cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 20 results. Next

A119259 Central terms of the triangle in A119258.

Original entry on oeis.org

1, 3, 17, 111, 769, 5503, 40193, 297727, 2228225, 16807935, 127574017, 973168639, 7454392321, 57298911231, 441739706369, 3414246490111, 26447737520129, 205272288591871, 1595964714385409, 12427568655368191, 96905907580960769, 756583504975757311, 5913649000782757889
Offset: 0

Views

Author

Reinhard Zumkeller, May 11 2006

Keywords

Comments

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 06 2022

References

  • R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Crossrefs

Programs

  • Haskell
    a119259 n = a119258 (2 * n) n  -- Reinhard Zumkeller, Aug 06 2014
    
  • Mathematica
    Table[Binomial[2k - 1, k] Hypergeometric2F1[-2k, -k, 1 - 2k, -1], {k, 0, 10}] (* Vladimir Reshetnikov, Feb 16 2011 *)
  • Python
    from itertools import count, islice
    def A119259_gen(): # generator of terms
        yield from (1,3)
        a, c = 2, 1
        for n in count(1):
            yield (a<>1
    A119259_list = list(islice(A119259_gen(),20)) # Chai Wah Wu, Apr 26 2023

Formula

a(n) = A119258(2*n,n).
a(n) = Sum_{k=0..n} C(2*n,k)*C(2*n-k-1,n-k). - Paul Barry, Sep 28 2007
a(n) = Sum_{k=0..n} C(n+k-1,k)*2^k. - Paul Barry, Sep 28 2007
2*a(n) = A064062(n)+A178792(n). - Joseph Abate, Jul 21 2010
G.f.: (4*x^2+3*sqrt(1-8*x)*x-5*x)/(sqrt(1-8*x)*(2*x^2+x-1)-8*x^2-7*x+1). - Vladimir Kruchinin, Aug 19 2013
a(n) = (-1)^n - 2^(n+1)*binomial(2*n,n-1)*hyper2F1([1,2*n+1],[n+2],2). - Peter Luschny, Jul 25 2014
a(n) = (-1)^n + 2^(n+1)*A014300(n). - Peter Luschny, Jul 25 2014
a(n) = [x^n] ( (1 + x)^2/(1 - x) )^n. Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 13*x^2 + 67*x^3 + ... is essentially the o.g.f. for A064062. - Peter Bala, Oct 01 2015
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^2/(1 - x)) and hence is algebraic by Stanley 1999, Theorem 6.33, p.197. - Peter Bala, Aug 21 2016
n*(3*n-4)*a(n) +(-21*n^2+40*n-12)*a(n-1) -4*(3*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Aug 09 2017
From Peter Bala, Mar 23 2020: (Start)
a(p) == 3 ( mod p^3 ) for prime p >= 5. Cf. A002003, A103885 and A156894.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. (End)
G.f.: (8*x)/(sqrt(1-8*x)*(1+4*x)-1+8*x). - Fabian Pereyra, Jul 20 2024
a(n) = 2^(n+1)*binomial(2*n,n) - A178792(n). - Akiva Weinberger, Dec 06 2024
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(2*n,k). - Seiichi Manyama, Jul 31 2025

A145661 Triangle T(n,k) = (-1)^k * A119258(n,k) read by rows, 0 <= k <= n.

Original entry on oeis.org

1, 1, -1, 1, -3, 1, 1, -5, 7, -1, 1, -7, 17, -15, 1, 1, -9, 31, -49, 31, -1, 1, -11, 49, -111, 129, -63, 1, 1, -13, 71, -209, 351, -321, 127, -1, 1, -15, 97, -351, 769, -1023, 769, -255, 1, 1, -17, 127, -545, 1471, -2561, 2815, -1793, 511, -1, 1, -19, 161, -799, 2561
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Mar 16 2009

Keywords

Comments

Row sums are (-1)^(n+1)*(n-1) for n >= 1.
A145661, A119258 and A118801 are all essentially the same (see the Shattuck and Waldhauser paper). - Tamas Waldhauser, Jul 25 2011

Examples

			Triangle begins
  1;
  1,  -1;
  1,  -3,   1;
  1,  -5,   7,   -1;
  1,  -7,  17,  -15,    1;
  1,  -9,  31,  -49,   31,    -1;
  1, -11,  49, -111,  129,   -63,    1;
  1, -13,  71, -209,  351,  -321,  127,    -1;
  1, -15,  97, -351,  769, -1023,  769,  -255,    1;
  1, -17, 127, -545, 1471, -2561, 2815, -1793,  511,    -1;
  1, -19, 161, -799, 2561, -5503, 7937, -7423, 4097, -1023, 1;
		

Crossrefs

A193844 is an essentially identical triangle.

Programs

  • Maple
    A119258 := proc(n,k)
            if k=0 or k = n then
                    1;
            elif k<0 or k> n then
                    0;
            else
                    2*procname(n-1,k-1)+procname(n-1,k) ;
            end if;
    end proc:
    seq(seq(A119258(n,k),k=0..n),n=0..10) ;
    A145661 := proc(n,k)
            (-1)^k*A119258(n,k) ;
    end proc: # R. J. Mathar, Oct 21 2011
  • Mathematica
    Clear[M, T, d, a, x, a0];
    T[n_, m_, d_] := If[ m == n + 1, 1, If[n == d, 1, 0]];
    M[d_] := MatrixPower[Table[T[n, m, d], {n, 1, d}, {m, 1, d}], d];
    Table[M[d], {d, 1, 10}];
    Table[Det[M[d]], {d, 1, 10}];
    Table[CharacteristicPolynomial[M[d], x], {d, 1, 10}];
    a = Join[{{1}}, Table[CoefficientList[Expand[CharacteristicPolynomial[M[n], x]], x], {n, 1, 10}]];
    Flatten[a]
    Join[{1}, Table[Apply[ Plus, CoefficientList[Expand[CharacteristicPolynomial[M[n], x]], x]], {n, 1, 10}]];

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

Original entry on oeis.org

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

Views

Author

Keywords

Comments

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

Examples

			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
		

References

  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 75.
  • Ralph P. Grimaldi, Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition, Addison-Wesley, 2004, p. 134.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, p. 79.
  • Johann Peter Hebel, Gesammelte Werke in sechs Bänden, Herausgeber: Jan Knopf, Franz Littmann und Hansgeorg Schmidt-Bergmann unter Mitarbeit von Ester Stern, Wallstein Verlag, 2019. Band 3, S. 20-21, Loesung, S. 36-37. See also the link below.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 60, 75-83.
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 141.
  • D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, "Tower of Hanoi", Penguin Books, 1987, pp. 112-113.

