A003463 -id:A003463 - cp's OEIS frontend

A000225 a(n) = 2^n - 1. (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.)

0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095, 8191, 16383, 32767, 65535, 131071, 262143, 524287, 1048575, 2097151, 4194303, 8388607, 16777215, 33554431, 67108863, 134217727, 268435455, 536870911, 1073741823, 2147483647, 4294967295, 8589934591

Offset: 0

N. J. A. Sloane

This is the Gaussian binomial coefficient [n,1] for q=2.
Number of rank-1 matroids over S_n.
Numbers k such that the k-th central binomial coefficient is odd: A001405(k) mod 2 = 1. - Labos Elemer, Mar 12 2003
This gives the (zero-based) positions of odd terms in the following convolution sequences: A000108, A007460, A007461, A007463, A007464, A061922.
Also solutions (with minimum number of moves) for the problem of Benares Temple, i.e., three diamond needles with n discs ordered by decreasing size on the first needle to place in the same order on the third one, without ever moving more than one disc at a time and without ever placing one disc at the top of a smaller one. - Xavier Acloque, Oct 18 2003
a(0) = 0, a(1) = 1; a(n) = smallest number such that a(n)-a(m) == 0 (mod (n-m+1)), for all m. - Amarnath Murthy, Oct 23 2003
Binomial transform of [1, 1/2, 1/3, ...] = [1/1, 3/2, 7/3, ...]; (2^n - 1)/n, n=1,2,3, ... - Gary W. Adamson, Apr 28 2005
Numbers whose binary representation is 111...1. E.g., the 7th term is (2^7) - 1 = 127 = 1111111 (in base 2). - Alexandre Wajnberg, Jun 08 2005
Number of nonempty subsets of a set with n elements. - Michael Somos, Sep 03 2006
For n >= 2, a(n) is the least Fibonacci n-step number that is not a power of 2. - Rick L. Shepherd, Nov 19 2007
Let P(A) be the power set of an n-element set A. Then a(n+1) = the number of pairs of elements {x,y} of P(A) for which x and y are disjoint and for which either x is a subset of y or y is a subset of x. - Ross La Haye, Jan 10 2008
A simpler way to state this is that it is the number of pairs (x,y) where at least one of x and y is the empty set. - Franklin T. Adams-Watters, Oct 28 2011
2^n-1 is the sum of the elements in a Pascal triangle of depth n. - Brian Lewis (bsl04(AT)uark.edu), Feb 26 2008
Sequence generalized: a(n) = (A^n -1)/(A-1), n >= 1, A integer >= 2. This sequence has A=2; A003462 has A=3; A002450 has A=4; A003463 has A=5; A003464 has A=6; A023000 has A=7; A023001 has A=8; A002452 has A=9; A002275 has A=10; A016123 has A=11; A016125 has A=12; A091030 has A=13; A135519 has A=14; A135518 has A=15; A131865 has A=16; A091045 has A=17; A064108 has A=20. - Ctibor O. Zizka, Mar 03 2008
a(n) is also a Mersenne prime A000668 when n is a prime number in A000043. - Omar E. Pol, Aug 31 2008
a(n) is also a Mersenne number A001348 when n is prime. - Omar E. Pol, Sep 05 2008
With offset 1, = row sums of triangle A144081; and INVERT transform of A009545 starting with offset 1; where A009545 = expansion of sin(x)*exp(x). - Gary W. Adamson, Sep 10 2008
Numbers n such that A000120(n)/A070939(n) = 1. - Ctibor O. Zizka, Oct 15 2008
For n > 0, sequence is equal to partial sums of A000079; a(n) = A000203(A000079(n-1)). - Lekraj Beedassy, May 02 2009
Starting with offset 1 = the Jacobsthal sequence, A001045, (1, 1, 3, 5, 11, 21, ...) convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 23 2009
Numbers n such that n=2*phi(n+1)-1. - Farideh Firoozbakht, Jul 23 2009
a(n) = (a(n-1)+1)-th odd numbers = A005408(a(n-1)) for n >= 1. - Jaroslav Krizek, Sep 11 2009
Partial sums of a(n) for n >= 0 are A000295(n+1). Partial sums of a(n) for n >= 1 are A000295(n+1) and A130103(n+1). a(n) = A006127(n) - (n+1). - Jaroslav Krizek, Oct 16 2009
If n is even a(n) mod 3 = 0. This follows from the congruences 2^(2k) - 1 ~ 2*2*...*2 - 1 ~ 4*4*...*4 - 1 ~ 1*1*...*1 - 1 ~ 0 (mod 3). (Note that 2*2*...*2 has an even number of terms.) - Washington Bomfim, Oct 31 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=2,(i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n)=det(A). - Milan Janjic, Jan 26 2010
This is the sequence A(0,1;1,2;2) = A(0,1;3,-2;0) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
a(n) = S(n+1,2), a Stirling number of the second kind. See the example below. - Dennis P. Walsh, Mar 29 2011
Entries of row a(n) in Pascal's triangle are all odd, while entries of row a(n)-1 have alternating parities of the form odd, even, odd, even, ..., odd.
Define the bar operation as an operation on signed permutations that flips the sign of each entry. Then a(n+1) is the number of signed permutations of length 2n that are equal to the bar of their reverse-complements and avoid the set of patterns {(-2,-1), (-1,+2), (+2,+1)}.  (See the Hardt and Troyka reference.) - Justin M. Troyka, Aug 13 2011
A159780(a(n)) = n and A159780(m) < n for m < a(n). - Reinhard Zumkeller, Oct 21 2011
This sequence is also the number of proper subsets of a set with n elements. - Mohammad K. Azarian, Oct 27 2011
a(n) is the number k such that the number of iterations of the map k -> (3k +1)/2 == 1 (mod 2) until reaching (3k +1)/2 == 0 (mod 2) equals n. (see the Collatz problem). - Michel Lagneau, Jan 18 2012
For integers a, b, denote by a<+>b the least c >= a such that Hd(a,c) = b (note that, generally speaking, a<+>b differs from b<+>a). Then a(n+1)=a(n)<+>1. Thus this sequence is the Hamming analog of nonnegative integers. - Vladimir Shevelev, Feb 13 2012
Pisano period lengths: 1, 1, 2, 1, 4, 2, 3, 1, 6, 4, 10, 2, 12, 3, 4, 1, 8, 6, 18, 4, ... apparently A007733. - R. J. Mathar, Aug 10 2012
Start with n. Each n generates a sublist {n-1,n-2,...,1}. Each element of each sublist also generates a sublist. Take the sum of all. E.g., 3->{2,1} and 2->{1}, so a(3)=3+2+1+1=7. - Jon Perry, Sep 02 2012
This is the Lucas U(P=3,Q=2) sequence. - R. J. Mathar, Oct 24 2012
The Mersenne numbers >= 7 are all Brazilian numbers, as repunits in base two. See Proposition 1 & 5.2 in Links: "Les nombres brésiliens". - Bernard Schott, Dec 26 2012
Number of line segments after n-th stage in the H tree. - Omar E. Pol, Feb 16 2013
Row sums of triangle in A162741. - Reinhard Zumkeller, Jul 16 2013
a(n) is the highest power of 2 such that 2^a(n) divides (2^n)!. - Ivan N. Ianakiev, Aug 17 2013
In computer programming, these are the only unsigned numbers such that k&(k+1)=0, where & is the bitwise AND operator and numbers are expressed in binary. - Stanislav Sykora, Nov 29 2013
Minimal number of moves needed to interchange n frogs in the frogs problem (see for example the NRICH 1246 link or the Britton link below). - N. J. A. Sloane, Jan 04 2014
a(n) !== 4 (mod 5); a(n) !== 10 (mod 11); a(n) !== 2, 4, 5, 6 (mod 7). - Carmine Suriano, Apr 06 2014
After 0, antidiagonal sums of the array formed by partial sums of integers (1, 2, 3, 4, ...). - Luciano Ancora, Apr 24 2015
a(n+1) equals the number of ternary words of length n avoiding 01,02. - Milan Janjic, Dec 16 2015
With offset 0 and another initial 0, the n-th term of 0, 0, 1, 3, 7, 15, ... is the number of commas required in the fully-expanded von Neumann definition of the ordinal number n. For example, 4 := {0, 1, 2, 3} := {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}}, which uses seven commas. Also, for n>0, a(n) is the total number of symbols required in the fully-expanded von Neumann definition of ordinal n - 1, where a single symbol (as usual) is always used to represent the empty set and spaces are ignored. E.g., a(5) = 31, the total such symbols for the ordinal 4. - Rick L. Shepherd, May 07 2016
With the quantum integers defined by [n+1]A001045%20are%20given%20by%20q%20=%20i%20*%20sqrt(2)%20for%20i%5E2%20=%20-1.%20Cf.%20A239473.%20-%20_Tom%20Copeland">q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)), the Mersenne numbers are a(n+1) = q^n [n+1]_q with q = sqrt(2), whereas the signed Jacobsthal numbers A001045 are given by q = i * sqrt(2) for i^2 = -1. Cf. A239473. - _Tom Copeland, Sep 05 2016
For n>1: numbers n such that n - 1 divides sigma(n + 1). - Juri-Stepan Gerasimov, Oct 08 2016
This is also the second column of the Stirling2 triangle A008277 (see also A048993). - Wolfdieter Lang, Feb 21 2017
Except for the initial terms, the decimal representation of the x-axis of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 659", "Rule 721" and "Rule 734", based on the 5-celled von Neumann neighborhood initialized with a single on cell. - Robert Price, Mar 14 2017
a(n), n > 1, is the number of maximal subsemigroups of the monoid of order-preserving partial injective mappings on a set with n elements. - James Mitchell and Wilf A. Wilson, Jul 21 2017
Also the number of independent vertex sets and vertex covers in the complete bipartite graph K_{n-1,n-1}. - Eric W. Weisstein, Sep 21 2017
Sum_{k=0..n} p^k is the determinant of n X n matrix M_(i, j) = binomial(i + j - 1, j)*p + binomial(i+j-1, i), in this case p=2 (empirical observation). - Tony Foster III, May 11 2019
The rational numbers r(n) = a(n+1)/2^(n+1) = a(n+1)/A000079(n+1) appear also as root of the n-th iteration f^{[n]}(c; x) = 2^(n+1)*x - a(n+1)*c of f(c; x) = f^{[0]}(c; x) = 2*x - c as r(n)*c. This entry is motivated by a riddle of Johann Peter Hebel (1760 - 1826): Erstes Rechnungsexempel(Ein merkwürdiges Rechnungs-Exempel) from 1803, with c = 24 and n = 2, leading to the root r(2)*24 = 21 as solution. See the link and reference. For the second problem, also involving the present sequence, see a comment in A130330. - Wolfdieter Lang, Oct 28 2019
a(n) is the sum of the smallest elements of all subsets of {1,2,..,n} that contain n. For example, a(3)=7; the subsets of {1,2,3} that contain 3 are {3}, {1,3}, {2,3}, {1,2,3}, and the sum of smallest elements is 7. - Enrique Navarrete, Aug 21 2020
a(n-1) is the number of nonempty subsets of {1,2,..,n} which don't have an element that is the size of the set. For example, for n = 4, a(3) = 7 and the subsets are {2}, {3}, {4}, {1,3}, {1,4}, {3,4}, {1,2,4}. - Enrique Navarrete, Nov 21 2020
From Eric W. Weisstein, Sep 04 2021: (Start)
Also the number of dominating sets in the complete graph K_n.
Also the number of minimum dominating sets in the n-helm graph for n >= 3. (End)
Conjecture: except for a(2)=3, numbers m such that 2^(m+1) - 2^j - 2^k - 1 is composite for all 0 <= j < k <= m. - Chai Wah Wu, Sep 08 2021
a(n) is the number of three-in-a-rows passing through a corner cell in n-dimensional tic-tac-toe. - Ben Orlin, Mar 15 2022
From Vladimir Pletser, Jan 27 2023: (Start)
a(n) == 1 (mod 30) for n == 1 (mod 4);
a(n) == 7 (mod 120) for n == 3 (mod 4);
(a(n) - 1)/30 = (a(n+2) - 7)/120 for n odd;
(a(n) - 1)/30 = (a(n+2) - 7)/120 = A131865(m) for n == 1 (mod 4) and m >= 0 with A131865(0) = 0. (End)
a(n) is the number of n-digit numbers whose smallest decimal digit is 8. - Stefano Spezia, Nov 15 2023
Also, number of nodes in a perfect binary tree of height n-1, or: number of squares (or triangles) after the n-th step of the construction of a Pythagorean tree: Start with a segment. At each step, construct squares having the most recent segment(s) as base, and isosceles right triangles having the opposite side of the squares as hypotenuse ("on top" of each square). The legs of these triangles will serve as the segments which are the bases of the squares in the next step. - M. F. Hasler, Mar 11 2024
a(n) is the length of the longest path in the n-dimensional hypercube. - Christian Barrientos, Apr 13 2024
a(n) is the diameter of the n-Hanoi graph.  Equivalently, a(n) is the largest minimum number of moves between any two states of the Towers of Hanoi problem (aka problem of Benares Temple described above). - Allan Bickle, Aug 09 2024

