A000302 -id:A000302 - cp's OEIS frontend

A289780 p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.

1, 4, 14, 47, 156, 517, 1714, 5684, 18851, 62520, 207349, 687676, 2280686, 7563923, 25085844, 83197513, 275925586, 915110636, 3034975799, 10065534960, 33382471801, 110713382644, 367182309614, 1217764693607, 4038731742156, 13394504020957, 44423039068114

Offset: 0

Clark Kimberling, Aug 10 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x).
Taking p(S) = 1 - S gives the INVERT transform of s, so that p-INVERT is a generalization of the INVERT transform (e.g., A033453).
Guide to p-INVERT sequences using p(S) = 1 - S - S^2:
t(A000012) = t(1,1,1,1,1,1,1,...)    = A001906
t(A000290) = t(1,4,9,16,25,36,...)   = A289779
t(A000027) = t(1,2,3,4,5,6,7,8,...)  = A289780
t(A000045) = t(1,2,3,5,8,13,21,...)  = A289781
t(A000032) = t(2,1,3,4,7,11,14,...)  = A289782
t(A000244) = t(1,3,9,27,81,243,...)  = A289783
t(A000302) = t(1,4,16,64,256,...)    = A289784
t(A000351) = t(1,5,25,125,625,...)   = A289785
t(A005408) = t(1,3,5,7,9,11,13,...)  = A289786
t(A005843) = t(2,4,6,8,10,12,14,...) = A289787
t(A016777) = t(1,4,7,10,13,16,...)   = A289789
t(A016789) = t(2,5,8,11,14,17,...)   = A289790
t(A008585) = t(3,6,9,12,15,18,...)   = A289795
t(A000217) = t(1,3,6,10,15,21,...)   = A289797
t(A000225) = t(1,3,7,15,31,63,...)   = A289798
t(A000578) = t(1,8,27,64,625,...)    = A289799
t(A000984) = t(1,2,6,20,70,252,...)  = A289800
t(A000292) = t(1,4,10,20,35,56,...)  = A289801
t(A002620) = t(1,2,4,6,9,12,16,...)  = A289802
t(A001906) = t(1,3,8,21,55,144,...)  = A289803
t(A001519) = t(1,1,2,5,13,34,...)    = A289804
t(A103889) = t(2,1,4,3,6,5,8,7,,...) = A289805
t(A008619) = t(1,1,2,2,3,3,4,4,...)  = A289806
t(A080513) = t(1,2,2,3,3,4,4,5,...)  = A289807
t(A133622) = t(1,2,1,3,1,4,1,5,...)  = A289809
t(A000108) = t(1,1,2,5,14,42,...)    = A081696
t(A081696) = t(1,1,3,9,29,97,...)    = A289810
t(A027656) = t(1,0,2,0,3,0,4,0,5...) = A289843
t(A175676) = t(1,0,0,2,0,0,3,0,...)  = A289844
t(A079977) = t(1,0,1,0,2,0,3,...)    = A289845
t(A059841) = t(1,0,1,0,1,0,1,...)    = A289846
t(A000040) = t(2,3,5,7,11,13,...)    = A289847
t(A008578) = t(1,2,3,5,7,11,13,...)  = A289828
t(A000142) = t(1!, 2!, 3!, 4!, ...)  = A289924
t(A000201) = t(1,3,4,6,8,9,11,...)   = A289925
t(A001950) = t(2,5,7,10,13,15,...)   = A289926
t(A014217) = t(1,2,4,6,11,17,29,...) = A289927
t(A000045*) = t(0,1,1,2,3,5,...)     = A289975 (* indicates prepended 0's)
t(A000045*) = t(0,0,1,1,2,3,5,...)   = A289976
t(A000045*) = t(0,0,0,1,1,2,3,5,...) = A289977
t(A290990*) = t(0,1,2,3,4,5,...)     = A290990
t(A290990*) = t(0,0,1,2,3,4,5,...)   = A290991
t(A290990*) = t(0,0,01,2,3,4,5,...)  = A290992

			Example 1:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S.
S(x) = x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - (x + 2x^2 + 3x^3 + 4x^4 + ... )
- p(0) + 1/p(S(x)) = -1 + 1 + x + 3x^2 + 8x^3 + 21x^4 + ...
T(x) = 1 + 3x + 8x^2 + 21x^3 + ...
t(s) = (1,3,8,21,...) = A001906.
***
Example 2:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S - S^2.
S(x) =  x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - ( x + 2x^2 + 3x^3 + 4x^4 + ...) - ( x + 2x^2 + 3x^3 + 4x^4 + ...)^2
- p(0) + 1/p(S(x)) = -1 + 1 + x + 4x^2 + 14x^3 + 47x^4 + ...
T(x) = 1 + 4x + 14x^2 + 47x^3 + ...
t(s) = (1,4,14,47,...) = A289780.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5, -7, 5, -1)

Cf. A000027.

P:=[1,4,14,47];; for n in [5..10^2] do P[n]:=5*P[n-1]-7*P[n-2]+5*P[n-3]-P[n-4]; od; P; # Muniru A Asiru, Sep 03 2017

z = 60; s = x/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289780 *)

x='x+O('x^99); Vec((1-x+x^2)/(1-5*x+7*x^2-5*x^3+x^4)) \\ Altug Alkan, Aug 13 2017

G.f.: (1 - x + x^2)/(1 - 5 x + 7 x^2 - 5 x^3 + x^4).

a(n) = 5*a(n-1) - 7*a(n-2) + 5*a(n-3) - a(n-4).

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

Original entry on oeis.org

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

A027306 a(n) = 2^(n-1) + ((1 + (-1)^n)/4)*binomial(n, n/2).

