A051597 -id:A051597 - cp's OEIS frontend

A132370 Array read by antidiagonals: T(m,n) = number of spotlight tilings of a width 1 m X n frame.

16, 34, 34, 58, 68, 58, 88, 112, 112, 88, 124, 166, 180, 166, 124, 166, 230, 262, 262, 230, 166, 214, 304, 358, 376, 358, 304, 214, 268, 388, 468, 508, 508, 468, 388, 268, 328, 482, 592, 658, 680, 658, 592, 482, 328, 394, 586, 730, 826, 874, 874, 826, 730, 586, 394

Offset: 3

Bridget Tenner, Nov 09 2007

			A 3 X 3 frame with width 1 has 16 spotlight tilings.
Array begins:
===============================================
m/n  |   3   4   5    6    7    8    9   10 ...
-----+-----------------------------------------
   3 |  16  34  58   88  124  166  214  268 ...
   4 |  34  68 112  166  230  304  388  482 ...
   5 |  58 112 180  262  358  468  592  730 ...
   6 |  88 166 262  376  508  658  826 1012 ...
   7 | 124 230 358  508  680  874 1090 1328 ...
   8 | 166 304 468  658  874 1116 1384 1678 ...
   9 | 214 388 592  826 1090 1384 1708 2062 ...
  10 | 268 482 730 1012 1328 1678 2062 2480 ...
  ...

Andrew Howroyd, Table of n, a(n) for n = 3..1277 (first 50 antidiagonals)
B. E. Tenner, Spotlight tiling, Ann. Combin. 14 (4) (2010) 553; arXiv preprint, arXiv:0711.1819 [math.CO], 2007-2008.

Cf. A051597, A051601.

T(m,n) = 2*(m-2)*(n-2)*(m+n-2) + (m-2)*(m+1) + (n-2)*(n+1) \\ Andrew Howroyd, Jan 02 2023

T(m,n) = 2*(m-2)*(n-2)*(m+n-2) + (m-2)*(m+1) + (n-2)*(n+1).

Terms a(31) and beyond from Andrew Howroyd, Jan 02 2023

A177696 Symmetrical triangle read by rows: T(n, k) = m*(T(n-1, k-1) + T(n-1, k)), where T(n, 1) = T(n, n) = n, and m = 2.

Original entry on oeis.org

1, 2, 2, 3, 8, 3, 4, 22, 22, 4, 5, 52, 88, 52, 5, 6, 114, 280, 280, 114, 6, 7, 240, 788, 1120, 788, 240, 7, 8, 494, 2056, 3816, 3816, 2056, 494, 8, 9, 1004, 5100, 11744, 15264, 11744, 5100, 1004, 9, 10, 2026, 12208, 33688, 54016, 54016, 33688, 12208, 2026, 10

Offset: 1

Roger L. Bagula, May 11 2010

			Triangle begins as:
   1;
   2,    2;
   3,    8,     3;
   4,   22,    22,     4;
   5,   52,    88,    52,     5;
   6,  114,   280,   280,   114,     6;
   7,  240,   788,  1120,   788,   240,     7;
   8,  494,  2056,  3816,  3816,  2056,   494,     8;
   9, 1004,  5100, 11744, 15264, 11744,  5100,  1004,    9;
  10, 2026, 12208, 33688, 54016, 54016, 33688, 12208, 2026, 10;

G. C. Greubel, Rows n = 1..50 of the triangle, flattened

Cf. A051597 (m=1).

function T(n, k) // T = A177696
  if k lt 1 or k gt n then return 0;
  elif k eq 1 or k eq n then return n;
  else return 2*(T(n-1, k-1) + T(n-1, k));
  end if;
end function;
[T(n, k): k in [1..n], n in [1..12]]; // G. C. Greubel, Oct 02 2024

m = 2;T[n_, k_]:= T[n, k]= If[k==1 || k==n, n, m*(T[n-1, k-1] + T[n-1, k])];Table[T[n, k], {n, 10}, {k, n}]//Flatten

@CachedFunction
def T(n, k): # T = A177696
    if (k<0 or k>n): return 0
    elif (k==1 or k==n): return n
    else: return 2*(T(n-1, k-1) + T(n-1, k))
flatten([[T(n, k) for k in range(1,n+1)] for n in range(1,13)]) # G. C. Greubel, Oct 02 2024

T(n, k) = m*(T(n-1, k-1) + T(n-1, k)), where T(n, 1) = T(n, n) = n, and m = 2.
From G. C. Greubel, Oct 02 2024: (Start)
Sum_{k=1..n} T(n, k) = (1/9)*(7*4^n + 6*n + 2).
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = (1-(-1)^n)*(2-n) - [n=1]. (End)

Edited by G. C. Greubel, Oct 02 2024

A213081 Exclusive-or based Pascal triangle, read by rows: T(n,1)=T(n,n)=n and T(n,k) = T(n-1,k-1) XOR T(n-1,k), where XOR is the bitwise exclusive-or operator.