			For n=3, a(3)=S(4,2)=7, a Stirling number of the second kind, since there are 7 ways to partition {a,b,c,d} into 2 nonempty subsets, namely,
  {a}U{b,c,d}, {b}U{a,c,d}, {c}U{a,b,d}, {d}U{a,b,c}, {a,b}U{c,d}, {a,c}U{b,d}, and {a,d}U{b,c}. - _Dennis P. Walsh_, Mar 29 2011
From _Justin M. Troyka_, Aug 13 2011: (Start)
Since a(3) = 7, there are 7 signed permutations of 4 that are equal to the bar of their reverse-complements and avoid {(-2,-1), (-1,+2), (+2,+1)}. These are:
  (+1,+2,-3,-4),
  (+1,+3,-2,-4),
  (+1,-3,+2,-4),
  (+2,+4,-1,-3),
  (+3,+4,-1,-2),
  (-3,+1,-4,+2),
  (-3,-4,+1,+2). (End)
G.f. = x + 3*x^2 + 7*x^3 + 15*x^4 + 31*x^5 + 63*x^6 + 127*x^7 + ...
For the Towers of Hanoi problem with 2 disks, the moves are as follows, so a(2) = 3.
12|_|_ -> 2|1|_ -> _|1|2 -> _|_|12  - _Allan Bickle_, Aug 07 2024

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Index entries for "core" sequences
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A000043 (Mersenne exponents).
Cf. A000668 (Mersenne primes).
Cf. A001348 (Mersenne numbers with n prime).
Cf. A000079, A001045, A009545, A016189, A052955, A083329, A085104, A144081.
Cf. a(n)=A112492(n, 2). Rightmost column of A008969.
a(n) = A118654(n, 1) = A118654(n-1, 3), for n > 0.
Subsequence of A132781.
Smallest number whose base b sum of digits is n: this sequence (b=2), A062318 (b=3), A180516 (b=4), A181287 (b=5), A181288 (b=6), A181303 (b=7), A165804 (b=8), A140576 (b=9), A051885 (b=10).
Cf. A000203, A239473, A279396.
Cf. A008277, A048993 (columns k=2), A000918, A130330.
Cf. A000225, A029858, A058809, A375256 (Hanoi graphs).

a000225 = (subtract 1) . (2 ^)
a000225_list = iterate ((+ 1) . (* 2)) 0
-- Reinhard Zumkeller, Mar 20 2012

A000225 := n->2^n-1; [ seq(2^n-1,n=0..50) ];
A000225:=1/(2*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation, sequence starting at a(1)

a[n_] := 2^n - 1; Table[a[n], {n, 0, 30}] (* Stefan Steinerberger, Mar 30 2006 *)
Array[2^# - 1 &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
NestList[2 # + 1 &, 0, 32] (* Robert G. Wilson v, Feb 28 2011 *)
2^Range[0, 20] - 1 (* Eric W. Weisstein, Jul 17 2017 *)
LinearRecurrence[{3, -2}, {1, 3}, 20] (* Eric W. Weisstein, Sep 21 2017 *)
CoefficientList[Series[1/(1 - 3 x + 2 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)

A000225(n) = 2^n-1  \\ Michael B. Porter, Oct 27 2009

concat(0, Vec(x/((1-2*x)*(1-x)) + O(x^100))) \\ Altug Alkan, Oct 28 2015

def A000225(n): return (1<Chai Wah Wu, Jul 06 2022

def isMersenne(n): return n == sum([(1 - b) << s for (s, b) in enumerate((n+1).bits())]) # Peter Luschny, Sep 01 2019

