A062196 → reverse
nonnegative coefficients
Proved for all n: all roots real, in (−∞, 0] · coefficients of one sign · Hadamard kernel what this means
Rows ?
| n | x0 | x1 | x2 | x3 | x4 | x5 | x6 | x7 | Pn(1) |
|---|---|---|---|---|---|---|---|---|---|
| 0 | 1 | 1 | |||||||
| 1 | 3 | 1 | 4 | ||||||
| 2 | 6 | 8 | 1 | 15 | |||||
| 3 | 10 | 30 | 15 | 1 | 56 | ||||
| 4 | 15 | 80 | 90 | 24 | 1 | 210 | |||
| 5 | 21 | 175 | 350 | 210 | 35 | 1 | 792 | ||
| 6 | 28 | 336 | 1050 | 1120 | 420 | 48 | 1 | 3003 | |
| 7 | 36 | 588 | 2646 | 4410 | 2940 | 756 | 63 | 1 | 11440 |
Generated b-files: A062196 (reversed triangle) ?
Computed in your browser ?
Technical details
- Node id
n-3f65f4dc3d5d123d· revision 1 · JSON- Definition source
derived:reversal- Fingerprint
f6-5f98bd4aea48ed7d
Observed on the stored rows ?
- Degrees: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 (affine: slope 1, intercept 0)
- Palindromic rows: no
- γ-positive (gamma_positive): no
- Nonnegative coefficients (
nonn): yes - Only rational roots (
rr_by_factors: every nonzero row is a product of linear factors over ℚ): no
OEIS matches ?
- A062196: reversed triangle, 55 terms agree, 10 full rows row-aligned (tabl) OEIS nonn in polytool catalog · view at the OEIS · generated b-file 990 terms from n = 0 ?
- A001791: row sums, 15 terms agree OEIS nonn (row-sum match: finite agreement only) · view at the OEIS
A generated b-file is computed from the node's definition along the row alignment of its OEIS link and starts at the entry's offset. It is an identification on a finite prefix, not a proof that the node and the OEIS entry agree for every n.
OEIS submission text ?
Fields to paste into an OEIS submission: the example %e, the formula
%F and programs %t (Mathematica) and %o (PARI/GP) that list the same rows.
For an OEIS link the row numbers, the offset and the orientation are those of the entry, as matched;
the matched terms were compared with the published ones, which identifies the node with the entry on a
finite prefix, not for every n. The PARI/GP program is not machine-checked here.
The atlas's own rows: n = 0, …, ascending powers of x
%e ExampleTriangle begins: 0: 1; 1: 3, 1; 2: 6, 8, 1; 3: 10, 30, 15, 1; 4: 15, 80, 90, 24, 1; 5: 21, 175, 350, 210, 35, 1; 6: 28, 336, 1050, 1120, 420, 48, 1; 7: 36, 588, 2646, 4410, 2940, 756, 63, 1; 8: 45, 960, 5880, 14112, 14700, 6720, 1260, 80, 1; 9: 55, 1485, 11880, 38808, 58212, 41580, 13860, 1980, 99, 1;
%F FormulaThe row polynomials satisfy (n^2+2*n)*P(n,x) = (n^2+4*n+4+(3*n^2+2*n-2)*x)*P(n-1,x) + ((2*n+2)*x-(2*n+2)*x^2)*d/dx P(n-1,x) for n >= 1, with P(0,x) = 1.
%t MathematicaP[0, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((n^2 + 4*n + 4 + (3*n^2 + 2*n - 2)*x)*P[n - 1, x] + ((2*n + 2)*x - (2*n + 2)*x^2)*D[P[n - 1, x], x])/(n^2 + 2*n)]];
Table[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten%o PARI/GPP(N) = {my(v = vector(N + 1)); v[1] = 1; for(n = 1, N, my(i = n + 1); v[i] = ((n^2 + 4*n + 4 + (3*n^2 + 2*n - 2)*x)*v[i - 1] + ((2*n + 2)*x - (2*n + 2)*x^2)*deriv(v[i - 1], x))/(n^2 + 2*n)); v};
row(n, p) = {my(c = Vecrev(p)); c};
T(N) = {my(v = P(N)); concat(vector(#v, i, row(i - 1, v[i])))};
T(9)Generated with polytool c48629374597.
