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 1-9 of 9 results.

A152726 a(n) = n^7 - (n-1)^7 + (n-2)^7 - ... + ((-1)^n)*0^7.

Original entry on oeis.org

0, 1, 127, 2060, 14324, 63801, 216135, 607408, 1489744, 3293225, 6706775, 12780396, 23051412, 39697105, 65716399, 105142976, 163292480, 247046193, 365173839, 528697900, 751302100, 1049786441, 1444571447, 1960254000
Offset: 0

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Author

Keywords

Crossrefs

Cf. A152725 (6th powers).

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+120*x+1191*x^2+2416*x^3+1191*x^4 +120*x^5+x^6)/( (1+x)*(x-1)^8))); // G. C. Greubel, Sep 01 2018
  • Mathematica
    k=0;lst={k};Do[k=n^7-k;AppendTo[lst,k],{n,1,5!}];lst
    LinearRecurrence[{7, -20, 28, -14, -14, 28, -20, 7, -1}, {0, 1, 127, 2060, 14324, 63801, 216135, 607408, 1489744}, 50] (* G. C. Greubel, Sep 01 2018 *)
    Table[Total[(Times@@@Partition[Riffle[Range[n,1,-1],{1,-1},{2,-1,2}],2])^7],{n,0,30}] (* Harvey P. Dale, Mar 14 2023 *)
  • PARI
    x='x+O('x^50); concat([0], Vec(x*(1+120*x+1191*x^2 +2416*x^3 +1191*x^4+120*x^5+x^6)/((1+x)*(x-1)^8))) \\ G. C. Greubel, Sep 01 2018
    

Formula

G.f.: x*(1 + 120*x + 1191*x^2 + 2416*x^3 + 1191*x^4 + 120*x^5 + x^6)/((1+x)*(x-1)^8). - R. J. Mathar, Jul 08 2013
a(n) = (17*(-1)^n + 84*n^2 - 17 + 28*n^6 + 8*n^7 - 70*n^4)/16. - R. J. Mathar, Jul 08 2013

Extensions

Offset corrected by R. J. Mathar, Jul 08 2013

A062393 a(n) = n^5 - (n-1)^5 + (n-2)^5 - ... +(-1)^n*0^5.

Original entry on oeis.org

0, 1, 31, 212, 812, 2313, 5463, 11344, 21424, 37625, 62375, 98676, 150156, 221137, 316687, 442688, 605888, 813969, 1075599, 1400500, 1799500, 2284601, 2869031, 3567312, 4395312, 5370313, 6511063, 7837844, 9372524, 11138625, 13161375
Offset: 0

Views

Author

Henry Bottomley, Jun 21 2001

Keywords

Comments

The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-6)(P(5,1)-(-1)^k P(5,2k+1))|. - Peter Luschny, Jul 12 2009

Crossrefs

Cf. A000539, A000584. A062392 for 4th powers, A152725 for 6th powers.

Programs

  • Maple
    a := n -> (1-(-1)^n+n^2*(n^2*(2*n+5)-5))/4; # Peter Luschny, Jul 12 2009
  • Mathematica
    k=0;lst={k};Do[k=n^5-k;AppendTo[lst, k], {n, 1, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
    Table[Total[(Times@@@Partition[Riffle[Range[n,0,-1],{1,-1},{2,-1,2}],2])^5],{n,0,30}] (* or *) LinearRecurrence[ {5,-9,5,5,-9,5,-1},{0,1,31,212,812,2313,5463},40] (* Harvey P. Dale, Feb 01 2013 *)
  • PARI
    { a=0; for (n=0, 1000, write("b062393.txt", n, " ", a=n^5 - a) ) } \\ Harry J. Smith, Aug 07 2009

Formula

a(n) = (2*n^5 + 5*n^4 - 5*n^2 + 1 - (-1)^n)/4 = n^5 - a(n-1).
G.f.: x*(x^4 + 26*x^3 + 66*x^2 + 26*x + 1)/((x-1)^6*(x+1)). - Colin Barker, Sep 19 2012
a(0)=0, a(1)=1, a(2)=31, a(3)=212, a(4)=812, a(5)=2313, a(6)=5463, a(n) = 5*a(n-1) - 9*a(n-2) + 5*a(n-3) + 5*a(n-4) - 9*a(n-5) + 5*a(n-6) - a(n-7). - Harvey P. Dale, Feb 01 2013

A152729 a(n) = (n-2)^4 - a(n-1) - a(n-2), with a(1) = a(2) = 0.

