A372288 -id:A372288 - cp's OEIS frontend

A372282 Array read by upward antidiagonals: A(n, k) = A371094(A(n-1, k)) for n > 1, k >= 1; A(1, k) = 2*k-1.

1, 21, 3, 5461, 21, 5, 357913941, 5461, 341, 7, 1537228672809129301, 357913941, 1398101, 45, 9, 28356863910078205288614550619314017621, 1537228672809129301, 23456248059221, 1109, 117, 11, 9649340769776349618630915417390658987772498722136713669954798667326094136661, 28356863910078205288614550619314017621, 6602346876188694799461995861, 873813, 11605, 69, 13

Offset: 1

Antti Karttunen, Apr 28 2024

			Array begins:
n\k|    1     2        3     4      5     6        7     8      9     10
---+----------------------------------------------------------------------
1  |    1,    3,       5,    7,     9,   11,      13,   15,    17,    19,
2  |   21,   21,     341,   45,   117,   69,     341,   93,   213,   117,
3  | 5461, 5461, 1398101, 1109, 11605, 3413, 1398101, 2261, 87381, 11605,

Cf. A005408 (row 1), A372351 (row 2, bisection of A371094), A372444 (column 14).
Arrays derived from this one:
  A372283,
  A372285 the number of terms of A086893 in the interval [A(n, k), A(1+n, k)],
  A372287 the column index of A(n, k) in array A257852,
  A372288 the sum of digits of A(n, k) in "Jacobsthal greedy base",
  A372353 differences between A(n,k) and the largest term of A086893 <= A(n,k),
  A372354 floor(log_2(.)) of terms, A372356 (and their columnwise first differences),
  A372359 terms xored with binary words of the same length, either of the form 10101...0101 or 110101...0101, depending on whether the binary length is odd or even.
Cf. also arrays A371096, A371102 that give subsets of columns of this array, and array A371100 that gives the terms of the row 2 in different order.

up_to = 28;
A371094(n) = { my(m=1+3*n, e=valuation(m,2)); ((m*(2^e)) + (((4^e)-1)/3)); };
A372282sq(n,k) = if(1==n,2*k-1,A371094(A372282sq(n-1,k)));
A372282list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A372282sq((a-(col-1)),col))); (v); };
v372282 = A372282list(up_to);
A372282(n) = v372282[n];

A265745 a(n) is the number of Jacobsthal numbers (A001045) needed to sum to n using the greedy algorithm.

Original entry on oeis.org

0, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 4, 3, 4, 5, 4, 5, 4, 5, 6, 5, 6, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2

Offset: 0

Antti Karttunen, Dec 17 2015

Sum of digits in "Jacobsthal greedy base", A265747.
It would be nice to know for sure whether this sequence gives also the least number of Jacobsthal numbers that add to n, i.e., that there cannot be even better nongreedy solutions.
The integer 63=21+21+21 has 3 for its 'non-greedy' solution, and a(63) = 5 for its greedy solution 63=43+11+5+3+1. - Yuriko Suwa, Jul 11 2021
Positions where a(n) is different from A372555(n) are n=63, 84, 148, 169, 191, 212, 234, 255, etc. See A372557. - Antti Karttunen, May 07 2024

			a(0) = 0, because no numbers are needed to form an empty sum, which is zero.
For n=1 we need just A001045(2) = 1, thus a(1) = 1.
For n=2 we need A001045(2) + A001045(2) = 1 + 1, thus a(2) = 2.
For n=4 we need A001045(3) + A001045(2) = 3 + 1, thus a(4) = 2.
For n=6 we form the greedy sum as A001045(4) + A001045(2) = 5 + 1, thus a(6) = 2. Alternatively, we could form the sum as A001045(3) + A001045(3) = 3 + 3, but the number of summands in that case is no less.
For n=7 we need A001045(4) + A001045(2) + A001045(2) = 5 + 1 + 1, thus a(7) = 3.
For n=8 we need A001045(4) + A001045(3) = 5 + 3, thus a(8) = 2.
For n=10 we need A001045(4) + A001045(4) = 5 + 5, thus a(10) = 2.