Original entry on oeis.org

1, 1, 3, 4, 11, 16, 42, 64, 163, 256, 638, 1024, 2510, 4096, 9908, 16384, 39203, 65536, 155382, 262144, 616666, 1048576, 2449868, 4194304, 9740686, 16777216, 38754732, 67108864, 154276028, 268435456, 614429672, 1073741824, 2448023843

Offset: 0

Clark Kimberling

Inverse binomial transform of A027914. Hankel transform (see A001906 for definition) is {1, 2, 3, 4, ..., n, ...}. - Philippe Deléham, Jul 21 2005
Number of walks of length n on a line that starts at the origin and ends at or above 0. - Benjamin Phillabaum, Mar 05 2011
Number of binary integers (i.e., with a leading 1 bit) of length n+1 which have a majority of 1-bits. E.g., for n+1=4: (1011, 1101, 1110, 1111) a(3)=4. - Toby Gottfried, Dec 11 2011
Number of distinct symmetric staircase walks connecting opposite corners of a square grid of side n > 1. - Christian Barrientos, Nov 25 2018
From Gus Wiseman, Aug 20 2021: (Start)
Also the number of integer compositions of n + 1 with alternating sum > 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A345917. For example, the a(0) = 1 through a(4) = 11 compositions are:
  (1)  (2)  (3)    (4)    (5)
            (21)   (31)   (32)
            (111)  (112)  (41)
                   (211)  (113)
                          (122)
                          (212)
                          (221)
                          (311)
                          (1121)
                          (2111)
                          (11111)
The following relate to these compositions:
- The unordered version is A027193.
- The complement is counted by A058622.
- The reverse unordered version is A086543.
- The version for alternating sum >= 0 is A116406.
- The version for alternating sum < 0 is A294175.
- Ranked by A345917. (End)
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 07 2022

			From _Gus Wiseman_, Aug 20 2021: (Start)
The a(0) = 1 through a(4) = 11 binary numbers with a majority of 1-bits (Gottfried's comment) are:
  1   11   101   1011   10011
           110   1101   10101
           111   1110   10110
                 1111   10111
                        11001
                        11010
                        11011
                        11100
                        11101
                        11110
                        11111
The version allowing an initial zero is A058622.
(End)

A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992, Eq. (4.2.1.6)

Alois P. Heinz, Table of n, a(n) for n = 0..1000
F. Disanto, A. Frosini, and S. Rinaldi, Square involutions, J. Int. Seq. 14 (2011) # 11.3.5.
Zachary Hamaker and Eric Marberg, Atoms for signed permutations, arXiv:1802.09805 [math.CO], 2018.
Donatella Merlini and Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.3.
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.

a(n) = Sum{(k+1)T(n, m-k)}, 0<=k<=[ (n+1)/2 ], T given by A008315.
Column k=2 of A226873. - Alois P. Heinz, Jun 21 2013
Cf. A008949, A248574, A001405, A246437.
The even bisection is A000302.
The odd bisection appears to be A032443.
Cf. A000984, A001700, A001791, A008549, A011782, A088218, A097805, A163493, A182616, A345197.

List([0..35],n->Sum([0..Int(n/2)],k->Binomial(n,k))); # Muniru A Asiru, Nov 27 2018

a027306 n = a008949 n (n `div` 2)  -- Reinhard Zumkeller, Nov 14 2014

[2^(n-1)+(1+(-1)^n)/4*Binomial(n, n div 2): n in [0..40]]; // Vincenzo Librandi, Jun 19 2016

a:= proc(n) add(binomial(n, j), j=0..n/2) end:
seq(a(n), n=0..32); # Zerinvary Lajos, Mar 29 2009

Table[Sum[Binomial[n, k], {k, 0, Floor[n/2]}], {n, 1, 35}]
(* Second program: *)
a[0] = a[1] = 1; a[2] = 3; a[n_] := a[n] = (2(n-1)(2a[n-2] + a[n-1]) - 8(n-2) a[n-3])/n; Array[a, 33, 0] (* Jean-François Alcover, Sep 04 2016 *)

a(n)=if(n<0,0,(2^n+if(n%2,0,binomial(n, n/2)))/2)

a(n) = Sum_{k=0..floor(n/2)} binomial(n,k).
Odd terms are 2^(n-1). Also a(2n) - 2^(2n-1) is given by A001700. a(n) = 2^n + (n mod 2)*binomial(n, (n-1)/2).
E.g.f.: (exp(2x) + I_0(2x))/2.
O.g.f.: 2*x/(1-2*x)/(1+2*x-((1+2*x)*(1-2*x))^(1/2)). - Vladeta Jovovic, Apr 27 2003
a(n) = A008949(n, floor(n/2)); a(n) + a(n-1) = A248574(n), n > 0. - Reinhard Zumkeller, Nov 14 2014
From Peter Bala, Jul 21 2015: (Start)
a(n) = [x^n]( 2*x - 1/(1 - x) )^n.
O.g.f.: (1/2)*( 1/sqrt(1 - 4*x^2) + 1/(1 - 2*x) ).
Inverse binomial transform is (-1)^n*A246437(n).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + ... is the o.g.f. for A001405. (End)
a(n) = Sum_{k=1..floor((n+1)/2)} binomial(n-1,(2n+1-(-1)^n)/4 -k). - Anthony Browne, Jun 18 2016
D-finite with recurrence: n*a(n) + 2*(-n+1)*a(n-1) + 4*(-n+1)*a(n-2) + 8*(n-2)*a(n-3) = 0. - R. J. Mathar, Aug 09 2017

Better description from Robert G. Wilson v, Aug 30 2000 and from Yong Kong (ykong(AT)curagen.com), Dec 28 2000

A001025 Powers of 16: a(n) = 16^n.

