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 21-30 of 53 results. Next

A008616 Expansion of 1/((1-x^2)(1-x^5)).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 3, 3, 3, 4, 3, 4, 3, 4, 4, 4, 4, 4, 4, 5, 4, 5, 4, 5, 5, 5, 5, 5, 5, 6, 5, 6, 5, 6, 6, 6, 6, 6, 6, 7, 6, 7, 6, 7, 7, 7, 7, 7, 7, 8, 7, 8, 7, 8, 8, 8, 8, 8, 8, 9, 8, 9, 8, 9, 9, 9, 9, 9, 9, 10, 9, 10, 9, 10, 10, 10, 10, 10, 10
Offset: 0

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Author

Keywords

Comments

Number of partitions of n into parts of size two and five.
It appears that, for n >= 2, a(n-2) is also (1) the number of partitions of 3n that are 6-term arithmetic progressions and (2) floor(n/2) - floor(2n/5). - John W. Layman, Jun 29 2009

References

  • G. E. Andrews, K. Eriksson, Integer Partitions, Cambridge Univ. Press, 2004. page 30, Exercise 48.
  • D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 100.

Crossrefs

Cf. A008615. - John W. Layman, Jun 29 2009

Programs

  • Magma
    m:=100; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/((1-x^2)*(1-x^5)))); // Wesley Ivan Hurt, Dec 27 2021
  • Mathematica
    CoefficientList[Series[1 / ((1 - x^2) (1 - x^5)),{x, 0, 100}], x] (* Vincenzo Librandi, Jun 21 2013 *)
  • PARI
    {a(n) = if( n<-6, -a(-7 - n), polcoeff( 1 / (1 - x^2) / (1 - x^5) + x * O(x^n), n))} /* Michael Somos, Jan 25 2005 */
    
  • PARI
    a(n) = floor(n/10+(3+(-1)^n)/4) \\ Charles R Greathouse IV, Jun 19 2013
    

Formula

G.f.: 1/((1-x^2)(1-x^5)) = 1/((x-1)^2*(1+x)*(1+x+x^2+x^3+x^4)).
Euler transform of finite sequence [0, 1, 0, 0, 1].
a(n) = -a(-7 - n) = a(n - 10) + 1 = a(n - 2) + a(n - 5) - a(n - 7). - Michael Somos, Jan 25 2005
A000217(a(n)) = A025810(n). - Michael Somos, Dec 15 2002
a(n) = 7/20 + n/10 + (-1)^n/4 + (A105384(n) + 2*(A010891(n) + A105384(n+4)))/5. - R. J. Mathar, Jun 28 2009
a(n) = floor(n/10 + (3 + (-1)^n)/4). - Tani Akinari, Jun 20 2013

A165027 Number of n-digit fixed points under the base-4 Kaprekar map A165012.

Original entry on oeis.org

1, 0, 1, 1, 0, 3, 1, 3, 3, 5, 3, 8, 5, 9, 8, 12, 9, 16, 12, 18, 16, 22, 18, 27, 22, 30, 27, 35, 30, 41, 35, 45, 41, 51, 45, 58, 51, 63, 58, 70, 63, 78, 70, 84, 78, 92, 84, 101, 92, 108, 101, 117, 108, 127, 117, 135, 127, 145, 135, 156, 145, 165, 156, 176, 165, 188, 176, 198
Offset: 1

Views

Author

Joseph Myers, Sep 04 2009

Keywords

Crossrefs

In other bases: A004526 (base 2, adjusted to start 1, 0, 0, 1, 1, ...), A008615 (base 3), A008617 (base 5), A165066 (base 6), A008722 (base 7, conjecturally), A165105 (base 8), A165125 (base 9), A164733 (base 10).

