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.

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A141324 Sum of digits of A002452(n).

Original entry on oeis.org

0, 1, 1, 10, 10, 19, 19, 37, 28, 37, 37, 37, 37, 46, 73, 55, 55, 64, 73, 73, 64, 82, 73, 109, 100, 118, 91, 109, 109, 118, 127, 127, 109, 136, 145, 127, 145, 136, 145, 163, 145, 154, 172, 190, 127, 181, 199, 208, 217, 190, 181, 235, 235, 253, 226, 217, 226, 235, 262
Offset: 0

Views

Author

Paul Curtz, Aug 03 2008

Keywords

Crossrefs

Cf. A017173.

Programs

  • Maple
    A002452 := proc(n) (9^n-1)/8 ; end: A007953 := proc(n) local i ; add(i,i=convert(n,base,10)) ; end: A141324 := proc(n) A007953(A002452(n)) ; end: for n from 0 to 80 do printf("%d,",A141324(n)) ; od: # R. J. Mathar, Aug 09 2008
  • Mathematica
    Total[IntegerDigits[#]]&/@LinearRecurrence[{10,-9},{0,1},60] (* Harvey P. Dale, Sep 23 2018 *)

Formula

a(n)=A007953(A002452(n)).

Extensions

Extended by R. J. Mathar, Aug 09 2008

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

A003463 a(n) = (5^n - 1)/4.

Original entry on oeis.org

0, 1, 6, 31, 156, 781, 3906, 19531, 97656, 488281, 2441406, 12207031, 61035156, 305175781, 1525878906, 7629394531, 38146972656, 190734863281, 953674316406, 4768371582031, 23841857910156, 119209289550781, 596046447753906, 2980232238769531
Offset: 0

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Author

Keywords

Comments

5^a(n) is the highest power of 5 dividing (5^n)!. - Benoit Cloitre, Feb 04 2002
n such that A002294(n) is not divisible by 5. - Benoit Cloitre, Jan 14 2003
Without leading zero, i.e., sequence {a(n+1) = (5*5^n-1)/4}, this is the binomial transform of A003947. - Paul Barry, May 19 2003 [Edited by M. F. Hasler, Oct 31 2014]
Numbers n such that a(n) is prime are listed in A004061(n) = {3, 7, 11, 13, 47, 127, 149, 181, 619, 929, ...}. Corresponding primes a(n) are listed in A086122(n) = {31, 19531, 12207031, 305175781, 177635683940025046467781066894531, ...}. 3^(m+1) divides a(2*3^m*k). 31 divides a(3k). 13 divides a(4k). 11 divides a(5k). 71 divides a(5k). 7 divides a(6k). 19531 divides a(7k). 313 divides a(8k). 19 divides a(9k). 829 divides a(9k). 71 divides a(10k). 521 divides a(10k). 17 divides a(16k). p divides a(p-1) for all prime p except p = {2,5}. p^(m+1) divides a(p^m*(p-1)) for all prime p except p = {2,5}. p divides a((p-1)/2) for prime p = {11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, ...} = A045468, Primes congruent to {1, 4} mod 5. p divides a((p-1)/3) for prime p = {13, 67, 127, 163, 181, 199, 211, 241, 313, 337, 367, 379, 409, 457, ...}. p divides a((p-1)/4) for prime p = {101, 109, 149, 181, 269, 389, 401, 409, 449, 461, 521, 541, ...} = A107219, Primes of the form x^2+100y^2. p divides a((p-1)/5) for prime p = {31, 191, 251, 271, 601, 641, 761, 1091, 1861, ...}. p divides a((p-1)/6) for prime p = {181, 199, 211, 241, 379, 409, 631, 691, 739, 769, 1039, ...}. - Alexander Adamchuk, Jan 23 2007
Starting with 1 = convolution square of A026375: (1, 3, 11, 45, 195, 873, ...). - Gary W. Adamson, May 17 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)=det(A). - Milan Janjic, Jan 27 2010
This is the sequence A(0,1;4,5;2) = A(0,1;6,-5;0) of the family of sequences [a,b:c,d:k] considered by Gary Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
It is the Lucas sequence U(6,5). - Felix P. Muga II, Mar 21 2014
a(2*n+1) is the sum of the numerators and denominators of the reduced fractions 0 < b/5^n < 1 plus 1, with b < 5^n. - J. M. Bergot, Jul 24 2015
The sequence multiplied by 10 (0, 10, 60, 310, 1560, ...) is the maximum number of coins which can be decided by n weighings on 2 balances in the counterfeit coin problem with undecided under/overweight. [Halbeisen and Hungerbuhler, Disc. Math. 147 (1995) 139 Theorem 1]. - R. J. Mathar, Sep 10 2015
Order of the rank-n projective geometry PG(n-1,5) over the finite field GF(5). - Anthony Hernandez, Oct 05 2016
Number of zeros in the substitution system {0 -> 11100, 1 -> 11110} at step n from initial string "1" (1 -> 11110 -> 1111011110111101111011100 -> ...). - Ilya Gutkovskiy, Apr 10 2017
a(n) is the numerator of Sum_{k=1..n} 1/5^k, which approaches a limit of 1/4. The denominators are 5^n. In general, Sum_{k=1..n} 1/x^k approaches a limit of 1/(x-1). It is of interest to note that as x increases, so does the rate of convergence. See Crossrefs for numerators for other values of x which have the general form (x^n-1)/(x-1). - Gary Detlefs, Aug 31 2021

