A394480 Irregular triangle T(n,k) read by rows in which row n gives the congruences (mod 2^n) that satisfy A006577(T(n,k)+a*2^n) = A006577(1+T(n,k)+a*2^n), with a>=0, in the '3x+1' (or Collatz) problem.

12, 18, 22, 37, 14, 45, 29, 49, 94, 62, 99, 145, 189, 65, 125, 209, 291, 368, 382, 131, 193, 240, 254, 419, 472, 473, 586, 587, 593, 765, 314, 315, 387, 390, 509, 769, 842, 843, 849, 944, 947, 961, 1173, 1175, 1187, 1520, 1534, 1752, 1753, 257, 528, 629, 631

Offset

3,1

Comments

As you can see in Garner's paper, it is possible to use formulas to determine some terms of the different rows.

From theorem 1 you can find one of these formulas: for n>4 we have that 2^(n-1)+(-1)^(n-1)*2^(n-2)-2 are terms of T(n,k).

Links

V. Barbera, Table of n, a(n) for n = 3..8640 (rows 3..20 flattened)

Lynn E. Garner, On Heights in the Collatz 3n+1 Problem, Discrete Math. 55 (1985) 57-64.

Example

The triangle T(n,k) begins:
   n\k  1  2  3   4   ...
   3:   12
   4:   18
   5:   22 37
   6:   14 45
   7:   29 49 94
   8:   62 99 145 189
   ...
For n=5 and k=1: A006577(22)=A006577(23)=15; A006577(22+2^5)=A006577(23+2^5)=112; A006577(22+2*2^5)=A006577(23+2*2^5)=30; ...

Prog

(PARI) collatz_k(x, k_max)= {my(k=valuation(x, 2), m=0, temp); k=min(k_max, k); x>>=k; while(k<k_max&&x>1, x++; temp=valuation(x, 2); temp=min(temp, k_max-k); x>>=temp; x*=3^temp; x--; k+=temp; m+=temp; temp=valuation(x, 2); temp=min(temp, k_max-k); x>>=temp; k+=temp); [x, m, k]};
rows(n)={if(n>2, my(dim=2^(n+1)-1, ra=vector(dim, i, i), a_min=2, dim_i=6); for(n_i=3, n, dim_i*=2; dim_i+=2; for(a=a_min, dim_i, if(ra[a]>0, my(xt, mt, kt); [xt, mt, kt]=collatz_k(a, n_i); b=a+1; if([xt, mt, kt]==collatz_k(b, n_i), print1(a, ", "); forstep(i=a, dim, 2^n_i, ra[i]=0), if(kt<n_i, ra[a]=0)), if(a==a_min, a_min++))); print()))};

Crossrefs

Cf. A006577.

Sequence in context: A031207 A107794 A280014 * A396571 A359929 A162151

Adjacent sequences: A394477 A394478 A394479 * A394481 A394482 A394483

Keyword

nonn,tabf

Author

V. Barbera, Apr 19 2026

Status

approved