Formula

Conjectures from Chai Wah Wu, Apr 13 2024: (Start)
a(n) = 2*a(n-2) + a(n-3) - a(n-4) - 2*a(n-5) + a(n-7) for n > 8.
G.f.: x*(x^7 - x^6 - 2*x^5 + x^4 + x^2 - 1)/((x - 1)^3*(x + 1)^2*(x^2 + x + 1)). (End)

A165066 Number of n-digit fixed points under the base-6 Kaprekar map A165051.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 2, 1, 4, 2, 3, 2, 5, 4, 5, 3, 8, 4, 8, 6, 9, 7, 10, 8, 13, 9, 13, 10, 17, 12, 16, 14, 19, 16, 21, 16, 24, 19, 25, 21, 28, 23, 29, 26, 33, 27, 35, 29, 39, 33, 39, 35, 44, 38, 46, 40, 50, 43, 53, 46, 56, 50, 58, 53, 64, 55, 66, 59, 71, 63, 73, 66, 78, 71, 81, 73, 87
Offset: 1

Views

Author

Joseph Myers, Sep 04 2009

Keywords

Crossrefs

In other bases: A004526 (base 2, adjusted to start 1, 0, 0, 1, 1, ...), A008615 (base 3), A165027 (base 4), A008617 (base 5), A008722 (base 7, conjecturally), A165105 (base 8), A165125 (base 9), A164733 (base 10).

Formula

Conjectures from Chai Wah Wu, Apr 13 2024: (Start)
a(n) = - a(n-1) + a(n-3) + 2*a(n-4) + 2*a(n-5) + a(n-6) - a(n-7) - 2*a(n-8) - 2*a(n-9) - a(n-10) + a(n-12) + a(n-13) for n > 15.
G.f.: x*(-x^14 + x^8 - x^5 + x^4 - x^2 - x - 1)/((x - 1)^3*(x + 1)^2*(x^2 + 1)*(x^2 + x + 1)*(x^4 + x^3 + x^2 + x + 1)). (End)

A025795 Expansion of 1/((1-x^2)*(1-x^3)*(1-x^5)).

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 7, 7, 8, 9, 9, 11, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 21, 23, 24, 25, 27, 28, 29, 31, 32, 34, 35, 37, 38, 40, 42, 43, 45, 47, 48, 51, 52, 54, 56, 58, 60, 62, 64, 66, 68, 71, 72, 75, 77, 79, 82, 84, 86, 89, 91, 94, 96, 99, 101, 104
Offset: 0

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Author

Keywords

Comments

a(n) is the number of ways to pay n dollars with coins of two, three and five dollars. E.g., a(0)=1 because there is one way to pay: with no coin; a(1)=0 no possibility; a(2)=1 (2=1*2); a(3)=1 (3=1*3); a(4)=1 (4=2*2) a(5)=2 (5=3+2=1*5) ... - Richard Choulet, Jan 20 2008
a(n) is the number of partitions of n into parts 2, 3, and 5. See the preceding comment by R. Choulet. - Wolfdieter Lang, Mar 15 2012

Examples

			G.f. = 1 + x^2 + x^3 + x^4 + 2*x^5 + 2*x^6 + 2*x^7 + 3*x^8 + 3*x^9 + 4*x^10 + ...
		

Crossrefs

Programs

  • Mathematica
    a[ n_] := Quotient[n^2 + 10 n + 1 - 13 Mod[n, 2], 60] + 1; (* Michael Somos, Nov 17 2017 *)
  • PARI
    {a(n) = (n^2 + 10*n + 1 - n%2 * 13) \60 + 1} /* Michael Somos, Feb 05 2008 */

