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 1-2 of 2 results.

A121671 Real part of (1 + n*i)^5.

Original entry on oeis.org

1, -4, 41, 316, 1121, 2876, 6121, 11516, 19841, 31996, 49001, 71996, 102241, 141116, 190121, 250876, 325121, 414716, 521641, 647996, 796001, 967996, 1166441, 1393916, 1653121, 1946876, 2278121, 2649916, 3065441, 3527996, 4041001, 4607996, 5232641, 5918716
Offset: 0

Views

Author

Gary W. Adamson, Aug 14 2006

Keywords

Comments

The imaginary term considered as an unsigned real integer = A121672(n). The companion sequence A121672 uses the operation (n + i)^5.

Examples

			a(4) = 1121 since (1 + 4i)^5 = (1121 + 404i) where 404 = A121672(4).
		

Crossrefs

Cf. A121672.

Programs

  • Mathematica
    Table[Re[(1 + n*I)^5], {n, 0, 35}] (* T. D. Noe, Mar 01 2012 *)
    LinearRecurrence[{5,-10,10,-5,1},{1,-4,41,316,1121},40] (* Harvey P. Dale, Apr 21 2019 *)
  • PARI
    a(n) = real((1 + n*I)^5); \\ Michel Marcus, Dec 19 2020

Formula

G.f.: (1-9*x+71*x^2+61*x^3-4*x^4)/(1-x)^5. - Bruno Berselli, Mar 01 2012
a(n) = 5*n^4-10*n^2+1. - Bruno Berselli, Mar 01 2012
a(n) = (1+n^2)^(5/2)*cos(5*arctan(n)). - Gerry Martens, Apr 06 2024

Extensions

Corrected and extended by T. D. Noe, Mar 01 2012

A272870 Real part of (n + i)^4.

Original entry on oeis.org

1, -4, -7, 28, 161, 476, 1081, 2108, 3713, 6076, 9401, 13916, 19873, 27548, 37241, 49276, 64001, 81788, 103033, 128156, 157601, 191836, 231353, 276668, 328321, 386876, 452921, 527068, 609953, 702236, 804601, 917756, 1042433, 1179388, 1329401, 1493276
Offset: 0

Views

Author

Colin Barker, May 08 2016

Keywords

Comments

a(1) and a(2) are the only two negative terms in the sequence. Since (n + i)^2 = (n^2 - 1) + 2ni, it follows that (n + i)^4 = (n^2 - 1 + 2ni)^2 = (n^4 - 6n^2 + 1) + (4n^3 - 4n)i, so the real part of (n + i)^4 is n^4 - 6n^2 + 1. Then n^4 + 1 > 6n^2 for all n > 2, ensuring a(n) is positive for all n > 2. - Alonso del Arte, Jun 04 2016

Examples

			a(5) = 476 because (5 + i)^4 = 476 + 480*i.
		

Crossrefs

Cf. A272871 (imaginary part).
Cf. A121670 ((n+i)^3), A121672 ((n+i)^5).

Programs

  • Mathematica
    Table[Re[(n + I)^4], {n, 0, 35}] (* or *)
    Table[n^4 - 6 n^2 + 1, {n, 0, 35}] (* or *)
    CoefficientList[Series[(1 - 9 x + 23 x^2 + 13 x^3 - 4 x^4)/(1 - x)^5, {x, 0, 35}], x] (* Michael De Vlieger, May 08 2016 *)
  • PARI
    a(n) = n^4-6*n^2+1
    
  • PARI
    vector(50, n, n--; real((n+I)^4))
    
  • PARI
    Vec((1-9*x+23*x^2+13*x^3-4*x^4)/(1-x)^5 + O(x^50))

Formula

a(n) = n^4 - 6*n^2 + 1.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4.
G.f.: (1-9*x+23*x^2+13*x^3-4*x^4) / (1-x)^5.
E.g.f.: (1 - 5*x + x^2 + 6*x^3 + x^4)*exp(x). - Ilya Gutkovskiy, May 08 2016
Showing 1-2 of 2 results.