A002450 -id:A002450 - cp's OEIS frontend

A020988 a(n) = (2/3)*(4^n-1).

0, 2, 10, 42, 170, 682, 2730, 10922, 43690, 174762, 699050, 2796202, 11184810, 44739242, 178956970, 715827882, 2863311530, 11453246122, 45812984490, 183251937962, 733007751850, 2932031007402, 11728124029610, 46912496118442, 187649984473770, 750599937895082

Offset: 0

N. J. A. Sloane

Numbers whose binary representation is 10, n times (see A163662(n) for n >= 1). - Alexandre Wajnberg, May 31 2005
Numbers whose base-4 representation consists entirely of 2's; twice base-4 repunits. - Franklin T. Adams-Watters, Mar 29 2006
Expected time to finish a random Tower of Hanoi problem with 2n disks using optimal moves, so (since 2n is even and A010684(2n) = 1) a(n) = A060590(2n). - Henry Bottomley, Apr 05 2001
a(n) is the number of derangements of [2n + 3] with runs consisting of consecutive integers. E.g., a(1) = 10 because the derangements of {1, 2, 3, 4, 5} with runs consisting of consecutive integers are 5|1234, 45|123, 345|12, 2345|1, 5|4|123, 5|34|12, 45|23|1, 345|2|1, 5|4|23|1, 5|34|2|1 (the bars delimit the runs). - Emeric Deutsch, May 26 2003
For n > 0, also smallest numbers having in binary representation exactly n + 1 maximal groups of consecutive zeros: A087120(n) = a(n-1), see A087116. - Reinhard Zumkeller, Aug 14 2003
Number of walks of length 2n + 3 between any two diametrically opposite vertices of the cycle graph C_6. Example: a(0) = 2 because in the cycle ABCDEF we have two walks of length 3 between A and D: ABCD and AFED. - Emeric Deutsch, Apr 01 2004
From Paul Barry, May 18 2003: (Start)
Row sums of triangle using cumulative sums of odd-indexed rows of Pascal's triangle (start with zeros for completeness):
            0  0
            1  1
         1  4  4  1
      1  6 14 14  6  1
   1  8 27 49 49 27  8  1 (End)
a(n) gives the position of the n-th zero in A173732, i.e., A173732(a(n)) = 0 for all n and this gives all the zeros in A173732. - Howard A. Landman, Mar 14 2010
Smallest number having alternating bit sum -n. Cf. A065359. For n = 0, 1, ..., the last digit of a(n) is 0, 2, 0, 2, ... . - Washington Bomfim, Jan 22 2011
Number of toothpicks minus 1 in the toothpick structure of A139250 after 2^n stages. - Omar E. Pol, Mar 15 2012
For n > 0 also partial sums of the odd powers of 2 (A004171). - K. G. Stier, Nov 04 2013
Values of m such that binomial(4*m + 2, m) is odd. Cf. A002450. - Peter Bala, Oct 06 2015
For a(n) > 2, values of m such that m is two steps away from a power of 2 under the Collatz iteration. - Roderick MacPhee, Nov 10 2016
a(n) is the position of the first occurrence of 2^(n+1)-1 in A020986. See the Brillhart and Morton link, pp. 856-857. - John Keith, Jan 12 2021
a(n) is the number of monotone paths in the n-dimensional cross-polytope for a generic linear orientation. See the Black and De Loera link. - Alexander E. Black, Feb 15 2023

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..170 from Vincenzo Librandi)
Andrei Asinowski, Cyril Banderier, and Benjamin Hackl, On extremal cases of pop-stack sorting, Permutation Patterns (Zürich, Switzerland, 2019) [link is not very stable].
Andrei Asinowski, Cyril Banderier, and Benjamin Hackl, Flip-sort and combinatorial aspects of pop-stack sorting, arXiv:2003.04912 [math.CO], 2020.
Peter Bala, A characterization of A002450, A020988 and A080674.
Alexander E. Black, Monotone Paths on Polytopes: Combinatorics and Optimization, Ph. D. Dissertation, Univ. Calif. Davis (2024). See p. 59.
Alexander Black and Jesús De Loera, Monotone paths on cross-polytopes, arXiv:2102.01237 [math.CO], Feb 2021
John Brillhart and Peter Morton, A case study in mathematical research: the Golay-Rudin-Shapiro sequence, Amer. Math. Monthly, 103 (1996) 854-869.
Nobushige Kurokawa, Zeta functions over F_1, Proc. Japan Acad., 81, Ser. A (2005), 180-184. See Theorem 3 (3).
Jonathan L. Merzel, Ján Minač, Tung T. Nguyen, and Nguyên Duy Tân, On divisibility relation graphs, 2025. See p. 14.
Andrei K. Svinin, Tuenter polynomials and a Catalan triangle, arXiv:1603.05748 [math.CO], 2016. See p.3.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A020989, A047849, A108019, A295521, A020522.

[(2/3)*(4^n-1): n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011

A020988 := proc(n)
    2*(4^n-1)/3 ;
end proc: # R. J. Mathar, Feb 19 2015

(2(4^Range[0, 30] - 1))/3 (* or *) LinearRecurrence[{5, -4}, {0, 2}, 30] (* Harvey P. Dale, Sep 25 2013 *)

vector(100, n, n--; (2/3)*(4^n-1)) \\ Altug Alkan, Oct 06 2015

Vec(2*z/((1-z)*(1-4*z)) + O(z^30)) \\ Altug Alkan, Oct 11 2015

def A020988(n): return (2 * ((1 << (2 * n)) - 1)) // 3 # John Reimer Morales, Aug 05 2025

(((List.fill(20)(4: BigInt)).scanLeft(1: BigInt)( * )).map(2 * )).scanLeft(0: BigInt)( + ) // _Alonso del Arte, Sep 12 2019

