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.

Previous Showing 41-50 of 241 results. Next

A191896 Decimal expansion of the constant whose continued fraction is based on Padovan numbers A000931.

Original entry on oeis.org

6, 3, 0, 8, 2, 1, 9, 8, 2, 0, 6, 2, 2, 6, 3, 4, 9, 7, 9, 9, 9, 8, 8, 1, 5, 0, 6, 5, 6, 0, 1, 9, 5, 8, 4, 8, 9, 3, 5, 6, 5, 2, 8, 9, 1, 7, 3, 6, 0, 2, 3, 0, 0, 9, 6, 4, 5, 5, 4, 4, 0, 8, 1, 7, 8, 5, 3, 2, 1, 0, 3, 4, 2, 6, 1
Offset: 0

Views

Author

Fabrice Auzanneau, Jun 18 2011

Keywords

Comments

The Padovan numbers starting at A000931(5) define a continued fraction which is converted to its floating point representation here.

Examples

			1/(1+1/(1+1/(1+1/(2+1/(2+1/(3+1/(4+1/(5+1/(7+1/(9+1/(12+1/(16+1/(21+1/(28+... = 0.630821982062263497...
		

Crossrefs

Cf. A000931.

Extensions

Decimals corrected by R. J. Mathar, Jul 01 2011

A200220 Product of Fibonacci and Padovan numbers: a(n) = A000045(n+1)*A000931(n+5).

Original entry on oeis.org

1, 1, 2, 6, 10, 24, 52, 105, 238, 495, 1068, 2304, 4893, 10556, 22570, 48363, 103805, 222224, 476634, 1021515, 2189200, 4693415, 10058607, 21561120, 46215400, 99056688, 212327858, 455105352, 975492413, 2090916520, 4481729501, 9606342690, 20590584676, 44134642493, 94599971180
Offset: 0

Views

Author

Paul D. Hanna, Nov 16 2011

Keywords

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 10*x^4 + 24*x^5 + 52*x^6 + 105*x^7 +...
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{0,3,4,-1,-1,1},{1,1,2,6,10,24},40] (* Harvey P. Dale, Mar 05 2019 *)
  • PARI
    {a(n)=fibonacci(n+1)*polcoeff((1+x)/(1-x^2-x^3+x*O(x^n)),n)}

Formula

G.f.: (1 + x - x^2 - x^3 + x^4) / (1 - 3*x^2 - 4*x^3 + x^4 + x^5 - x^6).

A285187 a(n) = Sum(psi(k-1)*psi(n-k-1),k=0..n)+(1-(-1)^n)/2, where psi(k) = A000931(k+6).

Original entry on oeis.org

1, 3, 3, 7, 9, 15, 22, 33, 48, 71, 101, 147, 208, 297, 419, 591, 829, 1161, 1619, 2255, 3130, 4339, 6000, 8285, 11419, 15717, 21600, 29649, 40645, 55659, 76135, 104043, 142045, 193759, 264078, 359637, 489408, 665539, 904449, 1228343, 1667216, 2261593, 3066183
Offset: 0

Views

Author

N. J. A. Sloane, Apr 23 2017

Keywords

Crossrefs

Programs

  • Maple
    A000931 := proc(n) option remember; if n = 0 then 1 elif n <= 2 then 0 else procname(n-2)+procname(n-3); fi; end;
    psi:=n->A000931(n+6);
    f:=n->add(psi(k-1)*psi(n-k-1),k=0..n)+(1-(-1)^n)/2;
    [seq(f(n),n=0..40)];
  • Mathematica
    (* b is A000931 *)
    b[n_] := b[n] = Which[n == 0, 1, n <= 2, 0, True, b[n-2] + b[n-3]];
    psi[n_] := b[n+6];
    a[n_] := Sum[psi[k-1]*psi[n-k-1], {k, 0, n}] + (1-(-1)^n)/2;
    Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Mar 19 2023, after Maple code *)

Formula

From Colin Barker, Apr 25 2017: (Start)
G.f.: (1 + 3*x - 4*x^3 - 3*x^4 + x^5 + 2*x^6 + x^7) / ((1 - x)*(1 + x)*(1 - x^2 - x^3)^2).
a(n) = 3*a(n-2) + 2*a(n-3) - 3*a(n-4) - 4*a(n-5) + 2*a(n-7) + a(n-8) for n>7. (End)

A329178 Sum of the products of pairs of Padovan numbers which are two apart, starting from A000931(5).

Original entry on oeis.org

1, 3, 5, 11, 19, 34, 62, 107, 191, 335, 587, 1035, 1812, 3184, 5589, 9803, 17213, 30199, 52999, 93014, 163214, 286439, 502655, 882095, 1547991, 2716503, 4767160, 8365776, 14680889, 25763219, 45211237, 79340227, 139232411, 244335770, 428779502, 752455475
Offset: 0

Views

Author

David Nacin, Nov 08 2019

Keywords

Examples

			For n=3, a(3) = 1*1 + 1*2 + 1*2 + 2*3 = 11.
		

Crossrefs

Programs

  • Python
    p = lambda x:[1, 1, 1][x] if x<3 else p(x-2)+p(x-3)
    a = lambda x:sum(p(i)*p(i+2) for i in range(x+1))

Formula

a(n) = Sum_{i=5..n+5} A000931(i)*A000931(i+2).
a(n) = A329227(n+7) - 1.
Conjectures from Colin Barker, Nov 09 2019: (Start)
G.f.: (1 + x - x^2 + x^3 - x^4) / ((1 - x)*(1 - 2*x + x^2 - x^3)*(1 + x - x^3)).
a(n) = 2*a(n-1) - 2*a(n-4) + 2*a(n-5) - 2*a(n-6) + a(n-7) for n>6.
(End)

A000073 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = a(1) = 0 and a(2) = 1.

Original entry on oeis.org

0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080, 98950096, 181997601, 334745777, 615693474, 1132436852
Offset: 0

Views

Author

Keywords

Comments

The name "tribonacci number" is less well-defined than "Fibonacci number". The sequence A000073 (which begins 0, 0, 1) is probably the most important version, but the name has also been applied to A000213, A001590, and A081172. - N. J. A. Sloane, Jul 25 2024
Also (for n > 1) number of ordered trees with n+1 edges and having all leaves at level three. Example: a(4)=2 because we have two ordered trees with 5 edges and having all leaves at level three: (i) one edge emanating from the root, at the end of which two paths of length two are hanging and (ii) one path of length two emanating from the root, at the end of which three edges are hanging. - Emeric Deutsch, Jan 03 2004
a(n) is the number of compositions of n-2 with no part greater than 3. Example: a(5)=4 because we have 1+1+1 = 1+2 = 2+1 = 3. - Emeric Deutsch, Mar 10 2004
Let A denote the 3 X 3 matrix [0,0,1;1,1,1;0,1,0]. a(n) corresponds to both the (1,2) and (3,1) entries in A^n. - Paul Barry, Oct 15 2004
Number of permutations satisfying -k <= p(i)-i <= r, i=1..n-2, with k=1, r=2. - Vladimir Baltic, Jan 17 2005
Number of binary sequences of length n-3 that have no three consecutive 0's. Example: a(7)=13 because among the 16 binary sequences of length 4 only 0000, 0001 and 1000 have 3 consecutive 0's. - Emeric Deutsch, Apr 27 2006
Therefore, the complementary sequence to A050231 (n coin tosses with a run of three heads). a(n) = 2^(n-3) - A050231(n-3) - Toby Gottfried, Nov 21 2010
Convolved with the Padovan sequence = row sums of triangle A153462. - Gary W. Adamson, Dec 27 2008
For n > 1: row sums of the triangle in A157897. - Reinhard Zumkeller, Jun 25 2009
a(n+2) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 0, 0, 1; 1, 0, 0] or [1, 1, 0; 1, 0, 1; 1, 0, 0] or [1, 1, 1; 1, 0, 0; 0, 1, 0] or [1, 0, 1; 1, 0, 0; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 0, 1; 1, 1, 1; 0, 1, 0], [0, 1, 0; 0, 1, 1; 1, 1, 0], [0, 0, 1; 1, 0, 1; 0, 1, 1] or [0, 1, 0; 0, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
Also row sums of A082601 and of A082870. - Reinhard Zumkeller, Apr 13 2014
Least significant bits are given in A021913 (a(n) mod 2 = A021913(n)). - Andres Cicuttin, Apr 04 2016
The nonnegative powers of the tribonacci constant t = A058265 are t^n = a(n)*t^2 + (a(n-1) + a(n-2))*t + a(n-1)*1, for n >= 0, with a(-1) = 1 and a(-2) = -1. This follows from the recurrences derived from t^3 = t^2 + t + 1. See the example in A058265 for the first nonnegative powers. For the negative powers see A319200. - Wolfdieter Lang, Oct 23 2018
The term "tribonacci number" was coined by Mark Feinberg (1963), a 14-year-old student in the 9th grade of the Susquehanna Township Junior High School in Pennsylvania. He died in 1967 in a motorcycle accident. - Amiram Eldar, Apr 16 2021
Andrews, Just, and Simay (2021, 2022) remark that it has been suggested that this sequence is mentioned in Charles Darwin's Origin of Species as bearing the same relation to elephant populations as the Fibonacci numbers do to rabbit populations. - N. J. A. Sloane, Jul 12 2022

Examples

			G.f. = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + 24*x^8 + 44*x^9 + 81*x^10 + ...
		