Crossrefs

Cf. A000043 (Mersenne exponents).
Cf. A000668 (Mersenne primes).
Cf. A001348 (Mersenne numbers with n prime).
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. A008277, A048993 (columns k=2), A000918, A130330.
Cf. A000225, A029858, A058809, A375256 (Hanoi graphs).

Programs

  • Haskell
    a000225 = (subtract 1) . (2 ^)
    a000225_list = iterate ((+ 1) . (* 2)) 0
    -- Reinhard Zumkeller, Mar 20 2012
    
  • Maple
    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)
  • Mathematica
    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 *)
  • PARI
    A000225(n) = 2^n-1  \\ Michael B. Porter, Oct 27 2009
    
  • PARI
    concat(0, Vec(x/((1-2*x)*(1-x)) + O(x^100))) \\ Altug Alkan, Oct 28 2015
    
  • Python
    def A000225(n): return (1<Chai Wah Wu, Jul 06 2022
  • SageMath
    def isMersenne(n): return n == sum([(1 - b) << s for (s, b) in enumerate((n+1).bits())]) # Peter Luschny, Sep 01 2019
    

Formula

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)

Extensions

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

A007051 a(n) = (3^n + 1)/2.

Original entry on oeis.org

1, 2, 5, 14, 41, 122, 365, 1094, 3281, 9842, 29525, 88574, 265721, 797162, 2391485, 7174454, 21523361, 64570082, 193710245, 581130734, 1743392201, 5230176602, 15690529805, 47071589414, 141214768241, 423644304722, 1270932914165, 3812798742494, 11438396227481
Offset: 0

Views

Author

Keywords

Comments

Number of ordered trees with n edges and height at most 4.
Number of palindromic structures using a maximum of three different symbols. - Marks R. Nester
Number of compositions of all even natural numbers into n parts <= 2 (0 is counted as a part), see example. - Adi Dani, May 14 2011
Consider the mapping f(a/b) = (a + 2*b)/(2*a + b). Taking a = 1, b = 2 to start with, and carrying out this mapping repeatedly on each new (reduced) rational number gives the sequence 1/2, 4/5, 13/14, 40/41, ... converging to 1. The sequence contains the denominators = (3^n+1)/2. The same mapping for N, i.e., f(a/b) = (a + N*b)/(a+b) gives fractions converging to N^(1/2). - Amarnath Murthy, Mar 22 2003
Second binomial transform of the expansion of cosh(x). - Paul Barry, Apr 05 2003
The sequence (1, 1, 2, 5, ...) = 3^n/6 + 1/2 + 0^n/3 has binomial transform A007581. - Paul Barry, Jul 20 2003
Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2n+2, s(0) = 1, s(2n+2) = 1. - Herbert Kociemba, Jun 10 2004
Density of regular language L over {1,2,3}^* (i.e., number of strings of length n in L) described by regular expression 11*+11*2(1+2)*+11*2(1+2)*3(1+2+3)*. - Nelma Moreira, Oct 10 2004
Sums of rows of the triangle in A119258. - Reinhard Zumkeller, May 11 2006
Number of n-words from the alphabet A = {a,b,c} which contain an even number of a's. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006
Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which either 0) x and y are disjoint and for which x is not a subset of y and y is not a subset of x, or 1) x = y. - Ross La Haye, Jan 10 2008
a(n+1) gives the number of primitive periodic multiplex juggling sequences of length n with base state <2>. - Steve Butler, Jan 21 2008
a(n) is also the number of idempotent order-preserving and order-decreasing partial transformations (of an n-chain). - Abdullahi Umar, Oct 02 2008
Equals row sums of triangle A147292. - Gary W. Adamson, Nov 05 2008
Equals leftmost column of A071919^3. - Gary W. Adamson, Apr 13 2009
A010888(a(n))=5 for n >= 2, that is, the digital root of the terms >= 5 equals 5. - Parthasarathy Nambi, Jun 03 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^n*charpoly(A,2). - Milan Janjic, Jan 27 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=6, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=(-1)^(n-1)*charpoly(A,3). - Milan Janjic, Feb 21 2010
It appears that if s(n) is a rational sequence of the form s(1)=2, s(n)= (2*s(n-1)+1)/(s(n-1)+2), n>1 then s(n)=a(n)/(a(n-1)-1).
Form an array with m(1,n)=1 and m(i,j) = Sum_{k=1..i-1} m(k,j) + Sum_{k=1..j-1} m(i,k), which is the sum of the terms to the left of m(i,j) plus the sum above m(i,j). The sum of the terms in antidiagonal(n-1) = a(n). - J. M. Bergot, Jul 16 2013
From Peter Bala, Oct 29 2013: (Start)
An Engel expansion of 3 to the base b := 3/2 as defined in A181565, with the associated series expansion 3 = b + b^2/2 + b^3/(2*5) + b^4/(2*5*14) + .... Cf. A034472.
More generally, for a positive integer n >= 3, the sequence [1, n - 1, n^2 - n - 1, ..., ( (n - 2)*n^k + 1 )/(n - 1), ...] is an Engel expansion of n/(n - 2) to the base n/(n - 1). Cases include A007583 (n = 4), A083065 (n = 5) and A083066 (n = 6). (End)
Diagonal elements (and one more than antidiagonal elements) of the matrix A^n where A=(2,1;1,2). - David Neil McGrath, Aug 17 2014
From M. Sinan Kul, Sep 07 2016: (Start)
a(n) is equal to the number of integer solutions to the following equation when x is equal to the product of n distinct primes: 1/x = 1/y + 1/z where 0 < x < y <= z.
If z = k*y where k is a fraction >= 1 then the solutions can be given as: y = ((k+1)/k)*x and z = (k+1)*x.
Here k can be equal to any divisor of x or to the ratio of two divisors.
For example for x = 2*3*5 = 30 (product of three distinct primes), k would have the following 14 values: 1, 6/5, 3/2, 5/3, 2, 5/2, 3, 10/3, 5, 6, 15/2, 10, 15, 30.
As an example for k = 10/3, we would have y=39, z=130 and 1/39 + 1/130 = 1/30.
Here finding the number of fractions would be equivalent to distributing n balls (distinct primes) to two bins (numerator and denominator) with no empty bins which can be found using Stirling numbers of the second kind. So another definition for a(n) is: a(n) = 2^n + Sum_{i=2..n} Stirling2(i,2)*binomial(n,i).
(End)
a(n+1) is the smallest i for which the Catalan number C(i) (see A000108) is divisible by 3^n for n > 0. This follows from the rule given by Franklin T. Adams-Watters for determining the multiplicity with which a prime divides C(n). We need to find the smallest number in base 3 to achieve a given count. Applied to prime 3, 1 is the smallest digit that counts but requires to be followed by 2 which cannot be at the end to count. Therefore the number in base 3 of the form 1{n-1 times}20 = (3^(n+1) + 1)/2 + 1 = a(n+1)+1 is the smallest number to achieve count n which implies the claim. - Peter Schorn, Mar 06 2020
Let A be a Toeplitz matrix of order n, defined by: A[i,j]=1, if ij; A[i,i]=2. Then, for n>=1, a(n) = det A. - Dmitry Efimov, Oct 28 2021
a(n) is the least number k such that A065363(k) = -(n-1), for n > 0. - Amiram Eldar, Sep 03 2022