a(n) = 4*a(n-1) + 2, a(0) = 0.
a(n) = A026644(2*n).
a(n) = A007583(n) - 1 = A039301(n+1) - 2 = A083584(n-1) + 1.
E.g.f. : (2/3)*(exp(4*x)-exp(x)). - Paul Barry, May 18 2003
a(n) = A007583(n+1) - 1 = A039301(n+2) - 2 = A083584(n) + 1. - Ralf Stephan, Jun 14 2003
G.f.: 2*x/((1-x)*(1-4*x)). - R. J. Mathar, Sep 17 2008
a(n) = a(n-1) + 2^(2n-1), a(0) = 0. - Washington Bomfim, Jan 22 2011
a(n) = A193652(2*n). - Reinhard Zumkeller, Aug 08 2011
a(n) = 5*a(n-1) - 4*a(n-2) (n > 1), a(0) = 0, a(1) = 2. - L. Edson Jeffery, Mar 02 2012
a(n) = (2/3)*A024036(n). - Omar E. Pol, Mar 15 2012
a(n) = 2*A002450(n). - Yosu Yurramendi, Jan 24 2017
From Seiichi Manyama, Nov 24 2017: (Start)
Zeta_{GL(2)/F_1}(s) = Product_{k = 1..4} (s-k)^(-b(2,k)), where Sum b(2,k)*t^k = t*(t-1)*(t^2-1). That is Zeta_{GL(2)/F_1}(s) = (s-3)*(s-2)/((s-4)*(s-1)).
Zeta_{GL(2)/F_1}(s) = Product_{n > 0} (1 - (1/s)^n)^(-A295521(n)) = Product_{n > 0} (1 - x^n)^(-A295521(n)) = (1-3*x)*(1-2*x)/((1-4*x)*(1-x)) = 1 + Sum_{k > 0} a(k-1)*x^k (x=1/s). (End)
From Oboifeng Dira, May 29 2020: (Start)
a(n) = A078008(2n+1) (second bisection).
a(n) = Sum_{k=0..n} binomial(2n+1, ((n+2) mod 3)+3k). (End)
From John Reimer Morales, Aug 04 2025: (Start)
a(n) = A000302(n) - A047849(n).
a(n) = A020522(n) + A000079(n) - A047849(n). (End)

Edited by N. J. A. Sloane, Sep 06 2006

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

Original entry on oeis.org

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

A094028 Expansion of 1/((1-x)(1-100x)).

Original entry on oeis.org

1, 101, 10101, 1010101, 101010101, 10101010101, 1010101010101, 101010101010101, 10101010101010101, 1010101010101010101, 101010101010101010101, 10101010101010101010101, 1010101010101010101010101, 101010101010101010101010101, 10101010101010101010101010101

Offset: 0

Paul Barry, Apr 22 2004

Regarded as binary numbers and converted to decimal, these become 1,5,21,85,... the partial sums of 4^n (see A002450).
Partial sums of 100^n.
Odd terms of A056830. - Alexandre Wajnberg, May 31 2005
101 is the only term that is prime, since (100^k-1)/99 = (10^k+1)/11 * (10^k-1)/9. When k is odd and not 1, (10^k+1)/11 is an integer > 1 and thus (100^k-1)/99 is nonprime. When k is even and greater than 2, (100^k-1)/99 has the prime factor 101 and is nonprime. - Felix Fröhlich, Oct 17 2015
Previous comment is the answer to the problem A1 proposed during the 50th Putnam Competition in 1989 (link). - Bernard Schott, Mar 24 2023

			From _Omar E. Pol_, Dec 13 2008: (Start)
=======================
n ....... a(n)
0 ........ 1
1 ....... 101
2 ...... 10101
3 ..... 1010101
4 .... 101010101
5 ... 10101010101
======================
(End)

Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 60.
Stephen Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Robert Price, Table of n, a(n) for n = 0..999
Kiran S. Kedlaya, The 50th William Lowell Putnam Mathematical Competition, Problem A1, Dec 02 1989.
J. V. Leyendekkers and A.G. Shannon, Modular Rings and the Integer 3, Notes on Number Theory & Discrete Mathematics, 17 (2011), pp. 47-51.
Robert Price, Comments on A094028 concerning Elementary Cellular Automata, Feb 21 2016.
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton.
Stephen Wolfram, A New Kind of Science.
Index to Elementary Cellular Automata.
Index entries for sequences related to cellular automata.
Index entries for linear recurrences with constant coefficients, signature (101,-100).
Index to sequences related to Olympiads and other Mathematical Competitions.

Bisection of A147759. [Omar E. Pol, Nov 13 2008]
Cf. A002450, A056830, A094027.
Cf. similar sequences of the form (k^n-1)/(k-1) listed in A269025.

[1+100*(100^n-1)/99 : n in [0..15]]; // Wesley Ivan Hurt, Oct 17 2015

A094028:=n->1+100*(100^n-1)/99: seq(A094028(n), n=0..15); # Wesley Ivan Hurt, Oct 17 2015

CoefficientList[Series[1/((1-x)(1-100x)),{x,0,20}],x] (* or *) Table[ FromDigits[ PadRight[{},2n-1,{1,0}]],{n,20}] (* or *) LinearRecurrence[ {101,-100},{1,101},20] (* or *) NestList[100#+1&,1,20] (* Harvey P. Dale, Apr 27 2015 *)

A094028(n):=1+100*(100^n-1)/99$
makelist(A094028(n),n,0,30); /* Martin Ettl, Nov 06 2012 */

a(n) = 1+100*(100^n-1)/99 \\ Felix Fröhlich, Oct 17 2015

Vec(1/((1-x)*(1-100*x)) + O(x^100)) \\ Altug Alkan, Oct 17 2015

G.f.: 1/((1-x)*(1-100*x)).
a(n) = 1 + 100*(100^n-1)/99. - N. J. A. Sloane, Apr 20 2008
a(n) = 100^(n+1)/99 - 1/99.
a(n) = A094027(2*n+1).
a(n) = 100*a(n-1) + 1, a(0) = 1. - Philippe Deléham, Feb 22 2014
a(n) = 101*a(n-1) - 100*a(n-2) for n > 1. - Wesley Ivan Hurt, Oct 17 2015
a(n) = (100^(n+1) - 1)/99. - Bernard Schott, Apr 15 2021
E.g.f.: exp(x)*(100*exp(99*x) - 1)/99. - Elmo R. Oliveira, Mar 06 2025

A106566 Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, 1, 1, 1, 1, 1, 1, ... ] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ... ] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 5, 5, 3, 1, 0, 14, 14, 9, 4, 1, 0, 42, 42, 28, 14, 5, 1, 0, 132, 132, 90, 48, 20, 6, 1, 0, 429, 429, 297, 165, 75, 27, 7, 1, 0, 1430, 1430, 1001, 572, 275, 110, 35, 8, 1, 0, 4862, 4862, 3432, 2002, 1001, 429, 154, 44, 9, 1

Offset: 0

Philippe Deléham, May 30 2005

Catalan convolution triangle; g.f. for column k: (x*c(x))^k with c(x) g.f. for A000108 (Catalan numbers).
Riordan array (1, xc(x)), where c(x) the g.f. of A000108; inverse of Riordan array (1, x*(1-x)) (see A109466).
Diagonal sums give A132364. - Philippe Deléham, Nov 11 2007

