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 1-3 of 3 results.

A261540 a(n) = n^7 + 7*n^5 + 14*n^3 + 7*n.

Original entry on oeis.org

0, 29, 478, 4287, 24476, 101785, 337434, 946043, 2333752, 5206581, 10714070, 20633239, 37597908, 65378417, 109216786, 176222355, 275832944, 420346573, 625528782, 911300591, 1302512140, 1829807049, 2530582538, 3450050347, 4642403496, 6172093925, 8115226054
Offset: 0

Views

Author

Raphael Ranna, Aug 24 2015

Keywords

Comments

Also numbers of the form (n-th metallic mean)^7 - 1/(n-th metallic mean)^7, see link to Wikipedia.

Links

Crossrefs

Cf. A001622, A014176, A098316, A098317, A098318, A176398, A176439, A176458, A176522, A261391.

Programs

  • Magma
    [n^7 + 7*n^5 + 14*n^3 + 7*n: n in [0..30]]; // Vincenzo Librandi, Aug 24 2015
  • Mathematica
    Table[n^7 + 7 n^5 + 14 n^3 + 7 n, {n, 0, 30}] (* Bruno Berselli, Aug 24 2015 *)
LinearRecurrence[{8, -28, 56, -70, 56, -28, 8, -1}, {0, 29, 478, 4287, 24476, 101785, 337434, 946043}, 30] (* Vincenzo Librandi, Aug 24 2015 *)
  • PARI
    a(n)=n^7+7*n^5+14*n^3+7*n \\ Charles R Greathouse IV, Aug 24 2015
  • Sage
    [n^7+7*n^5+14*n^3+7*n for n in (0..30)] # Bruno Berselli, Aug 24 2015

Formula

a(n) = -a(-n) = ( (n+sqrt(n^2+4))/2 )^7 - 1/( (n+sqrt(n^2+4))/2 )^7.
G.f.: x*(29 + 246*x + 1275*x^2 + 1940*x^3 + 1275*x^4 + 246*x^5 + 29*x^6)/(1 - x)^8. - Bruno Berselli, Aug 24 2015

Extensions

Offset changed from 1 to 0 and initial 0 added by Bruno Berselli, Aug 25 2015

A261574 a(n) = n*(n^2 + 3)*(n^6 + 6*n^4 + 9*n^2 + 3).

Original entry on oeis.org

0, 76, 2786, 46764, 439204, 2744420, 12813606, 48229636, 153992264, 432083484, 1092730090, 2537720636, 5489037036, 11179326964, 21624372014, 40001698260, 71163830416, 122319408236, 203920464114, 330799604044, 523606640180, 810600392196, 1229857906486
Offset: 0

Views

Author

Raphael Ranna, Aug 24 2015

Keywords

Comments

Also numbers of the form (n-th metallic mean)^9 - 1/(n-th metallic mean)^9, see link to Wikipedia.

Links

Crossrefs

Cf. A001622, A014176, A098316, A098317, A098318, A176398, A176439, A176458, A176522, A261391, A261540.

Programs

  • Magma
    [n*(n^2+3)*(n^6+6*n^4+9*n^2+3): n in [0..25]]; // Bruno Berselli, Aug 25 2015
  • Mathematica
    Table[n (n^2 + 3) (n^6 + 6 n^4 + 9 n^2 + 3), {n, 0, 25}] (* Bruno Berselli, Aug 25 2015 *)
  • PARI
    concat(0, Vec(2*x*(38*x^8 +1013*x^7 +11162*x^6 +43907*x^5 +69200*x^4 +43907*x^3 +11162*x^2 +1013*x +38) / (x -1)^10 + O(x^50))) \\ Colin Barker, Aug 25 2015

Formula

a(n) = -a(-n) = ( (n+sqrt(n^2+4))/2 )^9-1/( (n+sqrt(n^2+4))/2 )^9.
G.f.: 2*x*(38*x^8 +1013*x^7 +11162*x^6 +43907*x^5 +69200*x^4 +43907*x^3 +11162*x^2 +1013*x +38) / (x -1)^10. - Colin Barker, Aug 25 2015

Extensions

Formula in Name by Colin Barker, Aug 25 2015
Offset changed from 1 to 0 and initial 0 added by Bruno Berselli, Aug 25 2015

A272775 Squares of the form P(n, 5) + n, where P(x,k) is the Pochhammer function and n = square (A000290).

Original entry on oeis.org

121, 6724, 154449, 1860496, 14250625, 78960996, 344362249, 1250895424, 3936182121, 11035502500, 28143753121, 66322731024, 146186169649, 304278004996, 602680505625, 1143051786496, 2086600473049, 3681862517124, 6302555019121, 10498248010000, 17061121121121
Offset: 1

Views

Author

Jaroslav Krizek, May 06 2016

Keywords

Comments

Theorem: Only for a square n is the number M(n) = P(n, 5) + n also square, where P(x,k) = x*(x+1)*...*(x+k-1) is the Pochhammer function (rising factorial).
This sequence contains squares M(n) for the squares n from A000290.

Links

Crossrefs

Cf. A000290, A261391.

Programs

  • Magma
    [n*(n+1)*(n+2)*(n+3)*(n+4) + n: n in [1..7000] | IsSquare(n*(n+1)*(n+2)*(n+3)*(n+4) + n)];
  • PARI
    Vec(x*(1 +x)*(121 +5272*x +81868*x^2 +429544*x^3 +780790*x^4 +429544*x^5 +81868*x^6 +5272*x^7 +121*x^8)/(1-x)^11 + O(x^50)) \\ Colin Barker, May 06 2016

Formula

a(n) = (A261391(n))^2 = ((n-th metallic mean)^5 - 1/(n-th metallic mean)^5)^2.
a(n) = n^10 + 10*n^8 + 35*n^6 + 50*n^4 + 25*n^2 = (n^5 + 5*n^3 + 5*n)^2.
G.f.: x*(1 +x)*(121 +5272*x +81868*x^2 +429544*x^3 +780790*x^4 +429544*x^5 +81868*x^6 +5272*x^7 +121*x^8) / (1-x)^11. - Colin Barker, May 06 2016
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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