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A376276
Table T(n, k) n > 0, k > 2 read by upward antidiagonals. The sequences in each column k is a triangle read by rows (blocks), where each row is a permutation of the numbers of its constituents. The length of the row number n in column k is equal to the n-th k-gonal number A086270.
1
1, 3, 1, 4, 4, 1, 2, 3, 4, 1, 8, 5, 5, 5, 1, 7, 2, 3, 4, 5, 1, 9, 10, 6, 6, 6, 6, 1, 6, 11, 2, 3, 4, 5, 6, 1, 10, 9, 13, 7, 7, 7, 7, 7, 1, 5, 12, 12, 2, 3, 4, 5, 6, 7, 1, 16, 8, 14, 15, 8, 8, 8, 8, 8, 8, 1, 15, 13, 11, 16, 2, 3, 4, 5, 6, 7, 8, 1, 17, 7, 15, 14, 18, 9, 9, 9, 9, 9, 9, 9, 1, 14, 14, 10, 17, 17, 2, 3, 4, 5, 6, 7, 8, 9, 1, 18, 6, 16, 13, 19, 20, 10, 10, 10
OFFSET
1,2
COMMENTS
A209278 presents an algorithm for generating permutations.
The sequence is an intra-block permutation of integer positive numbers.
REFERENCES
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 45.
FORMULA
T(n,k) = P(n,k) + ((L(n,k)-1)^3*(k-2)+3*(L(n,k)-1)^2-(L(n,k)-1)*(k-5))/6, where L(n,k) = ceiling(x(n,k)), x(n,k) is largest real root of the equation x^3*(k - 2) + 3*x^2 - x*(k - 5) - 6*n = 0. R(n,k) = n - ((L(n,k) - 1)^3*(k-2)+3*(L(n,k)-1)^2-(L(n,k)-1)*(k-5))/6. P(n,k) = ((k * L(n,k) * (L(n,k) - 1) / 2) - L(n,k)^2 + 2 * L(n,k) + 2 - R(n,k)) / 2 if R is odd and (k * L(n,k) * (L(n,k) - 1) / 2) - L(n,k)^2 + 2 * L(n,k) is odd, P(n,k) = (R(n,k) + (k * L(n,k) * (L(n,k) - 1) / 2) - L(n,k)^2 + 2 * L(n,k) + 1) / 2 if R is odd and (k * L(n,k) * (L(n,k) - 1) / 2) - L(n,k)^2 + 2 * L(n,k) is even, P(n,k) = ceiling(((k * L(n,k) * (L(n,k) - 1) / 2) - L(n,k)^2 + 2 * L(n,k) + 1) / 2) + (R(n,k) / 2) if R is even and (k * L(n,k) * (L(n,k) - 1) / 2) - L(n,k)^2 + 2 * L(n,k) is odd, P(n,k) = ceiling(((k * L(n,k) * (L(n,k) - 1) / 2) - L(n,k)^2 + 2 * L(n,k) + 1) / 2) - (R(n,k) / 2) if R is even and (k * L(n,k) * (L(n,k) - 1) / 2) - L(n,k)^2 + 2 * L(n,k) is even.
T(1,n) = A000012(n). T(2,n) = A004526(n+7). T(3,n) = A028242(n+6). T(4,n) = A084964(n+5). T(n-2,n) = A000027(n) for n > 3. L(n,3) = A360010(n). L(n,4) = A074279(n).
EXAMPLE
Table begins:
k = 3 4 5 6 7 8
--------------------------------------
n = 1: 1, 1, 1, 1, 1, 1, ...
n = 2: 3, 4, 4, 5, 5, 6, ...
n = 3: 4, 3, 5, 4, 6, 5, ...
n = 4: 2, 5, 3, 6, 4, 7, ...
n = 5: 8, 2, 6, 3, 7, 4, ...
n = 6: 7, 10, 2, 7, 3, 8, ...
n = 7: 9, 11, 13, 2, 8, 3, ...
n = 8: 6, 9, 12, 15, 2, 9, ...
n = 9: 10, 12, 14, 16, 18, 2, ...
n =10: 5, 8, 11, 14, 17, 20, ...
n =11: 16, 13, 15, 17, 19, 21, ...
n =12: 15, 7, 10, 13, 16, 19, ...
n =13: 17, 14, 16, 18, 20, 22, ...
n =14: 14, 6, 9, 12, 15, 18, ...
n =15: 18, 23, 17, 19, 21, 23, ...
n =16: 13, 22, 8, 11, 14, 17, ...
n =17: 19, 24, 18, 20, 22, 24, ...
n =18: 12, 21, 7, 10, 13, 16, ...
n =19: 20, 25, 30, 21, 23, 25, ...
n =20: 11, 20, 29, 9, 12, 15, ...
... .
For k = 3 the first 4 blocks have lengths 1,3,6 and 10.
For k = 4 the first 3 blocks have lengths 1,4, and 9.
For k = 5 the first 3 blocks have lengths 1,5, and 12.
Each block is a permutation of the numbers of its constituents.
The first 6 antidiagonals are:
1;
3, 1;
4, 4, 1;
2, 3, 4, 1;
8, 5, 5, 5, 1;
7, 2, 3, 4, 5, 1;
MATHEMATICA
T[n_, k_]:=Module[{L, R, P, Res, result}, L=Ceiling[Max[x/.NSolve[x^3*(k-2)+3*x^2-x*(k-5)-6*n==0, x, Reals]]];
R=n-(((L-1)^3)*(k-2)+3*(L-1)^2-(L-1)*(k-5))/6; P=Which[OddQ[R]&&OddQ[k*L*(L-1)/2-L^2+2*L], ((k*L*(L-1)/2-L^2+2*L+1-R)+1)/2, OddQ[R]&&EvenQ[k*L*(L-1)/2-L^2+2*L], (R+k*L*(L-1)/2-L^2+2*L+1)/2, EvenQ[R]&&OddQ[k*L*(L-1)/2-L^2+2*L], Ceiling[(k*L*(L-1)/2-L^2+2*L+1)/2]+R/2, EvenQ[R]&&EvenQ[k*L*(L-1)/2-L^2+2*L], Ceiling[(k*L*(L-1)/2-L^2+2*L+1)/2]-R/2];
Res=P+((L-1)^3*(k-2)+3*(L-1)^2-(L-1)*(k-5))/6; result=Res; result]
Nmax=6; Table[T[n, k], {n, 1, Nmax}, {k, 3, Nmax+2}]
KEYWORD
nonn,tabl
AUTHOR
Boris Putievskiy, Sep 18 2024
STATUS
approved