Examples

			Base 5...........decimal
0......................0
1......................1
11.....................6
111...................31
1111.................156
11111................781
111111..............3906
1111111............19531
11111111...........97656
111111111.........488281
1111111111.......2441406
etc. ...............etc.
- _Zerinvary Lajos_, Jan 14 2007
		

References

  • Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
  • 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 282.

Crossrefs

Programs

Formula

Second binomial transform of A015518; binomial transform of A000302 (preceded by 0). - Paul Barry, Mar 28 2003
a(n) = Sum_{k=1..n} binomial(n,k)*4^(k-1). - Paul Barry, Mar 28 2003
a(n) = (-1)^n times the (i, j)-th element of M^n (for all i and j such that i is not equal to j), where M = ((1, -1, 1, -2), (-1, 1, -2, 1), (1, -2, 1, -1), (-2, 1, -1, 1)). - Simone Severini, Nov 25 2004
a(n) = A125118(n,4) for n>3. - Reinhard Zumkeller, Nov 21 2006
a(n) = ((3+sqrt(4))^n - (3-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008
a(n) = 6*a(n-1) - 5*a(n-2) n>1, a(0)=0, a(1)=1. - Philippe Deléham, Jan 01 2009
From Wolfdieter Lang, Oct 18 2010: (Start)
O.g.f.: x/((1-5*x)*(1-x)).
a(n) = 4*a(n-1) + 5*a(n-2) + 2, a(0)=0, a(1)=1.
a(n) = 5*a(n-1) + a(n-2) - 5*a(n-3) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3), a(0)=0, a(1)=1, a(2)=6. Observation by G. Detlefs. See the W. Lang comment and link. (End)
a(n) = 5*a(n-1) + 1 with n>0, a(0)=0. - Vincenzo Librandi, Nov 17 2010
a(n) = a(n-1) + A000351(n-1) n>0, a(0)=0. - Felix P. Muga II, Mar 19 2014
a(n) = a(n-1) + 20*a(n-2) + 5 for n > 1, a(0)=0, a(1)=1. - Felix P. Muga II, Mar 19 2014
a(n) = A060458(n)/2^(n+2), for n > 0. - R. J. Cano, Sep 25 2014
From Ilya Gutkovskiy, Oct 05 2016: (Start)
E.g.f.: (exp(4*x) - 1)*exp(x)/4.
Convolution of A000351 and A057427. (End)

A015518 a(n) = 2*a(n-1) + 3*a(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

Views

Author

Keywords

Comments

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

References

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

Crossrefs

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.