Original entry on oeis.org

1, 2, 2, 3, 0, 3, 4, 3, 3, 4, 5, 7, 0, 7, 5, 6, 2, 7, 7, 2, 6, 7, 4, 5, 0, 5, 4, 7, 8, 3, 1, 5, 5, 1, 3, 8, 9, 11, 2, 4, 0, 4, 2, 11, 9, 10, 2, 9, 6, 4, 4, 6, 9, 2, 10, 11, 8, 11, 15, 2, 0, 2, 15, 11, 8, 11, 12, 3, 3, 4, 13, 2, 2, 13, 4, 3, 3, 12, 13, 15

Offset: 1

Alex Ratushnyak, Jun 04 2012

			Table begins:
   1;
   2,  2;
   3,  0,  3;
   4,  3,  3,  4;
   5,  7,  0,  7,  5;
   6,  2,  7,  7,  2,  6;
   7,  4,  5,  0,  5,  4,  7;
   8,  3,  1,  5,  5,  1,  3,  8;
   9, 11,  2,  4,  0,  4,  2, 11,  9;
  10,  2,  9,  6,  4,  4,  6,  9,  2, 10;
  11,  8, 11, 15,  2,  0,  2, 15, 11,  8, 11;

Cf. A007318 - Pascal's triangle read by rows.
Cf. A051597 - Pascal's triangle, begin and end n-th row with n+1, read by rows.
Cf. A080046 - Multiplicative Pascal triangle, read by rows: T(n,1)=T(n,n)=n and T(n,k) = T(n-1,k-1) * T(n-1,k).

src = [0]*1024
dst = [0]*1024
for i in range(1,39):
    dst[0] = dst[i-1] = i
    for j in range(1,i-1):
        dst[j] = src[j-1]^src[j]
    for j in range(i):
        src[j] = dst[j]
        print(dst[j], end=',')

A336014 Irregular triangle read by rows: T(n,1) = T(n,2) = T(n,3n-2) = T(n,3n-1) = n for n >= 1 and T(n,k) = T(n-1,k-2) + T(n-1,k-1) for n > 1, 3 <= k <= 3*(n-1).

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 3, 3, 4, 4, 6, 7, 8, 8, 8, 7, 6, 4, 4, 5, 5, 8, 10, 13, 15, 16, 16, 15, 13, 10, 8, 5, 5, 6, 6, 10, 13, 18, 23, 28, 31, 32, 31, 28, 23, 18, 13, 10, 6, 6, 7, 7, 12, 16, 23, 31, 41, 51, 59, 63, 63, 59, 51, 41, 31, 23, 16, 12, 7, 7

Offset: 1

Lechoslaw Ratajczak, Jul 04 2020

The number of terms in row n is 3*n-1 = A016789(n-1).
The sum of row n is equal to 2*A094002(n-1) = 2*A188589(n).
Fibonacci(n) = T(n+k,n) - T(n+k-1,n) for n >= 1, k = 1,2,3,...
The elements b(k) of the main diagonal, superdiagonal 1 and all subdiagonals have the recursive formula: b(k) = 2*b(k-1) + b(k-2) - 2*b(k-3) - b(k-4) for k > 4.

			Triangle begins:
n\k 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20...
1   1  1
2   2  2  2  2  2
3   3  3  4  4  4  4  3  3
4   4  4  6  7  8  8  8  7  6  4  4
5   5  5  8 10 13 15 16 16 15 13 10  8  5  5
6   6  6 10 13 18 23 28 31 32 31 28 23 18 13 10  6  6
7   7  7 12 16 23 31 41 51 59 63 63 59 51 41 31 23 16 12  7  7
...

Superdiagonal 1 is A029907 for n >= 1.
The main diagonal is A208354 for n >= 1.
Subdiagonal 1 is A102702(n-1) for n >= 1.
Subdiagonal 2 is A206268(n+2) for n >= 1 (conjectured).
Subdiagonal 3 is A191830(n+3) for n >= 1.
Cf. A007318, A051597, A228196.

T(n,k) = T(n,3*k-n) for 1 <= k <= 3*n-1.
T(n,k) = Sum_{u=2*(n-k)+3..2*n-k+1} ceiling(u/2)*A065941(k-2,u-2*(n-k)-3) for n >= 3, 3 <= k <= n.
T(n,k) = Sum_{m1=1..k-n} A208354(m1)*binomial(n-m1-1, k-n-m1) + Sum_{m2=1..2*n-k} A208354(m2)*binomial(n-m2-1, 2*n-k-m2) for n >= 2, n+1 <= k <= 2*n-1.
T(n,k) = Sum_{u=2*(k-2*n)+3..k-n+1} ceiling(u/2)*A065941(3*n-k-2,u-2*(k-2*n)-3) for n>= 3, 2*n <= k <= 3*(n-1).
T(n,k) = A208354(k) + (n-k)*Fibonacci(k) for n >= 3, 3 <= k <= n.
T(n,k) = A029907(k-1) + (n-k+1)*Fibonacci(k) for n >= 2, 3 <= k <= n+1.

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A132370 Array read by antidiagonals: T(m,n) = number of spotlight tilings of a width 1 m X n frame.

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A177696 Symmetrical triangle read by rows: T(n, k) = m*(T(n-1, k-1) + T(n-1, k)), where T(n, 1) = T(n, n) = n, and m = 2.

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A213081 Exclusive-or based Pascal triangle, read by rows: T(n,1)=T(n,n)=n and T(n,k) = T(n-1,k-1) XOR T(n-1,k), where XOR is the bitwise exclusive-or operator.

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A336014 Irregular triangle read by rows: T(n,1) = T(n,2) = T(n,3*n-2) = T(n,3*n-1) = n for n >= 1 and T(n,k) = T(n-1,k-2) + T(n-1,k-1) for n > 1, 3 <= k <= 3*(n-1).

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A336014 Irregular triangle read by rows: T(n,1) = T(n,2) = T(n,3n-2) = T(n,3n-1) = n for n >= 1 and T(n,k) = T(n-1,k-2) + T(n-1,k-1) for n > 1, 3 <= k <= 3*(n-1).