A124861 -id:A124861 - cp's OEIS frontend

A291382 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.

2, 7, 22, 70, 222, 705, 2238, 7105, 22556, 71608, 227332, 721705, 2291178, 7273743, 23091762, 73308814, 232731578, 738846865, 2345597854, 7446508273, 23640235416, 75050038224, 238259397096, 756395887969, 2401310279090, 7623377054503, 24201736119310

Offset: 0

Clark Kimberling, Sep 04 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
In the following guide to p-INVERT sequences using s = (1,1,0,0,0,...) = A019590, in some cases t(1,1,0,0,0,...) is a shifted version of the cited sequence:
p(S)                     t(1,1,0,0,0,...)
1 - S                       A000045 (Fibonacci numbers)
1 - S^2                     A094686
1 - S^3                     A115055
1 - S^4                     A291379
1 - S^5                     A281380
1 - S^6                     A281381
1 - 2 S                     A002605
1 - 3 S                     A125145
(1 - S)^2                   A001629
(1 - S)^3                   A001628
(1 - S)^4                   A001629
(1 - S)^5                   A001873
(1 - S)^6                   A001874
1 - S - S^2                 A123392
1 - 2 S - S^2               A291382
1 - S - 2 S^2               A124861
1 - 2 S - S^2               A291383
(1 - 2 S)^2                 A073388
(1 - 3 S)^2                 A291387
(1 - 5 S)^2                 A291389
(1 - 6 S)^2                 A291391
(1 - S)(1 - 2 S)            A291393
(1 - S)(1 - 3 S)            A291394
(1 - 2 S)(1 - 3 S)          A291395
(1 - S)(1 - 2 S)            A291393
(1 - S)(1 - 2 S)(1 - 3 S)   A291396
1 - S - S^3                 A291397
1 - S^2 - S^3               A291398
1 - S - S^2 - S^3           A186812
1 - S - S^2 - S^3 - S^4     A291399
1 - S^2 - S^4               A291400
1 - S - S^4                 A291401
1 - S^3 - S^4               A291402
1 - 2 S^2 - S^4             A291403
1 - S^2 - 2 S^4             A291404
1 - 2 S^2 - 2 S^4           A291405
1 - S^3 - S^6               A291407
(1 - S)(1 - S^2)            A291408
(1 - S^2)(1 - S)^2          A291409
1 - S - S^2 - 2 S^3         A291410
1 - 2 S - S^2 + S^3         A291411
1 - S - 2 S^2 + S^3         A291412
1 - 3 S + S^2 + S^3         A291413
1 - 2 S + S^3               A291414
1 - 3 S + S^2               A291415
1 - 4 S + S^2               A291416
1 - 4 S + 2 S^2             A291417

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2, 3, 2, 1)

Cf. A019590, A290890, A291000, A291219, A291728, A292479, A292480.

z = 60; s = x + x^2; p = 1 - 2 s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291382 *)

G.f.: (-2 - 3 x - 2 x^2 - x^3)/(-1 + 2 x + 3 x^2 + 2 x^3 + x^4).

a(n) = 2*a(n-1) + 3*a(n-2) + 2*a(n-3) + a(n-4) for n >= 5.

A297682 T(n,k)=Number of nXk 0..1 arrays with every 1 horizontally, diagonally or antidiagonally adjacent to 0, 1 or 4 neighboring 1s.

Original entry on oeis.org

2, 4, 4, 7, 11, 8, 13, 29, 33, 16, 24, 80, 150, 98, 32, 44, 219, 629, 742, 291, 64, 81, 597, 2790, 4633, 3744, 865, 128, 149, 1632, 12110, 32911, 34872, 18840, 2570, 256, 274, 4459, 52889, 221420, 401678, 260924, 94891, 7637, 512, 504, 12181, 230406, 1519630

Offset: 1

R. H. Hardin, Jan 03 2018

Table starts
...2.....4.......7........13.........24...........44.............81
...4....11......29........80........219..........597...........1632
...8....33.....150.......629.......2790........12110..........52889
..16....98.....742......4633......32911.......221420........1519630
..32...291....3744.....34872.....401678......4202440.......45865837
..64...865...18840....260924....4870764.....78957968.....1368968852
.128..2570...94891...1955750...59210634...1487819051....41030621948
.256..7637..477850..14651847..719647644..28013761161..1229127412701
.512.22693.2406649.109783269.8748946600.527589764007.36837288191422

			Some solutions for n=4 k=4
..0..1..0..1. .1..1..0..0. .0..0..0..1. .0..0..0..1. .0..0..0..0
..0..0..1..1. .0..0..0..1. .1..0..1..0. .1..0..0..1. .1..0..0..0
..0..1..0..0. .0..0..1..0. .1..0..0..0. .1..0..0..1. .0..0..0..0
..0..1..0..0. .1..0..1..0. .0..1..0..0. .0..0..0..0. .0..0..1..1

R. H. Hardin, Table of n, a(n) for n = 1..287

Column 1 is A000079.
Column 2 is A282990.
Row 1 is A000073(n+3).
Row 2 is A124861(n+1).