Programs

  • Magma
    [Round(3^n/4): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011
    
  • Mathematica
    Table[(3^n-(-1)^n)/4,{n,0,30}] (* Alexander Adamchuk, Nov 19 2006 *)
  • Maxima
    a(n):= round(3^n/4)$ /* Dimitri Papadopoulos, Nov 28 2023 */
  • PARI
    a(n)=round(3^n/4)
    
  • Python
    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)]
    

Formula

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)

Extensions

More terms from Emeric Deutsch, Apr 01 2004
Edited by Ralf Stephan, Aug 30 2004

A136630 Triangular array: T(n,k) counts the partitions of the set [n] into k odd sized blocks.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 4, 0, 1, 0, 1, 0, 10, 0, 1, 0, 0, 16, 0, 20, 0, 1, 0, 1, 0, 91, 0, 35, 0, 1, 0, 0, 64, 0, 336, 0, 56, 0, 1, 0, 1, 0, 820, 0, 966, 0, 84, 0, 1, 0, 0, 256, 0, 5440, 0, 2352, 0, 120, 0, 1, 0, 1, 0, 7381, 0, 24970, 0, 5082, 0, 165, 0, 1, 0, 0, 1024, 0, 87296, 0
Offset: 0

Views

Author

Paul D. Hanna, Jan 14 2008

Keywords

Comments

For partitions into blocks of even size see A156289.
Essentially the unsigned matrix inverse of triangle A121408.
From Peter Bala, Jul 28 2014: (Start)
Define a polynomial sequence x_(n) by setting x_(0) = 1 and for n = 1,2,... setting x_(n) = x*(x + n - 2)*(x + n - 4)*...*(x + n - 2*(n - 1)). Then this table is the triangle of connection constants for expressing the monomial polynomials x^n in terms of the basis x_(k), that is, x^n = sum {k = 0..n} T(n,k)*x_(k) for n = 0,1,2,.... An example is given below.
Let M denote the lower unit triangular array A119467 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/ having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle, omitting the first row and column, equals the infinite matrix product M(0)*M(1)*M(2)*.... (End)
Also the Bell transform of A000035(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Triangle begins:
  1;
  0, 1;
  0, 0,   1;
  0, 1,   0,    1;
  0, 0,   4,    0,    1;
  0, 1,   0,   10,    0,     1;
  0, 0,  16,    0,   20,     0,    1;
  0, 1,   0,   91,    0,    35,    0,    1;
  0, 0,  64,    0,  336,     0,   56,    0,   1;
  0, 1,   0,  820,    0,   966,    0,   84,   0,   1;
  0, 0, 256,    0, 5440,     0, 2352,    0, 120,   0, 1;
  0, 1,   0, 7381,    0, 24970,    0, 5082,   0, 165, 0, 1;
T(5,3) = 10. The ten partitions of the set [5] into 3 odd-sized blocks are
(1)(2)(345), (1)(3)(245), (1)(4)(235), (1)(5)(234), (2)(3)(145),
(2)(4)(135), (2)(5)(134), (3)(4)(125), (3)(5)(124), (4)(5)(123).
Connection constants: Row 5 = [0,1,0,10,0,1]. Hence, with the polynomial sequence x_(n) as defined in the Comments section we have x^5 = x_(1) + 10*x_(3) + x_(5) = x + 10*x*(x+1)*(x-1) + x*(x+3)*(x+1)*(x-1)*(x-3).
		

References

  • L. Comtet, Analyse Combinatoire, Presses Univ. de France, 1970, Vol. II, pages 61-62.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 225-226.

Crossrefs

Cf. A121408; A136631 (antidiagonal sums), A003724 (row sums), A136632; A002452 (column 3), A002453 (column 5); A008958 (central factorial triangle), A156289. A185690, A196776.

