A193449 -id:A193449 - cp's OEIS frontend

A268766 T(n,k)=Number of nXk binary arrays with some element plus some horizontally, vertically, diagonally or antidiagonally adjacent neighbor totalling two exactly once.

0, 1, 1, 2, 6, 2, 5, 15, 15, 5, 10, 44, 56, 44, 10, 20, 105, 223, 223, 105, 20, 38, 258, 762, 1148, 762, 258, 38, 71, 595, 2607, 5170, 5170, 2607, 595, 71, 130, 1368, 8500, 23156, 32056, 23156, 8500, 1368, 130, 235, 3069, 27411, 99057, 193573, 193573, 99057

Offset: 1

R. H. Hardin, Feb 13 2016

Table starts
...0....1......2.......5........10.........20...........38............71
...1....6.....15......44.......105........258..........595..........1368
...2...15.....56.....223.......762.......2607.........8500.........27411
...5...44....223....1148......5170......23156........99057........418924
..10..105....762....5170.....32056.....193573......1129042.......6475898
..20..258...2607...23156....193573....1552272.....12111209......92571436
..38..595...8500...99057...1129042...12111209....127676872....1312123185
..71.1368..27411..418924...6475898...92571436...1312123185...18045771274
.130.3069..86622.1736105..36505596..696659613..13311824510..245588158242
.235.6830.270955.7122856.203462597.5178525870.133228716170.3292985469950

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

R. H. Hardin, Table of n, a(n) for n = 1..1404

Column 1 is A001629.

Column 2 is A193449.

Empirical for column k:
k=1: a(n) = 2*a(n-1) +a(n-2) -2*a(n-3) -a(n-4)
k=2: a(n) = 2*a(n-1) +3*a(n-2) -4*a(n-3) -4*a(n-4)
k=3: a(n) = 4*a(n-1) +2*a(n-2) -16*a(n-3) -a(n-4) +12*a(n-5) -4*a(n-6)
k=4: [order 8]
k=5: [order 12]
k=6: [order 16]
k=7: [order 28]

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)

A099429 A Jacobsthal-Lucas convolution.

Original entry on oeis.org

0, 0, 2, 3, 12, 25, 66, 147, 344, 765, 1710, 3751, 8196, 17745, 38234, 81915, 174768, 371365, 786438, 1660239, 3495260, 7340025, 15379122, 32156323, 67108872, 139810125, 290805086, 603979767, 1252698804, 2594876065, 5368709130, 11095332171, 22906492256

Offset: 0

Paul Barry, Oct 15 2004

Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 02 2014

If we concatenate the lexicographically ordered bit strings of length n, then a(n) is the number of times 11 appears as a substring, if overlapping substrings are not considered as being separate. - John M. Campbell, Jan 18 2019

			G.f. = 2*x^2 + 3*x^3 + 12*x^4 + 25*x^5 + 66*x^6 + 147*x^7 + 344*x^8 + ...
If we concatenate the lexicographically ordered bit strings of length 4, we obtain the expression 0000000100100011010001010110011110001001101010111100110111101111, and we see that the substring 11 appears a total of a(4) = 12 times, with overlapping substrings not being considered as being separate. - _John M. Campbell_, Jan 18 2019

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,3,-4,-4).

Cf. A001045, A014551, A193449.

m:=40; R:=PowerSeriesRing(Integers(), m); [0,0] cat Coefficients(R!( x^2*(2-x)/(1-x-2*x^2)^2 )); // G. C. Greubel, Feb 25 2019

CoefficientList[Series[x^2*(2-x)/(1-x-2x^2)^2, {x, 0, 32}], x] (* Michael De Vlieger, Jan 18 2019 *)

{a(n) = if( n>=0, polcoeff( x^2*(2-x)/((1+x)*(1-2*x))^2 + x*O(x^n), n), polcoeff( x*(1-2*x)/((1+x)*(2-x))^2 + x*O(x^-n), -n) )}; /* Michael Somos, Jun 02 2014 */

(x^2*(2-x)/(1-x-2*x^2)^2).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

G.f.: x^2*(2-x)/(1-x-2*x^2)^2. [Typo corrected by Colin Barker, Jun 16 2012]
a(n) = Sum_{k=0..n} J(n-k)*(2^(k-1) -(-1)^k +0^k/2).
a(n) = Sum_{k=0..n+1} J(n-k)*binomial(n-k+1, k)*binomial(1, (k+1)/2)*(1-(-1)^k)/2.
a(n) = A036289(n)/6 +(-1)^n*n/3. - R. J. Mathar, Sep 21 2012
a(-n) = (-2)^(-n-1) * A193449(n) for all n in Z. - Michael Somos, Jun 02 2014

A193450 Triangle of a binomial convolution sum related to Jacobsthal numbers.

Original entry on oeis.org

0, 1, 0, 2, 2, 2, 3, 6, 6, 0, 4, 12, 16, 8, 4, 5, 20, 35, 30, 15, 0, 6, 30, 66, 78, 54, 18, 6, 7, 42, 112, 168, 154, 84, 28, 0, 8, 56, 176, 320, 368, 272, 128, 32, 8, 9, 72, 261, 558, 774, 720, 450, 180, 45, 0, 10, 90, 370, 910, 1480, 1660, 1300, 700, 250, 50, 10

Offset: 0

Olivier Gérard, Jul 26 2011

Row sum is A193449(n) = A001045(n+1)*n.

			Triangle starts:
0;
1, 0;
2, 2, 2;
3, 6, 6, 0;
4, 12, 16, 8, 4;
5, 20, 35, 30, 15, 0;
...

Cf. A193451.

T(n,k) = sum(j=0, k, (-1)^j*n*binomial(n-j,k-j)); \\ Michel Marcus, Jun 04 2014

T(n,k) = sum( (-1)^j*n*C(n-j,k-j), j=0..k).

T(n,k) = n*C(n, k)*2F1( (1, -k); -n )(-1).

A193451 Triangle of a binomial convolution sum related to Jacobsthal numbers.

Original entry on oeis.org

0, 0, 1, 0, 3, 3, 0, 5, 8, 2, 0, 7, 17, 14, 6, 0, 9, 30, 39, 24, 3, 0, 11, 47, 83, 75, 33, 9, 0, 13, 68, 152, 184, 126, 48, 4, 0, 15, 93, 252, 384, 354, 198, 60, 12, 0, 17, 122, 389, 716, 830, 620, 290, 80, 5, 0, 19, 155, 569, 1229, 1718, 1610, 1010, 410, 95, 15

Offset: 0

Olivier Gérard, Jul 26 2011

Row sum is A193449(n) = n*A001045(n+1).

			Triangle starts:
0;
0, 1;
0, 3, 3;
0, 5, 8, 2;
0, 7, 17, 14, 6;
0, 9, 30, 39, 24, 3;
...

Cf. A193449, A193450, A001045.

T(n,k)= sum(j=0, k, (-1)^(j+k)*(j+k)*binomial(n-k+j,j)); \\ Michel Marcus, Jun 04 2014

T(n,k)= sum( (-1)^(j+k)*(j+k)*C(n-k+j,j), j=0..k).

cp's OEIS Frontend

A268766 T(n,k)=Number of nXk binary arrays with some element plus some horizontally, vertically, diagonally or antidiagonally adjacent neighbor totalling two exactly once.

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A124860 A Jacobsthal-Pascal triangle.

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A099429 A Jacobsthal-Lucas convolution.

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A193450 Triangle of a binomial convolution sum related to Jacobsthal numbers.

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A193451 Triangle of a binomial convolution sum related to Jacobsthal numbers.

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PARI

Formula