Antti Karttunen, Table of n, a(n) for n = 0..10923

Cf. A001045, A130249, A197911, A003158, A265746, A265747, A364377, A364379, A364380, A372288, A372555, A372557.
Cf. A054111 (apparently the positions of the first occurrence of each n > 0).
Cf. also A007895, A014420, A053610, A265404, A265743, A265744.

jacob[n_] := (2^n - (-1)^n)/3; maxInd[n_] := Floor[Log2[3*n + 1]]; A265745[n_] := A265745[n] = 1 + A265745[n - jacob[maxInd[n]]]; A265745[0] = 0; Array[A265745, 100, 0] (* Amiram Eldar, Jul 21 2023 *)

A130249(n) = floor(log(3*n + 1)/log(2));
A001045(n) = (2^n - (-1)^n) / 3;
A265745(n) = {if(n == 0, 0, my(d = n - A001045(A130249(n))); if(d == 0, 1, 1 + A265745(d)));} \\ Amiram Eldar, Jul 21 2023

def greedyJ(n): n1 = (3*n+1).bit_length() - 1; return (2**n1 - (-1)**n1)//3
def a(n): return 0 if n == 0 else 1 + a(n - greedyJ(n))
print([a(n) for n in range(107)]) # Michael S. Branicky, Jul 11 2021

a(0) = 0; for n >= 1, a(n) = 1 + a(n - A001045(A130249(n))). [This formula uses a simple greedy algorithm.]

A372283 Array read by upward antidiagonals: A(n, k) = R(A(n-1, k)) for n > 1, k >= 1; A(1, k) = 2*k-1, where Reduced Collatz function R(n) gives the odd part of 3n+1.

Original entry on oeis.org

1, 1, 3, 1, 5, 5, 1, 1, 1, 7, 1, 1, 1, 11, 9, 1, 1, 1, 17, 7, 11, 1, 1, 1, 13, 11, 17, 13, 1, 1, 1, 5, 17, 13, 5, 15, 1, 1, 1, 1, 13, 5, 1, 23, 17, 1, 1, 1, 1, 5, 1, 1, 35, 13, 19, 1, 1, 1, 1, 1, 1, 1, 53, 5, 29, 21, 1, 1, 1, 1, 1, 1, 1, 5, 1, 11, 1, 23, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 35, 25

Offset: 1

Antti Karttunen, Apr 28 2024

Collatz conjecture is equal to the claim that in each column 1 will eventually appear. See also arrays A372287 and A372288.

			Array begins:
n\k| 1  2  3   4   5   6   7   8   9  10  11  12  13   14  15   16  17  18
---+-----------------------------------------------------------------------
1  | 1, 3, 5,  7,  9, 11, 13, 15, 17, 19, 21, 23, 25,  27, 29,  31, 33, 35,
2  | 1, 5, 1, 11,  7, 17,  5, 23, 13, 29,  1, 35, 19,  41, 11,  47, 25, 53,
3  | 1, 1, 1, 17, 11, 13,  1, 35,  5, 11,  1, 53, 29,  31, 17,  71, 19,  5,
4  | 1, 1, 1, 13, 17,  5,  1, 53,  1, 17,  1,  5, 11,  47, 13, 107, 29,  1,
5  | 1, 1, 1,  5, 13,  1,  1,  5,  1, 13,  1,  1, 17,  71,  5, 161, 11,  1,
6  | 1, 1, 1,  1,  5,  1,  1,  1,  1,  5,  1,  1, 13, 107,  1, 121, 17,  1,
7  | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  5, 161,  1,  91, 13,  1,
8  | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 121,  1, 137,  5,  1,
9  | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  91,  1, 103,  1,  1,
10 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 137,  1, 155,  1,  1,
11 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 103,  1, 233,  1,  1,
12 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 155,  1, 175,  1,  1,
13 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 233,  1, 263,  1,  1,
14 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 175,  1, 395,  1,  1,
15 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 263,  1, 593,  1,  1,
16 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 395,  1, 445,  1,  1,

Cf. A005408 (row 1), A075677 (row 2), A372443 (column 14).
Cf. A258145 (A075680), A371093.
Arrays derived from this one or related to:
  A372282,
  A372287 the column index of A(n, k) in array A257852,
  A372361 terms xored with binary words of the same length, either of the form 10101...0101 or 110101...0101, depending on whether the binary length is odd or even,
  A372360 binary weights of A372361.
Cf. also array A371095 (giving every fourth column, 1, 5, 9, ...) and irregular array A256598 which gives the terms of each column, but only down to the first 1.