References

  • M. Agronomof, Sur une suite récurrente, Mathesis (Series 4), Vol. 4 (1914), pp. 125-126.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 47, ex. 4.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.2.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • 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

Cf. A000045, A000078, A000213, A000931, A001590 (first differences, also a(n)+a(n+1)), A001644, A008288 (tribonacci triangle), A008937 (partial sums), A021913, A027024, A027083, A027084, A046738 (Pisano periods), A050231, A054668, A062544, A063401, A077902, A081172, A089068, A118390, A145027, A153462, A230216.
A057597 is this sequence run backwards: A057597(n) = a(1-n).
Row 3 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
Partitions: A240844 and A117546.
Cf. also A092836 (subsequence of primes), A299399 = A092835 + 1 (indices of primes).

Programs

  • GAP
    a:=[0,0,1];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Oct 24 2018
  • Haskell
    a000073 n = a000073_list !! n
    a000073_list = 0 : 0 : 1 : zipWith (+) a000073_list (tail
                              (zipWith (+) a000073_list $ tail a000073_list))
    -- Reinhard Zumkeller, Dec 12 2011
    
  • Magma
    [n le 3 select Floor(n/3) else Self(n-1)+Self(n-2)+Self(n-3): n in [1..70]]; // Vincenzo Librandi, Jan 29 2016
    
  • Maple
    a:= n-> (<<0|1|0>, <0|0|1>, <1|1|1>>^n)[1,3]:
    seq(a(n), n=0..40);  # Alois P. Heinz, Dec 19 2016
    # second Maple program:
    A000073:=proc(n) option remember; if n <= 1 then 0 elif n=2 then 1 else procname(n-1)+procname(n-2)+procname(n-3); fi; end; # N. J. A. Sloane, Aug 06 2018
  • Mathematica
    CoefficientList[Series[x^2/(1 - x - x^2 - x^3), {x, 0, 50}], x]
    a[0] = a[1] = 0; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; Array[a, 36, 0] (* Robert G. Wilson v, Nov 07 2010 *)
    LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, May 24 2011 *)
    a[n_] := SeriesCoefficient[If[ n < 0, x/(1 + x + x^2 - x^3), x^2/(1 - x - x^2 - x^3)], {x, 0, Abs @ n}] (* Michael Somos, Jun 01 2013 *)
    Table[-RootSum[-1 - # - #^2 + #^3 &, -#^n - 9 #^(n + 1) + 4 #^(n + 2) &]/22, {n, 0, 20}] (* Eric W. Weisstein, Nov 09 2017 *)
  • Maxima
    A000073[0]:0$
    A000073[1]:0$
    A000073[2]:1$
    A000073[n]:=A000073[n-1]+A000073[n-2]+A000073[n-3]$
      makelist(A000073[n], n, 0, 40);  /* Emanuele Munarini, Mar 01 2011 */
    
  • PARI
    {a(n) = polcoeff( if( n<0, x / ( 1 + x + x^2 - x^3), x^2 / ( 1 - x - x^2 - x^3) ) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, Sep 03 2007 */
    
  • PARI
    my(x='x+O('x^99)); concat([0, 0], Vec(x^2/(1-x-x^2-x^3))) \\ Altug Alkan, Apr 04 2016
    
  • PARI
    a(n)=([0,1,0;0,0,1;1,1,1]^n)[1,3] \\ Charles R Greathouse IV, Apr 18 2016, simplified by M. F. Hasler, Apr 18 2018
    
  • Python
    def a(n, adict={0:0, 1:0, 2:1}):
        if n in adict:
            return adict[n]
        adict[n]=a(n-1)+a(n-2)+a(n-3)
        return adict[n] # David Nacin, Mar 07 2012
    from functools import cache
    @cache
    def A000073(n: int) -> int:
        if n <= 1: return 0
        if n == 2: return 1
        return A000073(n-1) + A000073(n-2) + A000073(n-3) # Peter Luschny, Nov 21 2022
    

Formula

G.f.: x^2/(1 - x - x^2 - x^3).
G.f.: x^2 / (1 - x / (1 - x / (1 + x^2 / (1 + x)))). - Michael Somos, May 12 2012
G.f.: Sum_{n >= 0} x^(n+2) *[ Product_{k = 1..n} (k + k*x + x^2)/(1 + k*x + k*x^2) ] = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + ... may be proved by the method of telescoping sums. - Peter Bala, Jan 04 2015
a(n+1)/a(n) -> A058265. a(n-1)/a(n) -> A192918.
a(n) = central term in M^n * [1 0 0] where M = the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 1 1]. (M^n * [1 0 0] = [a(n-1) a(n) a(n+1)].) a(n)/a(n-1) tends to the tribonacci constant, 1.839286755... = A058265, an eigenvalue of M and a root of x^3 - x^2 - x - 1 = 0. - Gary W. Adamson, Dec 17 2004
a(n+2) = Sum_{k=0..n} T(n-k, k), where T(n, k) = trinomial coefficients (A027907). - Paul Barry, Feb 15 2005
A001590(n) = a(n+1) - a(n); A001590(n) = a(n-1) + a(n-2) for n > 1; a(n) = (A000213(n+1) - A000213(n))/2; A000213(n-1) = a(n+2) - a(n) for n > 0. - Reinhard Zumkeller, May 22 2006
Let C = the tribonacci constant, 1.83928675...; then C^n = a(n)*(1/C) + a(n+1)*(1/C + 1/C^2) + a(n+2)*(1/C + 1/C^2 + 1/C^3). Example: C^4 = 11.444...= 2*(1/C) + 4*(1/C + 1/C^2) + 7*(1/C + 1/C^2 + 1/C^3). - Gary W. Adamson, Nov 05 2006
a(n) = j*C^n + k*r1^n + L*r2^n where C is the tribonacci constant (C = 1.8392867552...), real root of x^3-x^2-x-1=0, and r1 and r2 are the two other roots (which are complex), r1 = m+p*i and r2 = m-p*i, where i = sqrt(-1), m = (1-C)/2 (m = -0.4196433776...) and p = ((3*C-5)*(C+1)/4)^(1/2) = 0.6062907292..., and where j = 1/((C-m)^2 + p^2) = 0.1828035330..., k = a+b*i, and L = a-b*i, where a = -j/2 = -0.0914017665... and b = (C-m)/(2*p*((C-m)^2 + p^2)) = 0.3405465308... . - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n+1) = 3*c*((1/3)*(a+b+1))^n/(c^2-2*c+4) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3), c=(586+102*sqrt(33))^(1/3). Round to the nearest integer. - Al Hakanson (hawkuu(AT)gmail.com), Feb 02 2009
a(n) = round(3*((a+b+1)/3)^n/(a^2+b^2+4)) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3).. - Anton Nikonov
Another form of the g.f.: f(z) = (z^2-z^3)/(1-2*z+z^4). Then we obtain a(n) as a sum: a(n) = Sum_{i=0..floor((n-2)/4)} ((-1)^i*binomial(n-2-3*i,i)*2^(n-2-4*i)) - Sum_{i=0..floor((n-3)/4)} ((-1)^i*binomial(n-3-3*i,i)*2^(n-3-4*i)) with natural convention: Sum_{i=m..n} alpha(i) = 0 for m > n. - Richard Choulet, Feb 22 2010
a(n+2) = Sum_{k=0..n} Sum_{i=k..n, mod(4*k-i,3)=0} binomial(k,(4*k-i)/3)*(-1)^((i-k)/3)*binomial(n-i+k-1,k-1). - Vladimir Kruchinin, Aug 18 2010
a(n) = 2*a(n-2) + 2*a(n-3) + a(n-4). - Gary Detlefs, Sep 13 2010
Sum_{k=0..2*n} a(k+b)*A027907(n,k) = a(3*n+b), b >= 0 (see A099464, A074581).
a(n) = 2*a(n-1) - a(n-4), with a(0)=a(1)=0, a(2)=a(3)=1. - Vincenzo Librandi, Dec 20 2010
Starting (1, 2, 4, 7, ...) is the INVERT transform of (1, 1, 1, 0, 0, 0, ...). - Gary W. Adamson, May 13 2013
G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1 - x*(4*k+1 + x + x^2)/( x*(4*k+3 + x + x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 09 2013
a(n+2) = Sum_{j=0..floor(n/2)} Sum_{k=0..j} binomial(n-2*j,k)*binomial(j,k)*2^k. - Tony Foster III, Sep 08 2017
Sum_{k=0..n} (n-k)*a(k) = (a(n+2) + a(n+1) - n - 1)/2. See A062544. - Yichen Wang, Aug 20 2020
a(n) = A008937(n-1) - A008937(n-2) for n >= 2. - Peter Luschny, Aug 20 2020
From Yichen Wang, Aug 27 2020: (Start)
Sum_{k=0..n} a(k) = (a(n+2) + a(n) - 1)/2. See A008937.
Sum_{k=0..n} k*a(k) = ((n-1)*a(n+2) - a(n+1) + n*a(n) + 1)/2. See A337282. (End)
For n > 1, a(n) = b(n) where b(1) = 1 and then b(n) = Sum_{k=1..n-1} b(n-k)*A000931(k+2). - J. Conrad, Nov 24 2022
Conjecture: the congruence a(n*p^(k+1)) + a(n*p^k) + a(n*p^(k-1)) == 0 (mod p^k) holds for positive integers k and n and for all the primes p listed in A106282. - Peter Bala, Dec 28 2022
Sum_{k=0..n} k^2*a(k) = ((n^2-4*n+6)*a(n+1) - (2*n^2-2*n+5)*a(n) + (n^2-2*n+3)*a(n-1) - 3)/2. - Prabha Sivaramannair, Feb 10 2024
a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(3*r^2-2*r-1). - Fabian Pereyra, Nov 23 2024

Extensions

Minor edits by M. F. Hasler, Apr 18 2018
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A005251 a(0) = 0, a(1) = a(2) = a(3) = 1; thereafter, a(n) = a(n-1) + a(n-2) + a(n-4).

