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 49 results. Next

A241660 Indices of primes in A001630.

Original entry on oeis.org

3, 4, 7, 19, 62, 94, 722, 5197, 5262, 6182, 14007, 21579, 35354, 75592
Offset: 1

Views

Author

Robert Price, Apr 26 2014

Keywords

Comments

a(15) > 2*10^5.

Crossrefs

Programs

  • Mathematica
    a={0,0,1,2}; Print[3]; For[n=4, n<=1000, n++, sum=Plus@@a; If[PrimeQ[sum], Print[n]]; a=RotateLeft[a]; a[[4]]=sum]

Extensions

Prepended a(1)=3 and Mathematica program corrected by Robert Price, Sep 09 2014

A107242 Sum of squares of tetranacci numbers (A001630).

Original entry on oeis.org

0, 0, 1, 5, 14, 50, 194, 723, 2659, 9884, 36780, 136636, 507517, 1885793, 7006962, 26034006, 96728470, 359395319, 1335332919, 4961420008, 18434129192, 68491926888, 254481427113, 945524491213, 3513091674982, 13052875206698
Offset: 0

Views

Author

Jonathan Vos Post, May 18 2005

Keywords

Examples

			a(0) = 0 = 0^2,
a(1) = 0 = 0^2 + 0^2
a(2) = 1 = 0^2 + 0^2 + 1^2
a(3) = 5 = 0^2 + 0^2 + 1^2 + 2^2
a(4) = 14 = 0^2 + 0^2 + 1^2 + 2^2 + 3^2
a(5) = 50 = 0^2 + 0^2 + 1^2 + 2^2 + 3^2 + 6^2
a(6) = 194 = 0^2 + 0^2 + 1^2 + 2^2 + 3^2 + 6^2 + 12^2
a(7) = 723 = 0^2 + 0^2 + 1^2 + 2^2 + 3^2 + 6^2 + 12^2 + 23^2
a(8) = 2659 = 0^2 + 0^2 + 1^2 + 2^2 + 3^2 + 6^2 + 12^2 + 23^2 + 44^2
		

Crossrefs

Programs

  • Mathematica
    Accumulate[LinearRecurrence[{1,1,1,1},{0,0,1,2},40]^2] (* or *) LinearRecurrence[{3,2,2,6,-16,-2,6,-2,2,1,-1},{0,0,1,5,14,50,194,723,2659,9884,36780},40] (* Harvey P. Dale, Aug 25 2013 *)

Formula

a(n) = F_4(1)^2 + F_4(1)^2 + F_4(2)^2 + ... F_4(n)^2 where F_4(n) = A001630(n). a(0) = 0, a(n+1) = a(n) + A001630(n)^2.
a(n)= 3*a(n-1) +2*a(n-2) +2*a(n-3) +6*a(n-4) -16*a(n-5) -2*a(n-6) +6*a(n-7) -2*a(n-8) +2*a(n-9) +a(n-10) -a(n-11). G.f.: x^2*(1+x)*(x^6-x^5-4*x^2+x+1)/((x-1) *(x^4+x^3-3*x^2-3*x+1) *(x^6-x^5+2*x^4-\ 2*x^3-2*x^2-x-1)). [R. J. Mathar, Aug 11 2009]

Extensions

a(13) and a(23) corrected by R. J. Mathar, Aug 11 2009

A241661 Primes in A001630.

Original entry on oeis.org

2, 3, 23, 60217, 108412217573460833, 143003097309669584171480759
Offset: 1

Views

Author

Robert Price, Apr 26 2014

Keywords

Comments

a(7) is too large to display here. It has 206 digits and is the 722nd term in A001630.

Crossrefs

Programs

  • Mathematica
    a={0,0,1,2}; Print[2]; For[n=4, n<=1000, n++, sum=Plus@@a; If[PrimeQ[sum], Print[sum]]; a=RotateLeft[a]; a[[4]]=sum]

Extensions

a(1)=2 prepended and Mathematica program corrected by Robert Price, Sep 09 2014

A007283 a(n) = 3*2^n.

