%I A053218
%S A053218 1,2,3,3,5,8,4,7,12,20,5,9,16,28,48,6,11,20,36,64,112,7,13,24,44,80,144,
%T A053218 256,8,15,28,52,96,176,320,576,9,17,32,60,112,208,384,704,1280,10,19,36,
%U A053218 68,128,240,448,832,1536,2816,11,21,40,76,144,272,512,960,1792,3328
%N A053218 Triangle read by rows where the first element in row n is n, and for
k>=2 element k in row n is the sum of element k-1 in row n and element
k-1 in row n-1.
%C A053218 Last element in each row gives A001792. Difference between center element
of row 2n-1 and row sum of row n, (A053220(n+4) - A053221(n+4)) gives
A045618(n).
%C A053218 For all integers k>=2, if a sequence k,k-1,k+2,k-3,k+4,...,2,2k-2,1,2k-1,
b0(n) with offset 1, is written, the sequence b0(2)-b0(1), b0(3)-b0(2),
b0(4)-b0(3),..., b0(2k-1)-b0(2k-2), b1(n) with offset 1, is written
under it, the sequence b1(2)-b1(1), b1(3)-b1(2), b1(4)-b1(3),...,
b1(2k-2)-b1(2k-3), b2(n) with offset 1, is written under this, and
so on until the sequence b(2k-3)(2)-b(2k-3)(1), b(2k-2)(n) with offset
1 (which will contain only one term), is written, and then the sequence
b1(1);b1(2),b2(1);b1(3),b2(2),b3(1);...;b1(2k-2), b2(2k-3), b3(2k-4),
...,b(2k-2)(1) is obtained, then this sequence will be identical
to the first 2k^2-3k+1 terms of a(n), except that the first term
of this sequence will be negative, the next two terms will be positive,
the next three will be negative, the next four positive, and so on.
%e A053218 1; 2,3; 3,5,8; 4,7,12,20; 5,9,16,28,48; ...
%Y A053218 Cf. A053219 (reverse of this triangle), A053220 (center elements), A053221
(row sums), A001792, A045618.
%Y A053218 Sequence in context: A035066 A035068 A153643 this_sequence A154690 A046937
A069831
%Y A053218 Adjacent sequences: A053215 A053216 A053217 this_sequence A053219 A053220
A053221
%K A053218 easy,nonn,tabl
%O A053218 1,2
%A A053218 Asher Auel (asher.auel(AT)reed.edu) Jan 01 2000
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