Original entry on oeis.org

0, 1, 1, 1, 2, 4, 7, 12, 21, 37, 65, 114, 200, 351, 616, 1081, 1897, 3329, 5842, 10252, 17991, 31572, 55405, 97229, 170625, 299426, 525456, 922111, 1618192, 2839729, 4983377, 8745217, 15346786, 26931732, 47261895, 82938844, 145547525, 255418101, 448227521
Offset: 0

Views

Author

Keywords

Comments

a(n+3) is the number of n-bit sequences that avoid 010. Example: For n=4 the 12 sequences are all 4-bit sequences except 0100, 0101, 0010, 1010. - David Callan, Mar 25 2004
a(n+2) is the number of compositions (ordered partitions) of n where no two adjacent parts are != 1, see example. - Joerg Arndt, Jan 26 2013
a(n+1) is the number of compositions of n avoiding the part 2. - Joerg Arndt, Jul 13 2014
Number of different positive braids with n crossings of 3 strands.
This is a_2(n) in the Doroslovacki reference. Note that there is a typo in the paper in the formula for a_2(n): the upper bound in the inner sum should be "n-i" not "i-1". - Max Alekseyev, Jun 26 2007
a(n) is the number of peakless Motzkin paths of length n-1 with no UHH...HD's starting at level > 0 (here n > 0 and U=(1,1), H=(1,0), D=(1,-1)). Example: a(5)=7 because from all 8 peakless Motzkin paths of length 5 (see A004148) only UUHDD does not qualify. - Emeric Deutsch, Sep 13 2004
Equals the INVERT transform of (1, 0, 1, 1, 1, ...); equivalent to a(n) = a(n-1) + a(n-3) + a(n-4) + ... - Gary W. Adamson, Apr 27 2009
a(n) is the number of length n-1 words on {0,1} such that each string of 1's is followed by a string of at least two 0's. For example, a(5) = 4 because we have: 0000, 0100, 1000, and 1100. - Geoffrey Critzer, Aug 09 2013
a(n+1) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 0; 0, 1, 1; 1, 0, 0] or [1, 0, 1; 1, 1, 0; 0, 1, 0] or [1, 1, 0; 0, 0, 1; 1, 0, 1] or [1, 0, 1; 1, 0, 0; 0, 1, 1]. - R. J. Mathar, Feb 03 2014
For n >= 2, a(n) is the number of (n-2)-length binary words with no isolated zeros. - Milan Janjic, Mar 07 2015
Apart from the first three terms, the total number of bargraphs of semiperimeter n of height at most two for n >= 2 starts 1, 2, 4, 7, 12, ... - Arnold Knopfmacher, Nov 02 2016
Number of DD-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are DD-equivalent iff the positions of pattern DD are identical in these paths. - Sergey Kirgizov, Apr 08 2018
From Gus Wiseman, Nov 25 2019: (Start)
For n > 0, also the number of subsets of {1, ..., n - 3} such that if x and x + 2 are both in the subset, then so is x + 1. For example, the a(3) = 1 through a(7) = 12 subsets are:
{} {} {} {} {}
{1} {1} {1} {1}
{2} {2} {2}
{1,2} {3} {3}
{1,2} {4}
{2,3} {1,2}
{1,2,3} {1,4}
{2,3}
{3,4}
{1,2,3}
{2,3,4}
{1,2,3,4}
(End)
The two-dimensional version, which counts sets of pairs where, if two pairs are separated by graph-distance 2, then the intermediate pair or pairs are also in the set, is A329871. - Gus Wiseman, Nov 30 2019
a(n+1) is the number of ways to tile a strip of length n with squares, dominoes, and tetrominoes, where the first tile cannot be a domino. - Greg Dresden and Myanna Nash, Aug 18 2020
For n>=3, a(n) is the number of binary strings of length n-2 without any maximal runs of ones of length 1. - Félix Balado, Aug 25 2025

Examples

			From _Joerg Arndt_, Jan 26 2013: (Start)
The a(5+2) = 12 compositions of 5 where no two adjacent parts are != 1 are
  [ 1]  [ 1 1 1 1 1 ]
  [ 2]  [ 1 1 1 2 ]
  [ 3]  [ 1 1 2 1 ]
  [ 4]  [ 1 1 3 ]
  [ 5]  [ 1 2 1 1 ]
  [ 6]  [ 1 3 1 ]
  [ 7]  [ 1 4 ]
  [ 8]  [ 2 1 1 1 ]
  [ 9]  [ 2 1 2 ]
  [10]  [ 3 1 1 ]
  [11]  [ 4 1 ]
  [12]  [ 5 ]
(End)
G.f. = x + x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 12*x^7 + 21*x^8 + 37*x^9 + ...
		

References

  • S. Burckel, Efficient methods for three strand braids (submitted). [Apparently unpublished]
  • P. Chinn and S. Heubach, "Compositions of n with no occurrence of k", Congressus Numeratium, 2002, v. 162, pp. 33-51.
  • John H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, p. 205.
  • R. K. Guy, "Anyone for Twopins?" in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 2-15.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Bisection of Padovan sequence A000931.
Partial sums of A005314 shifted 3 times to the right, if we assume A005314(0) = 1.
Compositions without adjacent equal parts are A003242.
Compositions without isolated parts are A114901.
Row sums of A097230(n-2) for n>1.

Programs

  • Haskell
    a005251 n = a005251_list !! n
    a005251_list = 0 : 1 : 1 : 1 : zipWith (+) a005251_list
       (drop 2 $ zipWith (+) a005251_list (tail a005251_list))
    -- Reinhard Zumkeller, Dec 28 2011
    
  • Magma
    I:=[0,1,1,1]; [n le 4 select I[n] else Self(n-1)+Self(n-2)+Self(n-4): n in [1..45]]; // Vincenzo Librandi, Nov 30 2018
    
  • Magma
    R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1-x)/(1-2*x + x^2 - x^3) )); // Marius A. Burtea, Oct 24 2019
    
  • Maple
    A005251 := proc(n) option remember; if n <= 2 then n elif n = 3 then 4 else 2*A005251(n - 1) - A005251(n - 2) + A005251(n - 3); fi; end;
    A005251:=(-1+z)/(-1+2*z-z**2+z**3); # Simon Plouffe in his 1992 dissertation
    a := n -> `if`(n<=1, n, hypergeom([(2-n)/3, 1-n/3, (1-n)/3], [1/2, -n+1], 27/4)):
    seq(simplify(a(n)), n=0..36); # Peter Luschny, Apr 08 2018
  • Mathematica
    LinearRecurrence[{2,-1,1},{0,1,1},40]  (* Harvey P. Dale, May 05 2011 *)
    a[ n_]:= If[n<0, SeriesCoefficient[ -x(1-x)/(1 -x + 2x^2 -x^3), {x, 0, -n}], SeriesCoefficient[ x(1-x)/(1 -2x +x^2 -x^3), {x, 0, n}]] (* Michael Somos, Dec 13 2013 *)
    a[0] = 1; a[1] = a[2] = 0; a[n_] := a[n] = a[n-2] + a[n-3]; Table[a[2 n-1], {n, 1, 20}] (* Rigoberto Florez, Oct 15 2019 *)
    Table[If[n==0,0,Length[DeleteCases[Subsets[Range[n-3]],{_,x_,y_,_}/;x+2==y]]],{n,0,10}] (* Gus Wiseman, Nov 25 2019 *)
  • PARI
    Vec((1-x)/(1-2*x+x^2-x^3)+O(x^99)) /* Charles R Greathouse IV, Nov 20 2012 */
    
  • PARI
    {a(n) = if( n<0, polcoeff( -x*(1-x)/(1 -x +2*x^2 -x^3) + x*O(x^-n), -n), polcoeff( x*(1-x)/(1 -2*x +x^2 -x^3) + x*O(x^n), n))} /* Michael Somos, Dec 13 2013 */
    
  • SageMath
    [sum( binomial(n-j-1, 2*j) for j in (0..floor((n-1)/3)) ) for n in (0..50)] # G. C. Greubel, Apr 13 2022

