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A134658
Triangle read by rows, giving coefficients of extended Jacobsthal recurrence.
3
3, 1, 2, 2, -1, 2, 3, -3, 1, 2, 4, -6, 4, -1, 2, 5, -10, 10, -5, 1, 2, 6, -15, 20, -15, 6, -1, 2, 7, -21, 35, -35, 21, -7, 1, 2, 8, -28, 56, -70, 56, -28, 8, -1, 2, 9, -36, 84, -126, 126, -84, 36, -9, 1, 2, 10, -45, 120, -210, 252, -210, 120, -45, 10, -1, 2
OFFSET
0,1
COMMENTS
Sequence identical to half its p-th differences from the second term.
This sequence is the second of a family after A135356.
This triangle looks like a Pascal's triangle without first column, and with signs and with additional right diagonal consisting of 2's. - Michel Marcus, Apr 07 2019
FORMULA
Every row sums to 3. - Jean-François Alcover, Apr 04 2019 (further to a remark e-mailed by Paul Curtz).
EXAMPLE
Triangle begins
3; : A000244 = 1, 3, 9, 27, ... is the main sequence
1, 2; : A001045 = 0, 1, 1, 3, ... is the main sequence
2, -1, 2; : 0, 0, (A007910 = 1, 2, 3, ... ) is the main sequence
3, -3, 1, 2; : 0, 0, 0, 1, 3, 6, 10, 17, ... is the main sequence
4, -6, 4, -1, 2; : A134987 = 0, 0, 0, 0, 1, ... is the main sequence
...
See signatures of linear recurrence of corresponding sequences.
MATHEMATICA
T[n_, k_] := T[n, k] = Which[n == 0, 3, k == n, 2, k == 0, n, k == n-1, (-1)^k, True, T[n-1, k] - T[n-1, k-1]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 06 2019 *)
CROSSREFS
Cf. A000244, A001045, A007910, A134977 (sum of antidiagonals), A134987, A135356.
Sequence in context: A087282 A105973 A088429 * A296518 A111951 A222593
KEYWORD
sign,tabl
AUTHOR
Paul Curtz, Feb 01 2008
EXTENSIONS
In agreement with author, T(0, 0) = 3 and offset 0 by Michel Marcus, Apr 06 2019
STATUS
approved