Examples

			From _Adi Dani_, May 14 2011: (Start)
a(3)=14 because all compositions of even natural numbers into 3 parts <=2 are
for 0: (0,0,0)
for 2: (0,1,1), (1,0,1), (1,1,0), (0,0,2), (0,2,0), (2,0,0)
for 4: (0,2,2), (2,0.2), (2,2,0), (1,1,2), (1,2,1), (2,1,1)
for 6: (2,2,2).
(End)
		

References

  • J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 47.
  • Adi Dani, Quasicompositions of natural numbers, Proceedings of III congress of mathematicians of Macedonia, Struga Macedonia 29 IX -2 X 2005 pages 225-238.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E11.
  • 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]
  • P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 60.
  • P. Ribenboim, The Little Book of Big Primes, Springer-Verlag, NY, 1991, p. 53.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

Formula

a(n) = 3*a(n-1) - 1.
Binomial transform of Chebyshev coefficients A011782. - Paul Barry, Mar 16 2003
From Paul Barry, Mar 16 2003: (Start)
a(n) = 4*a(n-1) - 3*a(n-2) for n > 1, a(0)=1, a(1)=2.
G.f.: (1 - 2*x)/((1 - x)*(1 - 3*x)). (End)
E.g.f.: exp(2*x)*cosh(x). - Paul Barry, Apr 05 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k). - Paul Barry, May 08 2003
This sequence is also the partial sums of the first 3 Stirling numbers of second kind: a(n) = S(n+1, 1) + S(n+1, 2) + S(n+1, 3) for n >= 0; alternatively it is the number of partitions of [n+1] into 3 or fewer parts. - Mike Zabrocki, Jun 21 2004
For c=3, a(n) = (c^n)/c! + Sum_{k=1..c-2}((k^n)/k!*(Sum_{j=2..c-k}(((-1)^j)/j!))) or = Sum_{k=1..c} g(k, c)*k^n where g(1, 1) = 1, g(1, c) = g(1, c-1) + ((-1)^(c-1))/(c-1)! for c > 1, and g(k, c) = g(k-1, c-1)/k for c > 1 and 2 <= k <= c. - Nelma Moreira, Oct 10 2004
The i-th term of the sequence is the entry (1, 1) in the i-th power of the 2 X 2 matrix M = ((2, 1), (1, 2)). - Simone Severini, Oct 15 2005
If p[i]=fibonacci(2i-3) and if A is the Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)= det A. - Milan Janjic, May 08 2010
INVERT transform of A001519: [1, 1, 2, 5, 13, 34, ...]. - Gary W. Adamson, Jun 13 2011
a(n) = M^n*[1,1,1,0,0,0,...], leftmost column term; where M = an infinite bidiagonal matrix with all 1's in the superdiagonal and (1,2,3,...) in the main diagonal and the rest zeros. - Gary W. Adamson, Jun 23 2011
a(n) = M^n*[1,1,1,0,0,0,...], top entry term; where M is an infinite bidiagonal matrix with all 1's in the superdiagonal, (1,2,3,...) as the main diagonal, and the rest zeros. - Gary W. Adamson, Jun 24 2011
a(n) = A201730(n,0). - Philippe Deléham, Dec 05 2011
a(n) = A006342(n) + A006342(n-1). - Yuchun Ji, Sep 19 2018
From Dmitry Efimov, Oct 29 2021: (Start)
a(2*m+1) = Product_{k=-m..m} (2+i*tan(Pi*k/(2*m+1))),
a(2*m) = Product_{k=-m..m-1} (2+i*tan(Pi*(2*k+1)/(4*m))),
where i is the imaginary unit. (End)

A056220 a(n) = 2*n^2 - 1.

Original entry on oeis.org

-1, 1, 7, 17, 31, 49, 71, 97, 127, 161, 199, 241, 287, 337, 391, 449, 511, 577, 647, 721, 799, 881, 967, 1057, 1151, 1249, 1351, 1457, 1567, 1681, 1799, 1921, 2047, 2177, 2311, 2449, 2591, 2737, 2887, 3041, 3199, 3361, 3527, 3697, 3871, 4049, 4231, 4417, 4607, 4801
Offset: 0