Original entry on oeis.org

1, 16, 256, 4096, 65536, 1048576, 16777216, 268435456, 4294967296, 68719476736, 1099511627776, 17592186044416, 281474976710656, 4503599627370496, 72057594037927936, 1152921504606846976, 18446744073709551616, 295147905179352825856, 4722366482869645213696, 75557863725914323419136, 1208925819614629174706176

Offset: 0

N. J. A. Sloane

Same as Pisot sequences E(1, 16), L(1, 16), P(1, 16), T(1, 16). Essentially same as Pisot sequences E(16, 256), L(16, 256), P(16, 256), T(16, 256). See A008776 for definitions of Pisot sequences.
Convolution-square (auto-convolution) of A098430. - R. J. Mathar, May 22 2009
Subsequence of A161441: A160700(a(n)) = 1. - Reinhard Zumkeller, Jun 10 2009
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 16-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011

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).

Muniru A Asiru, Table of n, a(n) for n = 0..820 (terms n = 0..100 from T. D. Noe)
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 280
Tanya Khovanova, Recursive Sequences
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Index entries for linear recurrences with constant coefficients, signature (16).

Partial sums give A131865.

Cf. A000079, A000302, A001018, A005843, A008586.

List([0..20],n->16^n); # Muniru A Asiru, Nov 07 2018

a001025 = (16 ^)
a001025_list = iterate (* 16) 1  -- Reinhard Zumkeller, Nov 07 2012

A001025:=-1/(-1+16*z); # Simon Plouffe in his 1992 dissertation

Table[4^(2*n), {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Mar 01 2009 *)

A001025(n):=16^n$
makelist(A001025(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

a(n)=1<<(4*n) \\ Charles R Greathouse IV, Feb 01 2012

print([16**n for n in range(20)]) # Stefano Spezia, Nov 10 2018

[lucas_number1(n,16,0) for n in range(1, 18)] # Zerinvary Lajos, Apr 29 2009

G.f.: 1/(1-16*x).
E.g.f.: exp(16*x).
From Muniru A Asiru, Nov 07 2018: (Start)
a(n) = 16^n.
a(0) = 1, a(n) = 16*a(n-1). (End)
a(n) = 4^A005843(n) = 2^A008586(n) = A000302(n)^2 = A000079(n)*A001018(n). - Muniru A Asiru, Nov 10 2018
a(n) = ( Sum_{k = 0..n} (2*k + 1)*binomial(2*n + 1, n - k) ) * ( Sum_{k = 0..n} (-1)^k/(2*k + 1)*binomial(2*n + 1, n - k) ). - Peter Bala, Feb 12 2019
a(n) = Sum_{k = 0..2*n} A000984(k) * A000984(2*n-k). - Peter Bala, Aug 23 2025

A133494 Diagonal of the array of iterated differences of A047848.

Original entry on oeis.org

1, 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883

Offset: 0

Paul Barry, Paul Curtz, Dec 23 2007

a(n) is the number of ways to choose a composition C, and then choose a composition of each part of C. - Geoffrey Critzer, Mar 19 2012
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [1, 1, 1; 1, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
a(n) is the reptend length of 1/3^(n+1) in decimal. - Jianing Song, Nov 14 2018
Also the number of pairs of integer compositions, the first summing to n and the second with sum equal to the length of the first. If an integer composition is regarded as an arrow from sum to length, these are composable pairs, and the obvious composition operation founds a category of integer compositions. For example, we have (2,1,1,4) . (1,2,1) . (1,2) = (2,6), where dots represent the composition operation. The version without empty compositions is A000244. Composable triples are counted by 1 followed by A000302. The unordered version is A022811. - Gus Wiseman, Jul 14 2022

			From _Gus Wiseman_, Jul 15 2020: (Start)
The a(0) = 1 through a(3) = 9 ways to choose a composition of each part of a composition:
  ()  (1)  (2)      (3)
           (1,1)    (1,2)
           (1),(1)  (2,1)
                    (1,1,1)
                    (1),(2)
                    (2),(1)
                    (1),(1,1)
                    (1,1),(1)
                    (1),(1),(1)
(End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Bishal Deb, Cyclic sieving phenomena via combinatorics of continued fractions, arXiv:2508.13709 [math.CO], 2025. See p. 42.
Index entries for linear recurrences with constant coefficients, signature (3).

The strict version is A336139.
Splittings of partitions are A323583.
Multiset partitions of partitions are A001970.
Partitions of each part of a partition are A063834.
Compositions of each part of a partition are A075900.
Strict partitions of each part of a strict partition are A279785.
Compositions of each part of a strict partition are A304961.
Strict compositions of each part of a composition are A307068.
Compositions of each part of a strict composition are A336127.
Cf. A000012, A000244, A000302, A001906, A006951, A011782, A022811, A071919.
Cf. A072706, A078008, A112467, A182105, A316245, A336128, A336135.
Cf. A339006, A355382, A355384, A355387.

[n eq 0 select 1 else 3^(n-1): n in [0..30]]; // G. C. Greubel, Nov 20 2023

a:= n-> ceil(3^(n-1)):
seq(a(n), n=0..30);  # Alois P. Heinz, Jul 26 2020

CoefficientList[Series[(1 - 2 x)/(1 - 3 x), {x, 0, 50}], x] (* Vladimir Joseph Stephan Orlovsky, Jun 21 2011 *)
Join[{1}, 3^(Range[0, 30])] (* G. C. Greubel, Nov 20 2023 *)