G.f.: x/((1-2*x)*(1-x)).
E.g.f.: exp(2*x) - exp(x).
E.g.f. if offset 1: ((exp(x)-1)^2)/2.
a(n) = Sum_{k=0..n-1} 2^k. - Paul Barry, May 26 2003
a(n) = a(n-1) + 2*a(n-2) + 2, a(0)=0, a(1)=1. - Paul Barry, Jun 06 2003
Let b(n) = (-1)^(n-1)*a(n). Then b(n) = Sum_{i=1..n} i!*i*Stirling2(n,i)*(-1)^(i-1). E.g.f. of b(n): (exp(x)-1)/exp(2x). - Mario Catalani (mario.catalani(AT)unito.it), Dec 19 2003
a(n+1) = 2*a(n) + 1, a(0) = 0.
a(n) = Sum_{k=1..n} binomial(n, k).
a(n) = n + Sum_{i=0..n-1} a(i); a(0) = 0. - Rick L. Shepherd, Aug 04 2004
a(n+1) = (n+1)*Sum_{k=0..n} binomial(n, k)/(k+1). - Paul Barry, Aug 06 2004
a(n+1) = Sum_{k=0..n} binomial(n+1, k+1). - Paul Barry, Aug 23 2004
Inverse binomial transform of A001047. Also U sequence of Lucas sequence L(3, 2). - Ross La Haye, Feb 07 2005
a(n) = A099393(n-1) - A020522(n-1) for n > 0. - Reinhard Zumkeller, Feb 07 2006
a(n) = A119258(n,n-1) for n > 0. - Reinhard Zumkeller, May 11 2006
a(n) = 3*a(n-1) - 2*a(n-2); a(0)=0, a(1)=1. - Lekraj Beedassy, Jun 07 2006
Sum_{n>0} 1/a(n) = 1.606695152... = A065442, see A038631. - Philippe Deléham, Jun 27 2006
Stirling_2(n-k,2) starting from n=k+1. - Artur Jasinski, Nov 18 2006
a(n) = A125118(n,1) for n > 0. - Reinhard Zumkeller, Nov 21 2006
a(n) = StirlingS2(n+1,2). - Ross La Haye, Jan 10 2008
a(n) = A024036(n)/A000051(n). - Reinhard Zumkeller, Feb 14 2009
a(n) = A024088(n)/A001576(n). -Reinhard Zumkeller, Feb 15 2009
a(2*n) = a(n)*A000051(n); a(n) = A173787(n,0). - Reinhard Zumkeller, Feb 28 2010
For n > 0: A179857(a(n)) = A024036(n) and A179857(m) < A024036(n) for m < a(n). - Reinhard Zumkeller, Jul 31 2010
From Enrique Pérez Herrero, Aug 21 2010: (Start)
a(n) = J_n(2), where J_n is the n-th Jordan Totient function: (A007434, is J_2).
a(n) = Sum_{d|2} d^n*mu(2/d). (End)
A036987(a(n)) = 1. - Reinhard Zumkeller, Mar 06 2012
a(n+1) = A044432(n) + A182028(n). - Reinhard Zumkeller, Apr 07 2012
a(n) = A007283(n)/3 - 1. - Martin Ettl, Nov 11 2012
a(n+1) = A001317(n) + A219843(n); A219843(a(n)) = 0. - Reinhard Zumkeller, Nov 30 2012
a(n) = det(|s(i+2,j+1)|, 1 <= i,j <= n-1), where s(n,k) are Stirling numbers of the first kind. - Mircea Merca, Apr 06 2013
G.f.: Q(0), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 - 1/(2*4^k - 8*x*16^k/(4*x*4^k - 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013
E.g.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - (k+1)/Q(k+1))); (continued fraction).
G.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 23 2013
a(n) = A000203(2^(n-1)), n >= 1. - Ivan N. Ianakiev, Aug 17 2013
a(n) = Sum_{t_1+2*t_2+...+n*t_n=n} n*multinomial(t_1+t_2 +...+t_n,t_1,t_2,...,t_n)/(t_1+t_2 +...+t_n). - Mircea Merca, Dec 06 2013
a(0) = 0; a(n) = a(n-1) + 2^(n-1) for n >= 1. - Fred Daniel Kline, Feb 09 2014
a(n) = A125128(n) - A000325(n) + 1. - Miquel Cerda, Aug 07 2016
From Ilya Gutkovskiy, Aug 07 2016: (Start)
Binomial transform of A057427.
Sum_{n>=0} a(n)/n! = A090142. (End)
a(n) = A000918(n) + 1. - Miquel Cerda, Aug 09 2016
a(n+1) = (A095151(n+1) - A125128(n))/2. - Miquel Cerda, Aug 12 2016
a(n) = (A079583(n) - A000325(n+1))/2. - Miquel Cerda, Aug 15 2016
Convolution of binomial coefficient C(n,a(k)) with itself is C(n,a(k+1)) for all k >= 3. - Anton Zakharov, Sep 05 2016
a(n) = (A083706(n-1) + A000325(n))/2. - Miquel Cerda, Sep 30 2016
a(n) = A005803(n) + A005408(n-1). - Miquel Cerda, Nov 25 2016
a(n) = A279396(n+2,2). - Wolfdieter Lang, Jan 10 2017
a(n) = n + Sum_{j=1..n-1} (n-j)*2^(j-1). See a Jun 14 2017 formula for A000918(n+1) with an interpretation. - Wolfdieter Lang, Jun 14 2017
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} C(k,i). - Wesley Ivan Hurt, Sep 21 2017
a(n+m) = a(n)*a(m) + a(n) + a(m). - Yuchun Ji, Jul 27 2018
a(n+m) = a(n+1)*a(m) - 2*a(n)*a(m-1). - Taras Goy, Dec 23 2018
a(n+1) is the determinant of n X n matrix M_(i, j) = binomial(i + j - 1, j)*2 + binomial(i+j-1, i) (empirical observation). - Tony Foster III, May 11 2019
From Peter Bala, Jun 27 2025: (Start)
For n >= 1, a(3*n)/a(n) = A001576(n), a(4*n)/a(n) = A034496(n), a(5*n)/a(n) = A020514(n) a(6*n)/a(n) = A034665(n), a(7*n)/a(n) = A020516(n) and a(8*n)/a(n) = A034674(n).
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n.
Modulo differences in offsets, exp( Sum_{n >= 1} a(k*n)/a(n)*x^n/n ) is the o.g.f. of A006095 (k = 3), A006096 (k = 4), A006097 (k = 5), A006110 (k = 6), A022189 (k = 7), A022190 (k = 8), A022191 (k = 9) and A022192 (k = 10).
The following are all examples of telescoping series:
Sum_{n >= 1} 2^n/(a(n)*a(n+1)) = 1; Sum_{n >= 1} 2^n/(a(n)*a(n+1)*a(n+2)) = 1/9.
In general, for k >= 1, Sum_{n >= 1} 2^n/(a(n)*a(n+1)*...*a(n+k)) = 1/(a(1)*a(2)*...*a(k)*a(k)).
Sum_{n >= 1} 2^n/(a(n)*a(n+2)) = 4/9, since 2^n/(a(n)*a(n+2)) = b(n) - b(n+1), where b(n) = (2/3)*(3*2^(n-1) - 1)/((2^(n+1) - 1)*(2^n - 1)).
Sum_{n >= 1} (-2)^n/(a(n)*a(n+2)) = -2/9, since (-2)^n/(a(n)*a(n+2)) = c(n) - c(n+1), where c(n) = (1/3)*(-2)^n/((2^(n+1) - 1)*(2^n - 1)).
Sum_{n >= 1} 2^n/(a(n)*a(n+4)) = 18/175, since 2^n/(a(n)*a(n+4)) = d(n) - d(n+1), where d(n) = (120*8^n - 140*4^n + 45*2^n - 4)/(15*(2^n - 1)*(2^(n+1) - 1)*(2^(n+2) - 1)*(2^(n+3) - 1)).
Sum_{n >= 1} (-2)^n/(a(n)*a(n+4)) = -26/525, since (-2)^n/(a(n)*a(n+4)) = e(n) - e(n+1), where e(n) = (-1)^n*(40*8^n - 24*4^n + 5*2^n)/(15*(2^n - 1)*(2^(n+1) - 1)*(2^(n+2) - 1)*(2^(n+3) - 1)). (End)

Name partially edited by Eric W. Weisstein, Sep 04 2021

A000351 Powers of 5: a(n) = 5^n.

Original entry on oeis.org

1, 5, 25, 125, 625, 3125, 15625, 78125, 390625, 1953125, 9765625, 48828125, 244140625, 1220703125, 6103515625, 30517578125, 152587890625, 762939453125, 3814697265625, 19073486328125, 95367431640625, 476837158203125, 2384185791015625, 11920928955078125

Offset: 0

N. J. A. Sloane

Same as Pisot sequences E(1, 5), L(1, 5), P(1, 5), T(1, 5). Essentially same as Pisot sequences E(5, 25), L(5, 25), P(5, 25), T(5, 25). See A008776 for definitions of Pisot sequences.
a(n) has leading digit 1 if and only if n = A067497 - 1. - Lekraj Beedassy, Jul 09 2002
With interpolated zeros 0, 1, 0, 5, 0, 25, ... (g.f.: x/(1 - 5*x^2)) second inverse binomial transform of Fibonacci(3n)/Fibonacci(3) (A001076). Binomial transform is A085449. - Paul Barry, Mar 14 2004
Sums of rows of the triangles in A013620 and A038220. - Reinhard Zumkeller, May 14 2006
Sum of coefficients of expansion of (1 + x + x^2 + x^3 + x^4)^n. a(n) is number of compositions of natural numbers into n parts less than 5. a(2) = 25 there are 25 compositions of natural numbers into 2 parts less than 5. - Adi Dani, Jun 22 2011
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 5-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Numbers n such that sigma(5n) = 5n + sigma(n). In fact we have this theorem: p is a prime if and only if all solutions of the equation sigma(p*x) = p*x + sigma(x) are powers of p. - Jahangeer Kholdi, Nov 23 2013
From Doug Bell, Jun 22 2015: (Start)
Empirical observation: Where n is an odd multiple of 3, let x = (a(n) + 1)/9 and let y be the decimal expansion of x/a(n); then y*(x+1)/x + 1 = y rotated to the left.
Example:
  a(3) = 125;
  x = (125 + 1)/9 = 14;
  y = 112, which is the decimal expansion of 14/125 = 0.112;
  112*(14 + 1)/14 + 1 = 121 = 112 rotated to the left.
(End)
a(n) is the number of n-digit integers that contain only odd digits (A014261). - Bernard Schott, Nov 12 2022
Number of pyramids in the Sierpinski fractal square-based pyramid at the n-th step, while A279511 gives the corresponding number of vertices (see IREM link with drawings). - Bernard Schott, Nov 29 2022

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
O. M. Cain, The Exceptional Selfcondensability of Powers of Five, arXiv:1910.13829 [math.HO], 2019.
Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 270
IREM Paris-Nord, La pyramide de Sierpinski (in French).
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
Yash Puri and Thomas Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Eric Weisstein's World of Mathematics, Box Fractal
Index entries for linear recurrences with constant coefficients, signature (5).

Cf. A009969 (even bisection), A013710 (odd bisection), A005054 (first differences), A003463 (partial sums).
Cf. A006495, A006496, A159991, A001021.
Cf. A001076, A008776, A013620, A038220, A067497, A085449, A014261.
Sierpinski fractal square-based pyramid: A020858 (Hausdorff dimension), A279511 (number of vertices), this sequence (number of pyramids).

a000351 = (5 ^)
a000351_list = iterate (* 5) 1  -- Reinhard Zumkeller, Oct 31 2012

[5^n : n in [0..30]]; // Wesley Ivan Hurt, Sep 27 2016

[ seq(5^n,n=0..30) ];
A000351:=-1/(-1+5*z); # Simon Plouffe in his 1992 dissertation

Table[5^n, {n, 0, 30}] (* Stefan Steinerberger, Apr 06 2006 *)
5^Range[0, 30] (* Harvey P. Dale, Aug 22 2011 *)

makelist(5^n,n,0,20); /* Martin Ettl, Dec 27 2012 */

a(n)=5^n \\ Charles R Greathouse IV, Jun 10 2011

def a(n): return 5**n
print([a(n) for n in range(24)]) # Michael S. Branicky, Nov 12 2022

(List.fill(50)(5: BigInt)).scanLeft(1: BigInt)( * ) // Alonso del Arte, May 31 2019

a(n) = 5^n.
a(0) = 1; a(n) = 5*a(n-1) for n > 0.
G.f.: 1/(1 - 5*x).
E.g.f.: exp(5*x).
a(n) = A006495(n)^2 + A006496(n)^2.
a(n) = A159991(n) / A001021(n). - Reinhard Zumkeller, May 02 2009
From Bernard Schott, Nov 12 2022: (Start)
Sum_{n>=0} 1/a(n) = 5/4.
Sum_{n>=0} (-1)^n/a(n) = 5/6. (End)
a(n) = Sum_{k=0..n} C(2*n+1,n-k)*A000045(2*k+1). - Vladimir Kruchinin, Jan 14 2025

A015518 a(n) = 2a(n-1) + 3a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 2, 7, 20, 61, 182, 547, 1640, 4921, 14762, 44287, 132860, 398581, 1195742, 3587227, 10761680, 32285041, 96855122, 290565367, 871696100, 2615088301, 7845264902, 23535794707, 70607384120, 211822152361, 635466457082