Original entry on oeis.org

0, 0, 1, 15, 65, 176, 384, 736, 1281, 2079, 3201, 4720, 6720, 9296, 12545, 16575, 21505, 27456, 34560, 42960, 52801, 64239, 77441, 92576, 109824, 129376, 151425, 176175, 203841, 234640, 268800, 306560, 348161, 393855, 443905, 498576, 558144
Offset: 1

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Author

Keywords

Comments

a(n+2) - a(n-1) = n^4 - (n-1)^4 = A005917(n) for all n in Z. - Michael Somos, Sep 02 2018

Examples

			0 + 0 + 1 = 1^4; 0 + 1 + 15 = 2^4; 1 + 15 + 65 = 3^4; ...
G.f. = x^3 + 15*x^4 + 65*x^5 + 176*x^6 + 384*x^7 + 736*x^8 + 1281*x^9 + ... - _Michael Somos_, Sep 02 2018
		

Crossrefs

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Integers(), m); [0,0] cat Coefficients(R!(x^3*(x+1)*(x^2+10*x+1)/((1-x)^5*(x^2+x+1)))); // G. C. Greubel, Sep 01 2018
  • Mathematica
    k0=k1=0;lst={k0,k1};Do[kt=k1;k1=n^4-k1-k0;k0=kt;AppendTo[lst,k1],{n,1,4!}];lst
    LinearRecurrence[{4,-6,5,-5,6,-4,1}, {0,0,1,15,65,176,384}, 50] (* G. C. Greubel, Sep 01 2018 *)
    a[ n_] := With[ {m = Max[n, 2 - n]}, SeriesCoefficient[ x^3 (1 + x) (1 + 10 x + x^2) / ((1 - x)^5 (1 + x + x^2)), {x , 0, m}]]; (* Michael Somos, Sep 02 2018 *)
  • PARI
    concat([0,0], Vec(-x^3*(x+1)*(x^2+10*x+1)/((x-1)^5*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Oct 28 2014
    
  • PARI
    {a(n) = my(m = max(n, 2 - n)); polcoeff( x^3 * (1 + x) * (1 + 10*x + x^2) / ((1 - x)^5 * (1 + x + x^2)) + x * O(x^m), m)}; /* Michael Somos, Sep 02 2018 */
    

Formula

G.f.: -x^3*(x+1)*(x^2+10*x+1) / ((x-1)^5*(x^2+x+1)). - Colin Barker, Oct 28 2014
a(n) = a(2 - n) for all n in Z. - Michael Somos, Sep 02 2018

Extensions

Definition adapted to offset by Georg Fischer, Jun 18 2021

A152730 a(n) + a(n+1) + a(n+2) = n^5, with a(1) = a(2) = 0.

Original entry on oeis.org

0, 0, 1, 31, 211, 782, 2132, 4862, 9813, 18093, 31143, 50764, 79144, 118924, 173225, 245675, 340475, 462426, 616956, 810186, 1048957, 1340857, 1694287, 2118488, 2623568, 3220568, 3921489, 4739319, 5688099, 6782950, 8040100, 9476950
Offset: 1

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Author

Keywords

Examples

			0 + 0 + 1 = 1^5; 0 + 1 + 31 = 2^5; 1 + 31 + 211 = 3^5; ...
		

