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A351123
Irregular triangle read by rows: row n lists the partial sums of the number of divisions by 2 after each tripling step in the Collatz trajectory of 2n+1.
1
1, 5, 4, 1, 2, 4, 7, 11, 2, 3, 4, 6, 9, 13, 1, 3, 6, 10, 3, 7, 1, 2, 3, 8, 12, 2, 5, 9, 1, 4, 5, 7, 10, 14, 6, 1, 2, 7, 11, 2, 3, 6, 7, 9, 12, 16, 1, 3, 4, 5, 6, 7, 9, 11, 12, 14, 15, 16, 18, 19, 20, 21, 23, 26, 27, 28, 30, 31, 33, 34, 35, 36, 37, 38, 41, 42, 43, 44, 48, 50, 52, 56, 59, 60, 61, 66, 70
OFFSET
1,2
COMMENTS
The terms in row n are T(n,0), T(n,1), ..., T(n, A258145(n)-2), and are the partial sums of the terms in row n of A351122.
In each row n, the terms also satisfy the equation 3* (3* (3* (3* ... (3* (2n+1) +1) + 2^T(n,0)) + 2^T(n,1)) + 2^T(n,2)) + ... = 2^T(n, A258145(n)-2); e.g., for n=4, and A258145(4)-2=5: 3* (3* (3* (3* (3* (3*9+1) +2^2) +2^3) +2^4) +2^6) +2^9 = 2^13.
For row n, the right-hand side of the equation above is 2^A166549(n+1). E.g., for the above example (n=4), the right-hand side is 2^A166549(4+1) = 2^13.
EXAMPLE
Triangle starts at T(1,0):
n\k 0 1 2 3 4 5 6 7 8 9 10 ...
1: 1 5
2: 4
3: 1 2 4 7 11
4: 2 3 4 6 9 13
5: 1 3 6 10
6: 3 7
7: 1 2 3 8 12
8: 2 5 9
...
E.g., row 3 of A351122 is [1, 1, 2, 3, 4]; its partial sums are [1, 2, 4, 7, 11].
PROG
(PARI) orow(n) = my(m=2*n+1, list=List()); while (m != 1, if (m%2, m = 3*m+1, my(nb = valuation(m, 2)); m/=2^nb; listput(list, nb)); ); Vec(list); \\ A351122
row(n) = my(v = orow(n)); vector(#v, k, sum(i=1, k, v[i])); \\ Michel Marcus, Jul 18 2022
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
EXTENSIONS
Data corrected by Mohsen Maesumi, Jul 18 2022
Last row completed by Michel Marcus, Jul 18 2022
STATUS
approved