Offset: 0

Olivier Gérard

Number of walks of length n between any two distinct vertices of the complete graph K_4. - Paul Barry and Emeric Deutsch, Apr 01 2004
For n >= 1, a(n) is the number of integers k, 1 <= k <= 3^(n-1), whose ternary representation ends in an even number of zeros (see A007417). - Philippe Deléham, Mar 31 2004
Form the digraph with matrix A=[0,1,1,1;1,0,1,1;1,1,0,1;1,0,1,1]. A015518(n) corresponds to the (1,3) term of A^n. - Paul Barry, Oct 02 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is 2. - Cino Hilliard, Sep 25 2005
(A046717(n))^2 + (2*a(n))^2 = A046717(2n). E.g., A046717(3) = 13, 2*a(3) = 14, A046717(6) = 365. 13^2 + 14^2 = 365. - Gary W. Adamson, Jun 17 2006
For n >= 2, number of ordered partitions of n-1 into parts of sizes 1 and 2 where there are two types of 1 (singletons) and three types of 2 (twins). For example, the number of possible configurations of families of n-1 male (M) and female (F) offspring considering only single births and twins, where the birth order of M/F/pair-of-twins is considered and there are three types of twins; namely, both F, both M, or one F and one M - where birth order within a pair of twins itself is disregarded. In particular, for a(3)=7, two children could be either: (1) F, then M; (2) M, then F; (3) F,F; (4) M,M; (5) F,F twins; (6) M,M twins; or (7) M,F twins (emphasizing that birth order is irrelevant here when both/all children are the same gender and when two children are within the same pair of twins). - Rick L. Shepherd, Sep 18 2004
a(n) is prime for n = {2, 3, 5, 7, 13, 23, 43, 281, 359, ...}, where only a(2) = 2 corresponds to a prime of the form (3^k - 1)/4. All prime terms, except a(2) = 2, are the primes of the form (3^k + 1)/4. Numbers k such that (3^k + 1)/4 is prime are listed in A007658. Note that all prime terms have prime indices. Prime terms are listed in A111010. - Alexander Adamchuk, Nov 19 2006
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=charpoly(A,1). - Milan Janjic, Jan 26 2010
Select an odd size subset S from {1,2,...,n}, then select an even size subset from S. - Geoffrey Critzer, Mar 02 2010
a(n) is the number of ternary sequences of length n where the numbers of (0's, 1's) are (even, odd) respectively, and, by symmetry, the number of such sequences where those numbers are (odd, even) respectively. A122983 covers (even, even), and A081251 covers (odd, odd). - Toby Gottfried, Apr 18 2010
An elephant sequence, see A175654. For the corner squares just one A[5] vector, with decimal value 341, leads to this sequence (without the leading 0). For the central square this vector leads to the companion sequence A046717 (without the first leading 1). - Johannes W. Meijer, Aug 15 2010
Let R be the commutative algebra resulting from adjoining the elements of the Klein four-group to the integers (equivalently, K = Z[x,y,z]/{x*y - z, y*z - x, x*z - y, x^2 - 1, y^2 - 1, z^2 - 1}). Then a(n) is equal to the coefficients of x, y, and z in the expansion of (x + y + z)^n. - Joseph E. Cooper III (easonrevant(AT)gmail.com), Nov 06 2010
Pisano period lengths:  1, 2, 2, 4, 4, 2, 6, 8, 2, 4, 10, 4, 6, 6, 4, 16, 16, 2, 18, 4, ... - R. J. Mathar, Aug 10 2012
The ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 09 2014
This is a divisibility sequence, also the values of Chebyshev polynomials, and also the number of ways of packing a 2 X n-1 rectangle with dominoes and unit squares. - R. K. Guy, Dec 16 2016
For n>0, gcd(a(n),a(n+1))=1. - Kengbo Lu, Jul 02 2020

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
Jean-Paul Allouche, Jeffrey Shallit, Zhixiong Wen, Wen Wu, and Jiemeng Zhang, Sum-free sets generated by the period-k-folding sequences and some Sturmian sequences, arXiv:1911.01687 [math.CO], 2019.
Katerina Böhmová, Cristina Dalfó, and Clemens Huemer, On cyclic Kautz digraphs, Research Report, 2015.
Garry Bowlin and Matthew G. Brin, Coloring Planar Graphs via Colored Paths in the Associahedra, arXiv preprint arXiv:1301.3984 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 12 2013
AJ Bu and Doron Zeilberger, Using Symbolic Computation to Explore Generalized Dyck Paths and Their Areas, arXiv:2305.09030 [math.CO], 2023.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
Sergio Falcón, Binomial Transform of the Generalized k-Fibonacci Numbers, Communications in Mathematics and Applications (2019) Vol. 10, No. 3, 643-651.
Dale Gerdemann, Fractal generated from (2,3) recursion, YouTube Video, Dec 05 2014.
Toufik Mansour and Mark Shattuck, Pattern avoidance in flattened derangements, Disc. Math. Lett. (2025) Vol. 15, 67-74. See p. 74.
R. J. Mathar, Counting Walks on Finite Graphs, Nov 2020, Section 2.
Felix Pozon Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014.
Index entries for linear recurrences with constant coefficients, signature (2,3).
Index entries for sequences related to Chebyshev polynomials.

a(n) = A080926(n-1) + 1 = (1/3)*A054878(n+1) = (1/3)*abs(A084567(n+1)).
First differences of A033113 and A039300.
Partial sums of A046717.
The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.
Cf. A046717.
Cf. A007658, A111010.
Cf. A001045, A078008, A097073, A115341. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008
Cf. A007482, A059260.

[Round(3^n/4): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011

Table[(3^n-(-1)^n)/4,{n,0,30}] (* Alexander Adamchuk, Nov 19 2006 *)

a(n):= round(3^n/4)$ /* Dimitri Papadopoulos, Nov 28 2023 */

PARI
```
a(n)=round(3^n/4)
```

for n in range(0, 20): print(int((3**n-(-1)**n)/4), end=', ') # Stefano Spezia, Nov 30 2018

Sage
```
[round(3^n/4) for n in range(0,27)]
```

G.f.: x/((1+x)*(1-3*x)).
a(n) = (3^n - (-1)^n)/4 = floor(3^n/4 + 1/2).
a(n) = 3^(n-1) - a(n-1). - Emeric Deutsch, Apr 01 2004
E.g.f.: (exp(3*x) - exp(-x))/4. Second inverse binomial transform of (5^n-1)/4, A003463. Inverse binomial transform for powers of 4, A000302 (when preceded by 0). - Paul Barry, Mar 28 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k+1)*2^(2k). - Paul Barry, May 14 2003
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*4^(k-1). - Paul Barry, Apr 02 2003
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2*k)*3^k. - Paul Barry, Jul 13 2004
a(n) = U(n-1, i/sqrt(3))(-i*sqrt(3))^(n-1), i^2=-1. - Paul Barry, Nov 17 2003
G.f.: x*(1+x)^2/(1 - 6*x^2 - 8*x^3 - 3*x^4) = x(1+x)^2/characteristic polynomial(x^4*adj(K_4)(1/x)). - Paul Barry, Feb 03 2004
a(n) = sum_{k=0..3^(n-1)} A014578(k) = -(-1)^n*A014983(n) = A051068(3^(n-1)), for n > 0. - Philippe Deléham, Mar 31 2004
E.g.f.: exp(x)*sinh(2*x)/2. - Paul Barry, Oct 02 2004
a(2*n+1) = A054880(n) + 1. - M. F. Hasler, Mar 20 2008
2*a(n) + (-1)^n = A046717(n). - M. F. Hasler, Mar 20 2008
a(n) = ((1+sqrt(4))^n - (1-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008
a(n) = abs(A014983(n)). - Zerinvary Lajos, May 28 2009
a(n) = round(3^n/4). - Mircea Merca, Dec 28 2010
a(n) = Sum_{k=1,3,5,...} binomial(n,k)*2^(k-1). - Geoffrey Critzer, Mar 02 2010
From Sergei N. Gladkovskii, Jul 19 2012: (Start)
G.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - 1/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + 1/G(k+1)))))); (continued fraction).
E.g.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - (2*k+1)/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + (2*k+2)/G(k+1)))))); (continued fraction).  (End)
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(4*k-1)/(x*(4*k+3) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n+1) = Sum_{k = 0..n} A238801(n,k)*2^k. - Philippe Deléham, Mar 07 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-4)^k = (-1)^(n-1)*Sum_{k=0..n-1} (-3)^k. Equals (-1)^(n-1)*Phi(n,-3), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
a(n) = 2*A006342(n-1) - n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Nov 30 2018
a(n) = 2*A033113(n-2) + n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Aug 16 2019
a(2*k) = 2*A002452(k), a(2*k+1) = A066443(k). - Yuchun Ji, Aug 14 2019
a(n+1) = 2*Sum_{k=0..n} a(k) if n odd, and 1 + 2*Sum_{k=0..n} a(k) if n even. - Kengbo Lu, May 30 2020
a(n) = F(n) + Sum_{k=1..(n-1)} a(k)*L(n-k), for F(n) and L(n) the Fibonacci and Lucas numbers. - Kengbo Lu and Greg Dresden, Jun 05 2020
From Kengbo Lu, Jun 11 2020: (Start)
a(n) = A002605(n) + Sum_{k = 1..n-2} a(k)*A002605(n-k-1).
a(n) = A006130(n-1) + Sum_{k = 1..n-1} a(k)*A006130(n-k-1). (End)
a(2n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)* 2^(2n-2i-2j-1)* 3^(i+j). - Kengbo Lu, Jul 02 2020
a(n) = 3*a(n-1) - (-1)^n. - Dimitri Papadopoulos, Nov 28 2023
G.f.: x/((1 + x)*(1 - 3*x)) = Sum_{n >= 0} x^(n+1) * Product_{k = 1..n} (k + 3*x + 1)(1 + k*x) (a telescoping series). Cf. A007482. - Peter Bala, May 08 2024
From Peter Bala, Jun 29 2025: (Start)
For n >= 1, a(n+1) = 2^n * hypergeom([1/2 - (1/2)*n, -(1/2)*n], [-n], -3).
G.f. A(x) = x*exp(Sum_{n >= 1} a(2*n)/a(n)*x^n/n) =  x + 2*x^2 + 7*x^3 + 20*x^4 + ....
sqrt(A(x)/x) is the g.f. of A002426.
The following series telescope:
Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)) = -1; Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*a(n+2)*a(n+3)) = -1/98.
In general, for k >= 0, Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*...*a(n+2*k+1)) = -1/((a(1)*a(2)*...*a(2*k+1))*a(2*k+1)).
Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)) = 1/4; Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)* a(n+3)*a(n+4)) = 1/5600.
In general, for k >= 1, Sum_{n >= 1} 3^n/(a(n)*a(n+1)*...*a(n+2*k)) = 1/((a(1)*a(2)*...*a(2*k))*a(2*k)). (End)

More terms from Emeric Deutsch, Apr 01 2004

Edited by Ralf Stephan, Aug 30 2004

A003947 Expansion of (1+x)/(1-4*x).

