A376180 Triangle read by rows (blocks). Each row consists of a permutation of the numbers of its constituents. The length of row number n is the n-th pentagonal number n(3n-1)/2 = A000326(n); see Comments.
1, 4, 5, 3, 6, 2, 13, 12, 14, 11, 15, 10, 16, 9, 17, 8, 18, 7, 30, 29, 31, 28, 32, 27, 33, 26, 34, 25, 35, 24, 36, 23, 37, 22, 38, 21, 39, 20, 40, 19, 58, 59, 57, 60, 56, 61, 55, 62, 54, 63, 53, 64, 52, 65, 51, 66, 50, 67, 49, 68, 48, 69, 47, 70, 46, 71, 45, 72, 44, 73, 43, 74, 42, 75, 41, 101, 102, 100, 103, 99, 104, 98, 105, 97, 106, 96, 107
Offset: 1
Examples
Triangle begins:
k = 1 2 3 4 5 6 7 8 9 10 11 12
n=1: 1;
n=2: 4, 5, 3, 6, 2;
n=3: 13, 12, 14, 11, 15, 10, 16, 9, 17, 8, 18, 7;
Subtracting (n-1)^2*n/2 from each term in row n is a permutation of 1 .. n(3n-1)/2:
1;
3,4,2,5,1;
7,6,8,5,9,4,10,3,11,2,12,1;
Links
- Boris Putievskiy, Table of n, a(n) for n = 1..9126
- Boris Putievskiy, Integer Sequences: Irregular Arrays and Intra-Block Permutations, arXiv:2310.18466 [math.CO], 2023.
- Index entries for sequences that are permutations of the natural numbers.
Programs
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Mathematica
a[n_]:=Module[{L,R,P,Result},L=Ceiling[Max[x/.NSolve[x^2 (x+1)-2 n==0,x,Reals]]]; R=n-((L-1)^2)*L/2; P=Which[OddQ[R]&&OddQ[L*(3*L-1)/2],(L*(3*L-1)/2-R+2)/2,OddQ[R]&&EvenQ[L*(3*L-1)/2],(R+L*(3*L-1)/2+1)/2,EvenQ[R]&&OddQ[L*(3*L-1)/2],Ceiling[(L*(3*L-1)/2+1)/2]+R/2, EvenQ[R]&&EvenQ[L*(3*L-1)/2],Ceiling[(L*(3*L-1)/2+1)/2]-R/2 ]; Result=P+(L-1)^2*L/2; Result] Nmax=18; Table[a[n],{n,1,Nmax}]
Formula
Linear sequence:
a(n) = P(n) + (L(n)-1)^2*L(n)/2. a(n) = P(n) + A002411(L(n)-1), where P = (L(n)(3L(n) - 1)/2 - R(n) + 2)/2 if R(n) is odd and L(n)(3L(n) - 1)/2 is odd, P = (R(n) + L(n)(3L(n) - 1)/2 + 1)/2 if R(n) is odd and L(n)(3L(n) - 1)/2 is even, P = ceiling((L(n)(3L(n) - 1)/2 + 1)/2) + R(n)/2 if R(n) is even and L(n)(3L(n) - 1)/2 is odd, P = ceiling((L(n)(3L(n) - 1)/2 + 1)/2) - R(n)/2 if R(n) is even and L(n)(3L(n) - 1)/2 is even. L(n) = ceiling(x(n)), x(n) is largest real root of the equation x^2*(x+1)-2*n = 0.
Triangular array T(n,k) for 1 <= k <= n(3n-1)/2 (see Example):
T(n,k) = P(n,k) + (n-1)^2*n/2, T(n,k) = P(n,k) + A002411(n-1), where P(n,k) = (n(3n - 1)/2 - k + 2)/2 if k is odd and n(3n - 1)/2 is odd,
P(n,k) = (k + n(3n - 1)/2 + 1)/2 if k is odd and n(3n - 1)/2 is even, P(n,k) = ceiling((n(3n - 1)/2 + 1)/2) + k/2 if k is even and n(3n - 1)/2 is odd, P(n,k) = ceiling((n(3n - 1)/2 + 1)/2) - k/2 if k is even and n(3n - 1)/2 is even.
Comments