

A100216


Relates row sums of Pascal's triangle to expansion of cos(x)/exp(x).


3



1, 4, 9, 16, 26, 44, 84, 176, 376, 784, 1584, 3136, 6176, 12224, 24384, 48896, 98176, 196864, 393984, 787456, 1573376, 3144704, 6288384, 12578816, 25163776, 50335744, 100675584, 201342976, 402661376, 805289984, 1610563584, 3221159936
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OFFSET

0,2


COMMENTS

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).
This sequence is identical to its 4th differences.  JeanFrançois Alcover, Nov 07 2013


LINKS

Table of n, a(n) for n=0..31.
Index entries for linear recurrences with constant coefficients, signature (4,6,4).


FORMULA

a(n+3) = 4a(n)  6a(n+1) + 4a(n+2), a(0) = 1, a(1) = 4, a(2) = 9;
g.f. (x+1)(x1)/((2x1)(2x^22x+1))
(a(n)) = lesseq(.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e).
2*a(n) = 3*2^n A009545(n+1) +4*A009545(n).  R. J. Mathar, May 21 2019


EXAMPLE

Ex. a(2) = 9 because (.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e)^3 =
1'j + 1'k + 1j' + 1k' + 3'ii' + 2'jj' + 2'kk' + 1'jk' + 1'kj' + 1e
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).


MAPLE

a:= n> (<<010>, <001>, <464>>^n. <<1, 4, 9>>)[1, 1]:
seq(a(n), n=0..35); # Alois P. Heinz, Nov 07 2013


MATHEMATICA

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 *)


PROG

Floretion Algebra Multiplication Program, FAMP


CROSSREFS

Cf. A100213, A100214, A100216, A009116, A038503.
Sequence in context: A265044 A266336 A027365 * A180306 A138994 A195618
Adjacent sequences: A100213 A100214 A100215 * A100217 A100218 A100219


KEYWORD

nonn,easy


AUTHOR

Creighton Dement, Nov 11 2004


STATUS

approved



