A104563 A floretion-generated sequence relating to centered square numbers.

0, 1, 3, 5, 8, 13, 19, 25, 32, 41, 51, 61, 72, 85, 99, 113, 128, 145, 163, 181, 200, 221, 243, 265, 288, 313, 339, 365, 392, 421, 451, 481, 512, 545, 579, 613, 648, 685, 723, 761, 800, 841, 883, 925, 968, 1013, 1059, 1105, 1152, 1201, 1251

Offset: 0

Author

Creighton Dement, Mar 15 2005

Keywords

• nonn
• easy

Comments

Floretion Algebra Multiplication Program, FAMP Code: a(n) = 1vesrokseq[A*B] with A = - .5'i - .5i' + .5'ii' + .5e, B = + .5'ii' - .5'jj' + .5'kk' + .5e. RokType: Y[sqa.Findk()] = Y[sqa.Findk()] + Math.signum(Y[sqa.Findk()])*p (internal program code). Note: many slight variations of the "RokType" already exist, such that it has become difficult to assign them all names.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Crossrefs

Cf. A001844, A004525, A104564, A098180, A059100, A010000, A047599, A007877.

Programs

Mathematica

LinearRecurrence[{3, -4, 4, -3, 1}, {0, 1, 3, 5, 8}, 60] (* Amiram Eldar, Dec 14 2024 *)

PARI

concat(0, Vec(x*(1 + x)*(1 - x + x^2) / ((1 - x)^3*(1 + x^2)) + O(x^40))) \\ Colin Barker, Apr 29 2019

Formula

G.f.: x*(1 + x^3)/((1 + x^2)*(1 - x)^3).

FAMP result: 2*a(n) + 2*A004525(n+1) = A104564(n) + a(n+1).

Superseeker results:

a(2*n+1) = A001844(n) = 2*n*(n+1) + 1 (Centered square numbers);

a(n+1) - a(n) = A098180(n) (Odd numbers with two times the odd numbers repeated in order between them);

a(n) + a(n+2) = A059100(n+1) = A010000(n+1);

a(n+2) - a(n) = A047599(n+1) (Numbers that are congruent to {0, 3, 4, 5} mod 8);

a(n+2) - 2*a(n+1) + a(n) = A007877(n+3) (Period 4 sequence with initial period (0, 1, 2, 1));

Coefficients of g.f.*(1-x)/(1+x) = convolution of this with A280560 gives A004525;

Coefficients of g.f./(1+x) = convolution of this with A033999 gives A054925.

a(n) = (1/2)*(n^2 + 1 - cos(n*Pi/2)). - Ralf Stephan, May 20 2007

From Colin Barker, Apr 29 2019: (Start)

a(n) = (2 - (-i)^n - i^n + 2*n^2) / 4 where i=sqrt(-1).

a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) for n>4. (End)

a(n) = A011848(n-1)+A011848(n+2). - R. J. Mathar, Sep 11 2019

Sum_{n>=1} 1/a(n) = Pi^2/48 + (Pi/2) * tanh(Pi/2) + (Pi/(4*sqrt(2)) * tanh(Pi/(2*sqrt(2)))). - Amiram Eldar, Dec 14 2024

Extensions

Stephan's formula corrected by Bruno Berselli, Apr 29 2019