			Triangle begins:
  1;
  0,   1;
  0,   1,   1;
  0,   2,   2,  1;
  0,   5,   5,  3,  1;
  0,  14,  14,  9,  4,  1;
  0,  42,  42, 28, 14,  5, 1;
  0, 132, 132, 90, 48, 20, 6, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0, 1,
  0, 1, 1,
  0, 1, 1, 1,
  0, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1, 1, 1 (End)

Alois P. Heinz, Rows n = 0..140, flattened
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
Paul Barry, Chebyshev moments and Riordan involutions, arXiv:1912.11845 [math.CO], 2019.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
F. R. Bernhart, Catalan, Motzkin and Riordan numbers, Discr. Math., 204 (1999), 73-112.
D. Callan, A recursive bijective approach to counting permutations containing 3-letter patterns, arXiv:math/0211380 [math.CO], 2002.
E. Deutsch, Dyck path enumeration, Discrete Math., 204, 1999, 167-202.
FindStat - Combinatorial Statistic Finder, The number of touch points of a Dyck path, The number of initial rises of a Dyck paths, The number of nodes on the left branch of the tree, The number of subtrees.
R. K. Guy, Catwalks, sandsteps and Pascal pyramids, J. Integer Sequences, Vol. 3 (2000), Article #00.1.6.
A. Robertson, D. Saracino and D. Zeilberger, Refined restricted permutations, arXiv:math/0203033 [math.CO], 2002.
L. W. Shapiro, S. Getu, W.-J. Woan and L. C. Woodson, The Riordan group, Discrete Applied Math., 34 (1991), 229-239.

Column k for k = 0, 1, 2, ..., 13: A000007, A000108, A000108, A000245, A002057, A000344, A003517, A000588, A003517, A001392, A003518, A000589, A003519, A000590
The three triangles A059365, A106566 and A099039 are the same except for signs and the leading term.
Diagonals: A000012, A001477, A000096, A005586, A005587, A005557, A064059, A064061
See also A009766, A033184, A059365 for other versions.
Generalized Catalan numbers C(x, n) for -11 <= x <= 10: A064333, A064332, A064331, A064330, A064329, A064328, A064327, A064326, A064325, A064311, A064310, A000012, A000108, A064062, A064063, A064087, A064088, A064089, A064090, A064091, A064092, A064093.
The following are all versions of (essentially) the same Catalan triangle: A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.
Diagonals give A000108 A000245 A002057 A000344 A003517 A000588 A003518 A003519 A001392, ...

A106566:= func< n,k | n eq 0 select 1 else (k/n)*Binomial(2*n-k-1, n-k) >;
[A106566(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 06 2021

A106566 := proc(n,k)
    if n = 0 then
        1;
    elif k < 0 or k > n then
        0;
    else
        binomial(2*n-k-1,n-k)*k/n ;
    end if;
end proc: # R. J. Mathar, Mar 01 2015

T[n_, k_] := Binomial[2n-k-1, n-k]*k/n; T[0, 0] = 1; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 18 2017 *)
(* The function RiordanArray is defined in A256893. *)
RiordanArray[1&, #(1-Sqrt[1-4#])/(2#)&, 11] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

{T(n, k) = if( k<=0 || k>n, n==0 && k==0, binomial(2*n - k, n) * k/(2*n - k))}; /* Michael Somos, Oct 01 2022 */

def A106566(n, k): return 1 if (n==0) else (k/n)*binomial(2*n-k-1, n-k)
flatten([[A106566(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 06 2021

T(n, k) = binomial(2n-k-1, n-k)*k/n for 0 <= k <= n with n > 0; T(0, 0) = 1; T(0, k) = 0 if k > 0.
T(0, 0) = 1; T(n, 0) = 0 if n > 0; T(0, k) = 0 if k > 0; for k > 0 and n > 0: T(n, k) = Sum_{j>=0} T(n-1, k-1+j).
Sum_{j>=0} T(n+j, 2j) = binomial(2n-1, n), n > 0.
Sum_{j>=0} T(n+j, 2j+1) = binomial(2n-2, n-1), n > 0.
Sum_{k>=0} (-1)^(n+k)*T(n, k) = A064310(n). T(n, k) = (-1)^(n+k)*A099039(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000007(n), A000108(n), A000984(n), A007854(n), A076035(n), A076036(n), A127628(n), A126694(n), A115970(n) for x = 0,1,2,3,4,5,6,7,8 respectively.
Sum_{k>=0} T(n, k)*x^(n-k) = C(x, n); C(x, n) are the generalized Catalan numbers.
Sum_{j=0..n-k} T(n+k,2*k+j) = A039599(n,k).
Sum_{j>=0} T(n,j)*binomial(j,k) = A039599(n,k).
Sum_{k=0..n} T(n,k)*A000108(k) = A127632(n).
Sum_{k=0..n} T(n,k)*(x+1)^k*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x= 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Aug 25 2007
Sum_{k=0..n} T(n,k)*A000108(k-1) = A121988(n), with A000108(-1)=0. - Philippe Deléham, Aug 27 2007
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Oct 27 2007
T(n,k)*2^(n-k) = A110510(n,k); T(n,k)*3^(n-k) = A110518(n,k). - Philippe Deléham, Nov 11 2007
Sum_{k=0..n} T(n,k)*A000045(k) = A109262(n), A000045: Fibonacci numbers. - Philippe Deléham, Oct 28 2008
Sum_{k=0..n} T(n,k)*A000129(k) = A143464(n), A000129: Pell numbers. - Philippe Deléham, Oct 28 2008
Sum_{k=0..n} T(n,k)*A100335(k) = A002450(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A100334(k) = A001906(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A099322(k) = A015565(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A106233(k) = A003462(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A151821(k+1) = A100320(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A082505(k+1) = A144706(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A000045(2k+2) = A026671(n). - Philippe Deléham, Feb 11 2009
Sum_{k=0..n} T(n,k)*A122367(k) = A026726(n). - Philippe Deléham, Feb 11 2009
Sum_{k=0..n} T(n,k)*A008619(k) = A000958(n+1). - Philippe Deléham, Nov 15 2009
Sum_{k=0..n} T(n,k)*A027941(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
G.f.: Sum_{n>=0, k>=0} T(n, k)*x^k*z^n = 1/(1 - x*z*c(z)) where c(z) the g.f. of A000108. - Michael Somos, Oct 01 2022

Formula corrected by Philippe Deléham, Oct 31 2008
Corrected by Philippe Deléham, Sep 17 2009
Corrected by Alois P. Heinz, Aug 02 2012

A024036 a(n) = 4^n - 1.