a(n)=max(1,3^(n-1)) \\ Charles R Greathouse IV, Jul 07 2011

Vec((1-2*x)/(1-3*x) + O(x^100)) \\  Altug Alkan, Oct 30 2015

[(3^n + 2*int(n==0))//3 for n in range(31)] # G. C. Greubel, Nov 20 2023

Binomial transform of A078008. - Paul Curtz, Aug 04 2008
From R. J. Mathar, Nov 11 2008: (Start)
G.f.: (1 - 2*x)/(1 - 3*x).
a(n) = A000244(n-1), n > 0. (End)
From Philippe Deléham, Nov 13 2008: (Start)
a(n) = Sum_{k=0..n} A112467(n,k)*2^k.
a(n) = Sum_{k=0..n} A071919(n,k)*2^k. (End)
Let A(x) be the g.f. Then B(x) = x*A(x) satisfies B(x/(1-x)) = x/(1 - 2*B(x)). - Vladimir Kruchinin, Dec 05 2011
G.f.: 1/(1 - (Sum_{k>=1} (x/(1 - x))^k)). - Joerg Arndt, Sep 30 2012
For n > 0, a(n) = 2*(Sum_{k=0..n-1} a(k)) - 1 = 3^(n-1). - J. Conrad, Oct 29 2015
G.f.: 1 + x/(1 + x)*(1 + 4*x/(1 + 4*x)*(1 + 7*x/(1 + 7*x)*(1 + 10*x/(1 + 10*x)*(1 + .... - Peter Bala, May 27 2017
Invert transform of A011782(n) = 2^(n-1). Second invert transform of A000012. - Gus Wiseman, Jul 19 2020
a(n) = ceiling(3^(n-1)). - Alois P. Heinz, Jul 26 2020
From Elmo R. Oliveira, Mar 31 2025: (Start)
E.g.f.: (2 + exp(3*x))/3.
a(n) = 3*a(n-1) for n > 1. (End)

Definition clarified by R. J. Mathar, Nov 11 2008

A008549 Number of ways of choosing at most n-1 items from a set of size 2*n+1.

Original entry on oeis.org

0, 1, 6, 29, 130, 562, 2380, 9949, 41226, 169766, 695860, 2842226, 11576916, 47050564, 190876696, 773201629, 3128164186, 12642301534, 51046844836, 205954642534, 830382690556, 3345997029244, 13475470680616, 54244942336114, 218269673491780, 877940640368572

Offset: 0

N. J. A. Sloane

Area under Dyck excursions (paths ending in 0): a(n) is the sum of the areas under all Dyck excursions of length 2*n (nonnegative walks beginning and ending in 0 with jumps -1,+1).
Number of inversions in all 321-avoiding permutations of [n+1]. Example: a(2)=6 because the 321-avoiding permutations of [3], namely 123,132,312,213,231, have 0, 1, 2, 1, 2 inversions, respectively. - Emeric Deutsch, Jul 28 2003
Convolution of A001791 and A000984. - Paul Barry, Feb 16 2005
a(n) = total semilength of "longest Dyck subpath" starting at an upstep U taken over all upsteps in all Dyck paths of semilength n. - David Callan, Jul 25 2008
[1,6,29,130,562,2380,...] is convolution of A001700 with itself. - Philippe Deléham, May 19 2009
From Ran Pan, Feb 04 2016: (Start)
a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) bounce off the diagonal y = x to the right. This is related to paired pattern P_2 in Pan and Remmel's link and more details can be found in Section 3.2 in the link.
a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) horizontally cross the diagonal y = x. This is related to paired pattern P_3 in Pan and Remmel's link and more details can be found in Section 3.3 in the link.
2*a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) bounce off the diagonal y = x. This is related to paired pattern P_2 and P_4 in Pan and Remmel's link and more details can be found in Section 4.2 in the link.
2*a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) cross the diagonal y = x. This is related to paired pattern P_3 and P_4 in Pan and Remmel's link and more details can be found in Section 4.3 in the link. (End)
From Gus Wiseman, Jul 17 2021: (Start)
Also the number of integer compositions of 2*(n+1) with alternating sum < 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(3) = 29 compositions of 8 are:
  (1,7)  (1,5,2)  (1,1,1,5)  (1,1,1,4,1)  (1,1,1,1,1,3)
  (2,6)  (1,6,1)  (1,1,2,4)  (1,2,1,3,1)  (1,1,1,2,1,2)
  (3,5)  (2,5,1)  (1,2,1,4)  (1,3,1,2,1)  (1,1,1,3,1,1)
                  (1,2,2,3)  (1,4,1,1,1)  (1,2,1,1,1,2)
                  (1,3,1,3)               (1,2,1,2,1,1)
                  (1,3,2,2)               (1,3,1,1,1,1)
                  (1,4,1,2)
                  (1,4,2,1)
                  (1,5,1,1)
                  (2,1,1,4)
                  (2,2,1,3)
                  (2,3,1,2)
                  (2,4,1,1)
Also the number of integer compositions of 2*(n+1) with reverse-alternating sum < 0. For a bijection, keep the odd-length compositions and reverse the even-length ones.
Also the number of 2*(n+1)-digit binary numbers with more 0's than 1's. For example, the a(2) = 6 binary numbers are: 100000, 100001, 100010, 100100, 101000, 110000; or in decimal: 32, 33, 34, 36, 40, 48.
(End)

			a(2) = 6 because there are 6 ways to choose at most 1 item from a set of size 5: You can choose the empty set, or you can choose any of the five one-element sets.
G.f. = x + 6*x^2 + 29*x^3 + 130*x^4 + 562*x^5 + 2380*x^6 + 9949*x^7 + ...

D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.

