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 1-6 of 6 results.

A038201 5-wave sequence.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 3, 4, 5, 9, 12, 14, 15, 29, 41, 50, 55, 105, 146, 175, 190, 365, 511, 616, 671, 1287, 1798, 2163, 2353, 4516, 6314, 7601, 8272, 15873, 22187, 26703, 29056, 55759, 77946, 93819, 102091, 195910, 273856, 329615, 358671, 688286, 962142
Offset: 0

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Author

Keywords

Comments

This sequence is related to the hendecagon or 11-gon, see A120747.
Sequence of perfect distributions for a cascade merge with six tapes according to Knuth. - Michael Somos, Feb 07 2004

Examples

			The first few rows of the T(n,k) array are, n>=1, 1 <= k <=5:
  0,   0,   0,   0,   1
  1,   1,   1,   1,   1
  1,   2,   3,   4,   5
  5,   9,   12,  14,  15
  15,  29,  41,  50,  55
  55,  105, 146, 175, 190
  190, 365, 511, 616, 671
G.f. = 1 + x + x^2 + x^3 + x^4 + 2*x^5 + 3*x^6 + 4*x^7 + 9*x^8 + 12*x^9 + ...
		

References

  • D. E. Knuth, Art of Computer Programming, Vol. 3, Sect. 5.4.3, Eq. (1).

Crossrefs

The a(4*n) values (column 0) lead to A006358; the T(n,k) lead to A069006, A038342 and A120747.

Programs

  • Maple
    m:=5: nmax:=12: for k from 1 to m-1 do T(1,k):=0 od: T(1,m):=1: for n from 2 to nmax do for k from 1 to m do T(n,k):= add(T(n-1,k1), k1=m-k+1..m) od: od: for n from 1 to nmax/2 do seq(T(n,k), k=1..m) od; a(0):=1: Tx:=1: for n from 2 to nmax do for k from 2 to m do a(Tx):= T(n,k): Tx:=Tx+1: od: od: seq(a(n), n=0..Tx-1); # Johannes W. Meijer, Aug 03 2011
  • Mathematica
    LinearRecurrence[{0,0,0,3,0,0,0,3,0,0,0,-4,0,0,0,-1,0,0,0,1},{1,1,1,1,1,2,3,4,5,9,12,14,15,29,41,50,55,105,146,175},50] (* Harvey P. Dale, Dec 13 2012 *)
  • PARI
    {a(n) = local(m); if( n<=0, n==0, m = (n-1)\4 * 4; sum(k=2*m - n, m, a(k)))};

Formula

a(n) = a(n-1)+a(n-2) if n=4*m+1, a(n) = a(n-1)+a(n-4) if n=4*m+2, a(n) = a(n-1)+a(n-6) if n=4*m+3 and a(n) = a(n-1)+a(n-8) if n=4*m.
G.f.: -(1+x+x^2+x^3-2*x^4-x^5+x^7-x^8-x^11+x^12)/(-1+3*x^4+3*x^8-4*x^12-x^16+x^20).
a(n) = 3*a(n-4)+3*a(n-8)-4*a(n-12)-a(n-16)+a(n-20).
a(n-1) = sequence(sequence(T(n,k), k=2..5), n>=2) with a(0)=1; T(n,k) = sum(T(n-1,k1), k1 = 6-k..5) with T(1,1) = T(1,2) = T(1,3) = T(1,4) = 0 and T(1,5) = 1; n>=1 and 1 <= k <= 5. [Steinbach]

Extensions

Edited by Floor van Lamoen, Feb 05 2002
Edited and information added by Johannes W. Meijer, Aug 03 2011

A069006 Let M denote the 5 X 5 matrix with rows /1,1,1,1,1/1,1,1,1,0/1,1,1,0,0/1,1,0,0,0/1,0,0,0,0/ and A(n) = vector (x(n),y(n),z(n),t(n),u(n)) = M^n*A where A is the vector (1,1,1,1,1); then a(n) = t(n).

