login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A376353
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-pyramidal number A261720.
1
1, 4, 1, 3, 4, 1, 5, 5, 5, 1, 2, 3, 4, 5, 1, 11, 6, 6, 6, 6, 1, 10, 2, 3, 4, 5, 6, 1, 12, 14, 7, 7, 7, 7, 7, 1, 9, 13, 2, 3, 4, 5, 6, 7, 1, 13, 15, 17, 8, 8, 8, 8, 8, 8, 1, 8, 12, 16, 2, 3, 4, 5, 6, 7, 8, 1, 14, 16, 18, 20, 9, 9, 9, 9, 9, 9, 9, 1, 7, 11, 15, 19, 2, 3, 4, 5, 6, 7, 8, 9, 1, 15, 17, 19, 21, 23, 10, 10, 10, 10, 10, 10, 10, 10, 1, 6, 10, 14, 18, 22, 2, 3
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 93.
FORMULA
T(n,k) = P(n,k) + ((k-2)*(L(n,k)-1)^4+2*k*(L(n,k)-1)^3+(14-k)*(L(n,k)-1)^2+(12-2*k)*(L(n,k)-1))/24, where L(n,k) = ceiling(x(n,k)), x(n,k) is largest real root of the equation (k-2)*x^4+2*k*x^3+(14-k)*x^2+(12-2*k)*x-24*n = 0. R(n,k) = n - ((k-2)*(L(n,k)-1)^4+2*k*(L(n,k)-1)^3+(14-k)*(L(n,k)-1)^2+(12-2*k)*(L(n,k)-1))/24. P(n,k) = ((L(n,k)^3*(k-2)+3*L(n,k)^2-L(n,k)*(k-5))/6+2-R(n,k))/2 if R(n,k) is odd and (L(n,k)^3*(k-2)+3*L(n,k)^2-L(n,k)*(k-5))/6 is odd, P(n,k) = ((L(n,k)^3*(k-2)+3*L(n,k)^2-L(n,k)*(k-5))/6+1)+R(n,k))/2 if R(n,k) is odd and (L(n,k)^3*(k-2)+3*L(n,k)^2-L(n,k)*(k-5))/6 is even, P = ceiling(((L(n,k)^3*(k-2)+3*L(n,k)^2-L(n,k)*(k-5))/6+1)/2)+R(n,k)/2) if R(n,k) is even and (L(n,k)^3*(k-2)+3*L(n,k)^2-L(n,k)*(k-5))/6 is odd, P = ceiling(((L(n,k)^3*(k-2)+3*L(n,k)^2-L(n,k)*(k-5))/6+1)/2)-R(n,k)/2) if R(n,k) is even and (L(n,k)^3*(k-2)+3*L(n,k)^2-L(n,k)*(k-5))/6 is even.
T(1,n) = A000012(n). T(2,n) = A004526(n+8). T(3,n) = A028242(n+7). T(4,n) = A084964(n+6). T(5,n) = A168230(n+5). T(n-2,n) = 4*A000012(n) for n > 3. T(n-1,n) = A000027(n) for n > 2.
EXAMPLE
Table begins:
k = 3 4 5 6 7 8
--------------------------------------
n = 1: 1, 1, 1, 1, 1, 1, ...
n = 2: 4, 4, 5, 5, 6, 6, ...
n = 3: 3, 5, 4, 6, 5, 7, ...
n = 4: 5, 3, 6, 4, 7, 5, ...
n = 5: 2, 6, 3, 7, 4, 8, ...
n = 6: 11, 2, 7, 3, 8, 4, ...
n = 7: 10, 14, 2, 8, 3, 9, ...
n = 8: 12, 13, 17, 2, 9, 3, ...
n = 9: 9, 15, 16, 20, 2, 10, ...
n = 10: 13, 12, 18, 19, 23, 2, ...
n = 11: 8, 16, 15, 21, 22, 26, ...
n = 12: 14, 11, 19, 18, 24, 25, ...
n = 12: 7, 17, 14, 22, 21, 27, ...
n = 14: 15, 10, 20, 17, 25, 24, ...
n = 15: 6, 18, 13, 23, 20, 28, ...
... .
For k = 3 the first 3 blocks have lengths 1,4 and 10.
For k = 4 the first 2 blocks have lengths 1 and 5.
For k = 5 the first 2 blocks have lengths 1 and 6.
Each block is a permutation of the numbers of its constituents.
The first 6 antidiagonals are:
1;
4, 1;
3, 4, 1;
5, 5, 5, 1;
2, 3, 4, 5, 1;
11, 6, 6, 6, 6, 1;
MATHEMATICA
T[n_, k_]:=Module[{L, R, result}, L=Ceiling[Max[x/.NSolve[(k-2)*x^4+2*k*x^3+(14-k)*x^2+(12-2*k)*x-24*n==0, x, Reals]]]; R=n-((k-2)*(L-1)^4+2*k*(L-1)^3+(14-k)*(L-1)^2+(12-2*k)*(L-1))/24; P=Which[OddQ[R]&&OddQ[(L^3*(k-2)+3*L^2-L*(k-5))/6], ((L^3*(k-2)+3*L^2-L*(k-5))/6+2-R)/2, OddQ[R]&&EvenQ[(L^3*(k-2)+3*L^2-L*(k-5))/6], (R+(L^3*(k-2)+3*L^2-L*(k-5))/6+1)/2, EvenQ[R]&&OddQ[(L^3*(k-2)+3*L^2-L*(k-5))/6], Ceiling[((L^3*(k-2)+3*L^2-L*(k-5))/6+1)/2]+R/2, EvenQ[R]&&EvenQ[(L^3*(k-2)+3*L^2-L*(k-5))/6], Ceiling[((L^3*(k-2)+3*L^2-L*(k-5))/6+1)/2]-R/2]; Res= P +((k-2)*(L-1)^4+2*k*(L-1)^3+(14-k)*(L-1)^2+(12-2*k)*(L-1))/24; result=Res] Nmax=6; Table[T[n, k], {n, 1, Nmax}, {k, 3, Nmax+2}]
KEYWORD
nonn,tabl
AUTHOR
Boris Putievskiy, Sep 21 2024
STATUS
approved