Views

Author

N. J. A. Sloane, Aug 06 2000

Keywords

Comments

Image of squares (A000290) under "little Hankel" transform that sends [c_0, c_1, ...] to [d_0, d_1, ...] where d_n = c_n^2 - c_{n+1}*c_{n-1}. - Henry Bottomley, Dec 12 2000
Surround numbers of an n X n square. - Jason Earls, Apr 16 2001
Numbers n such that 2*n + 2 is a perfect square. - Cino Hilliard, Dec 18 2003, Juri-Stepan Gerasimov, Apr 09 2016
The sums of the consecutive integer sequences 2n^2 to 2(n+1)^2-1 are cubes, as 2n^2 + ... + 2(n+1)^2-1 = (1/2)(2(n+1)^2 - 1 - 2n^2 + 1)(2(n+1)^2 - 1 + 2n^2) = (2n+1)^3. E.g., 2+3+4+5+6+7 = 27 = 3^3, then 8+9+10+...+17 = 125 = 5^3. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 29 2005
X values (other than 0) of solutions to the equation 2*X^3 + 2*X^2 = Y^2. To find Y values: b(n) = 2n*(2*n^2 - 1). - Mohamed Bouhamida, Nov 06 2007
Average of the squares of two consecutive terms is also a square. In fact: (2*n^2 - 1)^2 + (2*(n+1)^2 - 1)^2 = 2*(2*n^2 + 2*n + 1)^2. - Matias Saucedo (solomatias(AT)yahoo.com.ar), Aug 18 2008
Equals row sums of triangle A143593 and binomial transform of [1, 6, 4, 0, 0, 0, ...] with n > 1. - Gary W. Adamson, Aug 26 2008
Start a spiral of square tiles. Trivially the first tile fits in a 1 X 1 square. 7 tiles fit in a 3 X 3 square, 17 tiles fit in a 5 X 5 square and so on. - Juhani Heino, Dec 13 2009
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) = coeff(charpoly(A,x),x^(n-2)). - Milan Janjic, Jan 26 2010
For each n > 0, the recursive series, formula S(b) = 6*S(b-1) - S(b-2) - 2*a(n) with S(0) = 4n^2-4n+1 and S(1) = 2n^2, has the property that every even term is a perfect square and every odd term is twice a perfect square. - Kenneth J Ramsey, Jul 18 2010
Fourth diagonal of A154685 for n > 2. - Vincenzo Librandi, Aug 07 2010
First integer of (2*n)^2 consecutive integers, where the last integer is 3 times the first + 1. As example, n = 2: term = 7; (2*n)^2 = 16; 7, 8, 9, ..., 20, 21, 22: 7*3 + 1 = 22. - Denis Borris, Nov 18 2012
Chebyshev polynomial of the first kind T(2,n). - Vincenzo Librandi, May 30 2014
For n > 0, number of possible positions of a 1 X 2 rectangle in a (n+1) X (n+2) rectangular integer lattice. - Andres Cicuttin, Apr 07 2016
This sequence also represents the best solution for Ripà's n_1 X n_2 X n_3 dots problem, for any 0 < n_1 = n_2 < n_3 = floor((3/2)*(n_1 - 1)) + 1. - Marco Ripà, Jul 23 2018

Examples

			a(0) = 0^2-1*1 = -1, a(1) = 1^2 - 4*0 = 1, a(2) = 2^2 - 9*1 = 7, etc.
a(4) = 31 = (1, 3, 3, 1) dot (1, 6, 4, 0) = (1 + 18 + 12 + 0). - _Gary W. Adamson_, Aug 29 2008
		

Crossrefs

Cf. A066049 (indices of prime terms)
Column 2 of array A188644 (starting at offset 1).

Programs

Formula

G.f.: (-1 + 4*x + x^2)/(1-x)^3. - Henry Bottomley, Dec 12 2000
a(n) = A119258(n+1,2) for n > 0. - Reinhard Zumkeller, May 11 2006
From Doug Bell, Mar 08 2009: (Start)
a(0) = -1,
a(n) = sqrt(A001844(n)^2 - A069074(n-1)),
a(n+1) = sqrt(A001844(n)^2 + A069074(n-1)) = sqrt(a(n)^2 + A069074(n-1)*2). (End)
a(n) + a(n+1) + 1 = (2n+1)^2. - Doug Bell, Mar 09 2009
a(n) = a(n-1) + 4*n - 2 (with a(0)=-1). - Vincenzo Librandi, Dec 25 2010
a(n) = A188653(2*n) for n > 0. - Reinhard Zumkeller, Apr 13 2011
a(n) = A162610(2*n-1,n) for n > 0. - Reinhard Zumkeller, Jan 19 2013
a(n) = ( Sum_{k=0..2} (C(n+k,3)-C(n+k-1,3))*(C(n+k,3)+C(n+k+1,3)) ) - (C(n+2,3)-C(n-1,3))*(C(n,3)+C(n+3,3)), for n > 3. - J. M. Bergot, Jun 16 2014
a(n) = j^2 + k^2 - 2 or 2*j*k if n >= 2 and j = n + sqrt(2)/2 and k = n - sqrt(2)/2. - Avi Friedlich, Mar 30 2015
a(n) = A002593(n)/n^2. - Bruce J. Nicholson, Apr 03 2017
a(n) = A000384(n) + n - 1. - Bruce J. Nicholson, Nov 12 2017
a(n)*a(n+k) + 2k^2 = m^2 (a perfect square), m = a(n) + (2n*k), for n>=1. - Ezhilarasu Velayutham, May 13 2019
From Amiram Eldar, Aug 10 2020: (Start)
Sum_{n>=1} 1/a(n) = 1/2 - sqrt(2)*Pi*cot(Pi/sqrt(2))/4.
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(2)*Pi*csc(Pi/sqrt(2))/4 - 1/2. (End)
From Amiram Eldar, Feb 04 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(2))*csc(Pi/sqrt(2)).
Product_{n>=2} (1 - 1/a(n)) = (Pi/(4*sqrt(2)))*csc(Pi/sqrt(2)). (End)
a(n) = A003215(n) - A000217(n-2)*2. - Leo Tavares, Jun 29 2021
Let T(n) = n*(n+1)/2. Then a(n)^2 = T(2n-2)*T(2n+1) + n^2. - Charlie Marion, Feb 12 2023
E.g.f.: exp(x)*(2*x^2 + 2*x - 1). - Stefano Spezia, Jul 08 2023

A000337 a(n) = (n-1)*2^n + 1.

