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 1-5 of 5 results.

A142883 a(n) = A142590(n)/3.

Original entry on oeis.org

0, 7, 5, 39, 4, 95, 33, 175, 14, 279, 85, 407, 30, 559, 161, 735, 52, 935, 261, 1159, 80, 1407, 385, 1679, 114, 1975, 533, 2295, 154, 2639, 705, 3007, 200, 3399, 901, 3815, 252, 4255, 1121, 4719, 310, 5207, 1365, 5719, 374, 6255, 1633, 6815, 444, 7399, 1925, 8007
Offset: 0

Views

Author

Paul Curtz, Sep 28 2008

Keywords

Programs

Extensions

Edited and extended by R. J. Mathar, Sep 07 2009

A165943 a(n) = A061037(7*n+2).

Original entry on oeis.org

0, 77, 63, 525, 56, 1365, 483, 2597, 210, 4221, 1295, 6237, 462, 8645, 2499, 11445, 812, 14637, 4095, 18221, 1260, 22197, 6083, 26565, 1806, 31325, 8463, 36477, 2450, 42021, 11235, 47957, 3192, 54285, 14399, 61005, 4032, 68117, 17955, 75621, 4970
Offset: 0

Views

Author

Paul Curtz, Oct 01 2009

Keywords

Comments

The (2k+1)-sections A061037((2*k+1)*n+2) are multiples of 2k+1:
0,...21,...15,..117,...12,..285,...99,..525,...42,..837,..255, k=1, A142590
0,...45,...35,..285,...30,..725,..255,.1365,..110,.2205,..675, k=2, A165248
0,...77,...63,..525,...56,.1365,..483,.2597,..210,.4221,.1295, k=3, here
0,..117,...99,..837,...90,.2205,..783,.4221,..342,.6885,.2115, k=4,
0,..165,..143,.1221,..132,.3245,.1155,.6237,..506,10197,.3135, k=5
0,..221,..195,.1677,..182,.4485,.1599,.8645,..702,14157,.4355, k=6
After division by 2k+1 these define a table T'(k,c) :
0,....7,....5,...39,....4,...95,...33,..175,...14,..279,...85, k=1, A142883
0,....9,....7,...57,....6,..145,...51,..273,...22,..441,..135, k=2
0,...11,....9,...75,....8,..195,...69,..371,...30,..603,..185, k=3
0,...13,...11,...93,...10,..245,...87,..469,...38,..765,..235, k=4
0,...15,...13,..111,...12,..295,..105,..567,...46,..927,..285, k=5
0,...17,...15,..129,...14,..345,..123,..665,...54,.1089,..335, k=6
Differences downwards each second column in this second table are 2 = 7-5 = 9-7..; 18 = 57-39 = 75-57..; 50 = 145-95 = 195-145... = A077591(n+1) = 2*A016754.
The difference T'(k+1,c)-T'(k,c) is 0, 2, 2, 18, 2, 50, 18, 98, 8 ... = 2*A181318(c) =A061037(c-2)+A061037(c+2). - Paul Curtz, Mar 12 2012
Let b(n)= a(n) mod 11. The sequence b(n) has the property b(n+44) = b(n) with the first 43 values being {0, 0 , 8, 1, 1, 10, 1, 1, 8, 8, 0, 0, 10, 5, 5, 9, 7, 1, 5, 6, 10, 7, 0, 2, 1, 1, 8, 1, 5, 8, 2, 0, 8, 10, 6, 5, 10, 7, 9, 5, 0, 10, 0}. - G. C. Greubel, Apr 18 2016

Links

Programs

  • Magma
    m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))); // G. C. Greubel, Sep 19 2018
  • Maple
    seq(numer(1/4 - 1/(7*n+2)^2), n=0..50); # Robert Israel, Apr 20 2016
  • Mathematica
    Table[Numerator[1/4 - 1/(7 n + 2)^2], {n, 0, 40}] (* Michael De Vlieger, Apr 19 2016 *)
CoefficientList[Series[7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 19 2018 *)
  • PARI
    a(n) = numerator(1/4 - 1/(7*n+2)^2); \\  Altug Alkan, Apr 18 2016
  • PARI
    x='x+O('x^50); concat([0], Vec(7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))) \\ G. C. Greubel, Sep 19 2018

Formula

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>12. - Conjectured by R. J. Mathar, Mar 02 2010, proved by Robert Israel, Apr 20 2016
From Ilya Gutkovskiy, Apr 19 2016: (Start)
G.f.: 7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3).
a(n) = -7*n*(7*n + 4)*(27*(-1)^n + 6*cos((Pi*n)/2) - 37)/64. (End)

Extensions

Partially edited and extended by R. J. Mathar, Mar 02 2010
Removed division by 7 in definition and formula - R. J. Mathar, Mar 23 2010

A165248 Quintisection A061037(5*n+2).

Original entry on oeis.org

0, 45, 35, 285, 30, 725, 255, 1365, 110, 2205, 675, 3245, 240, 4485, 1295, 5925, 420, 7565, 2115, 9405, 650, 11445, 3135, 13685, 930, 16125, 4355, 18765, 1260, 21605, 5775, 24645, 1640, 27885, 7395, 31325, 2070, 34965, 9215, 38805, 2550, 42845, 11235, 47085
Offset: 0

Views

Author

Paul Curtz, Sep 10 2009

Keywords

Comments

A trisection of A061037 is in A142590. These (2k+1)-sections A061037(2+n*(2k+1)) are multiples of 2k+1.

