A054925 -id:A054925 - cp's OEIS frontend

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

Creighton Dement, Mar 15 2005

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.

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).

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

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

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

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

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

A246659 a(n) = binomial(n-h,h)*hypergeometric([h-n/2,h-(n-1)/2],[1],4), h = floor(n/4).

Original entry on oeis.org

1, 1, 3, 7, 9, 28, 95, 306, 285, 1071, 3948, 14148, 11844, 47160, 182655, 690580, 547965, 2244385, 8961953, 35016345, 26885859, 112052304, 456606332, 1824478488, 1369818996, 5777515212, 23884958520, 97002706248, 71654875560, 304865648208, 1273989485439

Offset: 0

Peter Luschny, Sep 18 2014

nonn

Also middle column of A132885.

a(n) is the k-th term of n-th row of triangle of A132885 where k = floor(n/4). - Altug Alkan, Nov 29 2015

Cf. A132885.

a := proc(n) local h; h := iquo(n,4); binomial(n-h,h)*hypergeom([h-n/2, h-n/2+1/2],[1],4) end: seq(round(evalf(a(n),99)),n=0..30);

a[n_] := With[{h = Quotient[n, 4]}, Binomial[n-h, h]*Hypergeometric2F1[h-n/2, h-(n-1)/2, 1, 4]];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jun 18 2024 *)

a(n) = A132885(n, floor(n/4)), that is, a(n) = A132885(A054925(n+2) - 1). - Altug Alkan, Nov 29 2015

cp's OEIS Frontend

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

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