Original entry on oeis.org

0, 3, 15, 63, 255, 1023, 4095, 16383, 65535, 262143, 1048575, 4194303, 16777215, 67108863, 268435455, 1073741823, 4294967295, 17179869183, 68719476735, 274877906943, 1099511627775, 4398046511103, 17592186044415, 70368744177663, 281474976710655

Offset: 0

N. J. A. Sloane

This sequence is the normalized length per iteration of the space-filling Peano-Hilbert curve. The curve remains in a square, but its length increases without bound. The length of the curve, after n iterations in a unit square, is a(n)*2^(-n) where a(n) = 4*a(n-1)+3. This is the sequence of a(n) values. a(n)*(2^(-n)*2^(-n)) tends to 1, the area of the square where the curve is generated, as n increases. The ratio between the number of segments of the curve at the n-th iteration (A015521) and a(n) tends to 4/5 as n increases. - Giorgio Balzarotti, Mar 16 2006
Numbers whose base-4 representation is 333....3. - Zerinvary Lajos, Feb 03 2007
From Eric Desbiaux, Jun 28 2009: (Start)
It appears that for a given area, a square n^2 can be divided into n^2+1 other squares.
It's a rotation and zoom out of a Cartesian plan, which creates squares with side
= sqrt( (n^2) / (n^2+1) ) --> A010503|A010532|A010541... --> limit 1,
and diagonal sqrt(2*sqrt((n^2)/(n^2+1))) --> A010767|... --> limit A002193.
(End)
Also the total number of line segments after the n-th stage in the H tree, if 4^(n-1) H's are added at the n-th stage to the structure in which every "H" is formed by 3 line segments. A164346 (the first differences of this sequence) gives the number of line segments added at the n-th stage. - Omar E. Pol, Feb 16 2013
a(n) is the cumulative number of segment deletions in a Koch snowflake after (n+1) iterations. - Ivan N. Ianakiev, Nov 22 2013
Inverse binomial transform of A005057. - Wesley Ivan Hurt, Apr 04 2014
For n > 0, a(n) is one-third the partial sums of A002063(n-1). - J. M. Bergot, May 23 2014
Also the cyclomatic number of the n-Sierpinski tetrahedron graph. - Eric W. Weisstein, Sep 18 2017

			G.f. = 3*x + 15*x^2 + 63*x^3 + 255*x^4 + 1023*x^5 + 4095*x^6 + ...

Graham Everest, Alf van der Poorten, Igor Shparlinski, and Thomas Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.

Felix Fröhlich, Table of n, a(n) for n = 0..99
Alexander V. Kitaev, Meromorphic Solution of the Degenerate Third Painlevé Equation Vanishing at the Origin, Symmetry, Integrability and Geometry: Methods and Applications, Vol. 15 (2019), 046, 53 pages; arXiv preprint, arXiv:1809.00122 [math.CA], 2018-2019.
Eric Weisstein's World of Mathematics, Cyclomatic Number.
Eric Weisstein's World of Mathematics, Sierpinski Tetrahedron Graph.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A000051, A000120, A000225, A000302, A002001, A002063, A002193, A002450, A005057, A010503, A010532, A010541, A010767, A015521, A020988, A027637 (partial products), A078904 (partial sums), A079978, A080674, A164346 (first differences), A178789, A179857, A248721.

a024036 = (subtract 1) . a000302
a024036_list = iterate ((+ 3) . (* 4)) 0
-- Reinhard Zumkeller, Oct 03 2012

A024036:=n->4^n-1; seq(A024036(n), n=0..30); # Wesley Ivan Hurt, Apr 04 2014

Array[4^# - 1 &, 50, 0] (* Vladimir Joseph Stephan Orlovsky, Nov 03 2009 *)
(* Start from Eric W. Weisstein, Sep 19 2017 *)
Table[4^n - 1, {n, 0, 20}]
4^Range[0, 20] - 1
LinearRecurrence[{5, -4}, {0, 3}, 20]
CoefficientList[Series[3 x/(1 - 5 x + 4 x^2), {x, 0, 20}], x]
(* End *)

for(n=0, 100, print1(4^n-1, ", ")) \\ Felix Fröhlich, Jul 04 2014

[gaussian_binomial(2*n,1, 2) for n in range(21)] # Zerinvary Lajos, May 28 2009

[stirling_number2(2*n+1, 2) for n in range(21)] # Zerinvary Lajos, Nov 26 2009

a(n) = 3*A002450(n). - N. J. A. Sloane, Feb 19 2004
G.f.: 3*x/((-1+x)*(-1+4*x)) = 1/(-1+x) - 1/(-1+4*x). - R. J. Mathar, Nov 23 2007
E.g.f.: exp(4*x) - exp(x). - Mohammad K. Azarian, Jan 14 2009
a(n) = A000051(n)*A000225(n). - Reinhard Zumkeller, Feb 14 2009
A079978(a(n)) = 1. - Reinhard Zumkeller, Nov 22 2009
a(n) = A179857(A000225(n)), for n > 0; a(n) > A179857(m), for m < A000225(n). - Reinhard Zumkeller, Jul 31 2010
a(n) = 4*a(n-1) + 3, with a(0) = 0. - Vincenzo Librandi, Aug 01 2010
A000120(a(n)) = 2*n. - Reinhard Zumkeller, Feb 07 2011
a(n) = (3/2)*A020988(n). - Omar E. Pol, Mar 15 2012
a(n) = (Sum_{i=0..n} A002001(i)) - 1 = A178789(n+1) - 3. - Ivan N. Ianakiev, Nov 22 2013
a(n) = n*E(2*n-1,1)/B(2*n,1), for n > 0, where E(n,x) denotes the Euler polynomials and B(n,x) the Bernoulli polynomials. - Peter Luschny, Apr 04 2014
a(n) = A000302(n) - 1. - Sean A. Irvine, Jun 18 2019
Sum_{n>=1} 1/a(n) = A248721. - Amiram Eldar, Nov 13 2020
a(n) = A080674(n) - A002450(n). - Elmo R. Oliveira, Dec 02 2023

More terms Wesley Ivan Hurt, Apr 04 2014

A109466 Riordan array (1, x(1-x)).