Original entry on oeis.org

3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 1572864, 3145728, 6291456, 12582912, 25165824, 50331648, 100663296, 201326592, 402653184, 805306368, 1610612736, 3221225472, 6442450944, 12884901888
Offset: 0

Views

Author

Keywords

Comments

Same as Pisot sequences E(3, 6), L(3, 6), P(3, 6), T(3, 6). See A008776 for definitions of Pisot sequences.
Numbers k such that A006530(A000010(k)) = A000010(A006530(k)) = 2. - Labos Elemer, May 07 2002
Also least number m such that 2^n is the smallest proper divisor of m which is also a suffix of m in binary representation, see A080940. - Reinhard Zumkeller, Feb 25 2003
Length of the period of the sequence Fibonacci(k) (mod 2^(n+1)). - Benoit Cloitre, Mar 12 2003
The sequence of first differences is this sequence itself. - Alexandre Wajnberg and Eric Angelini, Sep 07 2005
Subsequence of A122132. - Reinhard Zumkeller, Aug 21 2006
Apart from the first term, a subsequence of A124509. - Reinhard Zumkeller, Nov 04 2006
Total number of Latin n-dimensional hypercubes (Latin polyhedra) of order 3. - Kenji Ohkuma (k-ookuma(AT)ipa.go.jp), Jan 10 2007
Number of different ternary hypercubes of dimension n. - Edwin Soedarmadji (edwin(AT)systems.caltech.edu), Dec 10 2005
For n >= 1, a(n) is equal to the number of functions f:{1, 2, ..., n + 1} -> {1, 2, 3} such that for fixed, different x_1, x_2,...,x_n in {1, 2, ..., n + 1} and fixed y_1, y_2,...,y_n in {1, 2, 3} we have f(x_i) <> y_i, (i = 1,2,...,n). - Milan Janjic, May 10 2007
a(n) written in base 2: 11, 110, 11000, 110000, ..., i.e.: 2 times 1, n times 0 (see A003953). - Jaroslav Krizek, Aug 17 2009
Subsequence of A051916. - Reinhard Zumkeller, Mar 20 2010
Numbers containing the number 3 in their Collatz trajectories. - Reinhard Zumkeller, Feb 20 2012
a(n-1) gives the number of ternary numbers with n digits with no two adjacent digits in common; e.g., for n=3 we have 010, 012, 020, 021, 101, 102, 120, 121, 201, 202, 210 and 212. - Jon Perry, Oct 10 2012
If n > 1, then a(n) is a solution for the equation sigma(x) + phi(x) = 3x-4. This equation also has solutions 84, 3348, 1450092, ... which are not of the form 3*2^n. - Farideh Firoozbakht, Nov 30 2013
a(n) is the upper bound for the "X-ray number" of any convex body in E^(n + 2), conjectured by Bezdek and Zamfirescu, and proved in the plane E^2 (see the paper by Bezdek and Zamfirescu). - L. Edson Jeffery, Jan 11 2014
If T is a topology on a set V of size n and T is not the discrete topology, then T has at most 3 * 2^(n-2) many open sets. See Brown and Stephen references. - Ross La Haye, Jan 19 2014
Comment from Charles Fefferman, courtesy of Doron Zeilberger, Dec 02 2014: (Start)
Fix a dimension n. For a real-valued function f defined on a finite set E in R^n, let Norm(f, E) denote the inf of the C^2 norms of all functions F on R^n that agree with f on E. Then there exist constants k and C depending only on the dimension n such that Norm(f, E) <= C*max{ Norm(f, S) }, where the max is taken over all k-point subsets S in E. Moreover, the best possible k is 3 * 2^(n-1).
The analogous result, with the same k, holds when the C^2 norm is replaced, e.g., by the C^1, alpha norm (0 < alpha <= 1). However, the optimal analogous k, e.g., for the C^3 norm is unknown.
For the above results, see Y. Brudnyi and P. Shvartsman (1994). (End)
Also, coordination sequence for (infinity, infinity, infinity) tiling of hyperbolic plane. - N. J. A. Sloane, Dec 29 2015
The average of consecutive powers of 2 beginning with 2^1. - Melvin Peralta and Miriam Ong Ante, May 14 2016
For n > 1, a(n) is the smallest Zumkeller number with n divisors that are also Zumkeller numbers (A083207). - Ivan N. Ianakiev, Dec 09 2016
Also, for n >= 2, the number of length-n strings over the alphabet {0,1,2,3} having only the single letters as nonempty palindromic subwords. (Corollary 21 in Fleischer and Shallit) - Jeffrey Shallit, Dec 02 2019
Also, a(n) is the minimum link-length of any covering trail, circuit, path, and cycle for the set of the 2^(n+2) vertices of an (n+2)-dimensional hypercube. - Marco Ripà, Aug 22 2022
The known fixed points of maps n -> A163511(n) and n -> A243071(n). [See comments in A163511]. - Antti Karttunen, Sep 06 2023
The finite subsequence a(3), a(4), a(5), a(6) = 24, 48, 96, 192 is one of only two geometric sequences that can be formed with all interior angles (all integer, in degrees) of a simple polygon. The other sequence is a subsequence of A000244 (see comment there). - Felix Huber, Feb 15 2024
A level 1 Sierpiński triangle is a triangle. Level n+1 is formed from three copies of level n by identifying pairs of corner vertices of each pair of triangles. For n>2, a(n-3) is the radius of the level n Sierpiński triangle graph. - Allan Bickle, Aug 03 2024