Formula

a(n) = 2*a(n-1) - a(n-2) + a(n-3).
G.f.: z*(1-z)/(1 - 2*z + z^2 - z^3). - Emeric Deutsch, Sep 13 2004
23*a_n = 3*P_{2n+1} + 7*P_{2n} - 2*P_{2n-1}, where P_n are the Perrin numbers, A001608. - Don Knuth, Dec 09 2008
a(n+1) = Sum_{k=0..n} binomial(n-k, 2k). - Richard L. Ollerton, May 12 2004
From Henry Bottomley, Feb 21 2001: (Start)
a(n) = (Sum_{j
a(n) = A005314(n) - A005314(n-1).
a(n) = A049853(n-1) - a(n-1).
a(n) = A005314(n) - a(n-2). (End)
Conjecture: a(n+1) + |A078065(n)| = 2*A005314(n+1). - Creighton Dement, Dec 21 2004
a(n+2) has g.f. (F_3(-x) + F_2(-x))/(F_4(-x) + F_3(-x)) = 1/(-x+1/(-x+1/(-x+1))) where F_n(x) is the n-th Fibonacci polynomial; see A011973. - Qiaochu Yuan (qchu(AT)mit.edu), Feb 19 2009
a(n) = A173022(2^(n-2) - 1) for n > 1. - Reinhard Zumkeller, Feb 07 2010
BINOMIAL transform of A176971 is a(n+1). - Michael Somos, Dec 13 2013
a(n) = hypergeom([(2-n)/3, 1-n/3, (1-n)/3], [1/2, -n+1], 27/4) for n > 1. - Peter Luschny, Apr 08 2018
G.f.: z/(1-z-z^3-z^4-z^5-...) for the compositions of n-1 avoiding 2. The g.f. for the number of compositions of n avoiding the part k is 1/(1-z-...-z^(k-1) - z^(k+1)-...). - Gregory L. Simay, Sep 09 2018
If p,q,r are the three solutions to x^3 = 2x^2 - x + 1, then a(n) = (p-1)*p^n/((p-q)*(p-r)) + (q-1)*q^n/((q-p)*(q-r)) + (r-1)*r^n/((r-p)*(r-q)). - Greg Dresden and AnXing Yang, Aug 12 2025

A001608 Perrin sequence (or Ondrej Such sequence): a(n) = a(n-2) + a(n-3) with a(0) = 3, a(1) = 0, a(2) = 2.

Original entry on oeis.org

3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39, 51, 68, 90, 119, 158, 209, 277, 367, 486, 644, 853, 1130, 1497, 1983, 2627, 3480, 4610, 6107, 8090, 10717, 14197, 18807, 24914, 33004, 43721, 57918, 76725, 101639, 134643, 178364, 236282, 313007, 414646, 549289
Offset: 0

Keywords

Comments

Has been called the skiponacci sequence or skiponacci numbers. - N. J. A. Sloane, May 24 2013
For n >= 3, also the numbers of maximal independent vertex sets, maximal matchings, minimal edge covers, and minimal vertex covers in the n-cycle graph C_n. - Eric W. Weisstein, Mar 30 2017 and Aug 03 2017
With the terms indexed as shown, has property that p prime => p divides a(p). The smallest composite n such that n divides a(n) is 521^2. For quotients a(p)/p, where p is prime, see A014981.
Asymptotically, a(n) ~ r^n, with r=1.3247179572447... the inverse of the real root of 1-x^2-x^3=0 (see A060006). If n>9 then a(n)=round(r^n). - Ralf Stephan, Dec 13 2002
The recursion can be used to compute a(-n). The result is -A078712(n). - T. D. Noe, Oct 10 2006
For n>=3, a(n) is the number of maximal independent sets in a cycle of order n. - Vincent Vatter, Oct 24 2006
Pisano period lengths are given in A104217. - R. J. Mathar, Aug 10 2012
From Roman Witula, Feb 01 2013: (Start)
Let r1, r2 and r3 denote the roots of x^3 - x - 1. Then the following identity holds:
a(k*n) + (a(k))^n - (a(k) - r1^k)^n - (a(k) - r2^k)^n - (a(k) - r3^k)^n
= 0 for n = 0, 1, 2,
= 6 for n = 3,
= 12*a(k) for n = 4,
= 10*[2*(a(k))^2 - a(-k)] for n = 5,
= 30*a(k)*[(a(k))^2 - a(-k)] for n = 6,
= 7*[6*(a(k))^4 - 9*a(-k)*(a(k))^2 + 2*(a(-k))^2 - a(k)] for n = 7,
= 56*a(k)*[((a(k))^2 - a(-k))^2 - a(k)/2] for n = 8,
where a(-k) = -A078712(k) and the formula (5.40) from the paper of Witula and Slota is used. (End)
The parity sequence of a(n) is periodic with period 7 and has the form (1,0,0,1,0,1,1). Hence we get that a(n) and a(2*n) are congruent modulo 2. Similarly we deduce that a(n) and a(3*n) are congruent modulo 3. Is it true that a(n) and a(p*n) are congruent modulo p for every prime p? - Roman Witula, Feb 09 2013
The trinomial x^3 - x - 1 divides the polynomial x^(3*n) - a(n)*x^(2*n) + ((a(n)^2 - a(2*n))/2)*x^n - 1 for every n>=1. For example, for n=3 we obtain the factorization x^9 - 3*x^6 + 2*x^3 - 1 = (x^3 - x - 1)*(x^6 + x^4 - 2*x^3 + x^2 - x + 1). Sketch of the proof: Let p,s,t be roots of the Perrin polynomial x^3 - x - 1. Then we have (a(n))^2 = (p^n + s^n + t^n)^2 = a(2*n) + 2*a(n)*x^n -2*x^n + 2/x^n for every x = p,s,t, i.e., x^(3*n) - a(n)*x^(2*n) + ((a(n)^2 - a(2*n))/2)*x^n - 1 = 0 for every x = p,s,t, which finishes the proof. By discussion of the power(a(n))^3 = (p^n + s^n + t^n)^3 it can be deduced that the trinomial x^3 - x - 1 divides the polynomial 2*x^(4*n) - a(n)*x^(3*n) - a(2*n)*x^(2*n) + ((a(n)^3 - a(3*n) - 3)/3)*x^n - a(n) = 0. Co-author of these divisibility relations is also my young student Szymon Gorczyca (13 years old as of 2013). - Roman Witula, Feb 09 2013
The sum of powers of the real root and complex roots of x^3-x-1=0 as expressed as powers of the plastic number r, (see A060006). Let r0=1, r1=r, r2=1+r^(-1) and c0=2, c1=-r and c3 = r^(-5) then a(n) = r(n-2)+r(n-3) + c(n-2)+c(n-3). Example: a(5) = 1 + r^(-1) + 1 + r + 2 - r + r^(-5) = 4 + r^(-1) + r^(-5) = 5. - Richard Turk, Jul 14 2016
Also the number of minimal total dominating sets in the n-sun graph. - Eric W. Weisstein, Apr 27 2018
Named after the French engineer François Olivier Raoul Perrin (1841-1910). - Amiram Eldar, Jun 05 2021
a(p) = p*A127687(p) for p prime. - Robert FERREOL, Apr 09 2024

Examples

			From _Roman Witula_, Feb 01 2013: (Start)
We note that if a + b + c = 0 then:
1) a^3 + b^3 + c^3 = 3*a*b*c,
2) a^4 + b^4 + c^4 = 2*((a^2 + b^2 + c^2)/2)^2,
3) (a^5 + b^5 + c^5)/5 = (a^3 + b^3 + c^3)/3 * (a^2 +
    b^2  + c^2)/2,
4) (a^7 + b^7 + c^7)/7 = (a^5 + b^5 + c^5)/5 * (a^2 + b^2 + c^2)/2 = 2*(a^3 + b^3 + c^3)/3 * (a^4 + b^4 + c^4)/4,
5) (a^7 + b^7 + c^7)/7 * (a^3 + b^3 + c^3)/3 = ((a^5 + b^5 + c^5)/5)^2.
Hence, by the Binet formula for a(n) we obtain the relations: a(3) = 3, a(4) = 2*(a(2)/2)^2 = 2, a(5)/5 = a(3)/3 * a(2)/2, i.e., a(5) = 5, and similarly that a(7) = 7. (End)
		

References

  • Olivier Bordellès, Thèmes d'Arithmétique, Ellipses, 2006, Exercice 4.11, p. 127.0
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.2.
  • Dmitry Fomin, On the properties of a certain recursive sequence, Mathematics and Informatics Quarterly, Vol. 3 (1993), pp. 50-53.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 70.
  • Manfred Schroeder, Number Theory in Science and Communication, 3rd ed., Springer, 1997.
  • A. G. Shannon, P. G. Anderson and A. F. Horadam, Properties of Cordonnier, Perrin and Van der Laan numbers, International Journal of Mathematical Education in Science and Technology, Volume 37:7 (2006), 825-831. See Q_n.
  • 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

Closely related to A182097.
Cf. A000931, bisection A109377.
Cf. A013998 (Unrestricted Perrin pseudoprimes).
Cf. A018187 (Restricted Perrin pseudoprimes).