Formula

G.f.: 1/((1-x^2)*(1-x^3)*(1-x^5)).
Let [b(1); b(2); ...; b(p)] denote a periodic sequence: e.g., [0; 1] defines the sequence c such that c(0)=c(2)=...=c(2*k)=0 and c(1)=c(3)=...=c(2*k+1)=1. Then a(n)=0.25*[0; 1] - (1/3)*[1; 0; 0] + (1/5)*[0; 1; 1; 0; 3] + ((n+1)*(n+2)/60) + (7*(n+1)/60). - Richard Choulet, Jan 20 2008
If ||A|| is the nearest number to A (A not a half-integer) we also have a(n) = ||((n+1)*(n+9)/60) + (1/5)[0; 1; 1; 0; 3]. - Richard Choulet, Jan 20 2008
a(n) = 77/360 + 7*(n+1)/60 + (n+2)*(n+1)/60 + (-1)^n/8 - (2/9)*cos(2*(n+2)*Pi/3) + (4/(5*sqrt(5)+25))*cos(2*n*Pi/5) - (4/(5*sqrt(5)-25))*cos(4*n*Pi/5). - Richard Choulet, Jan 20 2008
Euler transform of length 5 sequence [0, 1, 1, 0, 1]. - Michael Somos, Feb 05 2008
a(n) = a(-10-n) for all n in Z. - Michael Somos, Feb 25 2008
a(n) - a(n-2) = A008676(n). a(n) - a(n-5) = A103221(n) = A008615(n+2). A078495(n) = 2^(a(n-7) + a(n-9)) * 3^a(n-8) for all n in Z. - Michael Somos, Nov 17 2017, corrected Jun 23 2021
a(n)-a(n-3) = A008616(n). - R. J. Mathar, Jun 23 2021
a(n) = floor((n^2 + 10*n + 6*(9+(-1)^n))/60). - Hoang Xuan Thanh, Jun 15 2025

A165105 Number of n-digit fixed points under the base-8 Kaprekar map A165090.

Original entry on oeis.org

1, 1, 1, 0, 0, 2, 1, 1, 1, 1, 0, 4, 0, 4, 2, 2, 2, 4, 2, 3, 6, 5, 2, 7, 2, 6, 5, 10, 5, 8, 5, 8, 7, 9, 11, 12, 7, 11, 9, 12, 9, 21, 10, 14, 12, 15, 13, 18, 20, 18, 15, 20, 15, 23, 16, 30, 20, 23, 20, 26, 21, 27, 32, 29, 23, 32, 25, 32, 28, 43
Offset: 1

Views

Author

Joseph Myers, Sep 04 2009

Keywords

Crossrefs

In other bases: A004526 (base 2, adjusted to start 1, 0, 0, 1, 1, ...), A008615 (base 3), A165027 (base 4), A008617 (base 5), A165066 (base 6), A008722 (base 7, conjecturally), A165125 (base 9), A164733 (base 10).

Formula

Conjectures from Chai Wah Wu, Apr 13 2024: (Start)
a(n) = - a(n-1) + a(n-3) + 2*a(n-4) + 2*a(n-5) + a(n-6) - a(n-7) - 2*a(n-8) - 2*a(n-9) - a(n-10) + a(n-12) + a(n-13) for n > 15.
G.f.: x*(-x^14 + x^8 - x^5 + x^4 - x^2 - x - 1)/((x - 1)^3*(x + 1)^2*(x^2 + 1)*(x^2 + x + 1)*(x^4 + x^3 + x^2 + x + 1)). (End)

A165125 Number of n-digit fixed points under the base-9 Kaprekar map A165110.

Original entry on oeis.org

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

Views

Author

Joseph Myers, Sep 04 2009

Keywords

Crossrefs

In other bases: A004526 (base 2, adjusted to start 1, 0, 0, 1, 1, ...), A008615 (base 3), A165027 (base 4), A008617 (base 5), A165066 (base 6), A008722 (base 7, conjecturally), A165105 (base 8), A164733 (base 10).

Formula

Conjectures from Chai Wah Wu, Apr 13 2024: (Start)
a(n) = a(n-2) - a(n-3) + a(n-5) - a(n-6) + a(n-8) + a(n-15) - a(n-17) + a(n-18) - a(n-20) + a(n-21) - a(n-23) for n > 24.
G.f.: x*(x^23 + x^22 - x^21 + x^20 + 2*x^19 - x^18 + x^17 + 3*x^16 - 2*x^15 + 3*x^13 - x^12 + 2*x^10 - x^9 + 2*x^7 - x^5 + x^4 + x^3 - x^2 + 1)/(x^23 - x^21 + x^20 - x^18 + x^17 - x^15 - x^8 + x^6 - x^5 + x^3 - x^2 + 1). (End)

A321201 Irregular triangle T with the nontrivial solutions of 2*e2 + 3*e3 = n, for n >= 2, with nonnegative e2 and e3, ordered as pairs with increasing e2 values.