References

  • Jason I. Brown, Discrete Structures and Their Interactions, CRC Press, 2013, p. 71.
  • T. Ito, Method, equipment, program and storage media for producing tables, Publication number JP2004-272104A, Japan Patent Office (written in Japanese, a(2)=12, a(3)=24, a(4)=48, a(5)=96, a(6)=192, a(7)=384 (a(7)=284 was corrected)).
  • Kenji Ohkuma, Atsuhiro Yamagishi and Toru Ito, Cryptography Research Group Technical report, IT Security Center, Information-Technology Promotion Agency, JAPAN.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Subsequence of the following sequences: A029744, A029747, A029748, A029750, A362804 (after 3), A364494, A364496, A364289, A364291, A364292, A364295, A364497, A364964, A365422.
Essentially same as A003945 and A042950.
Row sums of (5, 1)-Pascal triangle A093562 and of (1, 5) Pascal triangle A096940.
Cf. Latin squares: A000315, A002860, A003090, A040082, A003191; Latin cubes: A098843, A098846, A098679, A099321.

Programs

Formula

G.f.: 3/(1-2*x).
a(n) = 2*a(n - 1), n > 0; a(0) = 3.
a(n) = Sum_{k = 0..n} (-1)^(k reduced (mod 3))*binomial(n, k). - Benoit Cloitre, Aug 20 2002
a(n) = A118416(n + 1, 2) for n > 1. - Reinhard Zumkeller, Apr 27 2006
a(n) = A000079(n) + A000079(n + 1). - Zerinvary Lajos, May 12 2007
a(n) = A000079(n)*3. - Omar E. Pol, Dec 16 2008
From Paul Curtz, Feb 05 2009: (Start)
a(n) = b(n) + b(n+3) for b = A001045, A078008, A154879.
a(n) = abs(b(n) - b(n+3)) with b(n) = (-1)^n*A084247(n). (End)
a(n) = 2^n + 2^(n + 1). - Jaroslav Krizek, Aug 17 2009
a(n) = A173786(n + 1, n) = A173787(n + 2, n). - Reinhard Zumkeller, Feb 28 2010
A216022(a(n)) = 6 and A216059(a(n)) = 7, for n > 0. - Reinhard Zumkeller, Sep 01 2012
a(n) = (A000225(n) + 1)*3. - Martin Ettl, Nov 11 2012
E.g.f.: 3*exp(2*x). - Ilya Gutkovskiy, May 15 2016
A020651(a(n)) = 2. - Yosu Yurramendi, Jun 01 2016
a(n) = sqrt(A014551(n + 1)*A014551(n + 2) + A014551(n)^2). - Ezhilarasu Velayutham, Sep 01 2019
a(A048672(n)) = A225546(A133466(n)). - Michel Marcus and Peter Munn, Nov 29 2019
Sum_{n>=1} 1/a(n) = 2/3. - Amiram Eldar, Oct 28 2020

A000078 Tetranacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) for n >= 4 with a(0) = a(1) = a(2) = 0 and a(3) = 1.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 4, 8, 15, 29, 56, 108, 208, 401, 773, 1490, 2872, 5536, 10671, 20569, 39648, 76424, 147312, 283953, 547337, 1055026, 2033628, 3919944, 7555935, 14564533, 28074040, 54114452, 104308960, 201061985, 387559437, 747044834, 1439975216, 2775641472
Offset: 0

