%I
%S 1,4,9,16,26,44,84,176,376,784,1584,3136,6176,12224,24384,48896,98176,
%T 196864,393984,787456,1573376,3144704,6288384,12578816,25163776,
%U 50335744,100675584,201342976,402661376,805289984,1610563584,3221159936
%N Relates row sums of Pascal's triangle to expansion of cos(x)/exp(x).
%C A100215(n) (ves) = ((1)^n)*A009116(n+3) (jes) + a(n) (les) + A038503(n+1) (tes) (Sn, below, corresponds to the generating function from above). Coefficients of Sn(z)*(1z)/(1+z) gives match to A038504 (Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 1"). Coefficients of Sn(z)/(1+z) gives match to A038505 (Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 2"). Coefficients of Sn(z)/(1z^2) gives match to A000749 (Number of strings over Z_2 of length n with trace 1 and subtrace 1). The elements 'i, 'j, 'k, i', j', k', 'ii', 'jj', 'kk', 'ij', 'ik', 'ji', 'jk', 'ki', 'kj', e ("floretions") are members of the quaternion product factor space Q x Q /{(1,1), (1,1)}. "les" sums over coefficients belonging to basis vectors which squared give the unit e (excluding e itself).
%C This sequence is identical to its 4th differences.  _JeanFrançois Alcover_, Nov 07 2013
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,6,4).
%F a(n+3) = 4a(n)  6a(n+1) + 4a(n+2), a(0) = 1, a(1) = 4, a(2) = 9;
%F G.f.: (x+1)(x1)/((2x1)(2x^22x+1)).
%F (a(n)) = lesseq(.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e).
%F 2*a(n) = 3*2^n  A009545(n+1) + 4*A009545(n).  _R. J. Mathar_, May 21 2019
%e a(2) = 9 because (.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e)^3 =
%e 1'j + 1'k + 1j' + 1k' + 3'ii' + 2'jj' + 2'kk' + 1'jk' + 1'kj' + 1e
%e and the sum of the coefficients belonging to basis vectors which squared give the unit e (excluding e itself) is 3+2+2+1+1 = 9 (see comment).
%p a:= n> (<<010>, <001>, <464>>^n. <<1, 4, 9>>)[1, 1]:
%p seq(a(n), n=0..35); # _Alois P. Heinz_, Nov 07 2013
%t d = 4; nmax = 31; a[n_ /; n < d] := (n + 1)^2; seq = Table[a[n], {n, 0, nmax}]; seq /. Solve[ Thread[ Take[seq, nmax  d + 1] == Differences[seq, d]]] // First (* _JeanFrançois Alcover_, Nov 07 2013 *)
%o Floretion Algebra Multiplication Program, FAMP
%Y Cf. A100213, A100214, A100216, A009116, A038503.
%K nonn,easy
%O 0,2
%A _Creighton Dement_, Nov 11 2004