Original entry on oeis.org

1, 0, 0, 1, 2, 0, 1, 1, 0, 2, 3, 0, 2, 1, 1, 2, 4, 0, 0, 3, 3, 1, 2, 2, 5, 0, 1, 3, 4, 1, 0, 4, 3, 2, 6, 0, 2, 3, 5, 1, 1, 4, 4, 2, 7, 0, 0, 5, 3, 3, 6, 1, 2, 4, 5, 2, 8, 0, 1, 5, 4, 3, 7, 1, 0, 6, 3, 4, 6, 2, 9, 0, 2, 5, 5, 3, 8, 1, 1, 6, 4, 4, 7, 2, 10, 0
Offset: 2

Views

Author

Wolfdieter Lang, Nov 05 2018

Keywords

Comments

The length of row n is 2*A(n), with A(n) = A008615(n+2) for n >= 2: 2*[1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 3, 3, 3, 4, 3, 4, ...].
The trivial solution for n = 0 is [0, 0]. There is no solution for n = 1.
The row sums are given in A321202.
If a partition of n with parts 2 or 3 (with inclusive or) is written as 2^{e2} 3^{e3}, where e2 and e3 are nonnegative numbers, then in row n, all pairs [e2, e3] are given, for n >= 2, ordered with increasing values of e2.
The corresponding irregular triangle with the multinomial numbers n!/((n - (e2 + e3))!*e2!*e3!) is given in A321203. It gives the coefficients of x^n = x^{2*{e2} + 3*{e3}} of (1 + x^2 + x^3)^n, for n >= 2.

Examples

			The triangle T(n, k) begins (pairs are separated by commas):
  n\k  0  1   2  3   4  5   6  7 ...
  2:   1  0
  3:   0  1
  4:   2  0
  5:   1  1
  6:   0  2,  3  0
  7:   2  1
  8:   1  2,  4  0
  9:   0  3,  3  1
  10:  2  2,  5  0
  11:  1  3,  4  1
  12:  0  4,  3  2,  6  0
  13:  2  3,  5  1,
  14:  1  4,  4  2,  7  0
  15:  0  5,  3  3,  6  1
  16:  2  4,  5  2,  8  0
  17:  1  5,  4  3,  7  1
  18:  0  6,  3  4,  6  2,  9  0
  19:  2  5,  5  3,  8  1
  20:  1  6,  4  4,  7  2, 10  0
  ...
n=8: the two solutions of 2*e2 + 3*e3 = 8 are [e2, e3] = [1, 2] and = [4, 0], and 1 < 4, therefore row 8 is 1  2  4  0, with a comma after the first pair.
		

Crossrefs

Programs

  • Mathematica
    row[n_] := Reap[Do[If[2 e2 + 3 e3 == n, Sow[{e2, e3}]], {e2, 0, n/2}, {e3, 0, n/3}]][[2, 1]];
    Table[row[n], {n, 2, 20}] // Flatten (* Jean-François Alcover, Nov 23 2018 *)

Formula

T(n, k) gives all pairs [e2, e3] solving 2*e2 + 3*e3 = n, ordered with increasing value of e2, for n >= 2. The trivial solution [0, 0] for n = 0 is not recorded. There is no solution for n = 1.

Extensions

Missing row 2 inserted by Jean-François Alcover, Nov 23 2018

A060548 a(n) is the number of D3-symmetric patterns that may be formed with a top-down equilateral triangular arrangement of closely packed black and white cells satisfying the local matching rule of Pascal's triangle modulo 2, where n is the number of cells in each edge of the arrangement. The matching rule is such that any elementary top-down triangle of three neighboring cells in the arrangement contains either one or three white cells.

