|
%I #29 Mar 28 2024 09:04:02
%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)*(1-z)/(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)/(1-z^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. - _Jean-François Alcover_, Nov 07 2013
%H G. C. Greubel, <a href="/A100216/b100216.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4).
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), with a(0) = 1, a(1) = 4, a(2) = 9.
%F G.f.: (1-x^2)/((1-2*x)*(1-2*x+2*x^2)).
%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
%F E.g.f.: (1/2)*exp(x)*(3*sin(x) - cos(x) + 3*exp(x)). - _G. C. Greubel_, Mar 28 2024
%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-> (<<0|1|0>, <0|0|1>, <4|-6|4>>^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 (* _Jean-François Alcover_, Nov 07 2013 *)
%t LinearRecurrence[{4,-6,4}, {1,4,9}, 41] (* _G. C. Greubel_, Mar 28 2024 *)
%o (Magma) [n le 3 select n^2 else 4*Self(n-1) -6*Self(n-2) +4*Self(n-3): n in [1..40]]; // _G. C. Greubel_, Mar 28 2024
%o (SageMath)
%o @CachedFunction
%o def a(n): # a = A100216
%o if n<3: return (n+1)^2
%o else: return 4*a(n-1) -6*a(n-2) +4*a(n-3)
%o [a(n) for n in range(41)] # _G. C. Greubel_, Mar 28 2024
%Y Cf. A000749, A009116, A009545, A038503, A038504, A038505, A100213, A100214, A100215.
%K nonn,easy
%O 0,2
%A _Creighton Dement_, Nov 11 2004
|