With[{dmax = 15}, Table[#[[k, n-k+1]], {n, dmax}, {k, n}] & [Array[NestList[(3*# + 1)/2^IntegerExponent[3*# + 1, 2] &, 2*# - 1, dmax - #] &, dmax]]] (* Paolo Xausa, Apr 29 2024 *)

up_to = 91;
R(n) = { n = 1+3*n; n>>valuation(n, 2); };
A372283sq(n,k) = if(1==n,2*k-1,R(A372283sq(n-1,k)));
A372283list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A372283sq((a-(col-1)),col))); (v); };
v372283 = A372283list(up_to);
A372283(n) = v372283[n];

For n > 1, A(n, k) = R(A372282(n-1, k)), where R(n) = (3*n+1)/2^A371093(n).

For all k >= 1, A(A258145(k-1), k) = 1 [which is the topmost 1 in each column].

A372287 Array read by upward antidiagonals: A(n, k) = A371092(A372283(n, k)), n,k >= 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 3, 2, 3, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 1, 3, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 6, 3, 1, 1, 1, 1, 1, 1, 1, 9, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 9, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 7

Offset: 1

Antti Karttunen, Apr 28 2024

A(n, k) gives the column index of A372282(n, k) [or equally, of A372283(n, k)] in array A257852.

Collatz conjecture is equal to the claim that in each column 1 will eventually appear.

			Array begins:
n\k| 1  2  3  4  5  6  7  8  9 10 11 12 13  14 15  16 17 18 19  20
---+---------------------------------------------------------------
1  | 1, 1, 1, 2, 2, 3, 1, 4, 3, 5, 1, 6, 4,  7, 2,  8, 5, 9, 2, 10,
2  | 1, 1, 1, 3, 2, 3, 1, 6, 1, 2, 1, 9, 5,  6, 3, 12, 4, 1, 2, 15,
3  | 1, 1, 1, 3, 3, 1, 1, 9, 1, 3, 1, 1, 2,  8, 3, 18, 5, 1, 3, 12,
4  | 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 1, 3, 12, 1, 27, 2, 1, 3, 17,
5  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 18, 1, 21, 3, 1, 1,  4,
6  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 27, 1, 16, 3, 1, 1,  5,
7  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 21, 1, 23, 1, 1, 1,  2,
8  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 18, 1, 1, 1,  3,
9  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 23, 1, 26, 1, 1, 1,  3,
10 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 18, 1, 39, 1, 1, 1,  1,
11 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 26, 1, 30, 1, 1, 1,  1,
12 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 39, 1, 44, 1, 1, 1,  1,
13 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 30, 1, 66, 1, 1, 1,  1,
14 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 44, 1, 99, 1, 1, 1,  1,
15 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 66, 1, 75, 1, 1, 1,  1,
16 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 99, 1, 28, 1, 1, 1,  1,
17 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 75, 1, 42, 1, 1, 1,  1,
18 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 28, 1, 63, 1, 1, 1,  1,
19 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 42, 1, 48, 1, 1, 1,  1,
20 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 63, 1, 71, 1, 1, 1,  1,

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A257852, A371092, A371094, A372282, A372283, A372288.

Cf. also A371097 (array giving every fourth column, 1, 5, 9, ...), A371103 (array giving every even numbered column), also array A371101.

up_to = 105;
A000265(n) = (n>>valuation(n,2));
A371092(n) = floor((A000265(1+(3*n))+5)/6);
R(n) = { n = 1+3*n; n>>valuation(n, 2); };
A372283sq(n,k) = if(1==n,2*k-1,R(A372283sq(n-1,k)));
A372287sq(n,k) = A371092(A372283sq(n,k));
A372287list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A372287sq((a-(col-1)),col))); (v); };
v372287 = A372287list(up_to);
A372287(n) = v372287[n];

A(n, k) = A371092(A372282(n,k)) = A371092(A372283(n,k)).

A372360 Array read by upward antidiagonals: A(n, k) = A000120(A372361(n, k)), n,k >= 1; Binary weights of terms of arrays A372359 and A372361.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 0, 1, 1, 2, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 3, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 3, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 3

Offset: 1

Antti Karttunen, May 01 2024

Entry A(n, k) at row n and column k tells how many bits needs to be flipped in the binary expansion of the (n-1)-th iterate of Reduced Collatz function R, when started from 2*k-1, to obtain the unique term of A086893 with the same binary length as that (n-1)-th iterate. That is, A(n, k) gives the Hamming distance between A372283(n, k) and A086893(1+A000523(A372283(n, k))).