Views

Author

Keywords

Comments

a(n) is the number of compositions of n-3 with no part greater than 4. Example: a(7) = 8 because we have 1+1+1+1 = 2+1+1 = 1+2+1 = 3+1 = 1+1+2 = 2+2 = 1+3 = 4. - Emeric Deutsch, Mar 10 2004
In other words, a(n) is the number of ways of putting stamps in one row on an envelope using stamps of denominations 1, 2, 3 and 4 cents so as to total n-3 cents [Pólya-Szegő]. - N. J. A. Sloane, Jul 28 2012
a(n+4) is the number of 0-1 sequences of length n that avoid 1111. - David Callan, Jul 19 2004
a(n) is the number of matchings in the graph obtained by a zig-zag triangulation of a convex (n-3)-gon. Example: a(8) = 15 because in the triangulation of the convex pentagon ABCDEA with diagonals AD and AC we have 15 matchings: the empty set, seven singletons and {AB,CD}, {AB,DE}, {BC,AD}, {BC,DE}, {BC,EA}, {CD,EA} and {DE,AC}. - Emeric Deutsch, Dec 25 2004
Number of permutations satisfying -k <= p(i)-i <= r, i=1..n-3, with k = 1, r = 3. - Vladimir Baltic, Jan 17 2005
For n >= 0, a(n+4) is the number of palindromic compositions of 2*n+1 into an odd number of parts that are not multiples of 4. In addition, a(n+4) is also the number of Sommerville symmetric cyclic compositions (= bilaterally symmetric cyclic compositions) of 2*n+1 into an odd number of parts that are not multiples of 4. - Petros Hadjicostas, Mar 10 2018
a(n) is the number of ways to tile a hexagonal double-strip (two rows of adjacent hexagons) containing (n-4) cells with hexagons and double-hexagons (two adjacent hexagons). - Ziqian Jin, Jul 28 2019
The term "tetranacci number" was coined by Mark Feinberg (1963; see A000073). - Amiram Eldar, Apr 16 2021
a(n) is the number of ways to tile a skew double-strip of n-3 cells using squares and all possible "dominos", as seen in Ziqian Jin's article, below. Here is the skew double-strip corresponding to n=15, with 12 cells:
_ ___ _ ___ _ ___
| | | | | | |
|__|___|_|___| |___|
| | | | | | |
|_|___|_|___|_|___|,
and here are the three possible "domino" tiles:
_ _
| | | |
| | | | | |
|_|, |_|, |_____|.
As an example, here is one of the a(15) = 1490 ways to tile the skew double-strip of 12 cells:
_ ___ _____ _______
| | | | | | |
|__|_ |_|_ | _| _|
| | | | | |
|_____|___|_|___|_|. - Greg Dresden, Jun 05 2024

Examples

			From _Petros Hadjicostas_, Mar 10 2018: (Start)
For n = 3, we get a(3+4) = a(7) = 8 palindromic compositions of 2*n+1 = 7 into an odd number of parts that are not a multiple of 4. They are the following: 7 = 1+5+1 = 3+1+3 = 2+3+2 = 1+2+1+2+1 = 2+1+1+1+2 = 1+1+3+1+1 = 1+1+1+1+1+1+1. If we put these compositions on a circle, they become bilaterally symmetric cyclic compositions of 2*n+1 = 7.
For n = 4, we get a(4+4) = a(8) = 15 palindromic compositions of 2*n + 1 = 9 into an odd number of parts that are not a multiple of 4. They are the following: 9 = 3+3+3 = 2+5+2 = 1+7+1 = 1+1+5+1+1 = 2+1+3+1+2 = 1+2+3+2+1 = 1+3+1+3+1 = 3+1+1+1+3 = 2+2+1+2+2 = 2+1+1+1+1+1+2 = 1+2+1+1+1+2+1 = 1+1+2+1+2+1+1 = 1+1+1+3+1+1+1 = 1+1+1+1+1+1+1+1+1.
As _David Callan_ points out in the comments above, for n >= 1, a(n+4) is also the number of 0-1 sequences of length n that avoid 1111. For example, for n = 5, a(5+4) = a(9) = 29 is the number of binary strings of length n that avoid 1111. Out of the 2^5 = 32 binary strings of length n = 5, the following do not avoid 1111: 11111, 01111, and 11110. (End)
		

References

  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, NY, 2 vols., 1972, Vol. 1, p. 1, Problems 3 and 4.
  • J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, Princeton, NJ, 1978.
  • 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

Row 4 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
First differences are in A001631.
Cf. A008287 (quadrinomial coefficients) and A073817 (tetranacci with different initial conditions).

Programs

  • GAP
    a:=[0,0,0,1];; for n in [5..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]+a[n-4]; od; a; # Muniru A Asiru, Mar 11 2018
  • Haskell
    import Data.List (tails, transpose)
    a000078 n = a000078_list !! n
    a000078_list = 0 : 0 : 0 : f [0,0,0,1] where
       f xs = y : f (y:xs) where
         y = sum $ head $ transpose $ take 4 $ tails xs
    -- Reinhard Zumkeller, Jul 06 2014, Apr 28 2011
    
  • Magma
    [n le 4 select Floor(n/4) else Self(n-1)+Self(n-2)+Self(n-3)+Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jan 29 2016
    