Original entry on oeis.org

0, 1, 5, 17, 49, 129, 321, 769, 1793, 4097, 9217, 20481, 45057, 98305, 212993, 458753, 983041, 2097153, 4456449, 9437185, 19922945, 41943041, 88080385, 184549377, 385875969, 805306369, 1677721601, 3489660929, 7247757313, 15032385537, 31138512897, 64424509441
Offset: 0

Views

Author

Keywords

Comments

a(n) also gives number of 0's in binary numbers 1 to 111..1 (n+1 bits). - Stephen G Penrice, Oct 01 2000
Numerator of m(n) = (m(n-1)+n)/2, m(0)=0. Denominator is A000079. - Reinhard Zumkeller, Feb 23 2002
a(n) is the number of directed column-convex polyominoes of area n+2 having along the lower contour exactly one vertical step that is followed by a horizontal step (a reentrant corner). - Emeric Deutsch, May 21 2003
a(n) is the number of bits in binary numbers from 1 to 111...1 (n bits). Partial sums of A001787. - Emeric Deutsch, May 24 2003
Genus of graph of n-cube = a(n-3) = 1+(n-4)*2^(n-3), n>1.
Sum of ordered partitions of n where each element is summed via T(e-1). See A066185 for more information. - Jon Perry, Dec 12 2003
a(n-2) is the number of Dyck n-paths with exactly one peak at height >= 3. For example, there are 5 such paths with n=4: UUUUDDDD, UUDUUDDD, UUUDDUDD, UDUUUDDD, UUUDDDUD. - David Callan, Mar 23 2004
Permutations in S_{n+2} avoiding 12-3 that contain the pattern 13-2 exactly once.
a(n) is prime for n = 2, 3, 7, 27, 51, 55, 81. a(n) is semiprime for n = 4, 5, 6, 8, 9, 10, 11, 13, 15, 19, 28, 32, 39, 57, 63, 66, 75, 97. - Jonathan Vos Post, Jul 18 2005
A member of the family of sequences defined by a(n) = Sum_{i=1..n} i*[c(1)*...*c(r)]^(i-1). This sequence has c(1)=2, A014915 has c(1)=3. - Ctibor O. Zizka, Feb 23 2008
Starting with 1 = row sums of A023758 as a triangle by rows: [1; 2,3; 4,6,7; 8,12,14,15; ...]. - Gary W. Adamson, Jul 18 2008
Equivalent formula given in Brehm: for each q >= 3 there exists a polyhedral map M_q of type {4, q} with [number of vertices] f_0 = 2^q and [genus] g = (2^(q-3))*(q-4) + 1 such that M_q and its dual have polyhedral embeddings in R^3 [McMullen et al.]. - Jonathan Vos Post, Jul 25 2009
Sums of rows of the triangle in A173787. - Reinhard Zumkeller, Feb 28 2010
This sequence is related to A000079 by a(n) = n*A000079(n)-Sum_{i=0..n-1} A000079(i). - Bruno Berselli, Mar 06 2012
(1 + 5*x + 17*x^2 + 49*x^3 + ...) = (1 + 2*x + 4*x^2 + 8*x^3 + ...) * (1 + 3*x + 7*x^2 + 15*x^3 + ...). - Gary W. Adamson, Mar 14 2012
The first barycentric coordinate of the centroid of Pascal triangles, assuming that numbers are weights, is A000295(n+1)/A000337(n), no matter what the triangle sides are. See attached figure. - César Eliud Lozada, Nov 14 2014
a(n) is the n-th number that is a sum of n positive n-th powers for n >= 1. a(4) = 49 = A003338(4). - Alois P. Heinz, Aug 01 2020
a(n) is the sum of the largest elements of all subsets of {1,2,..,n}. For example, a(3)=17; the subsets of {1,2,3} are {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}, and the sum of the largest elements is 17. - Enrique Navarrete, Aug 20 2020
a(n-1) is the sum of the second largest elements of the subsets of {1,2,..,n} that contain n. For example, for n = 4, a(3)=17; the subsets of {1,2,3,4} that contain 4 are {4}, {1,4}, {2,4}, {3,4}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}, and the sum of the second largest elements is 17. - Enrique Navarrete, Aug 24 2020
a(n-1) is also the sum of diameters of all subsets of {1,2,...,n} that contain n. For example, for n = 4, a(3)=17; the subsets of {1,2,3,4} that contain 4 are {4}, {1,4}, {2,4}, {3,4}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}; the diameters of these sets are 0,3,2,1,3,3,2,3 and the sum is 17. - Enrique Navarrete, Sep 07 2020
a(n-1) is also the number of additions required to compute the permanent of general n X n matrices using trellis methods (see Theorems 5 and 6, pp. 10-11 in Kiah et al.). - Stefano Spezia, Nov 02 2021

References

  • F. Harary, Topological concepts in graph theory, pp. 13-17 of F. Harary and L. Beineke, editors, A seminar on Graph Theory, Holt, Rinehart and Winston, New York, 1967.
  • V. G. Gutierrez and S. L. de Medrano, Surfaces as complete intersections, in Riemann and Klein Surfaces, Automorphisms, Symmetries and Moduli Spaces, edited by Milagros Izquierdo, S. Allen Broughton, Antonio F. Costa, Contemp. Math. vol. 629, 2014, pp. 171-.
  • F. Harary, Graph Theory. Addison-Wesley, Reading, MA, 1969, p. 119.
  • G. H. Hardy, A Theorem Concerning the Infinite Cardinal Numbers, Quart. J. Math., 35 (1904), p. 90 = Collected Papers of G. H. Hardy, Vol. VII, p. 430.
  • 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).

Crossrefs

a(n) = T(3, n), array T given by A048472. A036799/2.
Cf. A003338.
Main diagonal of A336725.

Programs

  • GAP
    List([0..30],n->(n-1)*2^n+1); # Muniru A Asiru, Oct 24 2018
  • Magma
    [(n-1)*2^n + 1: n in [0..40]]; // Vincenzo Librandi, Nov 21 2014
    
  • Maple
    A000337 := proc(n) 1+(n-1)*2^n ; end proc: # R. J. Mathar, Oct 10 2011
  • Mathematica
    Table[Sum[(-1)^(n - k) k (-1)^(n - k) Binomial[n + 1, k + 1], {k, 0, n}], {n, 0, 28}] (* Zerinvary Lajos, Jul 08 2009 *)
    Table[(n - 1) 2^n + 1, {n, 0, 40}] (* Harvey P. Dale, Jun 21 2011 *)
    LinearRecurrence[{5, -8, 4}, {0, 1, 5}, 40] (* Harvey P. Dale, Jun 21 2011 *)
    CoefficientList[Series[x / ((1 - x) (1 - 2 x)^2), {x, 0, 50}], x] (* Vincenzo Librandi, Nov 21 2014 *)
  • PARI
    a(n)=if(n<0,0,(n-1)*2^n+1)
    