Original entry on oeis.org

1, 2, 9, 29, 105, 365, 1287, 4516, 15873, 55759, 195910, 688286, 2418195, 8495917, 29849041, 104869718, 368442700, 1294463368, 4547886208, 15978257251, 56137003923, 197228218022, 692929213991, 2434493909304, 8553197751125
Offset: 0

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Author

Benoit Cloitre, Apr 02 2002

Keywords

Comments

a(n-1) (with a(-1) = 0) appears in the formula for 1/rho(11)^n, n >= 0, with rho(11) = 2*cos(Pi/11) (the length ratio (smallest diagonal)/side in the regular 11-gon), when written in the power basis of the degree 5 number field Q(rho(11)): 1/rho(11)^n = A038342(n)*1 + A230080*rho(11) - A230081(n)*rho(11)^2 - a(n-1)*rho(11)^3 + A038342(n-1)* rho(11)^4, n >= 0, with A038342(-1) = 0. See A230080 with the example for n=4. - Wolfdieter Lang, Nov 04 2013

Crossrefs

Cf. A006359, A069007, A069008, A069009, A070778, A006359(offset), for x(n), y(n), z(n), t(n), u(n), v(n).
A038342, A230080, A230081 (for powers of 1/rho(11), see a comment above).

Programs

  • Mathematica
    LinearRecurrence[{3,3,-4,-1,1},{1,2,9,29,105},30] (* Harvey P. Dale, Apr 16 2015 *)

Formula

G.f.:(1-x)/(1-x^5+x^4+4*x^3-3*x^2-3*x). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 12 2009
a(n) = 3*a(n-1) + 3*a(n-2) - 4*a(n-3) - a(n-4) + a(n-5), n >= 0, with a(-5)=0, a(-4)=-1, a(-3)=a(-2)=a(-1)=0. - Wolfdieter Lang, Nov 04 2013

Extensions

Edited by Henry Bottomley, May 06 2002

A120747 Sequence relating to the 11-gon (or hendecagon).

Original entry on oeis.org

0, 1, 4, 14, 50, 175, 616, 2163, 7601, 26703, 93819, 329615, 1158052, 4068623, 14294449, 50221212, 176444054, 619907431, 2177943781, 7651850657, 26883530748, 94450905714, 331837870408, 1165858298498, 4096053203771, 14390815650209, 50559786403254
Offset: 1

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Author

Gary W. Adamson, Jul 01 2006

Keywords

Comments

The hendecagon is an 11-sided polygon. The preferred word in the OEIS is 11-gon.
The lengths of the diagonals of the regular 11-gon are r[k] = sin(k*Pi/11)/sin(Pi/11), 1 <= k <= 5, where r[1] = 1 is the length of the edge.
The value of limit(a(n)/a(n-1),n=infinity) equals the longest diagonal r[5].
The a(n) equal the matrix elements M^n[1,2], where M = Matrix([[1,1,1,1,1], [1,1,1,1,0], [1,1,1,0,0], [1,1,0,0,0], [1,0,0,0,0]]). The characteristic polynomial of M is (x^5 - 3x^4 - 3x^3 + 4x^2 + x - 1) with roots x1 = -r[4]/r[3], x2 = -r[2]/r[4], x3 = r[1]/r[2], x4 = r[3]/r[5] and x5 = r[5]/r[1].
Note that M^4*[1,0,0,0,0] = [55, 50, 41, 29, 15] which are all terms of the 5-wave sequence A038201. This is also the case for the terms of M^n*[1,0,0,0,0], n>=1.

Examples

			From _Johannes W. Meijer_, Aug 03 2011: (Start)
The lengths of the regular hendecagon edge and diagonals are:
  r[1] = 1.000000000, r[2] = 1.918985948, r[3] = 2.682507066,
  r[4] = 3.228707416, r[5] = 3.513337092.
The first few rows of the T(n,k) array are, n>=1, 1 <= k <=5:
    0,   0,   0,   0,   1, ...
    1,   1,   1,   1,   1, ...
    1,   2,   3,   4,   5, ...
    5,   9,  12,  14,  15, ...
   15,  29,  41,  50,  55, ...
   55, 105, 146, 175, 190, ...
  190, 365, 511, 616, 671, ... (End)
		

Crossrefs

From Johannes W. Meijer, Aug 03 2011: (Start)
Cf. A006358 (T(n+2,1) and T(n+1,5)), A069006 (T(n+1,2)), A038342 (T(n+1,3)), this sequence (T(n,4)) (m=5: hendecagon or 11-gon).
Cf. A000045 (m=2; pentagon or 5-gon); A006356, A006054 and A038196 (m=3: heptagon or 7-gon); A006357, A076264, A091024 and A038197 (m=4: enneagon or 9-gon); A006359, A069007, A069008, A069009, A070778 (m=6; tridecagon or 13-gon); A025030 (m=7: pentadecagon or 15-gon); A030112 (m=8: heptadecagon or 17-gon). (End)

