A364602 Triangle T(n,k) with rows of length 2*n-1, generated by T(1,1)=0, T(n,1)=T(n-1,1)+2, T(n,2)=4*(n-1)-1, and for k>=3, T(n,k)=4*T(n-1,k-2)+1.

0, 2, 3, 1, 4, 7, 9, 13, 5, 6, 11, 17, 29, 37, 53, 21, 8, 15, 25, 45, 69, 117, 149, 213, 85, 10, 19, 33, 61, 101, 181, 277, 469, 597, 853, 341, 12, 23, 41, 77, 133, 245, 405, 725, 1109, 1877, 2389, 3413, 1365, 14, 27, 49, 93, 165, 309, 533, 981, 1621, 2901

Offset: 1

Author

Ruud H.G. van Tol, Jul 29 2023

Keywords

• nonn
• tabf

Comments

The sequence is a permutation of all integers >= 0.

Each row of T contains n*2-1 terms; the terms in column k increase by 2^k.

T(1,1) = 0; T(2,2) = 3.

T(2,1) = T(1,1)+2 = 2; T(2,3) = 4*T(1,1)+1 = 1 ("knight jump").

In the context of the 3x+1 problem, when a term x is used to represent the odd 4*x+1, its successor is 3*x+1, and k-1 is the 2-adic valuation of 3*x+1.

Right diagonal is A002450.

The terms at the top of the columns are A096773(k), or (2^(k-1)*(3 + 2*(-1)^k) - 1)/3.

When the table is analytically continued upwards by subtracting 2^k, the first layer of values are -A255138(k), or -(2^k*(3 + 2*(-1)^k) + 1)/3.

Examples

			Triangle T(n,k) begins:
n/k 1| 2| 3| 4|  5|  6|  7|  8|  9| 10| 11|
1|  0
2|  2  3  1
3|  4  7  9 13   5
4|  6 11 17 29  37  53  21
5|  8 15 25 45  69 117 149 213  85
6| 10 19 33 61 101 181 277 469 597 853 341
7| 12 ...
		

Crossrefs

Cf. A002450, A007310, A071797, A087445, A096773, A255138, A257480.

Programs

PARI

my(N=8, v=Vec([0, 2, 3, 1], N^2), p=4); for(n=3, N, my(K=2*n-1); for(k=1, K, v[p+k]=if(k<=2, v[p-K+k+2]+2^k, 4*v[p-K+k]+1)); p+=K); v

PARI

T(n, k) = 2^k*(n-(6*k+3-(-1)^k)/12)-1/3;

PARI

n_of_x(x) = my(n=0); while(1==x%4, x>>=2; n++); n + if(x%2,(x+1)/4,  x/2) + 1;

PARI

k_of_x(x) = valuation(3*x+1,2) + 1;

Formula

For n>1, T(n,k) = T(n-1,k) + 2^k, so T(n,1) = 2*(n-1).

T(n,2) = 4*(n-1)-1 = 2*T(n,1)-1, so T(2,2) = 3.

For n>1 and k>2, T(n,k) = 4*T(n-1,k-2)+1, so T(2,3) = 1.

For i>=0, a(i^2+1) = T(i+1,1).

T(n, k) = 2^k * (n - (6*k + 3 - (-1)^k)/12) - 1/3.

T(n,1) == 0 (mod 2); T(n,2) == 3 (mod 4); T(n,k>=3) == 1 (mod 4).

k = v2(3*T(n,k)+1) + 1, where v2(x) = A007814(x) is the 2-adic valuation of x.