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.
LINKS
Boris Putievskiy, Table of n, a(n) for n = 1..9870
Boris Putievskiy, Integer Sequences: Irregular Arrays and Intra-Block Permutations, arXiv:2310.18466 [math.CO], 2023.
Eric Weisstein's World of Mathematics, Pyramidal Number.
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.
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}]
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Boris Putievskiy, Sep 21 2024
STATUS
approved