Programs

  • Maple
    A136630 := proc (n, k) option remember; if k < 0 or n < k then 0 elif k = n then 1 else procname(n-2, k-2) + k^2*procname(n-2, k) end if end proc: seq(seq(A136630(n, k), k = 1 .. n), n = 1 .. 12); # Peter Bala, Jul 27 2014
    # The function BellMatrix is defined in A264428.
    BellMatrix(n -> (n+1) mod 2, 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    t[n_, k_] := Coefficient[ x^k/Product[ 1 - (2*j + k - 2*Quotient[k, 2])^2*x^2, {j, 0, k/2}] + x*O[x]^n, x, n]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 22 2013, after Pari *)
    BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
    rows = 13;
    M = BellMatrix[Mod[#+1, 2]&, rows];
    Table[M[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 23 2018, after Peter Luschny *)
  • PARI
    {T(n,k)=polcoeff(x^k/prod(j=0,k\2,1-(2*j+k-2*(k\2))^2*x^2 +x*O(x^n)),n)}

Formula

G.f. for column k: x^k/Product_{j=0..floor(k/2)} (1 - (2*j + k-2*floor(k/2))^2 * x^2).
G.f. for column 2*k: x^(2*k)/Product_{j=0..k} (1 - (2*j)^2*x^2).
G.f. for column 2*k+1: x^(2*k+1)/Product_{j=0..k} (1 - (2*j+1)^2*x^2).
From Peter Bala, Feb 21 2011 (Start)
T(n,k) = 1/(2^k*k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j-k)^n,
Recurrence relation T(n+2,k) = T(n,k-2) + k^2*T(n,k).
E.g.f.: F(x,z) = exp(x*sinh(z)) = Sum_{n>=0} R(n,x)*z^n/n! = 1 + x*z + x^2*z^2/2! + (x+x^3)*z^3/3! + ....
The row polynomials R(n,x) begin
R(1,x) = x
R(2,x) = x^2
R(3,x) = x+x^3.
The e.g.f. F(x,z) satisfies the partial differential equation d^2/dz^2(F) = x^2*F + x*F' + x^2*F'' where ' denotes differentiation w.r.t. x.
Hence the row polynomials satisfy the recurrence relation R(n+2,x) = x^2*R(n,x) + x*R'(n,x) + x^2*R''(n,x) with R(0,x) = 1.
The recurrence relation for T(n,k) given above follows from this.
(End)
For the corresponding triangle of ordered partitions into odd-sized blocks see A196776. Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of exp(t*M) lists the row polynomials for the present triangle. - Peter Bala, Oct 06 2011
Row generating polynomials equal D^n(exp(x*t)) evaluated at x = 0, where D is the operator sqrt(1+x^2)*d/dx. Cf. A196776. - Peter Bala, Dec 06 2011
From Peter Bala, Jul 28 2014: (Start)
E.g.f.: exp(t*sinh(x)) = 1 + t*x + t^2*x^2/2! + (t + t^3)*x^3/3! + ....
Hockey-stick recurrence: T(n+1,k+1) = Sum_{i = 0..floor((n-k)/2)} binomial(n,2*i)*T(n-2*i,k).
Recurrence equation for the row polynomials R(n,t):
R(n+1,t) = t*Sum_{k = 0..floor(n/2)} binomial(n,2*k)*R(n-2*k,t) with R(0,t) = 1. (End)

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

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Author

Keywords

Comments

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

References

  • 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.

Crossrefs

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).

Programs

  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    a(n) = (2^n - 1)*(2^(n-1) - 1)/3 \\ Charles R Greathouse IV, Jul 25 2011
    
  • PARI
    concat([0, 0], Vec(x^2/((1-x)*(1-2*x)*(1-4*x)) + O(x^50))) \\ Altug Alkan, Nov 12 2015
  • Sage
    [gaussian_binomial(n,2,2) for n in range(0,25)] # Zerinvary Lajos, May 24 2009
    

Formula

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

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Author

Keywords

Comments

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

Examples

			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
		

References

  • 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).

