With 1 2 3 4 5 6 ... I obtained 1 2 1 3 4 1 4 10 6 1 5 20 21 8 1 6 35 56 36 10 1 7 56 126 120 55 12 1 whose row sums correspond to the first bissection of Fibonacci numbers. A few days after, I tried (1) 1 1 2 3 4 5 6 ... Hence 1 1 1 2 2 1 3 5 3 1 4 10 9 4 1 5 18 22 14 5 1 6 30 48 40 20 6 1 (of whom the 45 degrees diagonal sums (2) 1 1 3 5 10 18 34 63 are of an unknown sequence). The row sums and their successive differences (3) 1 2 5 12 28 65 151 351 1 3 7 16 37 86 200 2 4 9 21 49 114 2 5 12 28 65 consituted my first "suite en trios" . After, I chose arbitrarly (4) 1 2 3 5 10 21 43 86 171 1 1 2 5 11 22 43 85 0 1 3 6 11 21 42 1 2 3 5 10 21 I integrated the first line of this sequence in the "tableau en trios" (5) 1 1 1 0 0 0 1 (6) 1 0 0 0 1 1 1 2 3 5 10 21 43 1 2 6 20 61 183 1 3 7 20 60 182 547 1 3 12 51 205 820 1 4 13 51 204 819 3277 1 4 20 104 520 2605 1 5 21 104 520 2604 13021 The third one is formed with the third line of (4) 0 1 3 6 The first line of the (5) array generates the present sequence by the principal diagonal : (7) 1 2 4 7 11 16 23 37 74 175 431 1024 2291 1 2 3 4 5 7 14 37 101 256 593 1267 1 1 1 1 2 7 23 64 155 337 674 0 0 0 1 5 16 41 91 182 337 0 0 1 4 11 25 50 91 155 0 1 3 7 14 25 41 64 1 2 4 7 11 16 23 1 2 3 4 5 7 1 1 1 1 2 0 0 0 1 0 0 1 0 1 1 The recurrence is, from the sixth term, u(n)=6*u(n-1)-15*u(n-2)+20*u(n-3)-15*u(n-4)+6*u(n-5) The array contains many 2^n like the other sequences I presented last week A129818, and called twisted number sequences. Sequence A038504 presents an almost equivalent array (the difference is the 0 at the beginning of the sequence). Instead of 1 1 0 0 (equivalent to 1 1 1 0 0 0), the beginning of the diagonal is 0 1 1 0 0. It is a twist numbers sequence. Many formulas were discovered by Paul Barry. Naturally, I studied too delate sequences. Note that for 1 1 1 2 3 4 5 , the sequence and its principal diagonal : 1 2 4 9 20 44 96 209 455 991 1 1 1 2 1 3 1 5 0 9 It can be defined by u(n+1) = 2*u(n)+v(n) , v(n) = 0 0 1 2 4 8 17 37 81 177 n=0... diff = 0 1 1 2 4 9 20 44 96 = 0 1 u(n) I wrote what I exposed here on two difference manuscripts which can be analysed (paper and ink).