Empirical for column k:
k=1: a(n) = 2*a(n-1)
k=2: a(n) = 2*a(n-1) +3*a(n-2) -a(n-4)
k=3: a(n) = 5*a(n-1) -a(n-2) +8*a(n-3) -5*a(n-4) -30*a(n-5) +17*a(n-6)
k=4: [order 16]
k=5: [order 30]
k=6: [order 57]
Empirical for row n:
n=1: a(n) = a(n-1) +a(n-2) +a(n-3)
n=2: a(n) = a(n-1) +3*a(n-2) +4*a(n-3) +2*a(n-4)
n=3: [order 8]
n=4: [order 17]
n=5: [order 41]

A124860 A Jacobsthal-Pascal triangle.

Original entry on oeis.org

1, 1, 1, 3, 6, 3, 5, 15, 15, 5, 11, 44, 66, 44, 11, 21, 105, 210, 210, 105, 21, 43, 258, 645, 860, 645, 258, 43, 85, 595, 1785, 2975, 2975, 1785, 595, 85, 171, 1368, 4788, 9576, 11970, 9576, 4788, 1368, 171, 341, 3069, 12276, 28644, 42966, 42966, 28644, 12276, 3069, 341

Offset: 0

Paul Barry, Nov 10 2006

Triangle T(n, k) read by rows given by [1, 2, -2, 0, 0, 0, ...] DELTA [1, 2, -2, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 11 2006

			Triangle begins
   1;
   1,   1;
   3,   6,   3;
   5,  15,  15,   5;
  11,  44,  66,  44,  11;
  21, 105, 210, 210, 105,  21;
  43, 258, 645, 860, 645, 258, 43;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A001045, A003683 (row sums), A016095, A084938, A124862 (diagonal sums), A193449.

A124860:= func< n,k | Binomial(n,k)*(2^(n+1) - (-1)^(n+1))/3 >;
[A124860(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 17 2023

A := proc(n,k) ## n >= 0 and k = 0 .. n
    ((-1)^n+2^(n+1))/3*binomial(n, k)
end proc: # Yu-Sheng Chang, Jan 15 2020

jacobPascal[n_, k_]:= Binomial[n, k]*(2^(n+1) -(-1)^(n+1))/3; ColumnForm[Table[jacobPascal[n, k], {n,0,12}, {k,0,n}], Center] (* Alonso del Arte, Jan 16 2020 *)

def A124860(n,k): return binomial(n,k)*(2^(n+1) - (-1)^(n+1))/3
flatten([[A124860(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Feb 17 2023

G.f.: 1/(1 - x*(1+y) - 2*x^2*(1+y)^2).
T(n, k) = J(n+1) * C(n, k), where J(n) = A001045(n).
T(n, 0) = T(n, n) = A001045(n+1).
T(2*n, n) = A124862(n).
Sum_{k=0..n} T(n, k) = A003683(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A124861(n).
T(n, k) = T(n-1, k-1) + T(n-1, k) + 2*T(n-2, k-2) + 4*T(n-2, k-1) + 2*T(n-2, k), T(0, 0) = 1, T(n, k) = 0 if k < 0 or if k > n . - Philippe Deléham, Nov 11 2006
G.f.: T(0)/2, where T(k) = 1 + 1/(1 - (2*k + 1 + 2*x*(1+y))*x*(1 + y)/((2*k + 2 + 2*x*(1+y))*x*(1+y) + 1/T(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 06 2013
From G. C. Greubel, Feb 17 2023: (Start)
T(n, n-k) = T(n, k).
T(n, 1) = A193449(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n). (End)

cp's OEIS Frontend

A291382 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.

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A297682 T(n,k)=Number of nXk 0..1 arrays with every 1 horizontally, diagonally or antidiagonally adjacent to 0, 1 or 4 neighboring 1s.

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A124860 A Jacobsthal-Pascal triangle.

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