Indranil Ghosh, Table of n, a(n) for n = 0..1500 (terms 0..200 from T. D. Noe)
José Agapito, Ângela Mestre, Maria M. Torres, and Pasquale Petrullo, On One-Parameter Catalan Arrays, Journal of Integer Sequences, 18 (2015), Article 15.5.1.
Octavio Arizmendi, Daniel Perales, and Josue Vazquez-Becerra, Finite Free Convolution: Infinitesimal Distributions, arXiv:c [math.PR], 2025. See p. 34.
Jean Christophe Aval, Adrien Boussicault, Patxi Laborde-Zubieta, and Mathias Pétréolle, Generating series of Periodic Parallelogram polyominoes, arXiv:1612.03759, 2016.
Roland Bacher, On generating series of complementary plane trees, arXiv:math/0409050 [math.CO], 2004.
Vijay Balasubramanian, Javier M. Magan, and Qingyue Wu, A Tale of Two Hungarians: Tridiagonalizing Random Matrices, arXiv:2208.08452 [hep-th], 2022.
Cyril Banderier, Analytic combinatorics of random walks and planar maps, Ph. D. Thesis, 2001. [Broken link]
Adrien Boussicault and P. Laborde-Zubieta, Periodic Parallelogram Polyominoes, arXiv preprint arXiv:1611.03766 [math.CO], 2016.
AJ Bu, Explicit Generating Functions for the Sum of the Areas Under Dyck and Motzkin Paths (and for Their Powers), arXiv:2310.17026 [math.CO], 2023.
AJ Bu and Doron Zeilberger, Using Symbolic Computation to Explore Generalized Dyck Paths and Their Areas, arXiv:2305.09030 [math.CO], 2023.
Alexander Burstein and Sergi Elizalde, Total occurrence statistics on restricted permutations, arXiv:1305.3177 [math.CO], 2013.
Robin Chapman, Moments of Dyck paths, Discrete Math., 204 (1999), 113-117.
Nicolle González, Pamela E. Harris, Gordon Rojas Kirby, Mariana Smit Vega Garcia, and Bridget Eileen Tenner, Pinnacle sets of signed permutations, arXiv:2301.02628 [math.CO], 2023.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
Niklas G. Johansson, Efficient Simulation of the Deutsch-Jozsa Algorithm, Master's Project, Department of Electrical Engineering & Department of Physics, Chemistry and Biology, Linkoping University, April, 2015.
Miles Jones, Sergey Kitaev, and Jeffrey Remmel, Frame patterns in n-cycles, arXiv preprint arXiv:1311.3332 [math.CO], 2013.
James A. Mingo and Josue Vazquez-Becerra, The Asymptotic Infinitesimal Distribution of a Real Wishart Random Matrix, arXiv:2112.15231 [math.PR], 2021.
Henri Mühle, Symmetric Chain Decompositions and the Strong Sperner Property for Noncrossing Partition Lattices, arXiv:1509.06942v1 [math.CO], 2015.
Ran Pan and Jeffrey B. Remmel, Paired patterns in lattice paths, arXiv:1601.07988 [math.CO], 2016.
Elisa Pergola, Two bijections for the area of Dyck paths, Discrete Math., 241 (2001), 435-447.
Wen-jin Woan, Area of Catalan Paths, Discrete Math., 226 (2001), 439-444.

Cf. A038608, A045894, A057571, A109196, A153338.
Odd bisection of A294175 (even is A000346).
For integer compositions of 2*(n+1) with alternating sum k < 0 we have:
- The opposite (k > 0) version is A000302.
- The weak (k <= 0) version is (also) A000302.
- The k = 0 version is A001700 or A088218.
- The reverse-alternating version is also A008549 (this sequence).
- These compositions are ranked by A053754 /\ A345919.
- The complement (k >= 0) is counted by A114121.
- The case of reversed integer partitions is A344743(n+1).
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A345197 counts compositions by length and alternating sum.
Cf. A000070, A001791, A007318, A025047, A027306, A032443, A058622, A120452, A163493, A239830, A344611.

[4^n-Binomial(2*n+1, n): n in [0..30]]; // Vincenzo Librandi, Feb 04 2016

A008549:=n->4^n-binomial(2*n+1,n): seq(A008549(n), n=0..30);

Table[4^n-Binomial[2n+1,n],{n,0,30}] (* Harvey P. Dale, May 11 2011 *)
a[ n_] := If[ n<-4, 0, 4^n - Binomial[2 n + 2, n + 1] / 2] (* Michael Somos, Jan 25 2014 *)

{a(n)=if(n<0, 0, 4^n - binomial(2*n+1, n))} /* Michael Somos Oct 31 2006 */

{a(n) = if( n<-4, 0, n++; (4^n / 2 - binomial(2*n, n)) / 2)} /* Michael Somos, Jan 25 2014 */

import math
def C(n,r):
    f=math.factorial
    return f(n)/f(r)/f(n-r)
def A008549(n):
    return int((4**n)-C(2*n+1,n)) # Indranil Ghosh, Feb 18 2017

a(n) = 4^n - C(2*n+1, n).
a(n) = Sum_{k=1..n} Catalan(k)*4^(n-k): convolution of Catalan numbers and powers of 4.
G.f.: x*c(x)^2/(1 - 4*x), c(x) = g.f. of Catalan numbers. - Wolfdieter Lang
Note Sum_{k=0..2*n+1} binomial(2*n+1, k) = 2^(2n+1). Therefore, by the symmetry of Pascal's triangle, Sum_{k=0..n} binomial(2*n+1, k) = 2^(2*n) = 4^n. This explains why the following two expressions for a(n) are equal: Sum_{k=0..n-1} binomial(2*n+1, k) = 4^n - binomial(2*n+1, n). - Dan Velleman
G.f.: (2*x^2 - 1 + sqrt(1 - 4*x^2))/(2*(1 + 2*x)*(2*x - 1)*x^3).
a(n) = Sum_{k=0..n} C(2*k, k)*C(2*(n-k), n-k-1). - Paul Barry, Feb 16 2005
Second binomial transform of 2^n - C(n, floor(n/2)) = A045621(n). - Paul Barry, Jan 13 2006
a(n) = Sum_{0 < i <= k < n} binomial(n, k+i)*binomial(n, k-i). - Mircea Merca, Apr 05 2012
D-finite with recurrence (n+1)*a(n) + 2*(-4*n-1)*a(n-1) + 8*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 03 2012
0 = a(n) * (256*a(n+1) - 224*a(n+2) + 40*a(n+3)) + a(n+1) * (-32*a(n+1) + 56*a(n+2) - 14*a(n+3)) + a(n+2) * (-2*a(n+2) + a(n+3)) if n > -5. - Michael Somos, Jan 25 2014
Convolution square is A045894. - Michael Somos, Jan 25 2014
HANKEL transform is [0, -1, 2, -3, 4, -5, ...]. - Michael Somos, Jan 25 2014
BINOMIAL transform of [0, 0, 1, 3, 11, 35,...] (A109196) is [0, 0, 1, 6, 29, 130, ...]. - Michael Somos, Jan 25 2014
(n+1) * a(n) = A153338(n+1). - Michael Somos, Jan 25 2014
a(n) = Sum_{m = n+2..2*n+1} binomial(2*n+1,m), n >= 0. - Wolfdieter Lang, May 22 2015
E.g.f.: (exp(2*x) - BesselI(0,2*x) - BesselI(1,2*x))*exp(2*x). - Ilya Gutkovskiy, Aug 30 2016