Links

Programs

  • Magma
    m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 + 4*x^7 + 33*x^8 + 3*x^9 + x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))); // G. C. Greubel, Sep 19 2018
  • Mathematica
    CoefficientList[Series[5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 +4*x^7 + 33*x^8 + 3*x^9 +x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 19 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,45,35,285,30,725,255,1365,110,2205,675,3245},50] (* Harvey P. Dale, Sep 18 2021 *)
  • PARI
    a(n) = numerator(1/4 - 1/(5*n+2)^2); \\ Altug Alkan, Apr 19 2016
  • PARI
    x='x+O('x^50); concat([0], Vec(5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 + 4*x^7 + 33*x^8 + 3*x^9 + x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))) \\ G. C. Greubel, Sep 19 2018

Formula

Conjecture: a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>11. - R. J. Mathar, Mar 02 2010
The conjecture is equivalent to a(4n) = 5n*(5n+1), a(4n+1) = 5*(20n+9)*(4n+1), a(4n+2) = 5*(10n+7)*(2n+1) and a(4n+3) = 5*(20n+19)*(4n+3). - R. J. Mathar, Feb 13 2011
The conjectures can be proved by taking the closed form of A061037, and writing up the quadrisections case by case. - Bruno Berselli, Feb 20 2011
From Ilya Gutkovskiy, Apr 19 2016: (Start)
G.f.: 5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 + 4*x^7 + 33*x^8 + 3*x^9 + x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3).
a(n) = -5*n (5*n + 4)*(27*(-1)^n + 6*cos((Pi*n)/2) - 37)/64. (End)

Extensions

Extended by R. J. Mathar, Mar 02 2010

A174325 Trisection A061037(3*n-2) of the Balmer spectrum numerators extended to negative indices.

Original entry on oeis.org

0, -3, 3, 45, 6, 165, 63, 357, 30, 621, 195, 957, 72, 1365, 399, 1845, 132, 2397, 675, 3021, 210, 3717, 1023, 4485, 306, 5325, 1443, 6237, 420, 7221, 1935, 8277, 552, 9405, 2499, 10605, 702, 11877, 3135, 13221, 870, 14637, 3843, 16125, 1056, 17685, 4623, 19317
Offset: 0

Views

Author

Paul Curtz, Nov 27 2010

Keywords

Comments

All terms are multiples of 3.

Links

Crossrefs

Cf. A061037, A142590, A142600.

Programs

  • Magma
    I:=[0,-3,3,45,6,165,63,357,30,621,195,957]; [n le 12 select I[n] else 3*Self(n-4)-3*Self(n-8)+Self(n-12): n in [1..50]]; // Vincenzo Librandi, Oct 15 2014
  • Mathematica
    Table[Numerator[(n-2)*(n+2)/(4*n^2)],{n,-2,300,3}] (* Vaclav Kotesovec, Oct 15 2014 *)
  • PARI
    concat(0, Vec(-3*x*(7*x^10 +5*x^9 +39*x^8 +4*x^7 +74*x^6 +18*x^5 +58*x^4 +2*x^3 +15*x^2 +x -1) / ((x -1)^3*(x +1)^3*(x^2 +1)^3) + O(x^100))) \\ Colin Barker, Oct 15 2014

Formula

a(n) = A142600(n-1), n>1.
G.f.: -3*x*(7*x^10 +5*x^9 +39*x^8 +4*x^7 +74*x^6 +18*x^5 +58*x^4 +2*x^3 +15*x^2 +x -1) / ((x -1)^3*(x +1)^3*(x^2 +1)^3). - Colin Barker, Oct 15 2014
Sum_{n>=1} 1/a(n) = 11*log(3)/16 - 5*Pi/(48*sqrt(3)) - 1/4. - Amiram Eldar, Sep 11 2022

A142599 Second trisection of A061037.

Original entry on oeis.org

5, 2, 77, 35, 221, 20, 437, 143, 725, 56, 1085, 323, 1517, 110, 2021, 575, 2597, 182, 3245, 899, 3965, 272, 4757, 1295, 5621, 380, 6557, 1763, 7565, 506, 8645, 2303, 9797, 650, 11021, 2915, 12317, 812, 13685, 3599, 15125, 992, 16637, 4355, 18221, 1190, 19877, 5183, 21605, 1406, 23405
Offset: 1

Views

Author

Paul Curtz, Sep 23 2008

Keywords

Comments

From Balmer spectrum of hydrogen. First trisection is A142590.

Links

Crossrefs

Cf. A061037, A142590.

Programs

  • Mathematica
    A061037[n_] := Numerator[(n-2)*(n+2)/(4*n^2)]; Table[A061037[n], {n, 1, 300}][[3 ;;  ;; 3]] (* G. C. Greubel, Sep 19 2018 *)

Formula

Sum_{n>=1} 1/a(n) = 5*Pi/(24*sqrt(3)) + 1/2. - Amiram Eldar, Sep 11 2022

Extensions

Edited by N. J. A. Sloane, Sep 27 2008
Terms a(18) onward added by G. C. Greubel, Sep 19 2018
Showing 1-5 of 5 results.

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