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.

A248619 a(n) = (n*(n+1))^4.

Original entry on oeis.org

0, 16, 1296, 20736, 160000, 810000, 3111696, 9834496, 26873856, 65610000, 146410000, 303595776, 592240896, 1097199376, 1944810000, 3317760000, 5473632256, 8767700496, 13680577296, 20851360000, 31116960000, 45558341136, 65554433296, 92844527616, 129600000000
Offset: 0

Views

Author

Eugene Chong, Oct 09 2014

Keywords

Links

Crossrefs

Cf. A016744, A059977; A002378: n*(n+1); A035287: n^2 *(n-1)^2; A060459: n^3*(n+1)^3.
Cf. A327773.

Programs

  • Magma
    [(n*(n+1))^4: n in [0..30]]; // Vincenzo Librandi, Oct 16 2014
  • Maple
    [ seq(n^4*(n+1)^4, n = 0..100) ];
  • Mathematica
    Table[(n (n + 1))^4, {n, 0, 70}] (* or *) CoefficientList[Series[16 x (1 + 72 x + 603 x^2 + 1168 x^3 + 603 x^4 + 72 x^5 + x^6)/(1 - x)^9, {x, 0, 30}], x] (* Vincenzo Librandi, Oct 16 2014 *)
LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{0,16,1296,20736,160000,810000,3111696,9834496,26873856},30] (* Harvey P. Dale, Sep 09 2016 *)

Formula

a(n) = A002378(n)^4 = A016744(A000217(n)).
a(n) = 16*A059977(n) for n>0.
G.f.: 16*x*(1 + 72*x + 603*x^2 + 1168*x^3 + 603*x^4 + 72*x^5 + x^6)/(1 - x)^9. - Vincenzo Librandi, Oct 16 2014
Sum_{n>=1} 1/a(n) = A327773 = -35 + 10*Pi^2/3 + Pi^4/45. - Vaclav Kotesovec, Sep 25 2019

Extensions

Terms a(76) and beyond corrected by Andrew Howroyd, Feb 20 2018

A361803 Least k > 1 such that k^n - n > 1 is semiprime, or 0 if no such k exists.

Original entry on oeis.org

5, 4, 5, 3, 6, 2, 2, 5, 8, 3, 4, 11, 15, 5, 2, 0, 4, 2, 14, 7, 48, 42, 6, 35, 2, 7, 602, 3, 16, 13, 2, 3, 2, 6, 37, 3185, 6, 9, 2, 33, 28, 2, 20, 9, 2, 135, 6, 5, 2, 49, 100, 5, 166, 5, 4, 9, 98, 15, 4, 27, 24, 2, 4, 17343, 34, 19, 24, 15, 56, 6, 90, 5, 2, 85
Offset: 1

Views

Author

Kevin P. Thompson, Jun 12 2023

Keywords

Comments

For n = 16, k^16 - 16 = (k^8 - 4)(k^8 + 4) = (k^4 - 2)(k^4 + 2)(k^8 + 4) always has at least three factors, so a(16) = 0. Similarly for any n of the form (2m)^4, so a(A016744(n)) = 0.

Examples

			For n = 3:
k = 1: 1^3 - 3 = -2 < 0 so ignore.
k = 2: 2^3 - 3 = 5 which is not semiprime.
k = 3: 3^3 - 3 = 24 = 2 * 2 * 2 * 3 which is not semiprime.
k = 4: 4^3 - 3 = 61 which is not semiprime.
k = 5: 5^3 - 3 = 122 = 2 * 61 which is semiprime.
Therefore, a(3) = 5 since k = 5 is the first value for which k^3 - 3 is semiprime.

Links

Crossrefs

Cf. A001358, A016744, A130827.

A228105 a(n) = 432*n^6.

Original entry on oeis.org

0, 432, 27648, 314928, 1769472, 6750000, 20155392, 50824368, 113246208, 229582512, 432000000, 765314352, 1289945088, 2085181488, 3252759552, 4920750000, 7247757312, 10427429808, 14693280768, 20323820592, 27648000000, 37050964272, 48980118528
Offset: 0

Views

Author

Arkadiusz Wesolowski, Aug 10 2013

Keywords

Comments

For any n > 0, the equation y^2 = x^3 - a(n) has exactly one solution in natural numbers (x = 12*n^2 and y = 36*n^3).

Examples

			a(2) = 432*2^6 = 27648.