Programs

  • GAP
    a:=[3,0,2];; for n in [4..20] do a[n]:=a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Jul 12 2018
    
  • Haskell
    a001608 n = a000931_list !! n
    a001608_list = 3 : 0 : 2 : zipWith (+) a001608_list (tail a001608_list)
    -- Reinhard Zumkeller, Feb 10 2011
    
  • Magma
    I:=[3,0,2]; [n le 3 select I[n] else Self(n-2) +Self(n-3): n in [1..50]]; // G. C. Greubel, Mar 18 2019
    
  • Maple
    A001608 :=proc(n) option remember; if n=0 then 3 elif n=1 then 0 elif n=2 then 2 else procname(n-2)+procname(n-3); fi; end proc;
    [seq(A001608(n),n=0..50)]; # N. J. A. Sloane, May 24 2013
  • Mathematica
    LinearRecurrence[{0, 1, 1}, {3, 0, 2}, 50] (* Harvey P. Dale, Jun 26 2011 *)
    per = Solve[x^3 - x - 1 == 0, x]; f[n_] := Floor @ Re[N[ per[[1, -1, -1]]^n + per[[2, -1, -1]]^n + per[[3, -1, -1]]^n]]; Array[f, 46, 0] (* Robert G. Wilson v, Jun 29 2010 *)
    a[n_] := n*Sum[Binomial[k, n-2*k]/k, {k, 1, n/2}]; a[0]=3; Table[a[n] , {n, 0, 45}] (* Jean-François Alcover, Oct 04 2012, after Vladimir Kruchinin *)
    CoefficientList[Series[(3 - x^2)/(1 - x^2 - x^3), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 03 2015 *)
    Table[RootSum[-1 - # + #^3 &, #^n &], {n, 0, 20}] (* Eric W. Weisstein, Mar 30 2017 *)
    RootSum[-1 - # + #^3 &, #^Range[0, 20] &] (* Eric W. Weisstein, Dec 30 2017 *)
  • PARI
    a(n)=if(n<0,0,polsym(x^3-x-1,n)[n+1])
    
  • PARI
    A001608_list(n) = polsym(x^3-x-1,n) \\ Joerg Arndt, Mar 10 2019
    
  • Python
    A001608_list, a, b, c = [3, 0, 2], 3, 0, 2
    for _ in range(100):
        a, b, c = b, c, a+b
        A001608_list.append(c) # Chai Wah Wu, Jan 27 2015
    
  • Sage
    ((3-x^2)/(1-x^2-x^3)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Mar 18 2019

Formula

G.f.: (3 - x^2)/(1 - x^2 - x^3). - Simon Plouffe in his 1992 dissertation
a(n) = r1^n + r2^n + r3^n where r1, r2, r3 are three roots of x^3-x-1=0.
a(n-1) + a(n) + a(n+1) = a(n+4), a(n) - a(n-1) = a(n-5). - Jon Perry, Jun 05 2003
From Gary W. Adamson, Feb 01 2004: (Start)
a(n) = trace(M^n) where M is the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 1 0], the companion matrix of the characteristic polynomial of this sequence, P = X^3 - X - 1.
M^n * [3, 0, 2] = [a(n), a(n+1), a(n+2)]; e.g., M^7 * [3, 0, 2] = [7, 10, 12].
a(n) = 2*A000931(n+3) + A000931(n). (End)
a(n) = 3*p(n) - p(n-2) = 2*p(n) + p(n-3), with p(n) := A000931(n+3), n >= 0. - Wolfdieter Lang, Jun 21 2010
From Francesco Daddi, Aug 03 2011: (Start)
a(0) + a(1) + a(2) + ... + a(n) = a(n+5) - 2.
a(0) + a(2) + a(4) + ... + a(2*n) = a(2*n+3).
a(1) + a(3) + a(5) + ... + a(2*n+1) = a(2*n+4) - 2. (End)
From Francesco Daddi, Aug 04 2011: (Start)
a(0) + a(3) + a(6) + a(9) + ... + a(3*n) = a(3*n+2) + 1.
a(0) + a(5) + a(10) + a(15) + ... + a(5*n) = a(5*n+1)+3.
a(0) + a(7) + a(14) + a(21) + ... + a(7*n) = (a(7*n) + a(7*n+1) + 3)/2. (End)
a(n) = n*Sum_{k=1..floor(n/2)} binomial(k,n-2*k)/k, n > 0, a(0)=3. - Vladimir Kruchinin, Oct 21 2011
(a(n)^3)/2 + a(3n) - 3*a(n)*a(2n)/2 - 3 = 0. - Richard Turk, Apr 26 2017
2*a(4n) - 2*a(n) - 2*a(n)*a(3n) - a(2n)^2 + a(n)^2*a(2n) = 0. - Richard Turk, May 02 2017
a(n)^4 + 6*a(4n) - 4*a(3n)*a(n) - 3*a(2n)^2 - 12a(n) = 0. - Richard Turk, May 02 2017
a(n+5)^2 + a(n+1)^2 - a(n)^2 = a(2*(n+5)) + a(2*(n+1)) - a(2*n). - Aleksander Bosek, Mar 04 2019
From Aleksander Bosek, Mar 18 2019: (Start)
a(n+12) = a(n) + 2*a(n+4) + a(n+11);
a(n+16) = a(n) + 4*a(n+9) + a(n+13);
a(n+18) = a(n) + 2*a(n+6) + 5*a(n+12);
a(n+21) = a(n) + 2*a(n+12) + 6*a(n+14);
a(n+27) = a(n) + 3*a(n+9) + 4*a(n+22). (End)
a(n) = Sum_{j=0..floor((n-g)/(2*g))} 2*n/(n-2*(g-2)*j-(g-2)) * Hypergeometric2F1([-(n-2g*j-g)/2, -(2j+1)], [1], 1), g = 3 and n an odd integer. - Richard Turk, Oct 14 2019
E.g.f.: exp(r1*x) + exp(r2*x) + exp(r3*x), where r1, r2, r3 are three roots of x^3 - x - 1 = 0. - Fabian Pereyra, Nov 02 2024

Extensions

Additional comments from Mike Baker, Oct 11 2005
Definition edited by Chai Wah Wu, Jan 27 2015
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A065941 T(n,k) = binomial(n-floor((k+1)/2), floor(k/2)). Triangle read by rows, for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 4, 3, 3, 1, 1, 1, 5, 4, 6, 3, 1, 1, 1, 6, 5, 10, 6, 4, 1, 1, 1, 7, 6, 15, 10, 10, 4, 1, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1, 1, 1, 11, 10, 45, 36, 84, 56, 70, 35, 21, 6, 1
Offset: 0