Original entry on oeis.org

2, 1, 2, 2, 2, 2, 4, 2, 4, 4, 4, 4, 8, 4, 8, 8, 8, 8, 16, 8, 16, 16, 16, 16, 32, 16, 32, 32, 32, 32, 64, 32, 64, 64, 64, 64, 128, 64, 128, 128, 128, 128, 256, 128, 256, 256, 256, 256, 512, 256, 512, 512, 512, 512, 1024, 512, 1024, 1024, 1024, 1024, 2048, 1024, 2048
Offset: 1

Views

Author

André Barbé (Andre.Barbe(AT)esat.kuleuven.ac.be), Apr 02 2001

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = a[n-2]*a[n-3]/a[n-5]; a[1] = a[3] = a[4] = a[5] = 2; a[2] = 1; Table[a[n], {n, 1, 63}] (* Jean-François Alcover, Dec 27 2011, after second formula *)
    LinearRecurrence[{0,0,0,0,0,2},{2,1,2,2,2,2},70] (* Harvey P. Dale, Sep 19 2016 *)
  • PARI
    a(n)=if(n<1,0,2^((n+3)\6+(n%6==1)))

Formula

a(n) = 2^A008615(n+1) = 2^floor(A008611(n+2)/2) for n >= 1.
a(n) = 2^(floor((n+3)/6) + d(n)), where d(n)=1 if n mod 6=1, else d(n)=0.
a(n) = a(n-2)*a(n-3)/a(n-5), n>5.
From Colin Barker, Aug 29 2013: (Start)
a(n) = 2*a(n-6) for n>1.
G.f.: -x*(2*x^5+2*x^4+2*x^3+2*x^2+x+2) / (2*x^6-1). (End)
Sum_{n>=1} 1/a(n) = 7. - Amiram Eldar, Dec 10 2022

A260708 a(2n) = n*(2*n+1), a(2n+7) = a(2n+1) + 12*n + 28, with a(1)=1, a(3)=6, a(5)=16.

Original entry on oeis.org

0, 1, 3, 6, 10, 16, 21, 29, 36, 46, 55, 68, 78, 93, 105, 122, 136, 156, 171, 193, 210, 234, 253, 280, 300, 329, 351, 382, 406, 440, 465, 501, 528, 566, 595, 636, 666, 709, 741, 786, 820, 868, 903, 953, 990, 1042, 1081, 1136, 1176, 1233, 1275, 1334, 1378
Offset: 0

Views

Author

Paul Curtz, Nov 17 2015

Keywords

Comments

Conjecture: this sequence is 0 followed by A264041.
After 3, if a(n) is prime then n == 1 (mod 6).
a(n) is a square for n = 0, 1, 5, 8, 145, 288, 1777, 6533, 9800, 168097, 332928, 2051425, 7539845, ...

Examples

			a(0) = 0*1 = 0,
a(1) = 1,
a(2) = 1*3 = 3,
a(3) = 6,
a(4) = 2*5 = 10,
a(5) = 16,
a(6) = 3*7 = 21,
a(7) = a(1) +12*0 +28 = 29, etc.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{1, 1, -1, 0, 0, 1, -1, -1, 1}, {0, 1, 3, 6, 10, 16, 21, 29, 36}, 50] (* Bruno Berselli, Nov 18 2015 *)
  • PARI
    concat(0, Vec(-x*(x^6+x^5+3*x^4+2*x^3+2*x^2+2*x+1)/((x-1)^3*(x+1)^2*(x^2-x+1)*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Nov 18 2015
    
  • Sage
    [n*(n+1)/2+(1-(-1)^n)*floor(n/6+1/3)/2 for n in (0..60)] # Bruno Berselli, Nov 18 2015