  • Python
    a=lambda n:((n-1)<<(n))+1 # Indranil Ghosh, Jan 05 2017
    

Formula

Binomial transform of A004273. Binomial transform of A008574 if the leading zero is dropped.
G.f.: x/((1-x)*(1-2*x)^2). - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(x) - exp(2*x)*(1-2*x). a(n) = 4*a(n-1) - 4*a(n-2)+1, n>0. Series reversion of g.f. A(x) is x*A034015(-x). - Michael Somos
Binomial transform of n/(n+1) is a(n)/(n+1). - Paul Barry, Aug 19 2005
a(n) = A119258(n+1,n-1) for n>0. - Reinhard Zumkeller, May 11 2006
Convolution of "Number of fixed points in all 231-avoiding involutions in S_n" (A059570) with "The odd numbers" (A005408), treating the result as if offset=0. - Graeme McRae, Jul 12 2006
a(n) = Sum_{k=1..n} k*2^(k-1), partial sums of A001787. - Zerinvary Lajos, Oct 19 2006
a(n) = 5*a(n-1) - 8*a(n-2) + 4*a(n-3), n > 2. - Harvey P. Dale, Jun 21 2011
a(n) = Sum_{k=1..n} Sum_{i=1..n} i * C(k,i). - Wesley Ivan Hurt, Sep 19 2017
a(n) = A000295(n+1)^2 - A000295(n)*A000295(n+2). - Gregory Gerard Wojnar, Oct 23 2018

A112468 Riordan array (1/(1-x), x/(1+x)).

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, -1, 1, 1, 0, 2, -2, 1, 1, 1, -2, 4, -3, 1, 1, 0, 3, -6, 7, -4, 1, 1, 1, -3, 9, -13, 11, -5, 1, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1
Offset: 0

Views

Author

Paul Barry, Sep 06 2005

Keywords

Comments

Row sums are A040000. Diagonal sums are A112469. Inverse is A112467. Row sums of k-th power are 1, k+1, k+1, k+1, .... Note that C(n,k) = Sum_{j=0..n-k} C(n-j-1, n-k-j).
Equals row reversal of triangle A112555 up to sign, where log(A112555) = A112555 - I. Unsigned row sums equals A052953 (Jacobsthal numbers + 1). Central terms of even-indexed rows are a signed version of A072547. Sums of squared terms in rows yields A112556, which equals the first differences of the unsigned central terms. - Paul D. Hanna, Jan 20 2006
Sum_{k=0..n} T(n,k)*x^k = A000012(n), A040000(n), A005408(n), A033484(n), A048473(n), A020989(n), A057651(n), A061801(n), A238275(n), A238276(n), A138894(n), A090843(n), A199023(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 respectively (see the square array in A112739). - Philippe Deléham, Feb 22 2014

Examples

			Triangle starts
  1;
  1,  1;
  1,  0,  1;
  1,  1, -1,  1;
  1,  0,  2, -2,  1;
  1,  1, -2,  4, -3,  1;
  1,  0,  3, -6,  7, -4,  1;
Matrix log begins:
  0;
  1,  0;
  1,  0,  0;
  1,  1, -1,  0;
  1,  1,  1, -2,  0;
  1,  1,  1,  1, -3,  0; ...
Production matrix begins
  1,  1,
  0, -1,  1,
  0,  0, -1,  1,
  0,  0,  0, -1,  1,
  0,  0,  0,  0, -1,  1,
  0,  0,  0,  0,  0, -1,  1,
  0,  0,  0,  0,  0,  0, -1,  1.
- _Paul Barry_, Apr 08 2011
		

Crossrefs

Cf. A174294, A174295, A174296, A174297. - Mats Granvik, Mar 15 2010
Cf. A072547 (central terms), A112555 (reversed rows), A112465, A052953, A112556, A112739, A119258.
See A279006 for another version.

Programs

  • GAP
    T:= function(n,k)
        if k=0 or k=n then return 1;
        else return T(n-1,k-1) - T(n-1,k);
        fi;
      end;
    Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 13 2019
  • Haskell
    a112468 n k = a112468_tabl !! n !! k
    a112468_row n = a112468_tabl !! n
    a112468_tabl = iterate (\xs -> zipWith (-) ([2] ++ xs) (xs ++ [0])) [1]
    -- Reinhard Zumkeller, Jan 03 2014
    
  • Magma
    function T(n,k)
      if k eq 0 or k eq n then return 1;
      else return T(n-1,k-1) - T(n-1,k);
      end if;
      return T;
    end function;
    [T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 13 2019
    
  • Maple
    T := (n,k,m) -> (1-m)^(-n+k)-m^(k+1)*pochhammer(n-k,k+1)*hypergeom( [1,n+1],[k+2],m)/(k+1)!; A112468 := (n,k) -> T(n,n-k,-1);
    seq(print(seq(simplify(A112468(n,k)),k=0..n)),n=0..10); # Peter Luschny, Jul 25 2014
  • Mathematica
    T[n_, 0] = 1; T[n_, n_] = 1; T[n_, k_ ]:= T[n, k] = T[n-1, k-1] - T[n-1, k]; Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* Jean-François Alcover, Mar 06 2013 *)
  • PARI
    {T(n,k)=local(m=1,x=X+X*O(X^n),y=Y+Y*O(Y^k)); polcoeff(polcoeff((1+(m-1)*x)*(1+m*x)/(1+m*x-x*y)/(1-x),n,X),k,Y)} \\ Paul D. Hanna, Jan 20 2006
    
  • PARI
    T(n,k) = if(k==0 || k==n, 1, T(n-1, k-1) - T(n-1, k)); \\ G. C. Greubel, Nov 13 2019
    
  • Sage
    @CachedFunction
    def T(n, k):
        if (k<0 or n<0): return 0
        elif (k==0 or k==n): return 1
        else: return T(n-1, k-1) - T(n-1, k)
    [[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 13 2019
    

Formula

Triangle T(n,k) read by rows: T(n,0)=1, T(n,k) = T(n-1,k-1) - T(n-1,k). - Mats Granvik, Mar 15 2010
Number triangle T(n, k)= Sum_{j=0..n-k} C(n-j-1, n-k-j)*(-1)^(n-k-j).
G.f. of matrix power T^m: (1+(m-1)*x)*(1+m*x)/(1+m*x-x*y)/(1-x). G.f. of matrix log: x*(1-2*x*y+x^2*y)/(1-x*y)^2/(1-x). - Paul D. Hanna, Jan 20 2006
T(n, k) = R(n,n-k,-1) where R(n,k,m) = (1-m)^(-n+k)-m^(k+1)*Pochhammer(n-k,k+1)*hyper2F1([1,n+1],[k+2],m)/(k+1)!. - Peter Luschny, Jul 25 2014

A228576 A triangle formed like generalized Pascal's triangle. The rule is T(n,k) = 2*T(n-1,k-1) + T(n-1,k), the left border is n and the right border is n^2 instead of 1.