Better description from Dan Velleman (djvelleman(AT)amherst.edu), Dec 01 2000

A032443 a(n) = Sum_{i=0..n} binomial(2*n, i).

Original entry on oeis.org

1, 3, 11, 42, 163, 638, 2510, 9908, 39203, 155382, 616666, 2449868, 9740686, 38754732, 154276028, 614429672, 2448023843, 9756737702, 38897306018, 155111585372, 618679078298, 2468152192772, 9848142504068, 39301087452632, 156861290196878, 626155256640188

Offset: 0

J. H. Conway, Dec 11 1999

Array interpretation: first row is filled with 1's, first column with powers of 2, b(i,j) = b(i-1,j) + b(i,j-1); then a(n) = b(n,n). - Benoit Cloitre, Apr 01 2002
   1   1   1   1   1   1   1 ...
   2   3   4   5   6   7   8 ...
   4   7  11  16  22 ...
   8  15  26  42  64 ...
  16  31  ..  99 163 ...
From Gary W. Adamson, Dec 27 2008: (Start)
This is an analog of the Catalan sequence: Let M denote an infinite Cartan matrix (-1's in the super and subdiagonals and (2,2,2,...) in the main diagonal which we modify to (1,2,2,2,...)). Then A000108 can be generated by accessing the leftmost term in M^n * [1,0,0,0,...]. Change the operation to M^n * [1,2,3,...] to get this sequence. Or, take iterates M * [1,2,3,...] -> M * ANS, -> M * ANS,...; accessing the leftmost term. (End)
Convolved with the Catalan sequence, A000108: (1, 1, 2, 5, 14, ...) = powers of 4, A000302: (1, 4, 16, 64, ...). - Gary W. Adamson, May 15 2009
Row sums of A094527. - Paul Barry, Sep 07 2009
Hankel transform of the aeration of this sequence is C(n+2, 2). - Paul Barry, Sep 26 2009
Number of 4-ary words of length n in which the number of 1's does not exceed the number of 0's. - David Scambler, Aug 14 2012
Number of options available to a voter who has up to n (0-n) votes to allot among 2n candidates. - Lorraine Lee, Apr 27 2019
2*a(n-1) is the number of all triangulations that can be obtained from a Möbius strip with n chosen points on its edge. See Bazier-Matte et al. - Michel Marcus, Sep 15 2020

			G.f. = 1 + 3*x + 11*x^2 + 42*x^3 + 163*x^4 + 638*x^5 + 2510*x^6 + 9908*x^7 + ...
According to the second formula, we see the fourth row of Pascal's triangle has the terms 1,4,6,4,1 and the partial sums are 1,5,11,15,16. Using these we get 1*1 + 4*5 + 6*11 + 4*15*1*16 = 1 + 20 + 66 + 60 + 16 = 163 = a(4). - _J. M. Bergot_, Apr 29 2014

D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Véronique Bazier-Matte, Ruiyan Huang, and Hanyi Luo, Number of Triangulations of a Möbius Strip, arXiv:2009.05785 [math.CO], 2020.
A. Bernini, F. Disanto, R. Pinzani and S. Rinaldi, Permutations defining convex permutominoes, J. Int. Seq. 10 (2007) # 07.9.7.
Celeste Damiani, Paul Martin, and Eric C. Rowell, Generalisations of Hecke algebras from Loop Braid Groups, arXiv:2008.04840 [math.GT], 2020.
M. Klazar, Twelve countings with rooted plane trees, European Journal of Combinatorics 18 (1997), 195-210; Addendum, 18 (1997), 739-740.
Mircea Merca, A Note on Cosine Power Sums J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.

Binomial transform of A027914. Hankel transform is {1, 2, 3, 4, ..., n, ...}.

Cf. A000108, A000302, A110166.