Links

Crossrefs

Cf. A134109.

Programs

  • Magma
    [432*n^6 : n in [0..22]];
  • Maple
    seq(432*n^6, n=0..22);
  • Mathematica
    Table[432*n^6, {n, 0, 22}]
LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,432,27648,314928,1769472,6750000,20155392},40] (* Harvey P. Dale, Apr 06 2018 *)
  • PARI
    concat(0, Vec(432*x*(1 + x)*(1 + 56*x + 246*x^2 + 56*x^3 + x^4) / (1 - x)^7 + O(x^40))) \\ Colin Barker, Dec 11 2017

Formula

a(n) = A008585(n)*A008591(n)*A016744(n).
G.f.: 432*x*(1 + x)*(1 + 56*x + 246*x^2 + 56*x^3 + x^4) / (1 - x)^7.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n>6. - Colin Barker, Dec 11 2017

A330151 Partial sums of 4th powers of the even numbers.

Original entry on oeis.org

0, 16, 272, 1568, 5664, 15664, 36400, 74816, 140352, 245328, 405328, 639584, 971360, 1428336, 2042992, 2852992, 3901568, 5237904, 6917520, 9002656, 11562656, 14674352, 18422448, 22899904, 28208320, 34458320, 41769936, 50272992, 60107488, 71423984, 84383984
Offset: 0

Views

Author

Assoul Abdelkarim, Dec 03 2019

Keywords

Examples

			a(4) = 0^4 + 2^4 + 4^4 + 6^4 + 8^4 = 5664.

Links

Crossrefs

Cf. A000538, A002309.
Partial sums of A016744.

Programs

  • Mathematica
    a[n_] := (8/15)*n*(6*n^4 + 15*n^3 + 10*n^2 - 1); Array[a, 31, 0] (* Amiram Eldar, Dec 08 2019 *)
  • PARI
    a(n) = sum(i=0, n, 16*i^4); \\ Jinyuan Wang, Dec 07 2019
  • PARI
    concat(0, Vec(16*x*(1 + x)*(1 + 10*x + x^2) / (1 - x)^6 + O(x^30))) \\ Colin Barker, Dec 08 2019
  • Python
    def A330151(n): return 8*n*(n**2*(n*(6*n + 15) + 10) - 1)//15 # Chai Wah Wu, Dec 07 2021

Formula

a(n) = Sum_{k=1..n} (2*k)^4 = (8/15)*n*(6*n^4 + 15*n^3 + 10*n^2 - 1).
a(n) = 16*A000538(n).
From Colin Barker, Dec 08 2019: (Start)
G.f.: 16*x*(1 + x)*(1 + 10*x + x^2) / (1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>5.
(End)
E.g.f.: (8/15)*exp(x)*x*(30 + 225*x + 250*x^2 + 75*x^3 + 6*x^4). - Stefano Spezia, Dec 08 2019
a(n+1) = 12*A002299(n) + A002492(n+1). - Yasser Arath Chavez Reyes, Mar 07 2024

Extensions

More terms from Jinyuan Wang, Dec 07 2019

A375497 Numbers k such that 16*k^4 - 1 is squarefree.

Original entry on oeis.org

1, 2, 3, 6, 7, 8, 10, 11, 15, 17, 18, 20, 21, 26, 27, 28, 29, 30, 33, 36, 39, 42, 43, 44, 45, 46, 47, 48, 51, 52, 53, 54, 55, 56, 57, 64, 65, 69, 70, 71, 72, 75, 78, 79, 80, 81, 82, 83, 89, 90, 92, 93, 96, 97, 98, 99, 100, 101, 102, 105, 106, 107, 108, 110, 111, 114, 115, 117, 118, 119, 120
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Aug 18 2024

Keywords

Crossrefs

Cf. A005117, A016744.

Programs

  • Magma
    [k: k in [1..120] | IsSquarefree(16*k^4-1)];
  • Maple
    select(k -> numtheory:-issqrfree(16*k^4-1), [$1..200]); # Robert Israel, Aug 18 2024
  • Mathematica
    Select[Range[120], SquareFreeQ[16 * #^4 - 1] &] (* Amiram Eldar, Aug 18 2024 *)
  • PARI
    isok(k) = issquarefree(16*k^4 - 1); \\ Michel Marcus, Aug 18 2024
Showing 1-5 of 5 results.

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