  • Maple
    a:= n-> (<<1|1|0|0>, <1|0|1|0>, <1|0|0|1>, <1|0|0|0>>^n)[1, 4]: seq(a(n), n=0..50); # Alois P. Heinz, Jun 12 2008
  • Mathematica
    CoefficientList[Series[x^3/(1-x-x^2-x^3-x^4), {x, 0, 50}], x]
    LinearRecurrence[{1,1,1,1}, {0,0,0,1}, 50]  (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
    (* From Eric W. Weisstein, Nov 09 2017 *)
    Table[RootSum[-1 -# -#^2 -#^3 +#^4 &, 10#^n +157#^(n+1) -103 #^(n+2) +16#^(n+3) &]/563, {n, 0, 40}]
    Table[RootSum[#^4 -#^3 -#^2 -# -1 &, #^(n-2)/(-#^3 +6# -1) &], {n, 0, 40}] (* End *)
  • Maxima
    a(n):=sum(sum(if mod(5*k-i,4)>0 then 0 else binomial(k,(5*k-i)/4)*(-1)^((i-k)/4)*binomial(n-i+k-1,k-1),i,k,n),k,1,n); /* Vladimir Kruchinin, Aug 18 2010 */
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( x^3 / (1 - x - x^2 - x^3 - x^4) + x * O(x^n), n))}
    
  • Python
    A000078 = [0,0,0,1]
    for n in range(4, 100):
        A000078.append(A000078[n-1]+A000078[n-2]+A000078[n-3]+A000078[n-4])
    # Chai Wah Wu, Aug 20 2014
    

Formula

a(n) = A001630(n) - a(n-1). - Henry Bottomley
G.f.: x^3/(1 - x - x^2 - x^3 - x^4). - Simon Plouffe in his 1992 dissertation
G.f.: x^3 / (1 - x / (1 - x / (1 + x^3 / (1 + x / (1 - x / (1 + x)))))). - Michael Somos, May 12 2012
G.f.: Sum_{n >= 0} x^(n+3) * (Product_{k = 1..n} (k + k*x + k*x^2 + x^3)/(1 + k*x + k*x^2 + k*x^3)). - Peter Bala, Jan 04 2015
a(n) = term (1,4) in the 4 X 4 matrix [1,1,0,0; 1,0,1,0; 1,0,0,1; 1,0,0,0]^n. - Alois P. Heinz, Jun 12 2008
Another form of the g.f.: f(z) = (z^3 - z^4)/(1 - 2*z + z^5), then a(n) = Sum_{i=0..floor((n-3)/5)} (-1)^i*binomial(n-3-4*i, i)*2^(n - 3 - 5*i) - Sum_{i=0..floor((n-4)/5)} (-1)^i*binomial(n-4-4*i, i)*2^(n - 4 - 5*i) with natural convention Sum_{i=m..n} alpha(i) = 0 for m > n. - Richard Choulet, Feb 22 2010
a(n+3) = Sum_{k=1..n} Sum_{i=k..n} [(5*k-i mod 4) = 0] * binomial(k, (5*k-i)/4) *(-1)^((i-k)/4) * binomial(n-i+k-1,k-1), n > 0. - Vladimir Kruchinin, Aug 18 2010 [Edited by Petros Hadjicostas, Jul 26 2020, so that the formula agrees with the offset of the sequence]
Sum_{k=0..3*n} a(k+b) * A008287(n,k) = a(4*n+b), b >= 0 ("quadrinomial transform"). - N. J. A. Sloane, Nov 10 2010
G.f.: x^3*(1 + x*(G(0)-1)/(x+1)) where G(k) = 1 + (1+x+x^2+x^3)/(1-x/(x+1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 26 2013
Starting (1, 2, 4, 8, ...) = the INVERT transform of (1, 1, 1, 1, 0, 0, 0, ...). - Gary W. Adamson, May 13 2013
a(n) ~ c*r^n, where c = 0.079077767399388561146007... and r = 1.92756197548292530426195... = A086088 (One of the roots of the g.f. denominator polynomial is 1/r.). - Fung Lam, Apr 29 2014
a(n) = 2*a(n-1) - a(n-5), n >= 5. - Bob Selcoe, Jul 06 2014
From Ziqian Jin, Jul 28 2019: (Start)
a(2*n+5) = a(n+4)^2 + a(n+3)^2 + a(n+2)^2 + 2*a(n+3)*(a(n+2) + a(n+1)).
a(n) - 1 = a(n-2) + 2*a(n-3) + 3*(a(n-4) + a(n-5) + ... + a(2) + a(1)), n >= 4. (End)
a(n) = (Sum_{i=0..n-1} a(i)*A073817(n-i))/(n-3) for n > 3. - Greg Dresden and Advika Srivastava, Sep 28 2019
a(n) = Sum_{r root of x^4-x^3-x^2-x-1} r^n/(4*r^3-3*r^2-2*r-1). - Fabian Pereyra, Dec 06 2024

Extensions

Definition augmented (with 4 initial terms) by Daniel Forgues, Dec 02 2009
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A179070 a(1)=a(2)=a(3)=1, a(4)=3; thereafter a(n) = a(n-1) + a(n-3).