[(4^n+Binomial(2*n, n))/2:n in [0.. 30]]; // Vincenzo Librandi, Oct 18 2014

A032443:=n->(4^n + binomial(2*n, n))/2; seq(A032443(n), n=0..30); # Wesley Ivan Hurt, Apr 15 2014

a[ n_] := If[ n<-3, 0, (4^n + Binomial[2 n, n]) / 2]; Table[a[n], {n, 0, 30}] (* Michael Somos, Jan 25 2014 *)

A032443 = n->sum(i=0,n,binomial(2*n,i)) \\ M. F. Hasler, Jan 02 2014

A032443 = n->binomial(2*n-1,n)+4^n\2 \\ M. F. Hasler, Jan 02 2014

{a(n) = if( n<-3, 0, (4^n + binomial(2*n, n)) / 2)} /* Michael Somos, Jan 25 2014 */

a(n) = (4^n+binomial(2*n, n))/2. - David W. Wilson
a(n) = Sum_{0 <= i_1 <= i_2 <= n} binomial(n, i_2) * binomial(n, i_1+i_2). - Benoit Cloitre, Oct 14 2004
Sequence with interpolated zeros has a(n) = Sum_{k=0..floor(n/2)} (if((n-2k) mod 2)=0, C(n, k), 0). - Paul Barry, Jan 14 2005
a(n) = Sum_{k=0..n} C(n+k-1, k)*2^(n-k). - Paul Barry, Sep 28 2007
E.g.f.: exp(2*x)*(exp(2*x) + BesselI(0,2*x))/2. For BesselI see Abramowitz-Stegun (reference and link under A008277), p. 375, eq. 9.6.10. See also A000984 for its e.g.f. - Wolfdieter Lang, Jan 16 2012
From Sergei N. Gladkovskii, Aug 13 2012: (Start)
G.f.: (1/sqrt(1-4*x) + 1/(1-4*x))/2 = G(0)/2 where G(k) = 1 + ((2*k)!)/(k!)^2/(4^k - 4*x*(16^k)/( 4*x*(4^k) + ((2*k)!)/(k!)^2/G(k+1))); (continued fraction).
E.g.f.: G(0)/2 where G(k) = 1 + ((2*k)!)/(k!)^2/(4^k - 4*x*(16^k)/( 4*x*(4^k) + (k+1)*((2*k)!)/(k!)^2/G(k+1))); (continued fraction).
(End)
O.g.f.: (1 - x*(2 + c(x)))/(1 - 4*x)^(3/2), with c the o.g.f. of A000108 (Catalan). - Wolfdieter Lang, Nov 22 2012
D-finite with recurrence: n*a(n) + 2*(-4*n+3)*a(n-1) + 8*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Dec 04 2012
a(n) = binomial(2*n-1,n) + floor(4^n/2), or a(n+1) = A001700(n) + A004171(n), for all n >= 0. See A000346 for the difference. - M. F. Hasler, Jan 02 2014
0 = a(n) * (256*a(n+1) - 224*a(n+2) + 40*a(n+3)) + a(n+1) * (-32*a(n+1) + 56*a(n+2) - 14*a(n+3)) + a(n+2) * (-2*a(n+2) + a(n+3)) if n > -4. - Michael Somos, Jan 25 2014
a(n) = coefficient of x^n in (4*x + 1 / (1 + x))^n. - Michael Somos, Jan 25 2014
Binomial transform is A110166. - Michael Somos, Jan 25 2014
Asymptotics: a(n) ~ 2^(2*n-1)*(1+1/sqrt(Pi*n)). - Fung Lam, Apr 13 2014
Self-convolution is A240879. - Fung Lam, Apr 13 2014
a(0) = 1, a(n+1) = A001700(n) + 2^(2n+1). - Philippe Deléham, Oct 11 2014
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 10*x^2 + 35*x^3 + 126*x^4 + ... is the o.g.f. for A001700. - Peter Bala, Jul 21 2015
a(n) = 4*a(n-1) - A000108(n-1). - Bob Selcoe, Apr 02 2017
a(n) = [x^n] 1/((1 - x)^n*(1 - 2*x)). - Ilya Gutkovskiy, Oct 12 2017
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 08 2022
O.g.f.: ( hypergeom([1/4, 3/4], [1/2], 4*x) )^2. - Peter Bala, Mar 04 2022
a(n) = binomial(2*n, n) * hypergeom([1, -n], [n + 1], -1). - Peter Luschny, Oct 06 2023

A001021 Powers of 12.

Original entry on oeis.org

1, 12, 144, 1728, 20736, 248832, 2985984, 35831808, 429981696, 5159780352, 61917364224, 743008370688, 8916100448256, 106993205379072, 1283918464548864, 15407021574586368, 184884258895036416, 2218611106740436992

Offset: 0

N. J. A. Sloane

Same as Pisot sequences E(1, 12), L(1, 12), P(1, 12), T(1, 12). Essentially same as Pisot sequences E(12, 144), L(12, 144), P(12, 144), T(12, 144). See A008776 for definitions of Pisot sequences.
Central terms of the triangle in A100851. - Reinhard Zumkeller, Mar 04 2006
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=1, a(n) equals the number of 12-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Starting with 12, this sequence appears in the film "Vollmond" (1998, dir. Fredi Murer), when the children write it on the sidewalk at night. - Alonso del Arte, Dec 21 2011

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).

T. D. Noe, Table of n, a(n) for n = 0..100
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 276.
Tanya Khovanova, Recursive Sequences.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Index entries for linear recurrences with constant coefficients, signature (12).

Cf. A024140 and A178248, 12^n +/- 1; A016125.

Cf. A000005, A000244, A000302, A000351, A000384, A001222, A008585, A008776, A100851, A159991.

a001021 = (12 ^)
a001021_list = iterate (* 12) 1  -- Reinhard Zumkeller, Mar 31 2012

A001021:=-1/(-1+12*z); # Simon Plouffe in his 1992 dissertation

Table[12^n, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 15 2011 *)

a(n)=12^n \\ Charles R Greathouse IV, Dec 21 2011

G.f.: 1/(1-12*x).
E.g.f.: exp(12x).
a(n) = 12*a(n-1). - Zerinvary Lajos, Apr 27 2009
a(n) = A159991(n)/A000351(n). - Reinhard Zumkeller, May 02 2009
From Reinhard Zumkeller, Mar 31 2012: (Start)
a(n) = A000302(n) * A000244(n). - Reinhard Zumkeller, Mar 31 2012
A001222(a(n)) = A008585(n); A000005(a(n)) = A000384(a(n)). (End)
a(n) = det(|ps(i+2, j)|, 1 <= i, j <= n), where ps(n, k) are Legendre-Stirling numbers of the first kind. - Mircea Merca, Apr 04 2013

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

A006127 a(n) = 2^n + n.