Author

Len Smiley, Nov 29 2001

Keywords

Comments

Also the q-Stirling2 numbers at q = -1. - Peter Luschny, Mar 09 2020
Row sums give the Fibonacci sequence. So do the alternating row sums.
Triangle of coefficients of polynomials defined by p(-1,x) = p(0,x) = 1, p(n, x) = x*p(n-1, x) + p(n-2, x), for n >= 1. - Benoit Cloitre, May 08 2005 [rewritten with correct offset. - Wolfdieter Lang, Feb 18 2020]
Another version of triangle in A103631. - Philippe Deléham, Jan 01 2009
The T(n,k) coefficients appear in appendix 2 of Parks's remarkable article "A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov" if we assume that the b(n) coefficients are all equal to 1 and ignore the first column. The complete version of this triangle including the first column is A103631. - Johannes W. Meijer, Aug 11 2011
Signed ++--++..., the roots are chaotic using f(x) --> x^2 - 2 with cycle lengths shown in A003558 by n-th rows. Example: given row 3, x^3 + x^2 - 2x - 1; the roots are (a = 1.24697, ...; b = -0.445041, ...; c = -1.802937, ...). Then (say using seed b with x^2 - 2) we obtain the trajectory -0.445041, ... -> -1.80193, ... -> 1.24697, ...; matching the entry "3" in A003558(3). - Gary W. Adamson, Sep 06 2011
From Gary W. Adamson, Aug 25 2019: (Start)
Roots to the polynomials and terms in A003558 can all be obtained from the numbers below using a doubling series mod N procedure as follows: (more than one row may result). Any row ends when the trajectory produces a term already used. Then try the next higher odd term not used as the leftmost term, then repeat.
For example, for N = 11, we get: (1, 2, 4, 3, 5), showing that when confronted with two choices after the 4: (8 and -3), pick the smaller (abs) term, = 3. Then for the next row pick 7 (not used) and repeat the algorithm; succeeding only if the trajectory produces new terms. But 7 is also (-4) mod 11 and 4 was used. Therefore what I call the "r-t table" (for roots trajectory) has only one row: (1, 2, 4, 3, 5). Conjecture: The numbers of terms in the first row is equal to A003558 corresponding to N, i.e., 5 in this case with period 2.
Now for the roots to the polynomials. Pick N = 7. The polynomial is x^3 - x^2 - 2x + 1 = 0, with roots 1.8019..., -1.2469... and 0.445... corresponding to 2*cos(j*Pi/N), N = 7, and j = (1, 2, and 3). The terms (1, 2, 3) are the r-t terms for N = 7. For 11, the r-t terms are (1, 2, 4, 3, 5). This implies that given any roots of the corresponding polynomial, they are cyclic using f(x) --> x^2 - 2 with cycle lengths shown in A003558. The terms thus generated are 2*cos(j*Pi), with j = (1, 2, 4, 3, 5). Check: Begin with 2*j*Pi/N, with j = 1 (1.9189...). The other trajectory terms are: --> 1.6825..., --> 0.83083..., -1.3097...; 545...; (a 5 period and cyclic since we can begin with any of the constants). The r-t table for odd N begins as follows:
3...............1
5...............1, 2
7...............1, 2, 3
9...............1, 2, 4
...............3 (singleton terms reduce to "1") (9 has two rows)
11...............1, 2, 4, 3, 5
13...............1, 2, 4, 5, 3, 6
15...............1, 2, 4, 7
................3, 6 (dividing through by the gcd gives (1, 2))
................5. (singleton terms reduce to "1")
The result is that 15 has 3 factors (since 3 rows), and the values of those factors are the previous terms "N", corresponding to the r-t terms in each row. Thus, the first row is new, the second (1, 2), corresponds to N = 5, and the "1" in row 3 corresponds to N = 3. The factors are those values apart from 15 and 1. Note that all of the unreduced r-t terms in all rows for N form a complete set of the terms 1 through (N-1)/2 without duplication. (End)
From Gary W. Adamson, Sep 30 2019: (Start)
The 3 factors of the 7th degree polynomial for 15: (x^7 - x^6 - 6x^5 + 5x^4 + 10x^3 - 6x^2 - 4x + 1) can be determined by getting the roots for 2*cos(j*Pi/1), j = (1, 2, 4, 7) and finding the corresponding polynomial, which is x^4 + x^3 - 4x^2 - 4x + 1. This is the minimal polynomial for N = 15 as shown in Table 2, p. 46 of (Lang). The degree of this polynomial is 4, corresponding to the entry in A003558 for 15, = 4. The trajectories (3, 6) and (5) are j values for 2*cos(j*Pi/15) which are roots to x^2 - x - 1 (relating to the pentagon), and (x - 1), relating to the triangle. (End)
From Gary W. Adamson, Aug 21 2019: (Start)
Matrices M of the form: (1's in the main diagonal, -1's in the subdiagonal, and the rest zeros) are chaotic if we replace (f(x) --> x^2 - 2) with f(x) --> M^2 - 2I, where I is the Identity matrix [1, 0, 0; 0, 1, 0; 0, 0, 1]. For example, with the 3 X 3 matrix M: [0, 0, 1; 0, 1, -1; 1, -1, 0]; the f(x) trajectory is:
....M^2 - 2I: [-1, -1, 0; -1, 0, -1; 0, -1, 0], then for the latter,
....M^2 - 2I: [0, 1, 1; 1, 0, 0; 1, 0, -1]. The cycle ends with period 3 since the next matrix is (-1) * the seed matrix. As in the case with f(x) --> x^2 - 2, the eigenvalues of the 3 chaotic matrices are (abs) 1.24697, 0.44504... and 1.80193, ... Also, the characteristic equations of the 3 matrices are the same as or variants of row 4 of the triangle below: (x^3 + x - 2x - 1) with different signs. (End)
Received from Herb Conn, Jan 2004: (Start)
Let x = 2*cos(2A) (A = Angle); then
sin(A)/sin A = 1
sin(3A)/sin A = x + 1
sin(5A)/sin A = x^2 + x - 1
sin(7A)/sin A = x^3 + x - 2x - 1
sin(9A)/sin A = x^4 + x^3 - 3x^2 - 2x + 1
... (signed ++--++...). (End)
Or Pascal's triangle (A007318) with duplicated diagonals. Also triangle of coefficients of polynomials defined by P_0(x) = 1 and for n>=1, P_n(x) = F_n(x) + F_(n+1)(x), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 12 2012
The matrix inverse is given by
1;
1, 1;
0, -1, 1;
0, 1, -2, 1;
0, 0, 1, -2, 1;
0, 0, -1, 3, -3, 1;
0, 0, 0, -1, 3, -3, 1;
0, 0, 0, 1, -4, 6, -4, 1;
0, 0, 0, 0, 1, -4, 6, -4, 1;
... apart from signs the same as A124645. - R. J. Mathar, Mar 12 2013

Examples

			Triangle T(n, k) begins:
n\k 0  1  2  3   4   5  6   7  8  9 ...
---------------------------------------
[0] 1,
[1] 1, 1,
[2] 1, 1, 1,
[3] 1, 1, 2, 1,
[4] 1, 1, 3, 2,  1,
[5] 1, 1, 4, 3,  3,  1,
[6] 1, 1, 5, 4,  6,  3,  1,
[7] 1, 1, 6, 5, 10,  6,  4,  1,
[8] 1, 1, 7, 6, 15, 10, 10,  4,  1,
[9] 1, 1, 8, 7, 21, 15, 20, 10,  5, 1,
---------------------------------------
From _Gary W. Adamson_, Oct 23 2019: (Start)
Consider the roots of the polynomials corresponding to odd N such that for N=7 the polynomial is (x^3 + x^2 - 2x - 1) and the roots (a, b, c) are (-1.8019377..., 1.247697..., and -0.445041...). The discriminant of a polynomial derived from the roots is the square of the product of successive differences: ((a-b), (b-c), (c-a))^2 in this case, resulting in 49, matching the method derived from the coefficients of a cubic. For our purposes we use the product of the differences, not the square, resulting in (3.048...) * (1.69202...) * (1.35689...) = 7.0. Conjecture: for all polynomials in the set, the product of the differences of the roots = the corresponding N. For N = 7, we get x^3 - 7x + 7. It appears that for all prime N's, these resulting companion polynomials are monic (left coefficient is 1), and all other coefficients are N or multiples thereof, with the rightmost term = N. The companion polynomials for the first few primes are:
  N =  5:  x^2 - 5;
  N =  7:  x^3 - 7x + 7;
  N = 11:  x^5 - 11x^3 + 11x^2 + 11x - 11;
  N = 13:  x^6 - 13x^4 + 13x^3 + 26x^2 - 39x + 13;
  N = 17:  x^8 - 17x^6 + 17x^5 + 68x^4 - 119x^3 + 17x^2 + 51x - 17;
  N = 19:  x^9 - 19x^7 + 19x^6 + 95x^5 - 171x^4 - 19x^3 + 190x^2 - 114x + 19. (End)
		

Crossrefs

Cf. A065942 (central stalk sequence), A000045 (row sums), A108299.
Reflected version of A046854.
Some triangle sums (see A180662): A000045 (Fi1), A016116 (Kn21), A000295 (Kn23), A094967 (Fi2), A000931 (Ca2), A001519 (Gi3), A000930 (Ze3).

Programs

  • Haskell
    a065941 n k = a065941_tabl !! n !! k
    a065941_row n = a065941_tabl !! n
    a065941_tabl = iterate (\row ->
       zipWith (+) ([0] ++ row) (zipWith (*) (row ++ [0]) a059841_list)) [1]
    -- Reinhard Zumkeller, May 07 2012
    
  • Magma
    [Binomial(n - Floor((k+1)/2), Floor(k/2)): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 10 2019
    
  • Maple
    A065941 := proc(n,k): binomial(n-floor((k+1)/2),floor(k/2)) end: seq(seq(A065941(n,k), k=0..n), n=0..15); # Johannes W. Meijer, Aug 11 2011
    A065941 := proc(n,k) option remember: local j: if k=0 then 1 elif k=1 then 1: elif k>=2 then add(procname(j,k-2), j=k-2..n-2) fi: end: seq(seq(A065941(n,k), k=0..n), n=0..15);  # Johannes W. Meijer, Aug 11 2011
    # The function qStirling2 is defined in A333143.
    seq(print(seq(qStirling2(n, k, -1), k=0..n)), n=0..9);
    # Peter Luschny, Mar 09 2020
  • Mathematica
    Flatten[Table[Binomial[n-Floor[(k+1)/2],Floor[k/2]],{n,0,15},{k,0,n}]] (* Harvey P. Dale, Dec 11 2011 *)
  • PARI
    T065941(n, k) = binomial(n-(k+1)\2, k\2); \\ Michel Marcus, Apr 28 2014
    
  • Sage
    [[binomial(n - floor((k+1)/2), floor(k/2)) for k in (0..n)] for n in (0..15)] # G. C. Greubel, Jul 10 2019

Formula

T(n, k) = binomial(n-floor((k+1)/2), floor(k/2)).
As a square array read by antidiagonals, this is given by T1(n, k) = binomial(floor(n/2) + k, k). - Paul Barry, Mar 11 2003
Triangle is a reflection of that in A066170 (absolute values). - Gary W. Adamson, Feb 16 2004
Recurrences: T(k, 0) = 1, T(k, n) = T(k-1, n) + T(k-2, n-2), or T(k, n) = T(k-1, n) + T(k-1, n-1) if n even, T(k-1, n-1) if n odd. - Ralf Stephan, May 17 2004
G.f.: sum[n, sum[k, T(k, n)x^ky^n]] = (1+xy)/(1-y-x^2y^2). sum[n>=0, T(k, n)y^n] = y^k/(1-y)^[k/2]. - Ralf Stephan, May 17 2004
T(n, k) = A108299(n, k)*A087960(k) = abs(A108299(n, k)). - Reinhard Zumkeller, Jun 01 2005
From Johannes W. Meijer, Aug 11 2011: (Start)
T(n,k) = A046854(n, n-k) = abs(A066170(n, n-k)).
T(n+k, n-k) = A109223(n,k).
T(n, k) = sum(T(j, k-2), j=k-2..n-2), 2 <= k <= n, n>=2;
T(n, 0) =1, T(n+1, 1) = 1, n >= 0. (End)
For n > 1: T(n, k) = T(n-2, k) + T(n-1, k), 1 < k < n. - Reinhard Zumkeller, Apr 24 2013

A030528 Triangle read by rows: a(n,k) = binomial(k,n-k).