Original entry on oeis.org

1, 1, 1, 3, 4, 5, 8, 12, 17, 25, 37, 54, 79, 116, 170, 249, 365, 535, 784, 1149, 1684, 2468, 3617, 5301, 7769, 11386, 16687, 24456, 35842, 52529, 76985, 112827, 165356, 242341, 355168, 520524, 762865, 1118033, 1638557, 2401422, 3519455, 5158012, 7559434
Offset: 1

Views

Author

Mark Dols, Jun 27 2010

Keywords

Comments

Also (essentially), coordination sequence for (2,4,infinity) tiling of hyperbolic plane. - N. J. A. Sloane, Dec 29 2015
Column sums of shifted (1,2) Pascal array:
1 1 1 1 1 1 1 1 1
......2 3 4 5 6 7
............2 5 9
.................
----------------- +
1 1 1 3 4 5 8 ...
a(n+1) is the number of multus bitstrings of length n with no runs of 2 0's. - Steven Finch, Mar 25 2020
From Areebah Mahdia and Greg Dresden, Jun 13 2020: (Start)
For n >= 5, a(n) gives the number of ways to tile the following board of length n-3 with squares and trominos:
.
|||
|||_ _ _
|||_|||_|_| ... . (End)

Crossrefs

Programs

Formula

a(n) = A000930(n-1) + A000930(n-4).
G.f.: x - x^2*(1+2*x^2) / ( -1+x+x^3 ). - R. J. Mathar, Oct 30 2011
a(n) = A000930(n-2)+2*A000930(n-4) for n>3. - R. J. Mathar, May 19 2024

Extensions

Simpler definition from N. J. A. Sloane, Aug 29 2013

A073817 Tetranacci numbers with different initial conditions: a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) starting with a(0)=4, a(1)=1, a(2)=3, a(3)=7.

Original entry on oeis.org

4, 1, 3, 7, 15, 26, 51, 99, 191, 367, 708, 1365, 2631, 5071, 9775, 18842, 36319, 70007, 134943, 260111, 501380, 966441, 1862875, 3590807, 6921503, 13341626, 25716811, 49570747, 95550687, 184179871, 355018116, 684319421, 1319068095, 2542585503
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Aug 12 2002

Keywords

Comments

These tetranacci numbers follow the same pattern as Lucas and generalized tribonacci(A001644) numbers: Binet's formula is a(n) = r1^n + r2^n + r3^n + r4^n, with r1, r2, r3, r4 roots of the characteristic polynomial.
For n >= 4, a(n) is the number of cyclic sequences consisting of n zeros and ones that do not contain four consecutive ones provided the positions of the zeros and ones are fixed on a circle. This is proved in Charalambides (1991) and Zhang and Hadjicostas (2015). For example, a(4)=15 because only the sequences 1110, 1101, 1011, 0111, 0011, 0101, 1001, 1010, 0110, 1100, 0001, 0010, 0100, 1000, 0000 avoid four consecutive ones on a circle. (For n=1,2,3 the statement is still true provided we allow the sequence to wrap around itself on a circle. For example, a(2)=3 because only the sequences 00, 01, 10 avoid four consecutive ones when wrapped around on a circle.) - Petros Hadjicostas, Dec 18 2016

Crossrefs

Cf. A000078, A001630, A001644, A000032, A106295 (Pisano periods). Two other versions: A001648, A074081.

Programs

  • GAP
    a:=[4,1,3,7];; for n in [5..40] do a[n]:=a[n-1]+a[n-2]+a[n-3] +a[n-4]; od; a; # G. C. Greubel, Feb 19 2019
  • Magma
    I:=[4,1,3,7]; [n le 4 select I[n] else Self(n-1) +Self(n-2) +Self(n-3) +Self(n-4): n in [1..40]]; // G. C. Greubel, Feb 19 2019
    
  • Mathematica
    a[0]=4; a[1]=1; a[2]=3; a[3]=7; a[4]=15; a[n_]:= 2*a[n-1] -a[n-5]; Array[a, 34, 0]
    CoefficientList[Series[(4-3x-2x^2-x^3)/(1-x-x^2-x^3-x^4), {x, 0, 40}], x]
    LinearRecurrence[{1,1,1,1},{4,1,3,7},40] (* Harvey P. Dale, Jun 01 2015 *)
  • PARI
    Vec((4-3*x-2*x^2-x^3)/(1-x-x^2-x^3-x^4) + O(x^40)) \\ Michel Marcus, Jan 29 2016
    
  • Sage
    ((4-3*x-2*x^2-x^3)/(1-x-x^2-x^3-x^4)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Feb 19 2019
    