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 0, -2, 1, 0, 0, 1, -3, 1, 0, 0, 0, 3, -4, 1, 0, 0, 0, -1, 6, -5, 1, 0, 0, 0, 0, -4, 10, -6, 1, 0, 0, 0, 0, 1, -10, 15, -7, 1, 0, 0, 0, 0, 0, 5, -20, 21, -8, 1, 0, 0, 0, 0, 0, -1, 15, -35, 28, -9, 1, 0, 0, 0, 0, 0, 0, -6, 35, -56, 36, -10, 1, 0, 0, 0, 0, 0, 0, 1, -21, 70, -84, 45, -11, 1, 0, 0, 0, 0

Offset: 0

Philippe Deléham, Aug 28 2005

Inverse is Riordan array (1, xc(x)) (A106566).
Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, -1, 1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.
Modulo 2, this sequence gives A106344. - Philippe Deléham, Dec 18 2008
Coefficient array of the polynomials Chebyshev_U(n, sqrt(x)/2)*(sqrt(x))^n. - Paul Barry, Sep 28 2009

			Rows begin:
  1;
  0,  1;
  0, -1,  1;
  0,  0, -2,  1;
  0,  0,  1, -3,  1;
  0,  0,  0,  3, -4,   1;
  0,  0,  0, -1,  6,  -5,   1;
  0,  0,  0,  0, -4,  10,  -6,   1;
  0,  0,  0,  0,  1, -10,  15,  -7,  1;
  0,  0,  0,  0,  0,   5, -20,  21, -8,  1;
  0,  0,  0,  0,  0,  -1,  15, -35, 28, -9, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0,    1,
  0,   -1,    1,
  0,   -1,   -1,   1,
  0,   -2,   -1,  -1,   1,
  0,   -5,   -2,  -1,  -1,  1,
  0,  -14,   -5,  -2,  -1, -1,  1,
  0,  -42,  -14,  -5,  -2, -1, -1,  1,
  0, -132,  -42, -14,  -5, -2, -1, -1,  1,
  0, -429, -132, -42, -14, -5, -2, -1, -1, 1 (End)

Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Tom Copeland, Addendum to Elliptic Lie Triad

Cf. A026729 (unsigned version), A000108, A030528, A124644.

/* As triangle */ [[(-1)^(n-k)*Binomial(k, n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jan 14 2016

(* The function RiordanArray is defined in A256893. *)
RiordanArray[1&, #(1-#)&, 13] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

Number triangle T(n, k) = (-1)^(n-k)*binomial(k, n-k).
T(n, k)*2^(n-k) = A110509(n, k); T(n, k)*3^(n-k) = A110517(n, k).
Sum_{k=0..n} T(n,k)*A000108(k)=1. - Philippe Deléham, Jun 11 2007
From Philippe Deléham, Oct 30 2008: (Start)
Sum_{k=0..n} T(n,k)*A144706(k) = A082505(n+1).
Sum_{k=0..n} T(n,k)*A002450(k) = A100335(n).
Sum_{k=0..n} T(n,k)*A001906(k) = A100334(n).
Sum_{k=0..n} T(n,k)*A015565(k) = A099322(n).
Sum_{k=0..n} T(n,k)*A003462(k) = A106233(n). (End)
Sum_{k=0..n} T(n,k)*x^(n-k) = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = -12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12 respectively. - Philippe Deléham, Oct 27 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A010892(n), A099087(n), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n) for x = 0,1,2,3,4,5,6,7,8,9,10 respectively. - Philippe Deléham, Oct 28 2008
G.f.: 1/(1-y*x+y*x^2). - Philippe Deléham, Dec 15 2011
T(n,k) = T(n-1,k-1) - T(n-2,k-1), T(n,0) = 0^n. - Philippe Deléham, Feb 15 2012
Sum_{k=0..n} T(n,k)*x^(n-k) = F(n+1,-x) where F(n,x)is the n-th Fibonacci polynomial in x defined in A011973. - Philippe Deléham, Feb 22 2013
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Feb 26 2013
Sum_{k=0..n} T(n,k)*T(n+1,k) = -A110320(n). - Philippe Deléham, Feb 26 2013
For T(0,0) = 0, the signed triangle below has the o.g.f. G(x,t) = [t*x(1-x)]/[1-t*x(1-x)] = L[t*Cinv(x)] where L(x) = x/(1-x) and Cinv(x)=x(1-x) with the inverses Linv(x) = x/(1+x) and C(x)= [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108, so the inverse o.g.f. is Ginv(x,t) = C[Linv(x)/t] = [1-sqrt[1-4*x/(t(1+x))]]/2 (cf. A124644 and A030528). - Tom Copeland, Jan 19 2016

A131865 Partial sums of powers of 16.

Original entry on oeis.org

1, 17, 273, 4369, 69905, 1118481, 17895697, 286331153, 4581298449, 73300775185, 1172812402961, 18764998447377, 300239975158033, 4803839602528529, 76861433640456465, 1229782938247303441, 19676527011956855057, 314824432191309680913, 5037190915060954894609

Offset: 0

Reinhard Zumkeller, Jul 22 2007

16 = 2^4 is the growth measure for the Jacobsthal spiral (compare with phi^4 for the Fibonacci spiral). - Paul Barry, Mar 07 2008
Second quadrisection of A115451. - Paul Curtz, May 21 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=16, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n-1) = det(A). - Milan Janjic, Feb 21 2010
Partial sums are in A014899. Also, the sequence is related to A014931 by A014931(n+1) = (n+1)*a(n) - Sum_{i=0..n-1} a(i) for n>0. - Bruno Berselli, Nov 07 2012
a(n) is the total number of holes in a certain box fractal (start with 16 boxes, 1 hole) after n iterations. See illustration in links. - Kival Ngaokrajang, Jan 28 2015
Except for 1 and 17, all terms are Brazilian repunits numbers in base 16, and so belong to A125134. All terms >= 273 are composite because a(n) = ((4^(n+1) + 1) * (4^(n+1) - 1))/15. - Bernard Schott, Jun 06 2017
The sequence in binary is 1, 10001, 100010001, 1000100010001, 10001000100010001, ... cf. Plouffe link, A330135. - Frank Ellermann, Mar 05 2020

			a(3) = 1 + 16 + 256 + 4096 = 4369 = in binary: 1000100010001.
a(4) = (16^5 - 1)/15 = (4^5 + 1) * (4^5 - 1)/15 = 1025 * 1023/15 = 205 * 341 = 69905 = 11111_16. - _Bernard Schott_, Jun 06 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..800
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
Kival Ngaokrajang, Illustration of initial terms
Quynh Nguyen, Jean Pedersen, and Hien T. Vu, New Integer Sequences Arising From 3-Period Folding Numbers, Vol. 19 (2016), Article 16.3.1. See Table 1.
Simon Plouffe, Identities and approximations inspired from Ramanujan notebooks, III, 2009.
Index entries related to partial sums.
Index entries related to q-numbers.
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A091045, A218721, A218722, A064108, A218724-A218734, A132469, A218736-A218753, A133853, A094028, A218723. - M. F. Hasler, Nov 05 2012