Original entry on oeis.org

1, 1, 1, 0, 2, 1, 0, 1, 3, 1, 0, 0, 3, 4, 1, 0, 0, 1, 6, 5, 1, 0, 0, 0, 4, 10, 6, 1, 0, 0, 0, 1, 10, 15, 7, 1, 0, 0, 0, 0, 5, 20, 21, 8, 1, 0, 0, 0, 0, 1, 15, 35, 28, 9, 1, 0, 0, 0, 0, 0, 6, 35, 56, 36, 10, 1, 0, 0, 0, 0, 0, 1, 21, 70, 84, 45, 11, 1, 0, 0, 0, 0, 0, 0, 7, 56, 126, 120, 55, 12, 1
Offset: 1

Keywords

Comments

A convolution triangle of numbers obtained from A019590.
a(n,m) := s1(-1; n,m), a member of a sequence of triangles including s1(0; n,m)= A023531(n,m) (unit matrix) and s1(2; n,m)= A007318(n-1,m-1) (Pascal's triangle).
The signed triangular matrix a(n,m)*(-1)^(n-m) is the inverse matrix of the triangular Catalan convolution matrix A033184(n+1,m+1), n >= m >= 0, with A033184(n,m) := 0 if n
Riordan array (1+x, x(1+x)). The signed triangle is the Riordan array (1-x,x(1-x)), inverse to (c(x),xc(x)) with c(x) g.f. for A000108. - Paul Barry, Feb 02 2005 [with offset 0]
Also, a(n,k)=number of compositions of n into k parts of 1's and 2's. Example: a(6,4)=6 because we have 2211, 2121, 2112, 1221, 1212 and 1122. - Emeric Deutsch, Apr 05 2005 [see MacMahon and Riordan. - Wolfdieter Lang, Jul 27 2023]
Subtriangle of A026729. - Philippe Deléham, Aug 31 2006
a(n,k) is the number of length n-1 binary sequences having no two consecutive 0's with exactly k-1 1's. Example: a(6,4)=6 because we have 01011, 01101, 01110, 10101, 10110, 11010. - Geoffrey Critzer, Jul 22 2013
Mirrored, shifted Fibonacci polynomials of A011973. The polynomials (illustrated below) of this entry have the property that p(n,t) = t * [p(n-1,t) + p(n-2,t)]. The additive properties of Pascal's triangle (A007318) are reflected in those of these polynomials, as can be seen in the Example Section below and also when the o.g.f. G(x,t) below is expanded as the series x*(1+x) + t * [x*(1+x)]^2 + t^2 * [x*(1+x)]^3 + ... . See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014
Rows of this entry appear as columns of an array for an infinitesimal generator presented in the Copeland link. - Tom Copeland, Dec 23 2015
For n >= 2, the n-th row is also the coefficients of the vertex cover polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017
With an additional initial matrix element a_(0,0) = 1 and column of zeros a_(n,0) = 0 for n > 0, these are antidiagonals read from bottom to top of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper, which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Cf. A011973. And A169803. - Tom Copeland, Jul 02 2018

Examples

			Triangle starts:
  [ 1]  1
  [ 2]  1   1
  [ 3]  0   2   1
  [ 4]  0   1   3   1
  [ 5]  0   0   3   4   1
  [ 6]  0   0   1   6   5   1
  [ 7]  0   0   0   4  10   6   1
  [ 8]  0   0   0   1  10  15   7   1
  [ 9]  0   0   0   0   5  20  21   8   1
  [10]  0   0   0   0   1  15  35  28   9   1
  [11]  0   0   0   0   0   6  35  56  36  10   1
  [12]  0   0   0   0   0   1  21  70  84  45  11   1
  [13]  0   0   0   0   0   0   7  56 126 120  55  12   1
  ...
From _Tom Copeland_, Nov 04 2014: (Start)
For quick comparison to other polynomials:
  p(1,t) = 1
  p(2,t) = 1 + 1 t
  p(3,t) = 0 + 2 t + 1 t^2
  p(4,t) = 0 + 1 t + 3 t^2 + 1 t^3
  p(5,t) = 0 + 0   + 3 t^2 + 4 t^3 +  1 t^4
  p(6,t) = 0 + 0   + 1 t^2 + 6 t^3 +  5 t^4 +  1 t^5
  p(7,t) = 0 + 0   + 0     + 4 t^3 + 10 t^4 +  6 t^5 + 1 t^6
  p(8,t) = 0 + 0   + 0     + 1 t^3 + 10 t^4 + 15 t^5 + 7 t^6 + 1 t^7
  ...
Reading along columns gives rows for Pascal's triangle. (End)
		

References

  • P. A. MacMahon, Combinatory Analysis, Two volumes (bound as one), Chelsea Publishing Company, New York, 1960, Vol. I, nr. 124, p. 151.
  • John Riordan, An Introduction to Combinatorial Analysis, John Wiley & Sons, London, 1958. eq. (35), p.124, 11. p. 154.

Crossrefs

Row sums A000045(n+1) (Fibonacci). a(n, 1)= A019590(n) (Fermat's last theorem). Cf. A049403.

Programs

  • Magma
    /* As triangle */ [[Binomial(k, n-k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Nov 05 2014
  • Maple
    for n from 1 to 12 do seq(binomial(k,n-k),k=1..n) od; # yields sequence in triangular form - Emeric Deutsch, Apr 05 2005
  • Mathematica
    nn=10;CoefficientList[Series[(1+x)/(1-y x - y x^2),{x,0,nn}],{x,y}]//Grid (* Geoffrey Critzer, Jul 22 2013 *)
    Table[Binomial[k, n - k], {n, 13}, {k, n}] // Flatten (* Michael De Vlieger, Dec 23 2015 *)
    CoefficientList[Table[x^(n/2 - 1) Fibonacci[n + 1, Sqrt[x]], {n, 10}],
       x] // Flatten (* Eric W. Weisstein, Apr 10 2017 *)

Formula

a(n, m) = 2*(2*m-n+1)*a(n-1, m)/n + m*a(n-1, m-1)/n, n >= m >= 1; a(n, m) := 0, n
G.f. for m-th column: (x*(1+x))^m.
As a number triangle with offset 0, this is T(n, k) = Sum_{i=0..n} (-1)^(n+i)*binomial(n, i)*binomial(i+k+1, 2k+1). The antidiagonal sums give the Padovan sequence A000931(n+5). Inverse binomial transform of A078812 (product of lower triangular matrices). - Paul Barry, Jun 21 2004
G.f.: (1 + x)/(1 - y*x - y*x^2). - Geoffrey Critzer, Jul 22 2013 [offset 0] [with offset 1: g.f. of row polynomials in y: x*(1+x)*y/(1 - x*(1+x)*y). - Wolfdieter Lang, Jul 27 2023]
From Tom Copeland, Nov 04 2014: (Start)
O.g.f: G(x,t) = x*(1+x) / [1 - t*x*(1+x)] = -P[Cinv(-x),t], where P(x,t)= x / (1 + t*x) and Cinv(x)= x*(1-x) are the compositional inverses in x of Pinv(x,t) = -P(-x,t) = x / (1 - t*x) and C(x) = [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108.
Therefore, Ginv(x,t) = -C[Pinv(-x,t)] = {-1 + sqrt[1 + 4*x/(1+t*x)]}/2, which is -A124644(-x,t).
This places this array in a family of arrays related by composition of P and C and their inverses and interpolation by t, such as A091867 and A104597, and associated to the Catalan, Motzkin, Fine, and Fibonacci numbers. Cf. A104597 (polynomials shifted in t) A125145, A146559, A057078, A000045, A155020, A125145, A039717, A001792, A057862, A011973, A115139. (End)

Extensions

More terms from Emeric Deutsch, Apr 05 2005

A087897 Number of partitions of n into odd parts greater than 1.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 5, 6, 8, 8, 10, 12, 13, 15, 18, 20, 23, 27, 30, 34, 40, 44, 50, 58, 64, 73, 83, 92, 104, 118, 131, 147, 166, 184, 206, 232, 256, 286, 320, 354, 394, 439, 485, 538, 598, 660, 730, 809, 891, 984, 1088, 1196, 1318, 1454, 1596, 1756
Offset: 0