Original entry on oeis.org

1, 5, 20, 80, 320, 1280, 5120, 20480, 81920, 327680, 1310720, 5242880, 20971520, 83886080, 335544320, 1342177280, 5368709120, 21474836480, 85899345920, 343597383680, 1374389534720, 5497558138880, 21990232555520, 87960930222080, 351843720888320

Offset: 0

N. J. A. Sloane

Coordination sequence for infinite tree with valency 5.
For n>=1, a(n+1) is equal to the number of functions f:{1,2,...,n+1}->{1,2,3,4,5} such that for fixed, different x_1, x_2,...,x_n in {1,2,...,n+1} and fixed y_1, y_2,...,y_n in {1,2,3,4,5} we have f(x_i)<>y_i, (i=1,2,...,n). - Milan Janjic, May 10 2007
Number of length-n strings of 5 letters with no two adjacent letters identical. The general case (strings of r letters) is the sequence with g.f. (1+x)/(1-(r-1)*x). - Joerg Arndt, Oct 11 2012
Create a rectangular prism with edges of lengths 2^(n-2), 2^(n-1), and 2^(n) starting at n=2; then the surface area = a(n). - J. M. Bergot, Aug 08 2013

T. D. Noe, Table of n, a(n) for n = 0..200
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 306
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (4).
Index entries for sequences related to trees

Cf. A003948, A003949. Column 5 in A265583.

Concatenation([1], List([1..30], n-> 5*4^(n-1) )); # G. C. Greubel, Aug 10 2019

[1] cat [5*4^(n-1): n in [1..30]]; // G. C. Greubel, Aug 10 2019

k := 5; if n = 0 then 1 else k*(k-1)^(n-1); fi;

q = 5; Join[{a = 1}, Table[If[n != 0, a = q*a - a, a = q*a], {n, 0, 25}]] (* and *) Join[{1}, 5*4^Range[0, 25]] (* Vladimir Joseph Stephan Orlovsky, Jul 11 2011 *)
LinearRecurrence[{4},{1,5},30] (* Harvey P. Dale, Apr 19 2015 *)

a(n)=5*4^n\4 \\ Charles R Greathouse IV, Sep 08 2011

[1]+[5*4^(n-1) for n in (1..30)] # G. C. Greubel, Aug 10 2019

Binomial transform of A060925. Its binomial transform is A003463 (without leading zero). - Paul Barry, May 19 2003
From Paul Barry, May 19 2003: (Start)
a(n) = (5*4^n - 0^n)/4.
G.f.: (1+x)/(1-4*x).
E.g.f.: (5*exp(4*x) - exp(0))/4. (End)
a(n) = Sum_{k=0..n} A029653(n, k)*x^k for x = 3. - Philippe Deléham, Jul 10 2005
a(n) = A146523(n)*A011782(n). - R. J. Mathar, Jul 08 2009
a(n) = 5*A000302(n-1), n>0.
a(n) = 4*a(n-1), n>1. - Vincenzo Librandi, Dec 31 2010
G.f.: 2+x- 2/G(0), where G(k)= 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 04 2013

Edited by N. J. A. Sloane, Dec 04 2009

A006095 Gaussian binomial coefficient [n, 2] for q = 2.

Original entry on oeis.org

0, 0, 1, 7, 35, 155, 651, 2667, 10795, 43435, 174251, 698027, 2794155, 11180715, 44731051, 178940587, 715795115, 2863245995, 11453115051, 45812722347, 183251413675, 733006703275, 2932028910251, 11728119835307, 46912487729835

Offset: 0

N. J. A. Sloane

Number of 4-block coverings of an n-set where every element of the set is covered by exactly 3 blocks (if offset is 3), so a(n) = (1/4!)*(4^n-6*2^n+8). - Vladeta Jovovic, Feb 20 2001
Number of non-coprime pairs of polynomials (f,g) with binary coefficients where both f and g have degree n+1 and nonzero constant term. - Luca Mariot and Enrico Formenti, Sep 26 2016
Number of triplets found from the integers 1 to 2^n-1 by converting to binary and performing an XOR operation on the corresponding bits of each pair. Defining addition in this carryless way (0+0=1+1=0, 0+1=1+0=1), each triplet (A,B,C) has the property A+B=C, A+C=B and B+C=A. For example, n=3 gives the 7 triplets (1,2,3), (1,4,5), (1,6,7), (2,4,6), (2,5,7), (3,4,7) and (3,5,6). Each integer appears in the set of triplets 2^(n-1)-1 times, for example 3 for n=3. - Ian Duff, Oct 05 2019
Number of 2-dimensional vector subspaces of (Z_2)^n, so also number of Klein subgroups of the group (C_2)^n. - Robert FERREOL, Jul 28 2021

J. Goldman and G.-C. Rota, The number of subspaces of a vector space, pp. 75-83 of W. T. Tutte, editor, Recent Progress in Combinatorics. Academic Press, NY, 1969.
I. P. Goulden and D. M. Jackson, Combinatorial Enumeration. Wiley, NY, 1983, p. 99.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
M. Sved, Gaussians and binomials, Ars. Combinatoria, 17A (1984), 325-351.

T. D. Noe, Table of n, a(n) for n = 0..200
L. Mariot and E. Formenti, The number of coprime/non-coprime pairs of polynomials over F_2 with degree n and nonzero constant term.
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 11.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
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
A. I. Solomon, C.-L. Ho and G. H. E. Duchamp, Degrees of entanglement for multipartite systems, arXiv preprint arXiv:1205.4958 [quant-ph], 2012. - _N. J. A. Sloane_, Oct 23 2012
M. Sved, Gaussians and binomials, Ars. Combinatoria, 17A (1984), 325-351. (Annotated scanned copy)
Index entries for linear recurrences with constant coefficients, signature (7,-14,8).

First differences: A006516.
Gaussian binomial coefficient [n, k] for q = 2: A000225 (k = 1), this sequence (k = 2), A006096 (k = 3), A006097 (k = 4), A006110 (k = 5), A022189 - A022195 (k = 6 thru 12).
Cf. A000392, A002275, A002452, A003462, A003463, A003464, A016123, A016125, A016208, A016256, A023000, A023001, A075113, A130324, A203235.

a:= n-> add((4^(n-1-j) - 2^(n-1-j))/2, j=0..n-1):
seq(a(n), n=0..24); # Zerinvary Lajos, Jan 04 2007
A006095 := -z^2/(z-1)/(2*z-1)/(4*z-1); # Simon Plouffe in his 1992 dissertation. [adapted to offset 0 by Peter Luschny, Jul 20 2021]
a := n -> (2^n - 2)*(2^n - 1)/6:
seq(a(n), n = 0..24); # Peter Luschny, Jul 20 2021

Join[{a=0,b=0},Table[c=6*b-8*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011 *)
CoefficientList[Series[x^2/((1-x)(1-2x)(1-4x)),{x,0,30}],x] (* or *) LinearRecurrence[{7,-14,8},{0,0,1},30] (* Harvey P. Dale, Jul 22 2011 *)
(* Next, using elementary symmetric functions *)
f[k_] := 2^(k - 1); t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[2, t[n]]
Table[a[n], {n, 2, 32}]    (* A203235 *)
Table[a[n]/2, {n, 2, 32}]  (* A006095 *)
(* Clark Kimberling, Dec 31 2011 *)
Table[QBinomial[n, 2, 2], {n, 0, 24}] (* Arkadiusz Wesolowski, Nov 12 2015 *)

a(n) = (2^n - 1)*(2^(n-1) - 1)/3 \\ Charles R Greathouse IV, Jul 25 2011

concat([0, 0], Vec(x^2/((1-x)*(1-2*x)*(1-4*x)) + O(x^50))) \\ Altug Alkan, Nov 12 2015

[gaussian_binomial(n,2,2) for n in range(0,25)] # Zerinvary Lajos, May 24 2009

G.f.: x^2/((1-x)(1-2x)(1-4x)).
a(n) = (2^n - 1)*(2^(n-1) - 1)/3 = 4^n/6 - 2^(n-1) + 1/3.
Row sums of triangle A130324. - Gary W. Adamson, May 24 2007
a(n) = Stirling2(n+1,3) + Stirling2(n+1,4). - Zerinvary Lajos, Oct 04 2007; corrected by R. J. Mathar, Mar 19 2011
a(n) = A139250(2^(n-1) - 1), n >= 1. - Omar E. Pol, Mar 03 2011
a(n) = 4*a(n-1) + 2^(n-1) - 1, n >= 2. - Vincenzo Librandi, Mar 19 2011
a(0) = 0, a(1) = 0, a(2) = 1, a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3). - Harvey P. Dale, Jul 22 2011
a(n) = Sum_{k=0..n-2} 2^k*C(2*n-k-2, k), n >= 2. - Johannes W. Meijer, Aug 19 2013
a(n) = Sum_{i=0..n-2, j=i..n-2} 2^{i+j} = 2^0 * (2^0 + 2^1 + ... + 2^(n-2)) + 2^1 * (2^1 + 2^2 + ... + 2^(n-2)) + ... + 2^(n-2) * 2^(n-2), n>1. - J. M. Bergot, May 08 2017
a(n) = a(n-1) + A000217(A000225(n-1)), n > 0. - Ivan N. Ianakiev, Dec 11 2017
E.g.f.: (2*exp(x)-3*exp(2*x)+exp(4*x))/6. - Paul Weisenhorn, Aug 22 2021
From Peter Bala, Jul 01 2025: (Start)
G.f. assuming an offset of 0: exp( Sum_{n >= 1} b(3*n)/b(n)*x^n/n ) = 1 + 7*x + 35*x^2 + ..., where b(n) = A000225(n) = 2^n - 1.
The following are examples of telescoping series:
Sum_{n >= 2} 2^n/a(n) = 6, follows from 1 - (1/6)*Sum_{k = 2..n} 2^k/a(k) = 1/(2^n - 1).
Sum_{n >= 2} 2^n/(a(n)*a(n+2)) = 6/49, follows from 1 - (49/6)*Sum_{k = 2..n} 2^k/(a(k)*a(k+2)) = 1/A006096(n+2);
Sum_{n >= 2} 4^n/(a(n)*a(n+2)) = 26/49, follows from 13 - (49/2)*Sum_{k = 2..n} 4^k/(a(k)*a(k+2)) = A086224(n)/A006096(n+2);
Sum_{n >= 2} 8^n/(a(n)*a(n+2)) = 129/49, follows from 43 - (49/3)*Sum_{k = 2..n} 8^k/(a(k)*a(k+2)) = A171479(n+1)/A006096(n+2). (End)

A000392 Stirling numbers of second kind S(n,3).