Crossrefs

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); [0,0] cat Coefficients(R!(x^3*(x^4+26*x^3+66*x^2+26*x+1)/((x-1)^6*(x^2+x+1)))); // G. C. Greubel, Sep 01 2018
  • Mathematica
    k0=k1=0;lst={k0,k1};Do[kt=k1;k1=n^5-k1-k0;k0=kt;AppendTo[lst,k1],{n,1,5!}];lst
    LinearRecurrence[{5,-10,11,-10,11,-10,5,-1}, {0,0,1,31,211,782,2132, 4862}, 50] (* G. C. Greubel, Sep 01 2018 *)
    CoefficientList[Series[x^2*(x^4 + 26*x^3 + 66*x^2 + 26*x + 1) / ((x - 1)^6*(x^2 + x + 1)),{x, 0, 50}], x] (* Stefano Spezia, Sep 02 2018 *)
  • PARI
    concat([0,0], Vec(x^3*(x^4+26*x^3+66*x^2+26*x+1)/((x-1)^6*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Oct 28 2014
    

Formula

G.f.: x^3*(x^4 + 26*x^3 + 66*x^2 + 26*x + 1) / ((x-1)^6*(x^2 + x + 1)). - Colin Barker, Oct 28 2014

A152731 a(n) + a(n+1) + a(n+2) = n^6, a(1)=a(2)=0.

Original entry on oeis.org

0, 0, 1, 63, 665, 3368, 11592, 31696, 74361, 156087, 300993, 542920, 927648, 1515416, 2383745, 3630375, 5376505, 7770336, 10990728, 15251160, 20803993, 27944847, 37017281, 48417776, 62600832, 80084368, 101455425, 127375983
Offset: 1

Views

Author

Keywords

Comments

0 + 0 + 1 = 1^6; 0 + 1 + 63 = 2^6; ...

Crossrefs

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); [0,0] cat Coefficients(R!(x^3*(1+x)*(x^4+56*x^3+246*x^2+56*x+1)/((1-x)^7*(1 +x+ x^2)))); // G. C. Greubel, Sep 01 2018
  • Mathematica
    k0=k1=0;lst={k0,k1};Do[kt=k1;k1=n^6-k1-k0;k0=kt;AppendTo[lst,k1],{n,1,5!}];lst
    LinearRecurrence[{6, -15, 21, -21, 21, -21, 15, -6, 1}, {0, 0, 1, 63, 665, 3368, 11592, 31696, 74361}, 5000]
    CoefficientList[Series[x^2*(1+x)*(x^4 + 56*x^3 + 246*x^2 + 56*x + 1)/((1-x)^7*(1+x+x^2)),{x, 0, 5000}], x] (* Stefano Spezia, Sep 02 2018 *)
  • PARI
    x='x+O('x^30); concat([0,0], Vec(x^3*(1+x)*(x^4+56*x^3 +246*x^2 +56*x+1)/((1-x)^7*(1+x+x^2)))) \\ G. C. Greubel, Sep 01 2018
    

Formula

From R. J. Mathar, Dec 12 2008: (Start)
a(n) = -26*n/3 + 20*n^3/3 - 5*n^2 + 7/3 - 2*n^5 + n^6/3 + 5*n^4/3 - 7*A131713(n)/3.
G.f.: x^3*(1+x)*(x^4 + 56*x^3 + 246*x^2 + 56*x + 1)/((1-x)^7*(1+x+x^2)). (End)

A152732 a(n) + a(n+1) + a(n+2) = 2^n.