[(16^(n+1)-1)/15: n in [0..20]]; // Vincenzo Librandi, Sep 17 2011

A131865:=n->(16^(n+1)-1)/15: seq(A131865(n), n=0..30); # Wesley Ivan Hurt, Apr 29 2017

Table[(2^(4 n) - 1)/15, {n, 16}] (* Robert G. Wilson v, Aug 22 2007 *)
Accumulate[16^Range[0,20]] (* or *) LinearRecurrence[{17,-16},{1,17},20] (* Harvey P. Dale, Jul 19 2019 *)

a[0]:0$
a[n]:=16*a[n-1]+1$
A131865(n):=a[n]$
makelist(A131865(n),n,1,30); /* Martin Ettl, Nov 05 2012 */

A131865(n)=16^n\15  \\ M. F. Hasler, Nov 05 2012

def A131865(n): return (1<<(n+1<<2))//15 # Chai Wah Wu, Nov 10 2022

[gaussian_binomial(n,1,16) for n in range(1,18)] # Zerinvary Lajos, May 28 2009

a(n) = if n=0 then 1 else a(n-1) + A001025(n).
for n > 0: A131851(a(n)) = n and abs(A131851(m)) < n for m < a(n).
a(n) = A098704(n+2)/2.
a(n) = (16^(n+1) - 1)/15. - Bernard Schott, Jun 06 2017
a(n) = (A001025(n+1) - 1)/15.
a(n) = 16*a(n-1) + 1. - Paul Curtz, May 20 2008
G.f.: 1 / ( (16*x-1)*(x-1) ). - R. J. Mathar, Feb 06 2011
E.g.f.: exp(x)*(16*exp(15*x) - 1)/15. - Stefano Spezia, Mar 06 2020

A075677 Reduced Collatz function R applied to the odd integers: a(n) = R(2n-1), where R(k) = (3k+1)/2^r, with r as large as possible.

Original entry on oeis.org

1, 5, 1, 11, 7, 17, 5, 23, 13, 29, 1, 35, 19, 41, 11, 47, 25, 53, 7, 59, 31, 65, 17, 71, 37, 77, 5, 83, 43, 89, 23, 95, 49, 101, 13, 107, 55, 113, 29, 119, 61, 125, 1, 131, 67, 137, 35, 143, 73, 149, 19, 155, 79, 161, 41, 167, 85, 173, 11, 179, 91, 185, 47, 191, 97, 197

Offset: 1

T. D. Noe, Sep 25 2002

The even-indexed terms a(2i+2) = 6i+5 = A016969(i), i >= 0 [Comment corrected by Bob Selcoe, Apr 06 2015]. The odd-indexed terms are the same as A067745. Note that this sequence is A016789 with all factors of 2 removed from each term. Also note that a(4i-1) = a(i). No multiple of 3 is in this sequence. See A075680 for the number of iterations of R required to yield 1.
From Bob Selcoe, Apr 06 2015: (Start)
All numbers in this sequence appear infinitely often.
From Eq. 1 and Eq. 2 in Formulas: Eq. 1 is used with 1/3 of the numbers in this sequence, Eq. 2 is used with 2/3 of the numbers.
(End)
Empirical: For arbitrary m, Sum_{n=2..A007583(m)} (a(n) - a(n-1)) = 0. - Fred Daniel Kline, Nov 23 2015
From Wolfdieter Lang, Dec 07 2021: (Start)
Only positive numbers congruent to 1 or 5 modulo 6 appear.
i) For the sequence entry with value A016921(m), for m >= 0, that is, a value from {1, 7, 13, ...}, the indices n are given by the row of array A178415(2*m+1, k), for k >= 1.
ii) For the sequence entry with value A007528(m), for m >= 1, that is, a value from {5, 11, 17, ...}, the indices n are given by the row of array A178415(2*m, k), for k >= 1.
See also the array A347834 with permuted row numbers and columns k >= 0. (End)

			a(11) = 1 because 21 is the 11th odd number and R(21) = 64/64 = 1.
From _Wolfdieter Lang_, Dec 07 2021: (Start)
i) 1 (mod 6) entry 1 = A016921(0) appears for n = A178415(1, k) = A347834(1, k-1) (the arrays), for k >= 1, that is, for {1, 5, 21, ..} = A002450.
ii) 5 (mod 6) entry 11 = A007528(2) appears for n = A178415(4, k) = A347835(3, k-1) (the arrays), for k >= 1, that is, for {7, 29, 117, ..} = A072261. (End)

Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section E16, pp. 330-336.
Victor Klee and Stan Wagon, Old and new unsolved problems in plane geometry and number theory, The Mathematical Association of America, 1991, p. 225, C(2n+1) = a(n+1), n >= 0.
Jeffrey C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010; see p. 57, also (90-9), p. 306.

T. D. Noe, Table of n, a(n) for n = 1..1000
Marc Chamberland, Una actualizacio del problema 3x + 1, Butl. Soc. Catalana Mat. (18), 19-45, 2003.
Jeffrey C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.
Fabian S. Reid, The Visual Pattern in the Collatz Conjecture and Proof of No Non-Trivial Cycles, arXiv:2105.07955 [math.GM], 2021.
Eric Weisstein's World of Mathematics, Collatz Problem.
Index entries for sequences related to 3x+1 (or Collatz) problem.

Cf. A000265, A000302, A002450, A007528, A016789, A016921, A016969, A065677, A072197, A072261, A075680, A081294, A178415, A191669, A329480, A347834.
Cf. A006370, A014682 (for non-reduced Collatz maps), A087230 (A371093), A371094.
Odd bisection of A139391.
Even bisection of A067745, which is also the odd bisection of this sequence.
After the initial 1, the second leftmost column of A256598.
Row 2 of A372283.