Original entry on oeis.org

0, 0, 0, 1, 6, 25, 90, 301, 966, 3025, 9330, 28501, 86526, 261625, 788970, 2375101, 7141686, 21457825, 64439010, 193448101, 580606446, 1742343625, 5228079450, 15686335501, 47063200806, 141197991025, 423610750290, 1270865805301

Offset: 0

N. J. A. Sloane

Number of palindromic structures using exactly three different symbols; Mobius transform: A056279. - Marks R. Nester
Number of ways of placing n labeled balls into k=3 indistinguishable boxes. - Thomas Wieder, Nov 30 2004
With two leading zeros, this is the second binomial transform of cosh(x)-1 and the binomial transform of A000225 (with extra leading zero). - Paul Barry, May 13 2003
Let [m] denote the first m positive integers. Then a(n) is the number of functions f from [n] to [n+1] that satisfy (i) f(x) > x for all x, (ii) f(x) = n+1 for exactly 3 elements and (iii) f(f(x)) = n+1 for the remaining n-3 elements of [n]. For example, a(4)=6 since there are exactly 6 functions from {1,2,3,4} to {1,2,3,4,5} such that f(x) > x, f(x) = 5 for 3 elements and f(f(x)) = 5 for the remaining element. The functions are f1 = {(1,5), (2,5), (3,4), (4,5)}, f2 = {(1,5), (2,3), (3,5), (4,5)}, f3 = {(1,5), (2,4), (3,5), (4,5)}, f4 = {(1,2), (2,5), (3,5), (4,5)}, f5 = {(1,3), (2,5), (3,5), (4,5)}, f6 = {(1,4), (2,5), (3,5), (4,5)}. - Dennis P. Walsh, Feb 20 2007
Conjecture. Let S(1)={1} and, for n > 1, let S(n) be the smallest set containing x, 2x and 3x for each element x in S(n-1). Then a(n+2) is the sum of the elements in S(n). (It is easy to prove that the number of elements in S(n) is the n-th triangular number given by A001952.) See A122554 for a sequence defined in this way. - John W. Layman, Nov 21 2007; corrected (a(n) to a(n+2) due to offset change) by Fred Daniel Kline, Oct 02 2014
Let P(A) be the power set of an n-element set A. Then a(n+1) = the number of pairs of elements {x,y} of P(A) for which x and y are disjoint and for which x is not a subset of y and y is not a subset of x. Wieder calls these "disjoint strict 2-combinations". - Ross La Haye, Jan 11 2008; corrected by Ross La Haye, Oct 29 2008
Also, let P(A) be the power set of an n-element set A. Then a(n+2) = the number of pairs of elements {x,y} of P(A) for which either 0) x and y are disjoint and for which either x is a subset of y or y is a subset of x, or 1) x and y are disjoint and for which x is not a subset of y and y is not a subset of x, or 2) x and y are intersecting and for which either x is a proper subset of y or y is a proper subset of x. - Ross La Haye, Jan 11 2008
3 * a(n+1) = p(n+1) where p(x) is the unique degree-n polynomial such that p(k) = a(k+1) for k = 0, 1, ..., n. - Michael Somos, Apr 29 2012
John W. Layman's conjecture that a(n+2) is the sum of elements in S(n) follows from the identification of S(n) with the first n rows of A036561, whose row sums are A001047. - Fred Daniel Kline, Oct 02 2014
From M. Sinan Kul, Sep 08 2016: (Start)
Let m be equal to the product of n-1 distinct primes. Then a(n) is equal to the number of distinct fractions >=1 that may be created by dividing a divisor of m by another divisor. For example for m = 2*3*5 = 30, we would have the following 6 fractions: 6/5, 3/2, 5/3, 5/2, 10/3, 15/2.
Here finding the number of fractions would be equivalent to distributing n-1 balls (distinct primes) to two bins (numerator and denominator) with no empty bins which can be found by Stirling numbers of the second kind. So another definition for a(n) is a(n) = Sum_{i=2..n-1} Stirling2(i,2)*binomial(n-1,i).
Also for n > 0, a(n) = (d(m^2)+1)/2 - d(m) where m is equal to the product of n-1 distinct primes. Example for a(5): m = 2*3*5*7 = 210 (product of four distinct primes) so a(5) = (d(210^2)+1)/2 - d(210) = 41 - 16 = 25. (End)
6*a(n) is the number of ternary strings of length n that contain at least one of each of the 3 symbols on which they are defined. For example, for n=4, the strings are the 12 permutations of 0012, the 12 permutations of 0112, and the 12 permutations of 0122. - Enrique Navarrete, Aug 23 2021
A simpler form of La Haye's first comment is: a(n+1) is the number of ways we can form disjoint unions of two nonempty subsets of [n] (see example below). Cf. A001047 for the requirement that the union contains n. - Enrique Navarrete, Aug 24 2021
As partial sums of the Nicomachus triangle's rows and the differences of the powers of 3 and 2 (A001047), each iteration corresponds to two figurate variations of the Sierpinski triangle (3^n) with cross-correlation to the Nicomachus triangle, see illustrations in links. The Sierpinski half-hexagons of (A001047) stack and conform to the footprint of 2^n - 1 triangular numbers. The 3^n Sierpinski triangle minus its 2^n bottom row, also correlates to the Nicomachus triangle according to each Sierpinski triangular sub-row. - John Elias, Oct 04 2021

			a(4) = 6. Let denote Z[i] the i-th labeled element = "ball". Then one has for n=4 six different ways to fill sets = "boxes" with the labeled elements:
Set(Set(Z[3], Z[4]), Set(Z[1]), Set(Z[2])), Set(Set(Z[3], Z[1]), Set(Z[4]), Set(Z[2])), Set(Set(Z[4], Z[1]), Set(Z[3]), Set(Z[2])), Set(Set(Z[4]), Set(Z[1]), Set(Z[3], Z[2])), Set(Set(Z[3]), Set(Z[1], Z[2]), Set(Z[4])), Set(Set(Z[3]), Set(Z[1]), Set(Z[4], Z[2])).
G.f. = x^3 + 6*x^4 + 25*x^5 + 90*x^6 + 301*x^7 + 966*x^8 + 3025*x^9 + ...
For example, for n=3, a(4)=6 since the disjoint unions are: {1}U{2}, {1}U{3}, {1}U{2,3}, {2}U{3}, {2}U{1,3}, and {1,2}U{3}. - _Enrique Navarrete_, Aug 24 2021

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]
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..200
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Harry Crane, Left-right arrangements, set partitions, and pattern avoidance, Australasian Journal of Combinatorics, 61(1) (2015), 57-72.
John Elias, Illustration: Stirling-Sierpinski triangles, Nicomachus-Sierpinski towers
M. Griffiths and I. Mezo, A generalization of Stirling Numbers of the Second Kind via a special multiset, JIS 13 (2010) #10.2.5.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 346
Fred Kline and Peter Taylor, Partial sums of Nicomachus' Triangle rows produce Stirling numbers of the 2nd kind, Mathematics Stack Exchange. - _Fred Daniel Kline_, Sep 22 2014
Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
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.
Anthony G. Shannon, Hakan Akkuş, Yeşim Aküzüm, Ömür Deveci, and Engin Özkan, A partial recurrence Fibonacci link, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 530-537. See Table 1, p. 531.
Kai Wang, Girard-Waring Type Formula For A Generalized Fibonacci Sequence, Fibonacci Quarterly (2020) Vol. 58, No. 5, 229-235.
Eric Weisstein's World of Mathematics, Minimal Cover.
Thomas Wieder, The number of certain k-combinations of an n-set, Applied Mathematics Electronic Notes, vol. 8 (2008).
Index entries for linear recurrences with constant coefficients, signature (6,-11,6).

Cf. A008277 (Stirling2 triangle), A007051, A056509, A000225.
Cf. A003462, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A016256.
Cf. A028243, A122554.

List([0..400], n->Stirling2(n,3)); # Muniru A Asiru, Feb 04 2018

A000392 := n -> 9/2*3^n-4*2^n+1/2;  [ seq(9/2*3^n-4*2^n+1/2,n=0..30) ]; # Thomas Wieder
A000392:=-1/(z-1)/(3*z-1)/(2*z-1); # Simon Plouffe in his 1992 dissertation

StirlingS2[Range[0,30],3] (* Harvey P. Dale, Dec 29 2011 *)

PARI
```
{a(n) = 3^(n-1) / 2 - 2^(n-1) + 1/2};
```

[stirling_number2(i,3) for i in (0..40)] # Zerinvary Lajos, Jun 26 2008

G.f.: x^3/((1-x)*(1-2*x)*(1-3*x)).
E.g.f.: ((exp(x) - 1)^3) / 3!.
Recurrence: a(n+3) = 6*a(n+2) - 11*a(n+1) + 6*a(n), a(3) = 1, a(4) = 6, a(5) = 25. - Thomas Wieder, Nov 30 2004
With offset 0, this is 9*3^n/2 - 4*2^n + 1/2, the partial sums of 3*3^n - 2*2^n = A001047(n+1). - Paul Barry, Jun 26 2003
a(n) = (1 + 3^(n-1) - 2^n)/2, n > 0. - Dennis P. Walsh, Feb 20 2007
For n >= 3, a(n) = 3*a(n-1) + 2^(n-2) - 1. - Geoffrey Critzer, Mar 03 2009
a(n) = 5*a(n-1) - 6*a(n-2) + 1, for n > 3. - Vincenzo Librandi Nov 25 2010
a(n) = det(|s(i+3,j+2)|, 1 <= i,j <= n-3), where s(n,k) are Stirling numbers of the first kind. - Mircea Merca, Apr 06 2013
G.f.:  x^3 + 12*x^4/(G(0)-12*x), where G(k) = x + 1 + 9*(3*x+1)*3^k - 8*(2*x+1)*2^k - x*(9*3^k+1-8*2^k)*(81*3^k+1-32*2^k)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Feb 01 2014
a(n + 2) = (1 - 2^(2 + n) + 3^(1 + n))/2 for n > 0. - Fred Daniel Kline, Oct 02 2014
For n > 0, a(n) = (1/2) * Sum_{k=1..n-1} Sum_{i=1..n-1} C(n-k-1,i) * C(n-1,k). - Wesley Ivan Hurt, Sep 22 2017
a(n) = Sum_{k=0..n-3} 2^(k-1)*(3^(n-2-k) - 1).  - J. M. Bergot, Feb 05 2018