Zeros occur in the same locations as where they occur in A372359, etc.

			Array begins:
n\k| 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
---+-------------------------------------------------------------------------
1  | 0, 0, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 2, 3, 1, 2, 2, 3, 1, 2, 3, 4, 2, 3,
2  | 0, 0, 0, 2, 1, 1, 0, 1, 0, 1, 0, 3, 2, 3, 2, 3, 2, 0, 1, 3, 2, 2, 1, 2,
3  | 0, 0, 0, 1, 2, 0, 0, 3, 0, 2, 0, 0, 1, 2, 1, 2, 2, 0, 2, 2, 3, 1, 0, 5,
4  | 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 2, 3, 0, 5, 1, 0, 1, 3, 2, 1, 0, 4,
5  | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 4, 2, 0, 0, 2, 5, 1, 0, 3,
6  | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 3, 1, 0, 0, 2, 4, 2, 0, 3,
7  | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 3, 0, 0, 0, 1, 3, 1, 0, 4,
8  | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 4, 0, 0, 0, 2, 3, 0, 0, 3,
9  | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 3, 0, 0, 0, 1, 4, 0, 0, 4,
10 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 3, 0, 0, 4,
11 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 4, 0, 0, 0, 0, 4, 0, 0, 5,
12 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 5, 0, 0, 0, 0, 4, 0, 0, 3,
13 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 3, 0, 0, 0, 0, 5, 0, 0, 6,
14 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 6, 0, 0, 0, 0, 3, 0, 0, 2,
15 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 2, 0, 0, 0, 0, 6, 0, 0, 4,
16 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 4, 0, 0, 0, 0, 2, 0, 0, 4,
17 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 4, 0, 0, 0, 0, 4, 0, 0, 4,
18 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 4, 0, 0, 3,
19 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 3, 0, 0, 0, 0, 4, 0, 0, 4,
20 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 3, 0, 0, 6,
21 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 6, 0, 0, 0, 0, 4, 0, 0, 4,
We have A372283(5, 14) = 71, and when we compare the binary expansion of 71 = 1000111_2 with the term of A086893 that has a binary expansion of the same length, which in this case is 85 = 1010101_2, we see that only the bits at positions 1 and 4 (indexed from the right hand end, with 0 being the least significant bit position at right) need to be toggled to obtain the 71 from 85 or vice versa, therefore A(5, 14) = 2.
We have A372283(6, 14) = 107 = 1101011_2, and when xored with A086893(7) = 85 = 1010101_2, we obtain A372361(6, 14) = 62 = 111110_2, with five 1-bits, therefore A(6, 14) = 5. I.e., five bits (all except the least and the most significant bit) need to be flipped to change 85 to 107 or vice versa.

Cf. A000120, A003987, A086893, A372282, A372283, A372287, A372354, A372358, A372360.
Binary weights of A372359 and A372361.
Cf. also A372288.

up_to = 105;
R(n) = { n = 1+3*n; n>>valuation(n, 2); };
A372283sq(n,k) = if(1==n,2*k-1,R(A372283sq(n-1,k)));
A000523(n) = logint(n,2);
A086893(n) = (if(n%2, 2^(n+1), 2^(n+1)+2^(n-1))\3);
A372358(n) = bitxor(A086893(1+A000523(n)),n);
A372361sq(n,k) = A372358(A372283sq(n,k));
A372361list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A372361sq((a-(col-1)),col))); (v); };
v372361 = A372361list(up_to);
A372361(n) = v372361[n];

A(n, k) = A000120(A372361(n, k)) = A000120(A372358(A372283(n, k))).

A(n, k) = A000120(A372359(n, k)) = A000120(A372358(A372282(n, k))).

A372561 Array read by upward antidiagonals: A(n, k) = A265745(A372560(n, k)) for n > 1, k >= 1.

