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Search: id:A053218
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| A053218 |
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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. |
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7
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| 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, 256, 8, 15, 28, 52, 96, 176, 320, 576, 9, 17, 32, 60, 112, 208, 384, 704, 1280, 10, 19, 36, 68, 128, 240, 448, 832, 1536, 2816, 11, 21, 40, 76, 144, 272, 512, 960, 1792, 3328
(list; table; graph; listen)
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OFFSET
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1,2
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COMMENT
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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).
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.
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EXAMPLE
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1; 2,3; 3,5,8; 4,7,12,20; 5,9,16,28,48; ...
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CROSSREFS
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Cf. A053219 (reverse of this triangle), A053220 (center elements), A053221 (row sums), A001792, A045618.
Sequence in context: A035066 A035068 A153643 this_sequence A154690 A046937 A069831
Adjacent sequences: A053215 A053216 A053217 this_sequence A053219 A053220 A053221
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KEYWORD
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easy,nonn,tabl
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AUTHOR
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Asher Auel (asher.auel(AT)reed.edu) Jan 01 2000
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