A062196: reversed triangle, offset 0, from the row P0
%e ExampleTriangle begins: 0: 1; 1: 1, 3; 2: 1, 8, 6; 3: 1, 15, 30, 10; 4: 1, 24, 90, 80, 15; 5: 1, 35, 210, 350, 175, 21; 6: 1, 48, 420, 1120, 1050, 336, 28; 7: 1, 63, 756, 2940, 4410, 2646, 588, 36; 8: 1, 80, 1260, 6720, 14700, 14112, 5880, 960, 45; 9: 1, 99, 1980, 13860, 41580, 58212, 38808, 11880, 1485, 55;
%F FormulaRow n lists the coefficients of P(n,x) in descending powers of x, where (n^2+2*n)*P(n,x) = (n^2+4*n+4+(3*n^2+2*n-2)*x)*P(n-1,x) + ((2*n+2)*x-(2*n+2)*x^2)*d/dx P(n-1,x) for n >= 1, with P(0,x) = 1.
%t MathematicaP[0, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((n^2 + 4*n + 4 + (3*n^2 + 2*n - 2)*x)*P[n - 1, x] + ((2*n + 2)*x - (2*n + 2)*x^2)*D[P[n - 1, x], x])/(n^2 + 2*n)]];
Table[Reverse[PadRight[CoefficientList[P[n, x], x], n + 1]], {n, 0, 9}] // Flatten%o PARI/GPP(N) = {my(v = vector(N + 1)); v[1] = 1; for(n = 1, N, my(i = n + 1); v[i] = ((n^2 + 4*n + 4 + (3*n^2 + 2*n - 2)*x)*v[i - 1] + ((2*n + 2)*x - (2*n + 2)*x^2)*deriv(v[i - 1], x))/(n^2 + 2*n)); v};
row(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n + 1 - #c)); c = Vecrev(c); c};
T(N) = {my(v = P(N)); concat(vector(#v, i, row(i - 1, v[i])))};
T(9)Generated with polytool c48629374597.
Proved properties
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Real-rooted. Every row P_n, n >= 0, is zero or real-rooted with all roots in (-inf, 0]
Every polynomial of the sequence has only real roots, inside the stated interval.by the cited steps:
e-b6d3c37eae668953 -
Coefficients of one sign. Every nonzero row, n >= 0: the coefficients of P_n have one sign
All coefficients of each polynomial have the same sign (after x -> -x when alternating).derived from the cited steps by a stated theorem:
e-b6d3c37eae668953 -
Hadamard kernel (Malo). Rows P_n, n >= 0, are Hadamard kernels (preserve real-rootedness)
Multiplying the coefficients of any real-rooted polynomial termwise by those of a row (Hadamard product) gives a real-rooted polynomial again.derived from the cited steps by a stated theorem:
e-b6d3c37eae668953
Proof chains
1 incoming proof step. Each step is expanded back to the roots of the proof DAG, the theorem checks of a node's own rule. JSON
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preserver_image reversal, shift 0
e-b6d3c37eae668953Affine, reversal and Möbius substitutions (elementary) polytool ?
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premise (sequence):
Triangle read by rows, T(n, k) = binomial(n, k)*binomial(n + 2, k).
uses: Every row P_n, n >= 0, is zero or real-rooted with all roots in (-inf, 0]
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theorem_check liu_wang
e-32e7e43d5a9debdaroot: theorem checkLiu–Wang theorem [LW07, Thm 2.1, Cor 2.4] ?
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premise (sequence):
Triangle read by rows, T(n, k) = binomial(n, k)*binomial(n + 2, k).
Related sequences ?
Same fingerprint (1)
Equal up to content, sign, reversal and a small index shift.
References
- [LW07] Lily L. Liu and Yi Wang. A unified approach to polynomial sequences with only real zeros. Advances in Applied Mathematics, 38(4):542–560, May 2007. doi:10.1016/j.aam.2006.02.003