Original entry on oeis.org

0, 1, 1, 2, 3, 4, 3, 7, 10, 9, 4, 13, 24, 29, 16, 5, 21, 50, 77, 74, 25, 6, 31, 92, 177, 228, 173, 36, 7, 43, 154, 361, 582, 629, 382, 49, 8, 57, 240, 669, 1304, 1793, 1640, 813, 64, 9, 73, 354, 1149, 2642, 4401, 5226, 4093, 1690, 81, 10, 91, 500, 1857, 4940, 9685, 14028, 14545, 9876, 3461, 100
Offset: 1

Views

Author

Boris Putievskiy, Aug 26 2013

Keywords

Examples

			The start of the sequence as triangle array read by rows:
  0;
  1,  1;
  2,  3,  4;
  3,  7, 10,  9;
  4, 13, 24, 29, 16;
  5, 21, 50, 77, 74, 25;
...
		

Crossrefs

Cf. We denote generalized Pascal's like triangle with coefficients a, b and with L(n) on the left border and R(n) on the right border by (a,b,L(n),R(n)). The list of sequences for (1,1,L(n),R(n)) see A228196;
A038207 (1,2,2^n,1), A105728 (1, 2, 1, n+1), A112468 (1,-1,1,1), A112626 (1,2,3^n,1), A119258 (2,1,1,1), A119673 (3,1,1,1), A119725 (3,2,1,1), A119726 (4,2,1,1), A119727 (5,2,1,1), A209705 (2,1,n+1,0);
A002061 (column 2), A000244 (sums of rows r of triangle array - (r-2)(r+1)/2).

Programs

  • GAP
    T:= function(n,k)
        if k=0 then return n;
        elif k=n then return n^2;
        else return 2*T(n-1,k-1) + T(n-1,k);
        fi;
      end;
    Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 13 2019
  • Magma
    function T(n,k)
      if k eq 0 then return n;
      elif k eq n then return n^2;
      else return 2*T(n-1,k-1) + T(n-1,k);
      end if;
      return T;
    end function;
    [T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 13 2019
    
  • Maple
    T := proc(n, k) option remember;
    if k = 0 then RETURN(n) fi;
    if k = n then RETURN(n^2) fi;
    2*T(n-1, k-1) + T(n-1, k) end:
    seq(seq(T(n,k),k=0..n),n=0..9);  # Peter Luschny, Aug 26 2013
  • Mathematica
    T[n_, 0]:= n; T[n_, n_]:= n^2; T[n_, k_]:= T[n, k] = 2*T[n-1, k-1]+T[n-1, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 25 2014 *)
  • PARI
    T(n,k) = if(k==0, n, if(k==n, n^2, 2*T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 13 2019
    
  • Sage
    @CachedFunction
    def T(n, k):
        if (k==0): return n
        elif (k==n): return n^2
        else: return 2*T(n-1,k-1) + T(n-1, k)
    [[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 13 2019
    

Formula

T(n, k) = 2*T(n-1, k-1) + T(n-1, k) for n,k >=0, with T(n,0) = n, T(n,n) = n^2.
Closed-form formula for generalized Pascal's triangle. Let a,b be any numbers. The rule is T(n, k) = a*T(n-1, k-1) + b*T(n-1, k) for n,k >0. Let L(m) and R(m) be the left border and the right border generalized Pascal's triangle, respectively.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} a^(n-m1) * b^k*R(m1)*C(n+k-m1-1,n-m1) + Sum_{m2=1..k} a^n*b^(k-m2)*L(m2)*C(n+k-m2-1,k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} a^(i-m1)*b^j*R(m1)*C(i+j-m1-1,i-m1) + Sum_{m2=1..j} a^i*b^(j-m2)*L(m2)*C(i+j-m2-1,j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If a=b=1, then the closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196.
If a=0, then as table read by antidiagonals T(n,k)=b*R(n), as linear sequence a(n)=b*R(i), where i=n-t*(t+1)/2-1, t=floor((-1+sqrt(8*n-7))/2); n>0. The sequence a(n) is the reluctant sequence of sequence b*R(n) - a(n) is triangle array read by rows: row number k coincides with first k elements of the sequence b*R(n). Similarly for b=0, we get T(n,k)=a*L(k).
For this sequence L(m)=m and R(m)=m^2, a=2, b=1. As table read by antidiagonals T(n,k) = Sum_{m1=1..n} 2^(n-m1)*m1^2*C(n+k-m1-1,n-m1) + Sum_{m2=1..k} 2^n*m2*C(n+k-m2-1,k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} 2^(i-m1)*m1^2*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} 2^i*m2*C(i+j-m2-1,j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.

A055580 Björner-Welker sequence: 2^n*(n^2 + n + 2) - 1.

Original entry on oeis.org

1, 7, 31, 111, 351, 1023, 2815, 7423, 18943, 47103, 114687, 274431, 647167, 1507327, 3473407, 7929855, 17956863, 40370175, 90177535, 200278015, 442499071, 973078527, 2130706431, 4647288831, 10099884031, 21877489663
Offset: 0

Views

Author

Wolfdieter Lang, May 26 2000; revised Feb 12 2001

Keywords

Comments

a(n) is the d=1 Betti number of the complement of '3-equal' arrangements in n-dimensional real space, see Björner-Welker reference, Table I, pp. 308-309, column '1' with k=3 and Th. 5.2, pp. 297-298.
Binomial transform of [1/2, 2/3, 3/4, 4/5, ...] = 1/2, 7/6, 31/12, 111/20, 351/30, 1023/42, ..., where 2, 6, 12, 20, ... = A002378 (deleting the zero). - Gary W. Adamson, Apr 28 2005
Number of three-dimensional block structures associated with n joint systems in the construction of stable underground structures. - Richard M. Green, Jul 26 2011
Number of monotone mappings from the chain with three points to the complete binary tree of height n (n+1 levels). For example, the seven monotone mappings from the chain with three points (denoted 1,2,3, in order) to the complete binary tree with two levels (with a the root of the tree, and b, c the atoms) are: f(1)=f(2)=f(3)=a; f(1)=f(2)=a, f(3)=b; f(1)=f(2)=a, f(3)=c; f(1)=a, f(2)=f(3)=b; f(1)=a, f(2)=f(3)=c; f(1)=f(2)=f(3)=b; f(1)=f(2)=f(3)=c. - Pietro Codara, Mar 26 2015

References

  • H. Barcelo and S. Smith, The discrete fundamental group of the order complex of B_n, Abstract 1020-05-141, 1020th Meeting Amer. Math. Soc., Cincinatti, Ohio, Oct 21-22, 2006.