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x^2*(1+x-x^2)/(1-3*x-3*x^2+4*x^3+x^4-x^5) )); // G. C. Greubel, Nov 13 2022
    
  • Maple
    nmax:=27: m:=5: for k from 1 to m-1 do T(1,k):=0 od: T(1,m):=1: for n from 2 to nmax do for k from 1 to m do T(n,k):= add(T(n-1,k1), k1=m-k+1..m) od: od: for n from 1 to nmax/3 do seq(T(n,k), k=1..m) od; for n from 1 to nmax do a(n):=T(n,4) od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Aug 03 2011
  • Mathematica
    LinearRecurrence[{3, 3, -4, -1, 1}, {0, 1, 4, 14, 50}, 41] (* G. C. Greubel, Nov 13 2022 *)
  • SageMath
    def A120747_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( x*(1+x-x^2)/(1-3*x-3*x^2+4*x^3+x^4-x^5) ).list()
    A120747_list(40) # G. C. Greubel, Nov 13 2022

Formula

a(n) = 3*a(n-1) + 3*a(n-2) - 4*a(n-3) - a(n-4) + a(n-5).
G.f.: x^2*(1+x-x^2)/(1-3*x-3*x^2+4*x^3+x^4-x^5). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 12 2009
From Johannes W. Meijer, Aug 03 2011: (Start)
a(n) = T(n,4) with T(n,k) = Sum_{k1 = 6-k..6} T(n-1, k1), T(1,1) = T(1,2) = T(1,3) = T(1,4) = 0 and T(1,5) = 1, n>=1 and 1 <= k <= 5. [Steinbach]
Sum_{k=1..5} T(n,k)*r[k] = r[5]^n, n>=1. [Steinbach]
r[k] = sin(k*Pi/11)/sin(Pi/11), 1 <= k <= 5. [Kappraff]
Sum_{k=1..5} T(n,k) = A006358(n-1).
Limit_{n -> 00} T(n,k)/T(n-1,k) = r[5], 1 <= k <= 5.
sequence(sequence( T(n,k), k=2..5), n>=1) = A038201(n-4).
G.f.: (x^2*(x - x1)*(x - x2))/((x - x3)*(x - x4)*(x - x5)*(x - x6)*(x - x7)) with x1 = phi, x2 = (1-phi), x3 = r[1] - r[3], x4 = r[3] - r[5], x5 = r[5] - r[4], x6 = r[4] - r[2], x7 = r[2], where phi = (1 + sqrt(5))/2 is the golden ratio A001622. (End)

Extensions

Edited and information added by Johannes W. Meijer, Aug 03 2011

A230080 Sequence needed for the nonpositive powers of rho(11) = 2*cos(Pi/11) in terms of the power basis of the degree 5 number field Q(rho(11)). Coefficients of the first power.

Original entry on oeis.org

0, 3, 5, 23, 73, 265, 920, 3245, 11385, 40018, 140574, 493911, 1735243, 6096533, 21419128, 75252674, 264387942, 928884046, 3263482673, 11465714843, 40282921096, 141527481021, 497233748352, 1746949771565, 6137623429414
Offset: 0

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Author

Wolfdieter Lang, Nov 04 2013

Keywords

Comments

The formula for the nonpositive powers of rho(11) := 2*cos(Pi/11) (the length ratio (smallest diagonal/side) in the regular 11-gon), when written in the power basis of the degree 5 algebraic number field Q(rho(11)) is: 1/rho(11)^n = A038342(n)*1 + a(n)*rho(11) - A230081(n)*rho(11)^2 - A069006(n-1)*rho(11)^3 + A038342(n-1)*rho(11)^4, n >= 0, with A069006(-1) = 0 = A038342(-1).

