-1), n > 0. - Yosu Yurramendi, Jan 17 2017
G.f.: -x^2*(-3+2*x) / ( (x-1)*(2*x+1)*(2*x-1) ). - R. J. Mathar, Mar 07 2017
a(2n) = A072197(n-1), n > 0; a(2n+1) = A002450(n), n >= 0. - Yosu Yurramendi, Mar 07 2017
a(2n) = (A266753(n) + A004171(n-1))/2, a(2n+1) = (A266753(n) - A004171(n-1))/2, n > 0. - Yosu Yurramendi, Mar 07 2017
a(n) = least residue 2*3^(2^(n-4)-1) - 1 (mod 2^n), n >= 5. - Bob Selcoe, Jul 26 2018
a(n) = 2*A176965(n-1) + 1 for n > 1. - Loren M. Pearson, Dec 06 2024

cp's OEIS Frontend

A083420 a(n) = 2*4^n - 1.

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A039770 Numbers k such that phi(k) is a perfect square.

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A000980 Number of ways of writing 0 as Sum_{k=-n..n} e(k)*k, where e(k) is 0 or 1.

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A002063 a(n) = 9*4^n.

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A087289 a(n) = 2^(2*n+1) + 1.

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A152978 a(n) = A139251(n+2)/4 = A152968(n+1)/2.

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