Crossrefs

Fourth column of triangle A055252.

Programs

  • Magma
    [2^n*(n^2+n+2)-1: n in [0..35]]; // Vincenzo Librandi, Jul 28 2011
    
  • Mathematica
    Table[ n*(n+1)*2^(n-2), {n, 0, 26}] // Accumulate // Rest (* Jean-François Alcover, Jul 09 2013, after Paul Barry *)
    LinearRecurrence[{7,-18,20,-8},{1,7,31,111},30] (* Harvey P. Dale, Nov 27 2014 *)
  • PARI
    a(n)=(n^2+n+2)<Charles R Greathouse IV, Jul 28 2011

Formula

a(n) = A055252(n+3, 3).
a(n) = Sum_{j=0..n-1} a(j) + A045618(n), n >= 1.
G.f.: 1/((1-2*x)^3*(1-x)).
Partial sums of A001788 (without leading zero). - Paul Barry, Jun 26 2003
a(n) = A001788(n) - A000337(n). - Jon Perry, Dec 12 2003
a(n) = A119258(n+4,n). - Reinhard Zumkeller, May 11 2006
E.g.f.: 2*(1 + 2*x + 2*x^2)*exp(2*x) - exp(x). - G. C. Greubel, Oct 28 2016
a(n) = Sum_{k=0..n+1} Sum_{i=0..n+1} i^2 * C(k,i). - Wesley Ivan Hurt, Sep 21 2017

Extensions

Edited (for consistency with change of offset) by M. F. Hasler, Nov 03 2012

A112857 Triangle T(n,k) read by rows: number of Green's R-classes in the semigroup of order-preserving partial transformations (of an n-element chain) consisting of elements of height k (height(alpha) = |Im(alpha)|).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 15, 17, 7, 1, 1, 31, 49, 31, 9, 1, 1, 63, 129, 111, 49, 11, 1, 1, 127, 321, 351, 209, 71, 13, 1, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1
Offset: 0

Views

Author

Abdullahi Umar, Aug 25 2008

Keywords

Comments

Sum of rows of T(n, k) is A007051; T(n,k) = |A118801(n,k)|.
Row-reversed variant of A119258. - R. J. Mathar, Jun 20 2011
Pairwise sums of row terms starting from the right yields triangle A038207. - Gary W. Adamson, Feb 06 2012
Riordan array (1/(1 - x), x/(1 - 2*x)). - Philippe Deléham, Jan 17 2014
Appears to coincide with the triangle T(n,m) (n >= 1, 1 <= m <= n) giving number of set partitions of [n], avoiding 1232, with m blocks [Crane, 2015]. See also A250118, A250119. - N. J. A. Sloane, Nov 25 2014
(A007318)^2 = A038207 = T*|A167374|. See A118801 for other relations to the Pascal matrix. - Tom Copeland, Nov 17 2016

Examples

			T(3,2) = 5 because in a regular semigroup of transformations the Green's R-classes coincide with convex partitions of subsets of {1,2,3} with convex classes (modulo the subsets): {1}, {2}/{1}, {3}/{2}, {3}/{1,2}, {3}/{1}, {2,3}
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,    1;
  1,    3,    1;
  1,    7,    5,    1;
  1,   15,   17,    7,    1;
  1,   31,   49,   31,    9,    1;
  1,   63,  129,  111,   49,   11,    1;
  1,  127,  321,  351,  209,   71,   13,   1;
  1,  255,  769, 1023,  769,  351,   97,  15,   1;
  1,  511, 1793, 2815, 2561, 1471,  545, 127,  17,  1;
  1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1;
  ...
As to matrix M, top row of M^3 = (1, 7, 5, 1, 0, 0, 0, ...)
		

Crossrefs

Programs

  • Maple
    A112857 := proc(n,k) if k=0 or k=n then 1; elif k <0 or k>n then 0; else 2*procname(n-1,k)+procname(n-1,k-1) ; end if; end proc: # R. J. Mathar, Jun 20 2011
  • Mathematica
    Table[Abs[1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]] - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

Formula

T(n,k) = Sum_{j = k..n} C(n,j)*C(j-1,k-1).
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) for n >= 2 and 1 <= k <= n-1 with T(n,0) = 1 = T(n,n) for n >= 0.
n-th row = top row of M^n, deleting the zeros, where M is an infinite square production matrix with (1,1,1,...) as the superdiagonal and (1,2,2,2,...) as the main diagonal. - Gary W. Adamson, Feb 06 2012
From Peter Bala, Mar 05 2018 (Start):
The following remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,2*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(2*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,2*x). For example, when n = 3 we have exp(x)*(1 + 7*(2*x) + 5*(2*x)^2/2! + (2*x)^3/3!) = 1 + 15*x + 49*x^2/2! + 111*x^3/3! + 209*x^4/4! + ....
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then P(n,x) is the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = x^4 + 15*x^3 + 17*x^2 + 7*x + 1 + O(x^5).
See A118801 for a signed version of this triangle and A145661 for a signed version of the row reversed triangle. (End)
Bivariate o.g.f.: Sum_{n,k>=0} T(n,k)*x^n*y^k = (1 - 2*x)/((1 - x)*(1 - 2*x - x*y)). - Petros Hadjicostas, Feb 14 2021
The matrix inverse of the Lucas triangle A029635 is -T(n, k)/(-2)^(n-k+1). - Peter Luschny, Dec 22 2024
Showing 1-10 of 20 results. Next