Offset changed by N. J. A. Sloane, Feb 08 2008

A027868 Number of trailing zeros in n!; highest power of 5 dividing n!.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 18, 18, 18, 18, 18, 19

Offset: 0

Warut Roonguthai

Also the highest power of 10 dividing n! (different from A054899). - Hieronymus Fischer, Jun 18 2007
Alternatively, a(n) equals the expansion of the base-5 representation A007091(n) of n (i.e., where successive positions from right to left stand for 5^n or A000351(n)) under a scale of notation whose successive positions from right to left stand for (5^n - 1)/4 or A003463(n); for instance, n = 7392 has base-5 expression 2*5^5 + 1*5^4 + 4*5^3 + 0*5^2 + 3*5^1 + 2*5^0, so that a(7392) = 2*781 + 1*156 + 4*31 + 0*6 + 3*1 + 2*0 = 1845. - Lekraj Beedassy, Nov 03 2010
Partial sums of A112765. - Hieronymus Fischer, Jun 06 2012
Also the number of trailing zeros in A000165(n) = (2*n)!!. - Stefano Spezia, Aug 18 2024

			a(100)  = 24.
a(10^3) = 249.
a(10^4) = 2499.
a(10^5) = 24999.
a(10^6) = 249998.
a(10^7) = 2499999.
a(10^8) = 24999999.
a(10^9) = 249999998.
a(10^n) = 10^n/4 - 3 for 10 <= n <= 15 except for a(10^14) = 10^14/4 - 2. - _M. F. Hasler_, Dec 27 2019

M. Gardner, "Factorial Oddities." Ch. 4 in Mathematical Magic Show: More Puzzles, Games, Diversions, Illusions and Other Mathematical Sleight-of-Mind from Scientific American. New York: Vintage, 1978, pp. 50-65.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000 (first 1000 terms from T. D. Noe)
David S. Hart, James E. Marengo, Darren A. Narayan, and David S. Ross, On the number of trailing zeros in n!, College Math. J., 39(2):139-145, 2008.
Enrique Pérez Herrero, Trailing Zeros in n!, Psychedelic Geometry Blogspot.
Soichi Ikeda and Kaneaki Matsuoka, On transcendental numbers generated by certain integer sequences, Siauliai Math. Semin., 8 (16) 2013, 63-69.
Shyam Sunder Gupta, Fascinating Factorials, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 16, 411-442.
Shu-Chung Liu and Jean C.-C. Yeh, Catalan numbers modulo 2^k, J. Int. Seq. 13 (2010), 10.5.4, eq (5).
Antonio M. Oller-Marcén, A new look at the trailing zeros of n!, arXiv:0906.4868v1 [math.NT], 2009.
Antonio M. Oller-Marcén and J. Maria Grau, On the Base-b Expansion of the Number of Trailing Zeros of b^k!, J. Int. Seq. 14 (2011) 11.6.8
Eric Weisstein's World of Mathematics, Factorial.
Index entries for sequences related to factorial numbers

See A000966 for the missing numbers. See A011371 and A054861 for analogs involving powers of 2 and 3.
Cf. A054899, A007953, A112765, A067080, A098844, A132027, A067080, A098844, A132029, A054999, A112765, A191610, A000351.
Cf. also A000142, A004154.
Cf. A000165, A008904.

a027868 n = sum $ takeWhile (> 0) $ map (n `div`) $ tail a000351_list
-- Reinhard Zumkeller, Oct 31 2012

[Valuation(Factorial(n), 5): n in [0..80]]; // Bruno Berselli, Oct 11 2021

0, seq(add(floor(n/5^i),i=1..floor(log[5](n))), n=1..100); # Robert Israel, Nov 13 2014

Table[t = 0; p = 5; While[s = Floor[n/p]; t = t + s; s > 0, p *= 5]; t, {n, 0, 100} ]
Table[ IntegerExponent[n!], {n, 0, 80}] (* Robert G. Wilson v *)
zOF[n_Integer?Positive]:=Module[{maxpow=0},While[5^maxpow<=n,maxpow++];Plus@@Table[Quotient[n,5^i],{i,maxpow-1}]]; Attributes[zOF]={Listable}; Join[{0},zOF[ Range[100]]] (* Harvey P. Dale, Apr 11 2022 *)

a(n)={my(s);while(n\=5,s+=n);s} \\ Charles R Greathouse IV, Nov 08 2012, edited by M. F. Hasler, Dec 27 2019

a(n)=valuation(n!,5) \\ Charles R Greathouse IV, Nov 08 2012

apply( A027868(n)=(n-sumdigits(n,5))\4, [0..99]) \\ M. F. Hasler, Dec 27 2019

from sympy import multiplicity
A027868, p5 = [0,0,0,0,0], 0
for n in range(5,10**3,5):
    p5 += multiplicity(5,n)
    A027868.extend([p5]*5) # Chai Wah Wu, Sep 05 2014

def A027868(n): return 0 if n<5 else n//5 + A027868(n//5) # David Radcliffe, Jun 26 2016

from sympy.ntheory.factor_ import digits
def A027868(n): return n-sum(digits(n,5)[1:])>>2 # Chai Wah Wu, Oct 18 2024

a(n) = Sum_{i>=1} floor(n/5^i).
a(n) = (n - A053824(n))/4.
From Hieronymus Fischer, Jun 25 2007 and Aug 13 2007, edited by M. F. Hasler, Dec 27 2019: (Start)
G.f.: g(x) = Sum_{k>0} x^(5^k)/(1-x^(5^k))/(1-x).
a(n) = Sum_{k=5..n} Sum_{j|k, j>=5} (floor(log_5(j)) - floor(log_5(j-1))).
G.f.: g(x) = L[b(k)](x)/(1-x) where L[b(k)](x) = Sum_{k>=0} b(k)*x^k/(1-x^k) is a Lambert series with b(k) = 1, if k>1 is a power of 5, else b(k) = 0.
G.f.: g(x) = Sum_{k>0} c(k)*x^k/(1-x), where c(k) = Sum_{j>1, j|k} floor(log_5(j)) - floor(log_5(j - 1)).
Recurrence:
a(n) = floor(n/5) + a(floor(n/5));
a(5*n) = n + a(n);
a(n*5^m) = n*(5^m-1)/4 + a(n).
a(k*5^m) = k*(5^m-1)/4, for 0 <= k < 5, m >= 0.
Asymptotic behavior:
a(n) = n/4 + O(log(n)),
a(n+1) - a(n) = O(log(n)), which follows from the inequalities below.
a(n) <= (n-1)/4; equality holds for powers of 5.
a(n) >= n/4 - 1 - floor(log_5(n)); equality holds for n = 5^m-1, m > 0.
lim inf (n/4 - a(n)) = 1/4, for n -> oo.
lim sup (n/4 - log_5(n) - a(n)) = 0, for n -> oo.
lim sup (a(n+1) - a(n) - log_5(n)) = 0, for n -> oo.
(End)
a(n) <= A027869(n). - Reinhard Zumkeller, Jan 27 2008
10^a(n) = A000142(n) / A004154(n). - Reinhard Zumkeller, Nov 24 2012
a(n) = Sum_{k=1..floor(n/2)} floor(log_5(n/k)). - Ammar Khatab, Feb 01 2025

Examples added by Hieronymus Fischer, Jun 06 2012

A016123 a(n) = (11^(n+1) - 1)/10.

Original entry on oeis.org

1, 12, 133, 1464, 16105, 177156, 1948717, 21435888, 235794769, 2593742460, 28531167061, 313842837672, 3452271214393, 37974983358324, 417724816941565, 4594972986357216, 50544702849929377, 555991731349223148

Offset: 0

N. J. A. Sloane

11^a(n) is highest power of 11 dividing (11^(n+1))!.
Partial sums of powers of 11 (A001020).
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=11, (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
Let A be the Hessenberg matrix of the order n, defined by: A[1,j]=1, A[i,i]:=12, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=2, a(n-1)=(-1)^n*charpoly(A,1). - Milan Janjic, Feb 21 2010

G. C. Greubel, Table of n, a(n) for n = 0..950
D. C. Santos, E. A. Costa, and P. M. M. C. Catarino, On Gersenne Sequence: A Study of One Family in the Horadam-Type Sequence, Axioms 14, 203, (2025). See p. 4.
Eric Weisstein's World of Mathematics, Repunit
Index entries for linear recurrences with constant coefficients, signature (12,-11).

For analogs with primes 2, 3, 5, 7, 13 and 17 see: A000225, A003462, A003463, A023000, A091030 and A091045, respectively.

Cf. A001020, A004191.

[(11^(n+1)-1)/10: n in [0..30]]; // G. C. Greubel, Feb 21 2024

(11^Range[0,20]-1)/10 (* or *) LinearRecurrence[{12,-11},{0,1},20] (* Harvey P. Dale, Apr 05 2012 *)

A016123(n):=(11^(n+1)-1)/10$
makelist(A016123(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

a(n)=(11^(n+1)-1)/10 \\ Charles R Greathouse IV, Sep 24 2015

[lucas_number1(n,12,11) for n in range(1, 19)] # Zerinvary Lajos, Apr 27 2009

[gaussian_binomial(n,1,11) for n in range(1,19)] # Zerinvary Lajos, May 28 2009

a(n) = Sum_{k=0..n} 11^k = (11^(n+1) - 1)/10.
G.f.: (1/(1-11*x) - 1/(1-x))/(10*x) = 1/((1-11*x)*(1-x)).
a(0)=1, a(n) = 11*a(n-1) + 1. - Vincenzo Librandi, Feb 05 2011
a(0)=0, a(1)=1, a(n) = 12*a(n-1) - 11*a(n-2). - Harvey P. Dale, Apr 05 2012
E.g.f.: exp(x)*(11*exp(10*x) - 1)/10. - Stefano Spezia, Mar 11 2023

Title edited by Daniel Forgues, Jul 08 2011

A026375 a(n) = Sum_{k=0..n} binomial(n,k)binomial(2k,k).