Formula

From Colin Barker, Nov 17 2015: (Start)
G.f.: x*(1 + 2*x + 2*x^2 + 2*x^3 + 3*x^4 + x^5 + x^6) / ((1 - x)^3*(1 + x)^2*(1 - x + x^2)*(1 + x + x^2)).
a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-6) - a(n-7) - a(n-8) + a(n-9) for n>8. (End)
a(2*k) = A000217(2*k) by definition; for odd indices:
a(6*k+1) = 18*k^2 + 10*k + 1,
a(6*k+3) = 2*(9*k^2 + 11*k + 3),
a(6*k+5) = 2*(k + 1)*(9*k + 8), that is A178574.
a(n) = A260699(n) + A008615(n).
a(n) = n*(n + 1)/2 + (1 - (-1)^n)*floor(n/6 + 1/3)/2. [Bruno Berselli, Nov 18 2015]

Extensions

Edited by Bruno Berselli, Nov 18 2015

A029143 Expansion of 1/((1-x^2)*(1-x^3)*(1-x^5)*(1-x^6)). Molien series for u.g.g.r. #31 of order 46080. Poincaré series [or Poincare series] for ring of even weight Siegel modular forms of genus 2.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 3, 2, 4, 4, 5, 6, 8, 7, 10, 11, 12, 14, 17, 16, 21, 22, 24, 27, 31, 31, 37, 39, 42, 46, 52, 52, 60, 63, 67, 73, 80, 81, 91, 95, 101, 108, 117, 119, 131, 137, 144, 153, 164, 167, 182, 189, 198, 209, 222
Offset: 0

Views

Author

Keywords

Comments

a(k) for k>0 is the dimension of the space of Siegel modular forms of genus 2 and weight 2k (for the full modular group Gamma_2). Also: Number of solutions of 4x+6y+10z+12w=k in nonnegative integers x,y,z,w. - Kilian Kilger (kilian(AT)nihilnovi.de), Sep 26 2009
Number of partitions of n into parts 2, 3, 5, and 6. - Joerg Arndt, Jun 21 2014

References

  • H. Klingen, Intro. lectures on Siegel modular forms, Cambridge, p. 123, Corollary.
  • L. Smith, Polynomial Invariants of Finite Groups, Peters, 1995, p. 199 (No. 31).

Crossrefs

Cf. A027640 for the dimension of even and odd weight Siegel modular forms. See A165684 (resp. A165685) for the corresponding space of cusp forms. - Kilian Kilger (kilian(AT)nihilnovi.de), Sep 26 2009

Programs

  • Maple
    M := Matrix(16, (i,j)-> if (i=j-1) or (j=1 and member(i, [2, 3, 6, 10, 13, 14])) then 1 elif j=1 and member(i, [7, 9, 16]) then -1 elif j=1 and i=8 then -2 else 0 fi): a:= n -> (M^(n))[1,1]: seq(a(n), n=0..54); # Alois P. Heinz, Jul 25 2008
  • Mathematica
    CoefficientList[Series[1/((1-x^2)*(1-x^3)*(1-x^5)*(1-x^6)),{x,0,54}],x] (* Jean-François Alcover, Mar 20 2011 *)
    LinearRecurrence[{0,1,1,0,0,1,-1,-2,-1,1,0,0,1,1,0,-1},{1,0,1,1,1,2,3,2,4,4,5,6,8,7,10,11},60] (* Harvey P. Dale, May 12 2015 *)

Formula

a(n) = A165684(n) + A008615(n+2). - Kilian Kilger (kilian(AT)nihilnovi.de), Sep 26 2009
a(n) ~ 1/1080*n^3. - Ralf Stephan, Apr 29 2014
a(0)=1, a(1)=0, a(2)=1, a(3)=1, a(4)=1, a(5)=2, a(6)=3, a(7)=2, a(8)=4, a(9)=4, a(10)=5, a(11)=6, a(12)=8, a(13)=7, a(14)=10, a(15)=11, a(n)= a(n-2)+ a(n-3)+a(n-6)-a(n-7)- 2*a(n-8)-a(n-9)+a(n-10)+a(n-13)+ a(n-14)- a(n-16). - Harvey P. Dale, May 12 2015

Extensions

Definition corrected by Kilian Kilger (kilian(AT)nihilnovi.de), Sep 25 2009
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