Crossrefs

Programs

  • GAP
    List([0..400], n->Stirling2(n,3)); # Muniru A Asiru, Feb 04 2018
  • Maple
    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
  • Mathematica
    StirlingS2[Range[0,30],3] (* Harvey P. Dale, Dec 29 2011 *)
  • PARI
    {a(n) = 3^(n-1) / 2 - 2^(n-1) + 1/2};
    
  • Sage
    [stirling_number2(i,3) for i in (0..40)] # Zerinvary Lajos, Jun 26 2008
    

Formula

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

Extensions

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

A003951 Expansion of g.f.: (1+x)/(1-8*x).

Original entry on oeis.org

1, 9, 72, 576, 4608, 36864, 294912, 2359296, 18874368, 150994944, 1207959552, 9663676416, 77309411328, 618475290624, 4947802324992, 39582418599936, 316659348799488, 2533274790395904, 20266198323167232, 162129586585337856, 1297036692682702848
Offset: 0

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Keywords

Comments

Coordination sequence for infinite tree with valency 9.
Binomial transform is {1, 10, 91, 820, 7381, ...}, see A002452. - Philippe Deléham, Jul 22 2005
a(n) equals the number of words of length n on alphabet {0,1,...,8} with no two adjacent letters identical. - Milan Janjic, Jan 31 2015 [Corrected by David Nacin, May 31 2017]

Crossrefs

Cf. A003945.

Programs

Formula

a(n) = Sum_{k=0..n} A029653(n, k)*x^k for x = 7. - Philippe Deléham, Jul 10 2005
a(0) = 1; for n>0, a(n) = 9*8^(n-1). - Vincenzo Librandi, Nov 18 2010
a(0) = 1, a(1) = 9, a(n) = 8*a(n-1). - Vincenzo Librandi, Dec 10 2012
E.g.f.: (9*exp(8*x) -1)/8. - G. C. Greubel, Sep 24 2019

Extensions

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

A131865 Partial sums of powers of 16.

Original entry on oeis.org

1, 17, 273, 4369, 69905, 1118481, 17895697, 286331153, 4581298449, 73300775185, 1172812402961, 18764998447377, 300239975158033, 4803839602528529, 76861433640456465, 1229782938247303441, 19676527011956855057, 314824432191309680913, 5037190915060954894609
Offset: 0

Views

Author

Reinhard Zumkeller, Jul 22 2007

Keywords

Comments

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

Examples

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

Crossrefs

Programs

Formula

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

A091030 Partial sums of powers of 13 (A001022).

Original entry on oeis.org

1, 14, 183, 2380, 30941, 402234, 5229043, 67977560, 883708281, 11488207654, 149346699503, 1941507093540, 25239592216021, 328114698808274, 4265491084507563, 55451384098598320, 720867993281778161
Offset: 1

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Author

Wolfdieter Lang, Jan 23 2004

Keywords

Comments

13^a(n) is highest power of 13 dividing (13^n)!.
For analogs with primes 2, 3, 5, 7 and 11 see A000225, A003462, A003463, A023000 and A016123 respectively.
Let A be the Hessenberg matrix of the order n, defined by: A[1,j]=1,A[i,i]:=13, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=14, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=(-1)^(n)*charpoly(A,1). - Milan Janjic, Feb 21 2010

Examples

			For n=6, a(6) = 1*6 + 12*15 + 144*20 + 1728*15 + 20736*6 + 248832*1 = 402234. - _Bruno Berselli_, Nov 12 2015
		

Crossrefs

Programs

Formula

G.f.: x/((1-13*x)*(1-x)) = (1/(1-13*x) - 1/(1-x))/12.
a(n) = Sum_{k=0..n-1} 13^k = (13^n-1)/12.
a(n) = 13*a(n-1)+1 for n>1, a(1)=1. - Vincenzo Librandi, Feb 05 2011
a(n) = Sum_{k=0...n-1} 12^k*binomial(n,n-1-k). - Bruno Berselli, Nov 12 2015
E.g.f.: exp(x)*(exp(12*x) - 1)/12. - Stefano Spezia, Mar 11 2023
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