Original entry on oeis.org

1, 3, 6, 11, 20, 37, 70, 135, 264, 521, 1034, 2059, 4108, 8205, 16398, 32783, 65552, 131089, 262162, 524307, 1048596, 2097173, 4194326, 8388631, 16777240, 33554457, 67108890, 134217755, 268435484, 536870941, 1073741854, 2147483679, 4294967328, 8589934625

Offset: 0

N. J. A. Sloane

For numbers m=n+2^n such that equation x=2^(m-x) has solution x=2^n, see A103354. - Zak Seidov, Mar 23 2005
Primes of the form x^x+1 must be of the form 2^2^(a(n))+1, that is, Fermat number F_(a(n)) (Sierpiński 1958). - David W. Wilson, May 08 2005
a(n) = n-th Mersenne number + n + 1 = A000225(n) + n + 1. Partial sums of a(n) are A132925(n+1). - Jaroslav Krizek, Oct 16 2009
Intersection of A188916 and A188917: A188915(a(n)) = (2^n)^2 = 2^(2*n) = A000302(n). - Reinhard Zumkeller, Apr 14 2011
a(n) is also the number of all connected subtrees of a star tree, having n leaves. The star tree is a tree, where all n leaves are connected to one parent P. - Viktar Karatchenia, Feb 29 2016

			From _Viktar Karatchenia_, Feb 29 2016: (Start)
a(0) = 1. There are n=0 leaves, it is a trivial tree consisting of a single parent node P.
a(1) = 3. There is n=1 leaf, the tree is P-A, the subtrees are: 2 singles: P, A; 1 pair: P-A; 2+1 = 3 subtrees in total.
a(2) = 6. When n=2, the tree is P-A P-B, the subtrees are: 3 singles: P, A, B; 2 pairs: P-A, P-B; 1 triple: A-P-B (the whole tree); 3+2+1 = 6.
a(3) = 11. For n=3 leaf nodes, the tree is P-A P-B P-C, the subtrees are: 4 singles: P, A, B, C; 3 pairs: P-A, P-B, P-C; 3 triples: A-P-B, A-P-C, B-P-C; 1 quad: P-A P-B P-C (the whole tree); 4+3+3+1 = 11 in total.
a(4) = 20. For n=4 leaves, the tree is P-A P-B P-C P-D, the subtrees are: 5 singles: P, A, B, C, D; 4 pairs: P-A, P-B, P-C, P-D; 6 triples: A-P-B, A-P-C, B-P-C, A-P-D, B-P-D, C-P-D; 4 quads: P-A P-B P-C, P-A P-B P-D, P-A P-C P-D, P-B P-C P-D; the whole tree counts as 1; 5+4+6+4+1 = 20.
In general, for n leaves, connected to the parent node P, there will be: (n+1) singles; (n, 1) pairs; (n, 2) triples; (n, 3) quads; ... ; (n, n-1) subtrees having (n-1) edges; 1 whole tree, having all n edges. Thus, the total number of all distinct subtrees is: (n+1) + (n, 1) + (n, 2) + (n, 3) + ... + (n, n-1) + 1 = (n + (n, 0)) + (n, 1) + (n, 2) + (n, 3) + ... + (n, n-1) + (n, n) = n + (sum of all binomial coefficients of size n) = n + (2^n). (End)

John H. Conway, R. K. Guy, The Book of Numbers, Copernicus Press, p. 84.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..100
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 435
C. L. Mallows & N. J. A. Sloane, Emails, May 1991
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Sierpiński Number of the First Kind
Eric Weisstein's World of Mathematics, Star Graph
Eric Weisstein's World of Mathematics, Vertex-Induced Subgraph
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A135227, A000079, A052944; A000051 (first differences).

Cf. A000325.

a006127 n = a000079 n + n
a006127_list = s [1] where
   s xs = last xs : (s $ zipWith (+) [1..] (xs ++ reverse xs))
Reinhard Zumkeller, May 19 2015, Feb 05 2011

A006127:=(-1+z+z**2)/(2*z-1)/(z-1)**2; # conjectured by Simon Plouffe in his 1992 dissertation

Table[2^n + n, {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, May 19 2011 *)
Table[BitXOr(i, 2^i), {i, 1, 100}] (* Peter Luschny, Jun 01 2011 *)
LinearRecurrence[{4,-5,2},{1,3,6},40] (* Harvey P. Dale, Feb 08 2023 *)

a(n)=1<Charles R Greathouse IV, Jul 19 2011

print([2**n + n for n in range(34)]) # Karl V. Keller, Jr., Aug 18 2020

def A006127(n): return (1<Chai Wah Wu, Jan 11 2023

Row sums of triangle A135227. - Gary W. Adamson, Nov 23 2007
Partial sums of A094373. G.f.: (1-x-x^2)/((1-x)^2(1-2x)). - Paul Barry, Aug 05 2004
Binomial transform of [1,2,1,1,1,1,1,...]. - Franklin T. Adams-Watters, Nov 29 2006
a(n) = 2*a(n-1) - n + 2 (with a(0)=1). - Vincenzo Librandi, Dec 30 2010
E.g.f.: exp(x)*(exp(x) + x). - Stefano Spezia, Dec 10 2021

cp's OEIS Frontend

A289780 p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.

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

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A027306 a(n) = 2^(n-1) + ((1 + (-1)^n)/4)*binomial(n, n/2).

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A001025 Powers of 16: a(n) = 16^n.

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A133494 Diagonal of the array of iterated differences of A047848.

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