Original entry on oeis.org

1, 3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261, 212601015, 1024642875, 4950790605, 23973456915, 116312293305, 565280386625, 2751474553575, 13411044301945, 65448142561035, 319756851757695

Offset: 0

Clark Kimberling

nonn

a(n) is the number of integer strings s(0),...,s(n) counted by array T in A026374 that have s(n)=0; also a(n)=T(2n,n).
Partial sums of A085362. Number of bilateral Schroeder paths (i.e., lattice paths consisting of steps U=(1,1), D=(1,-1) and H=(2,0)) from (0,0) to (2n,0) and with no H-steps at odd (positive or negative) levels. Example: a(2)=11 because we have HUD, UDH, UDUD, UUDD, UDDU, their reflections in the x-axis and HH. - Emeric Deutsch, Jan 30 2004
Largest coefficient of (1+3*x+x^2)^n; row sums of triangle in A124733. - Philippe Deléham, Oct 02 2007
Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), the H steps come in three colors. - N-E. Fahssi, Feb 05 2008
Equals INVERT transform of A109033: (1, 2, 6, 22, 88, ...), INVERTi transform of A111966, binomial transform of A000984, and inverse Binomial transform of A081671. Convolved with A002212: (1, 3, 10, 36, ...) = A026376: (1, 6, 30, 144, ...). Equals convolution square root of A003463: (1, 6, 31, 156, 781, 3906, ...). - Gary W. Adamson, May 17 2009
Diagonal of array with rational generating function 1/(1 - (x^2 + 3*x*y + y^2)). - Gheorghe Coserea, Jul 29 2018
a(n) == 0 (mod 3) if and only if n is in A081606. - Fabio Visonà, Aug 03 2023

			G.f. = 1 + 3*x + 11*x^2 + 45*x^3 + 195*x^4 + 873*x^5 + 3989*x^6 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
David Callan, A combinatorial interpretation for an identity of Barrucand, JIS 11 (2008) 08.3.4.
Shu-Chiuan Chang and Robert Shrock, Structure of the Partition Function and transfer matrices for the Potts model in a magnetic field on lattice strips, J. Stat. Phys. 137 (2009) 667.
D. E. Davenport, L. W. Shapiro and L. C. Woodson, The Double Riordan Group, The Electronic Journal of Combinatorics, 18(2) (2012), #P33. - From _N. J. A. Sloane_, May 11 2012
Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
Rui Duarte and António Guedes de Oliveira, Generating functions of lattice paths, Univ. do Porto (Portugal 2023).
Francesc Fite, Kiran S. Kedlaya, Victor Rotger and Andrew V. Sutherland, Sato-Tate distributions and Galois endomorphism modules in genus 2, arXiv preprint arXiv:1110.6638 [math.NT], 2011-2012.
Francesc Fite and Andrew V. Sutherland, Sato-Tate distributions of twists of y^2= x^5-x and y^2= x^6+1, arXiv preprint arXiv:1203.1476 [math.NT], 2012. - From _N. J. A. Sloane_, Sep 14 2012
J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.
Mathematics Stack Exchange, Proof that a(n) == 0 (mod 3) if and only if n is in A081606.
László Németh, Tetrahedron trinomial coefficient transform, arXiv:1905.13475 [math.CO], 2019.
H. D. Nguyen and D. Taggart, Mining the OEIS: Ten Experimental Conjectures, 2013. Mentions this sequence. - From _N. J. A. Sloane_, Mar 16 2014
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.
Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 96.

Column 3 of A292627. Column 1 of A110165. Central column of A272866.
Cf. A002893, A084610, A000172, A002212.
First differences are in A085362. Bisection of A026380.
m-th binomial transforms of A000984: A126869 (m = -2), A002426 (m = -1 and m = -3 for signed version), A000984 (m = 0 and m = -4 for signed version), A026375 (m = 1 and m = -5 for signed version), A081671 (m = 2 and m = -6 for signed version), A098409 (m = 3 and m = -7 for signed version), A098410 (m = 4 and m = -8 for signed version), A104454 (m = 5 and m = -9 for signed version).

List([0..25],n->Sum([0..n],k->Binomial(n,k)*Binomial(2*k,k))); # Muniru A Asiru, Jul 29 2018

a026375 n = a026374 (2 * n) n  -- Reinhard Zumkeller, Feb 22 2014

seq( add(binomial(n,k)*binomial(2*k,k), k=0..n), n=0..30 ); # Detlef Pauly (dettodet(AT)yahoo.de), Nov 08 2001
a := n -> simplify(GegenbauerC(n, -n, -3/2)):
seq(a(n), n=0..22); # Peter Luschny, May 09 2016

Table[SeriesCoefficient[1/Sqrt[1-6*x+5*x^2],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 08 2012 *)
(* From Michael Somos, May 11 2014: (Start) *)
a[ n_] := Sum[ Binomial[n, k] Binomial[2 k, k], {k, 0, n}];
a[ n_] := If[ n < 0, 0, Hypergeometric2F1[-n, 1/2, 1, -4]];
a[ n_] := If[ n < 0, 0, Coefficient[(1 + 3 x + x^2)^n, x, n]];
a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[Exp[3 x] BesselI[0,2 x], {x, 0, n}]];
(* (End) *)

A026375(n):=coeff(expand((1+3*x+x^2)^n),x,n);
makelist(A026375(n),n,0,12); /* Emanuele Munarini, Mar 02 2011 */

{a(n) = if( n<0, 0, polcoeff( (1 + 3*x + x^2)^n, n))}; /* Michael Somos, Sep 09 2002 */

a(n)={my(v=Vec((1-x-x^2)^n)); sum(k=1,#v, v[k]^2);} \\ Joerg Arndt, Jul 06 2011

{a(n) = sum(k=0, n, 5^(n-k)*(-1)^k*binomial(n, k)*binomial(2*k, k))} \\ Seiichi Manyama, Apr 22 2019

{a(n) = sum(k=0, n\2, 3^(n-2*k)*binomial(n, 2*k)*binomial(2*k, k))} \\ Seiichi Manyama, May 04 2019

Representation by Gauss's hypergeometric function, in Maple notation: a(n)=hypergeom([ -n, 1/2 ], [ 1 ], -4). - Karol A. Penson, Apr 20 2001
This sequence is the binomial transform of A000984. - John W. Layman, Aug 11 2000; proved by Emeric Deutsch, Oct 26 2002
E.g.f.: exp(3*x)*I_0(2x), where I_0 is Bessel function. - Michael Somos, Sep 17 2002
G.f.: 1/sqrt(1-6*x+5*x^2). - Emeric Deutsch, Oct 26 2002
D-finite with recurrence: n*a(n)-3*(2*n-1)*a(n-1)+5*(n-1)*a(n-2)=0 for n > 1. - Emeric Deutsch, Jan 24 2004
From Emeric Deutsch, Jan 30 2004: (Start)
a(n) = [t^n](1+3*t+t^2)^n;
a(n) = Sum_{j=ceiling(n/2)..n} 3^(2*j-n)*binomial(n, j)*binomial(j, n-j). (End)
a(n) = A026380(2*n-1) (n>0). - Emeric Deutsch, Feb 18 2004
G.f.: 1/(1-x-2*x/(1-x/(1-x-x/(1-x/(1-x-x/(1-x/(1-x-x/(1-x... (continued fraction). - Paul Barry, Jan 06 2009
a(n) = sum of squared coefficients of (1+x-x^2)^n - see triangle A084610. - Paul D. Hanna, Jul 18 2009
a(n) = sum of squares of coefficients of (1-x-x^2)^n. - Joerg Arndt, Jul 06 2011
a(n) = (1/Pi)*Integral_{x=-2..2} ((3+x)^n/sqrt((2-x)*(2+x))) dx. - Peter Luschny, Sep 12 2011
a(n) ~ 5^(n+1/2)/(2*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 08 2012
G.f.: G(0)/(1-x), where G(k) = 1 + 4*x*(4*k+1)/( (4*k+2)*(1-x) - 2*x*(1-x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1-x)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 24 2013
0 = a(n)*(+25*a(n+1) - 45*a(n+2) + 10*a(n+3)) + a(n+1)*(-15*a(n+1) + 36*a(n+2) - 9*a(n+3)) + a(n+2)*(-3*a(n+2) + a(n+3)) for all n in Z. - Michael Somos, May 11 2014
a(n) = GegenbauerC(n, -n, -3/2). - Peter Luschny, May 09 2016
a(n) = Sum_{k=0..n} 5^(n-k) * (-1)^k * binomial(n,k) * binomial(2*k,k). - Seiichi Manyama, Apr 22 2019
a(n) = Sum_{k=0..floor(n/2)} 3^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). - Seiichi Manyama, May 04 2019
a(n) = (1/Pi) * Integral_{x = -1..1} (1 + 4*x^2)^n/sqrt(1 - x^2) dx = (1/Pi) * Integral_{x = -1..1} (5 - 4*x^2)^n/sqrt(1 - x^2) dx. - Peter Bala, Jan 27 2020
From Peter Bala, Jan 10 2022: (Start)
1 + x*exp(Sum_{n >= 1} a(n)*x^n/n) = 1 + x + 3*x^2 + 10*x^3 + 36*x^4 + ... is the o.g.f. of A002212.
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. (End)
a(n) = (1/4)^n * Sum_{k=0..n} 5^k * binomial(2*k,k) * binomial(2*(n-k),n-k). - Seiichi Manyama, Aug 18 2025

Definition simplified by N. J. A. Sloane, Feb 16 2012

cp's OEIS Frontend

A000225 a(n) = 2^n - 1. (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.)

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A000351 Powers of 5: a(n) = 5^n.

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A015518 a(n) = 2*a(n-1) + 3*a(n-2), with a(0)=0, a(1)=1.

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A003947 Expansion of (1+x)/(1-4*x).

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A006095 Gaussian binomial coefficient [n, 2] for q = 2.

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A015518 a(n) = 2a(n-1) + 3a(n-2), with a(0)=0, a(1)=1.

A026375 a(n) = Sum_{k=0..n} binomial(n,k)binomial(2k,k).