Examples

			1/rho(11)^4 = 146*1 + 73*rho(11) - 173*rho(11)^2 - 29*rho(11)^3  + 41*rho(11)^4 (approximately 0.07374164519).
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{3,3,-4,-1,1},{0,3,5,23,73},30] (* Harvey P. Dale, May 19 2017 *)

Formula

G.f.: x*(3 - 4*x - x^2 + x^3)/(1 - 3*x - 3*x^2 + 4*x^3 + x^4 - x^5).
a(n) = 3*a(n-1) +3*a(n-2) -4*a(n-3) -a(n-4) +a(n-5) for n >= 0, with a(-5)=-3, a(-4)=a(-3)=a(-2)=0, a(-1)=1.

A230081 Sequence needed for the nonpositive powers of rho(11) = 2*cos(Pi/11) in the power basis of the degree 5 number field Q(rho(11)). Negative of the coefficients of the second power.

Original entry on oeis.org

0, 4, 13, 50, 173, 613, 2149, 7557, 26543, 93264, 327657, 1151183, 4044478, 14209634, 49923211, 175397097, 616229093, 2165020570, 7606447024, 26724012524, 93890464368, 329868851103, 1158940469863, 4071748539926, 14305425173111
Offset: 0

Views

Author

Wolfdieter Lang, Nov 04 2013

Keywords

Comments

The formula for the nonpositive powers of rho(11) := 2*cos(Pi/11) (the length ratio (smallest diagonal/side) in the regular 11-gon), when written in the power basis of the degree 5 algebraic number field Q(rho(11)) is: 1/rho(11)^n = A038342(n)*1 + A230080*rho(11) - a(n)*rho(11)^2 - A069006(n-1)*rho(11)^3 + A038342(n-1)*rho(11)^4, n >= 0, with A069006(-1) = 0 = A038342(-1).

Examples

			1/rho(11)^4 = 146*1 + 73*rho(11) - 173*rho(11)^2 - 29*rho(11)^3  + 41*rho(11)^4 (approximately 0.07374164519).
		

Crossrefs

Formula

G.f.: x*(4 + x - x^2) / (1 - 3*x - 3*x^2 + 4*x^3 + x^4 - x^5).
a(n) = 3*a(n-1) +3*a(n-2) -4*a(n-3) -a(n-4) +a(n-5) for n >= 0, with a(-5)=3, a(-4)=0, a(-3)=0, a(-2)=-1, a(-1)=0.

A231186 Decimal expansion of the ratio (longest diagonal)/side in a regular 11-gon (or hendecagon).

Original entry on oeis.org

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

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Author

Wolfdieter Lang, Nov 20 2013

Keywords

Comments

omega(11):= S(4, x) = 1 - 3*x^2 + x^4 with x = rho(11) := 2*cos(Pi/11). See A049310 for Chebyshev s-polynomials. rho(11) is the ratio (shortest diagonal)/side in a regular 11-gon. See the Q(2*cos(Pi/n)) link given in A187360. This is the power basis representation of omega(11) in the algebraic number field Q(2*cos(Pi/11)) of degree 5.
omega(11) = 1/(2*cos(Pi*5/11)) = 1/R(5, rho(11)) with the R-polynomial given in A127672. This follows from a computation of the power basis coefficients of the reciprocal of R(5, x) (mod C(11, x)) = 1+2*x-3*x^2-x^3+x^4, where C(11, x) is the minimal polynomial of rho(11) given in A187360. The result for this reciprocal (mod C(11, x)) is 1 - 3*x^2 + x^4 giving the power base coefficients [1,0,-3,0,1] for omega(11).
omega(11) is the analog in the regular 11-gon of the golden section in the regular 5-gon (pentagon), because it is the limit of a(n+1)/a(n) for n -> infinity of sequences like A038342, A069006, A230080 and A230081.
The ratio diagonal/side of the second and third shortest diagonals in a regular 11-gon are respectively x^2 - 1 and x^3 - 2*x, where x = 2*cos(Pi/11). - Mohammed Yaseen, Nov 03 2020

Examples

			3.51333709166613518878217159629798184207459481777014...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[Sin[5*Pi/11]/Sin[Pi/11], 10, 120][[1]] (* Amiram Eldar, Jun 01 2023 *)

Formula

omega(11) = 1 - 3*x^2 + x^4 with x = rho(11) := 2*cos(Pi/11) = 1/(2*cos(Pi*5/11)) = 3.5133370916661... See the comments above.
Equals sin(5*Pi/11)/sin(Pi/11). - Mohammed Yaseen, Nov 03 2020
Showing 1-6 of 6 results.