Original entry on oeis.org

0, 0, 2, 2, 4, 10, 18, 36, 74, 146, 292, 586, 1170, 2340, 4682, 9362, 18724, 37450, 74898, 149796, 299594, 599186, 1198372, 2396746, 4793490, 9586980, 19173962, 38347922, 76695844, 153391690, 306783378, 613566756, 1227133514, 2454267026, 4908534052
Offset: 1

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Author

Keywords

Comments

0 + 0 + 2 = 2^1; 0 + 2 + 2 = 2^2; 2 + 2 + 4 = 2^3; 2 + 4 + 10 = 2^4; ...
With a(0)=1, a(n) is the number of length n strings in the language over alphabet {0,1} generated by the regular expression: ((0+1)(0*(11)*)*10)*. - Geoffrey Critzer, Jan 25 2014

Crossrefs

Programs

  • Magma
    I:=[0,0,2]; [n le 3 select I[n] else Self(n-1) +Self(n-2) +2*Self(n-3): n in [1..30]]; // G. C. Greubel, Sep 01 2018
  • Mathematica
    k0=k1=0;lst={k0,k1};Do[kt=k1;k1=2^n-k1-k0;k0=kt;AppendTo[lst,k1],{n,1,5!}];lst
    LinearRecurrence[{1, 1, 2}, {0, 0, 2}, 70] (* Vladimir Joseph Stephan Orlovsky, Feb 24 2012 *)
  • PARI
    concat([0,0],Vec(2/(1-2*x)/(1+x+x^2)+O(x^99))) \\ Charles R Greathouse IV, Feb 24 2012
    

Formula

From R. J. Mathar, Dec 12 2008: (Start)
a(n) = 2*A077947(n-3).
G.f.: 2*x^3/((1-2*x)*(1+x+x^2)). (End)
a(n) = (1/21)*(3*2^n + 18*cos((2*n*Pi)/3) + 2*sqrt(3)*sin((2*n*Pi)/3)). - Zak Seidov, Dec 12 2008

A152733 a(n) + a(n+1) + a(n+2) = 3^n.

Original entry on oeis.org

0, 0, 3, 6, 18, 57, 168, 504, 1515, 4542, 13626, 40881, 122640, 367920, 1103763, 3311286, 9933858, 29801577, 89404728, 268214184, 804642555, 2413927662, 7241782986, 21725348961, 65176046880, 195528140640, 586584421923, 1759753265766, 5279259797298
Offset: 1

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Author

Keywords

Examples

			0 + 0 + 3 = 3^1; 0 + 3 + 6 = 3^2; 3 + 6 + 18 = 3^3; ...
		

Crossrefs

Programs

  • Magma
    [n le 2 select 0 else 3^(n-2) -Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 31 2014
    
  • Mathematica
    k0=k1=0;lst={k0,k1};Do[kt=k1;k1=3^n-k1-k0;k0=kt;AppendTo[lst,k1],{n,1,5!}];lst
    Rest[CoefficientList[Series[3x^3/((1-3x)(1+x+x^2)),{x,0,30}],x]] (* Harvey P. Dale, Aug 31 2014 *)
  • PARI
    x='x+O('x^30); concat([0,0], Vec(3*x^3/((1-3*x)*(1+x+x^2)))) \\ G. C. Greubel, Sep 01 2018

Formula

From R. J. Mathar, Dec 12 2008: (Start)
a(n) = 3*A077834(n-3).
G.f.: 3*x^3/((1-3*x)*(1+x+x^2)). (End)
a(n) = (1/13)*(3^n + 12*cos((2*n*Pi)/3) + 2*sqrt(3)*sin((2*n*Pi)/3)), n=1,2,... - Zak Seidov, Dec 12 2008

A187620 a(n) = n^6 - a(n-1), a(0)=1.

Original entry on oeis.org

1, 0, 64, 665, 3431, 12194, 34462, 83187, 178957, 352484, 647516, 1124045, 1861939, 2964870, 4564666, 6825959, 9951257, 14186312, 19825912, 27219969, 36780031, 48986090, 64393814, 83642075, 107460901
Offset: 0

Views

Author

Vincenzo Librandi, Mar 12 2011

Keywords

Comments

a(n)+a(n-1) is a perfect 6th power, hence a perfect square and a perfect cube.

Crossrefs

Cf. A152725.