a075677 = a000265 . subtract 2 . (* 6) -- Reinhard Zumkeller, Jan 08 2014

f:=proc(n) local t1;
if n=1 then RETURN(1) else
t1:=3*n+1;
while t1 mod 2 = 0 do t1:=t1/2; od;
RETURN(t1); fi;
end;
# N. J. A. Sloane, Jan 21 2011

nextOddK[n_] := Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; (* assumes odd n *) Table[nextOddK[n], {n, 1, 200, 2}]
v[x_] := IntegerExponent[x, 2]; f[x_] := (3*x + 1)/2^v[3*x + 1]; Table[f[2*n - 1], {n, 66}] (* L. Edson Jeffery, May 06 2015 *)

a(n)=n+=2*n-1;n>>valuation(n,2) \\ Charles R Greathouse IV, Jul 05 2013

from sympy import divisors
def a(n):
    return max(d for d in divisors(n) if d % 2)
print([a(6*n - 2) for n in range(1, 101)]) # Indranil Ghosh, Apr 15 2017, after formula by Reinhard Zumkeller

a(n) = A000265(6*n-2) = A000265(3*n-1). - Reinhard Zumkeller, Jan 08 2014
From Bob Selcoe, Apr 05 2015: (Start)
For all n>=1 and for every k, there exists j>=0 dependent upon n and k such that either:
Eq. 1: a(n) = (3n-1)/2^(2j+1) when k = ((4^(j+1)-1)/3) mod 2^(2j+3). Alternatively: a(n) = A016789(n-1)/A081294(j+1) when k = A002450(j+1) mod A081294(j+2). Example: n=51; k=101 == 5 mod 32, j=1. a(51) = 152/8 = 19.
  or
Eq. 2: a(n) = (3n-1)/4^j when k = (5*2^(2j+1) - 1)/3 mod 4^(j+1). Alternatively: a(n) = A016789(n-1)/A000302(j) when k = A072197(j) mod A000302(j+1). Example: n=91; k=181 == 53 mod 64, j=2. a(91) = 272/16 = 17.
(End) [Definition corrected by William S. Hilton, Jul 29 2017]
a(n) = a(n + g*2^r) - 6*g, n > -g*2^r. Examples: n=59; a(59)=11, r=5. g=-1: 11 = a(27) = 5 - (-1)*6; g=1: 11 = a(91) = 17 - 1*6; g=2: 11 = a(123) = 23 - 2*6; g=3: 11 = a(155) = 29 - 3*6; etc. - Bob Selcoe, Apr 06 2015
a(n) = a((1 + (3*n - 1)*4^(k-1))/3), k>=1 (cf. A191669). - L. Edson Jeffery, Oct 05 2015
a(n) = a(4n-1). - Bob Selcoe, Aug 03 2017
a(n) = A139391(2n-1). - Antti Karttunen, May 06 2024
Sum_{k=1..n} a(k) ~ n^2. - Amiram Eldar, Aug 26 2024
G.f.: Sum_{k>=1} ((3 + 2*(-1)^k)*x^(3*2^(k - 1) - (-2)^k/3 + 1/3) + (3 - 2*(-1)^k)*x^(2^(k - 1) - (-2)^k/3 + 1/3))/(x^(2^k) - 1)^2. - Miles Wilson, Oct 26 2024

A047849 a(n) = (4^n + 2)/3.

Original entry on oeis.org

1, 2, 6, 22, 86, 342, 1366, 5462, 21846, 87382, 349526, 1398102, 5592406, 22369622, 89478486, 357913942, 1431655766, 5726623062, 22906492246, 91625968982, 366503875926, 1466015503702, 5864062014806, 23456248059222, 93824992236886, 375299968947542

Offset: 0

Clark Kimberling

Counts closed walks of length 2n at a vertex of the cyclic graph on 6 nodes C_6. - Paul Barry, Mar 10 2004
The number of closed walks of odd length of the cyclic graph is zero. See the array w(N,L) and triangle a(K,N) given in A199571 for the general case. - Wolfdieter Lang, Nov 08 2011
A. A. Ivanov conjectures that the dimension of the universal embedding of the unitary dual polar space DSU(2n,4) is a(n). - J. Taylor (jt_cpp(AT)yahoo.com), Apr 02 2004
Permutations with two fixed points avoiding 123 and 132.
Related to A024495(6n), A131708(6n+2), A024493(6n+4). First differences give A000302. - Paul Curtz, Mar 25 2008
Also the number of permutations of length n which avoid 4321 and 4123 (or 4321 and 3412, or 4123 and 3214, or 4123 and 2143). - Vincent Vatter, Aug 17 2009; minor correction by Henning Ulfarsson, May 14 2017
This sequence is related to A014916 by A014916(n) = n*a(n)-Sum_{i=0..n-1} a(i). - Bruno Berselli, Jul 27 2010, Mar 02 2012
For n >= 2, a(n) equals 2^n times the permanent of the (2n-2) X (2n-2) tridiagonal matrix with 1/sqrt(2)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
For n > 0, counts closed walks of length (n) at a vertex of a triangle with two (x2) loops at each vertex. - David Neil McGrath, Sep 11 2014
From Michel Lagneau, Feb 26 2015: (Start)
a(n) is also the sum of the largest odd divisors of the integers 1,2,3, ..., 2^n.
Proof:
All integers of the set {2^(n-1)+1, 2^(n-1)+2, ..., 2^n} are of the form 2^k(2m+1) where k and m integers. The greatest odd divisor of a such integer is 2m+1. Reciprocally, if 2m+1 is an odd integer <= 2^n, there exists a unique integer in the set {2^(n-1)+1, 2^(n-1)+2, ..., 2^n} where 2m+1 is the greatest odd divisor. Hence the recurrence relation:
   a(n) = a(n+1) + (1 + 3 + ... + 2*2^(n-1) - 1) = a(n-1) + 4^(n-1) for n >= 2.
We obtain immediately: a(n) = a(1) + 4 + ... + 4^n = (4^n+2)/3. (End)
The number of Riordan graphs of order n+1. See Cheon et al., Proposition 2.8. - Peter Bala, Aug 12 2021
Let q = 2^(2n+1) and Omega_n be the Suzuki ovoid with q^2 + 1 points. Then a(n) is the number of orbits of the finite Suzuki group Sz(q) on 3-subsets of Omega_n. Link to result in References. - Paul M. Bradley, Jun 04 2023
Also the cogrowth sequence for the 8-element dihedral group D8 = . - Sean A. Irvine, Nov 04 2024

			a(2) = 6 for the number of round trips in C_6 from the six round trips from, say, vertex no. 1: 12121, 16161, 12161, 16121, 12321 and 16561. - _Wolfdieter Lang_, Nov 08 2011