Formula

G.f.: (4 - 3*x - 2*x^2 - x^3)/(1 - x - x^2 - x^3 - x^4).
a(n) = 2*a(n-1) - a(n-5), with a(0)=4, a(1)=1, a(2)=3, a(3)=7, a(4)=15. - Vincenzo Librandi, Dec 20 2010
a(n) = A000078(n+2) + 2*A000078(n+1) + 3*A000078(n) + 4*A000078(n-1). - Advika Srivastava, Aug 22 2019
a(n) = 8*a(n-3) - a(n-5) - 2*a(n-6) - 4*a(n-7). - Advika Srivastava, Aug 25 2019
a(n) = Trace(M^n), where M is the 4 X 4 matrix [0, 0, 0, 1; 1, 0, 0, 1; 0, 1, 0, 1; 0, 0, 1, 1], the companion matrix to the monic polynomial x^4 - x^3 - x^2 - x - 1. It follows that the sequence satisfies the Gauss congruences: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for positive integers n and r and all primes p. See Zarelua. - Peter Bala, Dec 31 2022
a(n) = Sum_{r root of x^4-x^3-x^2-x-1} r^n. - Fabian Pereyra, Feb 28 2025

Extensions

Typo in definition corrected by Vincenzo Librandi, Dec 20 2010

A054886 Layer counting sequence for hyperbolic tessellation by cuspidal triangles of angles (Pi/3,Pi/3,0) (this is the classical modular tessellation).

Original entry on oeis.org

1, 3, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050, 242786, 392836, 635622, 1028458, 1664080, 2692538, 4356618, 7049156, 11405774, 18454930, 29860704, 48315634, 78176338, 126491972
Offset: 0

Views

Author

Paolo Dominici (pl.dm(AT)libero.it), May 23 2000

Keywords

Comments

The layer sequence is the sequence of the cardinalities of the layers accumulating around a ( finite-sided ) polygon of the tessellation under successive side-reflections; see the illustration accompanying A054888.
Equivalently, coordination sequence for (3,3,infinity) tiling of hyperbolic plane. - N. J. A. Sloane, Dec 29 2015
Equivalently, spherical growth series for modular group.
Also, number of sequences of length n with terms 1, 2, and 3, with no adjacent terms equal, and no three consecutive terms (1, 2, 3) or (3, 2, 1). - Pontus von Brömssen, Jan 03 2022

References

  • P. de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 156.

Crossrefs

Essentially the same as A006355.

Programs

  • Mathematica
    Join[{1,3},2Fibonacci[Range[4,40]]] (* Harvey P. Dale, Jan 06 2012 *)
  • PARI
    my(x='x+O('x^50)); Vec((1+2*x+2*x^2+x^3)/(1-x-x^2)) \\ G. C. Greubel, Aug 06 2017

Formula

G.f.: (1+2*x+2*x^2+x^3)/(1-x-x^2) = (x^2+x+1)*(1+x)/(1-x-x^2).
a(n) = 2*F(n+2) for n >= 2, with F(n) the n-th Fibonacci number (cf. A000045).
E.g.f.: 2*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5 - 1 - x. - Stefano Spezia, Apr 18 2022

Extensions

Offset changed to 0 by N. J. A. Sloane, Jan 03 2022 at the suggestion of Pontus von Brömssen

A074048 Pentanacci numbers with initial conditions a(0)=5, a(1)=1, a(2)=3, a(3)=7, a(4)=15.

Original entry on oeis.org

5, 1, 3, 7, 15, 31, 57, 113, 223, 439, 863, 1695, 3333, 6553, 12883, 25327, 49791, 97887, 192441, 378329, 743775, 1462223, 2874655, 5651423, 11110405, 21842481, 42941187, 84420151, 165965647, 326279871, 641449337, 1261056193, 2479171199, 4873922247
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Aug 14 2002

Keywords

Comments

These pentanacci numbers follow the same pattern as Lucas, generalized tribonacci(A001644) and generalized tetranacci (A073817) numbers: Binet's formula is a(n)=r1^n+r^2^n+r3^n+r4^n+r5^n, with r1, r2, r3, r4, r5 roots of the characteristic polynomial. a(n) is also the trace of A^n, where A is the pentamatrix ((1,1,0,0,0),(1,0,1,0,0),(1,0,0,1,0),(1,0,0,0,1),(1,0,0,0,0)).
For n >= 5, a(n) is the number of cyclic sequences consisting of n zeros and ones that do not contain five consecutive ones provided the positions of the zeros and ones are fixed on a circle. This is proved in Charalambides (1991) and Zhang and Hadjicostas (2015). (For n=1,2,3,4 the statement is still true provided we allow the sequence to wrap around itself on a circle). - Petros Hadjicostas, Dec 18 2016
a(3407) has 1001 decimal digits. - Michael De Vlieger, Dec 28 2016

Crossrefs

Cf. A000078, A001630, A001644, A000032, A073817, A106297 (Pisano Periods).
Essentially the same as A023424.
Cf. A106273.