Programs

  • Magma
    [(-1)^n + n*(3-5*n^2+3*n^4+n^5)/2: n in [0..30]]; // Vincenzo Librandi, Oct 04 2013
  • Mathematica
    CoefficientList[Series[(- 1 - 78 x^2 - 267 x^3 - 337 x^4 - 36 x^5 - 8 x^6 + x^7 + 6 x)/((1 + x) (x - 1)^7), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 04 2013 *)
    LinearRecurrence[{6,-14,14,0,-14,14,-6,1},{1,0,64,665,3431,12194,34462,83187},30] (* Harvey P. Dale, Apr 30 2020 *)

Formula

From R. J. Mathar, Mar 15 2011: (Start)
a(n) = (-1)^n + n*(3-5*n^2+3*n^4+n^5)/2.
a(n) = (-1)^n + A152725(n).
G.f.: ( -1-78*x^2-267*x^3-337*x^4-36*x^5-8*x^6+x^7+6*x ) / ( (1+x)*(x-1)^7 ). (End)

A261032 a(n) = (-1)^n*(n^8 + 4*n^7 - 14*n^5 + 28*n^3 - 17*n)/2.

Original entry on oeis.org

0, -1, 255, -6306, 59230, -331395, 1348221, -4416580, 12360636, -30686085, 69313915, -145044966, 284936730, -530793991, 944995065, -1617895560, 2677071736, -4298685705, 6721274871, -10262288170, 15337711830, -22485147531, 32390726005, -45920259276, 64155054900, -88432835725
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 18 2015

Keywords

Comments

Alternating sum of eighth powers (A001016).
For n>0, a(n) is divisible by A000217(n).

Examples

			a(0) = 0^8 = 0,
a(1) = 0^8 -1^8 = -1,
a(2) = 0^8 -1^8 + 2^8 = 255,
a(3) = 0^8 -1^8 + 2^8 - 3^8 = -6306,
a(4) = 0^8 -1^8 + 2^8 - 3^8 + 4^8 = 59230,
a(5) = 0^8 -1^8 + 2^8 - 3^8 + 4^8 - 5^8 = -331395, etc.
		

Crossrefs

Programs

  • Magma
    [(-1)^n*(n^8+4*n^7-14*n^5+28*n^3-17*n)/2: n in [0..30]]; // Vincenzo Librandi, Nov 20 2015
    
  • Maple
    seq((-1)^n*(n^8 + 4*n^7 - 14*n^5 + 28*n^3 - 17*n)/2, n = 0 .. 100); # Robert Israel, Nov 18 2015
  • Mathematica
    Table[(1/2) (-1)^n n (n + 1) (n^6 + 3 n^5 - 3 n^4 - 11 n^3 + 11 n^2 + 17 n - 17), {n, 0, 25}]
  • PARI
    vector(100, n, n--; (-1)^n*(n^8+4*n^7-14*n^5+28*n^3-17*n)/2) \\ Altug Alkan, Nov 18 2015
    
  • Sage
    [(-1)^n*(n^8 +4*n^7 -14*n^5 +28*n^3 -17*n)/2 for n in (0..40)] # G. C. Greubel, Apr 02 2021

Formula

G.f.: -x*(1 - 246*x + 4047*x^2 - 11572*x^3 + 4047*x^4 - 246*x^5 + x^6)/(1 + x)^9.
a(n) = Sum_{k = 0..n} (-1)^k*k^8.
a(n) = (-1)^n*n*(n + 1)*(n^6 + 3*n^5 - 3*n^4 - 11*n^3 + 11*n^2 + 17*n - 17)/2.
Sum_{n>0} 1/a(n) = -0.9962225712723456482...
Sum_{j=0..9} binomial(9,j)*a(n-j) = 0. - Robert Israel, Nov 18 2015
E.g.f.: (x/2)*(-2 +253*x -1848*x^2 +2961*x^3 -1596*x^4 +350*x^5 -32*x^6 +x^7)*exp(-x). - G. C. Greubel, Apr 02 2021
Showing 1-9 of 9 results.