Author

N. J. A. Sloane, Dec 04 2003

Keywords

Comments

Also number of partitions of n into distinct parts which are not powers of 2.
Also number of partitions of n into distinct parts such that the two largest parts differ by 1.
Also number of partitions of n such that the largest part occurs an odd number of times that is at least 3 and every other part occurs an even number of times. Example: a(10) = 2 because we have [2,2,2,1,1,1,1] and [2,2,2,2,2]. - Emeric Deutsch, Mar 30 2006
Also difference between number of partitions of 1+n into distinct parts and number of partitions of n into distinct parts. - Philippe LALLOUET, May 08 2007
In the Berndt reference replace {a -> -x, q -> x} in equation (3.1) to get f(x). G.f. is 1 - x * (1 - f(x)).
Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
Also number of symmetric unimodal compositions of n+3 where the maximal part appears three times. - Joerg Arndt, Jun 11 2013
Let c(n) = number of palindromic partitions of n whose greatest part has multiplicity 3; then c(n) = a(n-3) for n>=3. - Clark Kimberling, Mar 05 2014
From Gus Wiseman, Aug 22 2021: (Start)
Also the number of integer partitions of n - 1 whose parts cover an interval of positive integers starting with 2. These partitions are ranked by A339886. For example, the a(6) = 1 through a(16) = 5 partitions are:
32 222 322 332 432 3322 3332 4332 4432 5432 43332
2222 3222 22222 4322 33222 33322 33332 44322
32222 222222 43222 43322 333222
322222 332222 432222
2222222 3222222
(End)

Examples

			1 + x^3 + x^5 + x^6 + x^7 + x^8 + 2*x^9 + 2*x^10 + 2*x^11 + 3*x^12 + 3*x^13 + ...
q + q^73 + q^121 + q^145 + q^169 + q^193 + 2*q^217 + 2*q^241 + 2*q^265 + ...
a(10)=2 because we have [7,3] and [5,5].
From _Joerg Arndt_, Jun 11 2013: (Start)
There are a(22)=13 symmetric unimodal compositions of 22+3=25 where the maximal part appears three times:
01:  [ 1 1 1 1 1 1 1 1 3 3 3 1 1 1 1 1 1 1 1 ]
02:  [ 1 1 1 1 1 1 2 3 3 3 2 1 1 1 1 1 1 ]
03:  [ 1 1 1 1 1 5 5 5 1 1 1 1 1 ]
04:  [ 1 1 1 1 2 2 3 3 3 2 2 1 1 1 1 ]
05:  [ 1 1 1 2 5 5 5 2 1 1 1 ]
06:  [ 1 1 2 2 2 3 3 3 2 2 2 1 1 ]
07:  [ 1 1 3 5 5 5 3 1 1 ]
08:  [ 1 1 7 7 7 1 1 ]
09:  [ 1 2 2 5 5 5 2 2 1 ]
10:  [ 1 4 5 5 5 4 1 ]
11:  [ 2 2 2 2 3 3 3 2 2 2 2 ]
12:  [ 2 3 5 5 5 3 2 ]
13:  [ 2 7 7 7 2 ]
(End)
From _Gus Wiseman_, Feb 16 2021: (Start)
The a(7) = 1 through a(19) = 8 partitions are the following (A..J = 10..19). The Heinz numbers of these partitions are given by A341449.
  7  53  9    55  B    75    D    77    F      97    H      99      J
         333  73  533  93    553  95    555    B5    755    B7      775
                       3333  733  B3    753    D3    773    D5      955
                                  5333  933    5533  953    F3      973
                                        33333  7333  B33    5553    B53
                                                     53333  7533    D33
                                                            9333    55333
                                                            333333  73333
(End)
		

References

  • J. W. L. Glaisher, Identities, Messenger of Mathematics, 5 (1876), pp. 111-112. see Eq. I

Crossrefs

The ordered version is A000931.
Partitions with no ones are counted by A002865, ranked by A005408.
The even version is A035363, ranked by A066207.
The version for factorizations is A340101.
Partitions whose only even part is the smallest are counted by A341447.
The Heinz numbers of these partitions are given by A341449.
A000009 counts partitions into odd parts, ranked by A066208.
A025147 counts strict partitions with no 1's.
A025148 counts strict partitions with no 1's or 2's.
A026804 counts partitions whose smallest part is odd, ranked by A340932.
A027187 counts partitions with even length/maximum, ranks A028260/A244990.
A027193 counts partitions with odd length/maximum, ranks A026424/A244991.
A058695 counts partitions of odd numbers, ranked by A300063.
A058696 counts partitions of even numbers, ranked by A300061.
A340385 counts partitions with odd length and maximum, ranked by A340386.

Programs

  • Haskell
    a087897 = p [3,5..] where
       p [] _ = 0
       p _  0 = 1
       p ks'@(k:ks) m | m < k     = 0
                      | otherwise = p ks' (m - k) + p ks m
    -- Reinhard Zumkeller, Aug 12 2011
    
  • Maple
    To get 128 terms: t4 := mul((1+x^(2^n)),n=0..7); t5 := mul((1+x^k),k=1..128): t6 := series(t5/t4,x,100); t7 := seriestolist(t6);
    # second Maple program:
    b:= proc(n, i) option remember; `if`(n=0, 1,
          `if`(i<3, 0, b(n, i-2)+`if`(i>n, 0, b(n-i, i))))
        end:
    a:= n-> b(n, n-1+irem(n, 2)):
    seq(a(n), n=0..80);  # Alois P. Heinz, Jun 11 2013
  • Mathematica
    max = 65; f[x_] := Product[ 1/(1 - x^(2k+1)), {k, 1, max}]; CoefficientList[ Series[f[x], {x, 0, max}], x] (* Jean-François Alcover, Dec 16 2011, after Emeric Deutsch *)
    b[n_, i_] := b[n, i] = If[n==0, 1, If[i<3, 0, b[n, i-2]+If[i>n, 0, b[n-i, i]]] ]; a[n_] := b[n, n-1+Mod[n, 2]]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Apr 01 2015, after Alois P. Heinz *)
    Flatten[{1, Table[PartitionsQ[n+1] - PartitionsQ[n], {n, 0, 80}]}] (* Vaclav Kotesovec, Dec 01 2015 *)
    Table[Length[Select[IntegerPartitions[n],FreeQ[#,1]&&OddQ[Times@@#]&]],{n,0,30}] (* Gus Wiseman, Feb 16 2021 *)
  • PARI
    {a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( (1 - x) * eta(x^2 + A) / eta(x + A), n))} /* Michael Somos, Nov 13 2011 */
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A087897_T(n,k):
        if n==0: return 1
        if k<3 or n<0: return 0
        return A087897_T(n,k-2)+A087897_T(n-k,k)
    def A087897(n): return A087897_T(n,n-(n&1^1)) # Chai Wah Wu, Sep 23 2023, after Alois P. Heinz

Formula

Expansion of q^(-1/24) * (1 - q) * eta(q^2) / eta(q) in powers of q.
Expansion of (1 - x) / chi(-x) in powers of x where chi() is a Ramanujan theta function.
G.f.: 1 + x^3 + x^5*(1 + x) + x^7*(1 + x)*(1 + x^2) + x^9*(1 + x)*(1 + x^2)*(1 + x^3) + ... [Glaisher 1876]. - Michael Somos, Jun 20 2012
G.f.: Product_{k >= 1} 1/(1-x^(2*k+1)).
G.f.: Product_{k >= 1, k not a power of 2} (1+x^k).
G.f.: Sum_{k >= 1} x^(3*k)/Product_{j = 1..k} (1 - x^(2*j)). - Emeric Deutsch, Mar 30 2006
a(n) ~ exp(Pi*sqrt(n/3)) * Pi / (8 * 3^(3/4) * n^(5/4)) * (1 - (15*sqrt(3)/(8*Pi) + 11*Pi/(48*sqrt(3)))/sqrt(n) + (169*Pi^2/13824 + 385/384 + 315/(128*Pi^2))/n). - Vaclav Kotesovec, Aug 30 2015, extended Nov 04 2016
G.f.: 1/(1 - x^3) * Sum_{n >= 0} x^(5*n)/Product_{k = 1..n} (1 - x^(2*k)) = 1/((1 - x^3)*(1 - x^5)) * Sum_{n >= 0} x^(7*n)/Product_{k = 1..n} (1 - x^(2*k)) = ..., extending Deutsch's result dated Mar 30 2006. - Peter Bala, Jan 15 2021
G.f.: Sum_{n >= 0} x^(n*(2*n+1))/Product_{k = 2..2*n+1} (1 - x^k). (Set z = x^3 and q = x^2 in Mc Laughlin et al., Section 1.3, Entry 7.) - Peter Bala, Feb 02 2021
a(2*n+1) = Sum{j>=1} A008284(n+1-j,2*j - 1) and a(2*n) = Sum{j>=1} A008284(n-j, 2*j). - Gregory L. Simay, Sep 22 2023
Previous Showing 41-50 of 241 results. Next