Programs

  • Mathematica
    CoefficientList[Series[(5-4*x-3*x^2-2*x^3-x^4)/(1-x-x^2-x^3-x^4-x^5), {x, 0, 30}], x]
    LinearRecurrence[{1, 1, 1, 1, 1}, {5, 1, 3, 7, 15}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)
  • PARI
    polsym(polrecip(1-x-x^2-x^3-x^4-x^5),33) \\ Joerg Arndt, Jan 28 2019

Formula

a(n) = a(n-1) +a(n-2) +a(n-3) +a(n-4) +a(n-5).
G.f.: (5-4*x-3*x^2-2*x^3-x^4) / (1-x-x^2-x^3-x^4-x^5).
a(n) = 2*a(n-1) -a(n-6), n>5. [Vincenzo Librandi, Dec 20 2010]
For k>0 and n>=0, a(n+5*k) = a(k)*a(n+4*k) - A123127(k-1)*a(n+3*k) + A123126(k-1)*a(n+2*k) - A074062(k)*a(n+k) + a(n). For example, if k=4, n=3, we have a(n+5*k) = a(23) = 5651423, a(4)*a(19) - A123127(3)*a(15) + A123126(3)*a(1695) - A074062(4)*a(7) + a(3) = (15)*(378329) - (1)*(25327) + (1)*(1695) - (-1)*(113) + (7) = 5651423. - Kai Wang, Sep 13 2020
From Kai Wang, Dec 16 2020: (Start)
For k >= 0,
| a(k+4) a(k+5) a(k+6) a(k+7) a(k+8) |
| a(k+3) a(k+4) a(k+5) a(k+6) a(k+7) |
det | a(k+2) a(k+3) a(k+4) a(k+5) a(k+6) | = 9584 = A106273(5).
| a(k+1) a(k+2) a(k+3) a(k+4) a(k+5) |
| a(k) a(k+1) a(k+2) a(k+3) a(k+4) |
(End)

A079262 Octanacci numbers: a(0)=a(1)=...=a(6)=0, a(7)=1; for n >= 8, a(n) = Sum_{i=1..8} a(n-i).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 255, 509, 1016, 2028, 4048, 8080, 16128, 32192, 64256, 128257, 256005, 510994, 1019960, 2035872, 4063664, 8111200, 16190208, 32316160, 64504063, 128752121, 256993248, 512966536, 1023897200, 2043730736
Offset: 0

Views

Author

Michael Joseph Halm, Feb 04 2003

Keywords

Comments

a(n+7) is the number of compositions of n into parts <= 8. - Joerg Arndt, Sep 24 2020

Examples

			a(16) = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255.
		

Crossrefs

Row 8 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
Cf. A253706, A253705. Primes and indices of primes in this sequence.

Programs

  • Maple
    for j from 0 to 6 do a[j]:=0 od: a[7]:=1: for n from 8 to 45 do a[n]:=sum(a[n-i],i=1..8) od:seq(a[n],n=0..45); # Emeric Deutsch, Apr 16 2005
  • Mathematica
    LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
    With[{nn=8},LinearRecurrence[Table[1,{nn}],Join[Table[0,{nn-1}],{1}],50]] (* Harvey P. Dale, Aug 17 2013 *)

Formula

G.f.: x^7/(1 - x - x^2 - x^3 - x^4 - x^5 - x^6 - x^7 - x^8). - Emeric Deutsch, Apr 16 2005
a(1)..a(9) = 1, 1, 2, 4, 8, 16, 32, 64, 128. a(10) and following are given by 63*2^(n-8)+(1/2+sqrt(5/4))^(n-6)/sqrt(5)-(1/2-sqrt(5/4))^(n-6)/sqrt(5). Offset 10. a(10)=255. - Al Hakanson (hawkuu(AT)gmail.com), Feb 14 2009
Another form of the g.f.: f(z) = (z^7 - z^8)/(1 - 2*z + z^9), then a(n) = Sum_{i=0..floor((n-7)/9)} (-1)^i*binomial(n-7-8*i,i)*2^(n-7-9*i) - Sum_{i=0..floor((n-8)/9)} (-1)^i*binomial(n-8-8*i,i)*2^(n-8-9*i) with Sum_{i=m..n} alpha(i) = 0 for m>n. - Richard Choulet, Feb 22 2010
Sum_{k=0..7*n} a(k+b)*A171890(n,k) = a(8*n+b), b>=0.
a(n) = 2*a(n-1) - a(n-9). - Vincenzo Librandi, Dec 20 2010

Extensions

Corrected by Joao B. Oliveira (oliveira(AT)inf.pucrs.br), Nov 25 2004
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