Bruno Berselli, Table of n, a(n) for n = 0..1000
Jeffrey M. Barnes, Georgia Benkart, and Tom Halverson, McKay centralizer algebras. Proc. Lond. Math. Soc. (3) 112, No. 2, 375-414 (2016).
Christian Bean, Finding structure in permutation sets, Ph.D. Dissertation, Reykjavík University, School of Computer Science, 2018.
Georgia Benkart and Tom Halverson, McKay Centralizer Algebras, hal-02173744 [math.CO], 2020.
Bruno Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
Paul Bradley and Peter Rowley, Orbits on k-subsets of 2-transitive Simple Lie-type Groups, Algebraic Combinatorics, Volume 5 (2022) no. 1, pp. 37-51.
Pascal Caron, Jean-Gabriel Luque, and Bruno Patrou, A combinatorial approach for the state complexity of the Shuffle product, arXiv:1905.08120 [cs.FL], 2019.
Gi-Sang Cheon, Ji-Hwan Jung, Sergey Kitaev, and Seyed Ahmad Mojallal, Riordan graphs I: structural properties, Linear Algebra and its Applications, 579. pp. 89-135, Prop. 2.8. (2019).
B. N. Cooperstein and E. E. Shult, A note on embedding and generating dual polar spaces. Adv. Geom. 1 (2001), 37-48. See Conjecture 5.5.
Kittitat Iamthong, Ji-Hwan Jung, and Sergey Kitaev, Encoding labelled p-Riordan graphs by words and pattern-avoiding permutations, arXiv:2009.01410 [math.CO], 2020.
D. Kremer and W. C. Shiu, Finite transition matrices for permutations avoiding pairs of length four patterns, Discrete Mathematics 268 (2003), 171-183.
T. Mansour and A. Robertson, Refined restricted permutations avoiding subsets of patterns of length three, arXiv:math/0204005 [math.CO], 2002.
Wikipedia, Permutation classes avoiding two patterns of length 4.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A000302, A001045, A002450, A007583, A024493, A047848, A078008, A121314, A131708, A178789, A199571.

Cf. A014916, A073724, A020988, A039301.

[(4^n+2)/3: n in [0..30]]; // Vincenzo Librandi, Dec 07 2015

(4^Range[0,30] +2)/3 (* or *) LinearRecurrence[{5,-4},{1,2},30] (* Harvey P. Dale, Nov 27 2015 *)

PARI
```
a(n)=(4^n+2)/3;
```

def A047849(n): return (pow(4, n) +2)//3
print([A047849(n) for n in range(41)]) # G. C. Greubel, Jan 17 2025

def A047849(n): return ((1 << (2 * n)) + 2) // 3 # John Reimer Morales, Aug 05 2025

n-th difference of a(n), a(n-1), ..., a(0) is 3^(n-1) for n >= 1.
From Henry Bottomley, Aug 29 2000: (Start)
a(n) = (4^n + 2)/3.
a(n) = 4*a(n-1) - 2.
a(n) = 5*a(n-1) - 4*a(n-2).
a(n) = 2*A007583(n-1) = A002450(n) + 1. (End)
a(n) = A047848(1,n).
With interpolated zeros, this is (-2)^n/6 + 2^n/6 + (-1)^n/3 + 1/3. - Paul Barry, Aug 26 2003
a(n) = A007583(n) - A002450(n) = A001045(2n+1) - A001045(2n) . - Philippe Deléham, Feb 25 2004
Second binomial transform of A078008. Binomial transform of 1, 1, 3, 9, 81, ... (3^n/3 + 2*0^n/3). a(n) = A078008(2n). - Paul Barry, Mar 14 2004
G.f.: (1-3*x)/((1-x)*(1-4*x)). - Herbert Kociemba, Jun 06 2004
a(n) = Sum_{k=0..n} 2^k*A121314(n,k). - Philippe Deléham, Sep 15 2006
a(n) = (A001045(2*n+1) + 1)/2. - Paul Barry, Dec 05 2007
From Bruno Berselli, Jul 27 2010: (Start)
a(n) = (A020988(n) + 2)/2 = A039301(n+1)/2.
Sum_{i=0..n} a(i) = A073724(n). (End)
For n >= 3, a(n) equals [2, 1, 1; 1, 2, 1; 1, 1, 2]^(n - 2)*{1, 1, 2}*{1, 1, 2}. - John M. Campbell, Jul 09 2011
a(n) = Sum_{k=0..n} binomial(2*n, mod(n,3) + 3*k). - Oboifeng Dira, May 29 2020
From Elmo R. Oliveira, Dec 21 2023: (Start)
E.g.f.: (exp(x)*(exp(3*x) + 2))/3.
a(n) = A178789(n+1)/3. (End)
a(n) = A000302(n) - A020988(n). - John Reimer Morales, Aug 03 2025

New name from Charles R Greathouse IV, Dec 22 2011

A218722 a(n) = (19^n-1)/18.

Original entry on oeis.org

0, 1, 20, 381, 7240, 137561, 2613660, 49659541, 943531280, 17927094321, 340614792100, 6471681049901, 122961939948120, 2336276859014281, 44389260321271340, 843395946104155461, 16024522975978953760, 304465936543600121441

Offset: 0

M. F. Hasler, Nov 04 2012

Partial sums of powers of 19 (A001029); q-integers for q=19: diagonal k=1 in triangle A022183.

Partial sums are in A014903. Also, the sequence is related to A014936 by A014936(n) = n*a(n)-sum(a(i), i=0..n-1) for n>0. - Bruno Berselli, Nov 06 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..700
Index entries related to partial sums
Index entries for linear recurrences with constant coefficients, signature (20,-19).

Cf. similar sequences of the form (k^n-1)/(k-1): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A064108, A218724-A218734, A132469, A218736-A218753, A133853, A094028, A218723.

Cf. A001029, A014903, A014936, A022183.

[n le 2 select n-1 else 20*Self(n-1)-19*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012

LinearRecurrence[{20, -19}, {0, 1}, 40] (* Vincenzo Librandi, Nov 07 2012 *)

A218722(n):=(19^n-1)/18$ makelist(A218722(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

PARI
```
A218722(n)=19^n\18
```

a(n) = floor(19^n/18).
G.f.: x/((1-x)*(1-19*x)). - Bruno Berselli, Nov 06 2012
a(n) = 20*a(n-1) - 19*a(n-2). - Vincenzo Librandi, Nov 07 2012
E.g.f.: exp(10*x)*sinh(9*x)/9. - Stefano Spezia, Mar 11 2023

cp's OEIS Frontend

A020988 a(n) = (2/3)*(4^n-1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Python

Scala

Formula

Extensions

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

Python

Formula

A094028 Expansion of 1/((1-x)*(1-100*x)).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Maxima

PARI

PARI

Formula

A106566 Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, 1, 1, 1, 1, 1, 1, ... ] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ... ] where DELTA is the operator defined in A084938.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A024036 a(n) = 4^n - 1.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

A094028 Expansion of 1/((1-x)(1-100x)).