Original entry on oeis.org

3, 5, 5, 5, 5, 3, 5, 5, 5, 3, 5, 5, 5, 5, 5, 3, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 3, 3, 3, 3, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 10347, 6251, 2155, 1131, 619, 363, 235, 107, 43, 27, 11, 7, 5

Offset: 1

Antti Karttunen, May 08 2024

In general, it seems that for n>2, k>1, A(n, k) = A(n-1, k+1) = A(k, n), except on those two anomalous antidiagonals, first on the thirteenth antidiagonal, where for n=1..13, A(n,14-n) obtains values 5, 7, 11, 27, 43, 107, 235, 363, 619, 1131, 2155, 6251, 10347, and then on the 30th antidiagonal, where for n=1.., A(n,31-n) obtains values 5, 11, 15, 23, 39, 71, 135, 391, 647, 1671, 2695, 4743, 17031, 33415, 49799, 82567, 148103, 410247, etc. The corresponding antidiagonals in A372560 begin as:
  233, 933, 14933, 978670933, 64138178286933, 1183140560213014108063589658350933, ..., and:
  911, 58325, 933205, 238900565, 15656587449685, 67244531063362552157525, etc. I conjecture that for the former sequence of numbers x, from 933 onward, A372555(x) = 7, and for the latter sequence of numbers y, from 58325 onward, A372555(y) = 9, and that the array A372555(A372560(n, k)) is symmetric apart from its borders, i.e, that for n, k > 1, A372555(A372560(n, k)) = A372555(A372560(k, n)).

			Array begins:
n\k| 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21
---+----------------------------------------------------------------
1  | 3, 5, 3, 3, 5, 3, 5, 5, 3, 5, 5, 5, 5, 3, 5, 3, 7, 5, 7, 5, 5,
2  | 5, 5, 5, 5, 3, 5, 5, 3, 5, 5, 5, 7, 5, 5, 5, 7, 5, 7, 7, 5, 5,
3  | 5, 5, 5, 3, 5, 5, 3, 5, 5, 5, 11, 5, 5, 5, 7, 5, 7, 7, 5, 5, 7,
4  | 5, 5, 3, 5, 5, 3, 5, 5, 5, 27, 5, 5, 5, 7, 5, 7, 7, 5, 5, 7, 9,
5  | 5, 3, 5, 5, 3, 5, 5, 5, 43, 5, 5, 5, 7, 5, 7, 7, 5, 5, 7, 9, 7,
6  | 3, 5, 5, 3, 5, 5, 5, 107, 5, 5, 5, 7, 5, 7, 7, 5, 5, 7, 9, 7, 7,
7  | 5, 5, 3, 5, 5, 5, 235, 5, 5, 5, 7, 5, 7, 7, 5, 5, 7, 9, 7, 7, 7,
8  | 5, 3, 5, 5, 5, 363, 5, 5, 5, 7, 5, 7, 7, 5, 5, 7, 9, 7, 7, 7, 9,
9  | 3, 5, 5, 5, 619, 5, 5, 5, 7, 5, 7, 7, 5, 5, 7, 9, 7, 7, 7, 9, 7,
10 | 5, 5, 5, 1131, 5, 5, 5, 7, 5, 7, 7, 5, 5, 7, 9, 7, 7, 7, 9, 7, 1671,
11 | 5, 5, 2155, 5, 5, 5, 7, 5, 7, 7, 5, 5, 7, 9, 7, 7, 7, 9, 7, 2695, 3,
12 | 5, 6251, 5, 5, 5, 7, 5, 7, 7, 5, 5, 7, 9, 7, 7, 7, 9, 7, 4743, 3, 5,
13 | 10347, 5, 5, 5, 7, 5, 7, 7, 5, 5, 7, 9, 7, 7, 7, 9, 7, 17031, 3, 5, 3,
14 | 5, 5, 5, 7, 5, 7, 7, 5, 5, 7, 9, 7, 7, 7, 9, 7, 33415, 3, 5, 3, 5,
15 | 5, 5, 7, 5, 7, 7, 5, 5, 7, 9, 7, 7, 7, 9, 7, 49799, 3, 5, 3, 5, 5,
etc.
From column 19 to column 41, the first 11 rows:
n\k|19 20 ........................................................... 40 41
---+-------------------------------------------------------------------------
1  | 7, 5, 5, 5, 7, 7, 5, 5, 5, 7, 7, 5,    3, 3, 3, 5, 5, 5, 5, 3, 3, 3, 1,
2  | 7, 5, 5, 7, 9, 7, 7, 7, 9, 7, 11,   3, 5, 3, 5, 5, 5, 5, 3, 5, 3, 1, 1,
3  | 5, 5, 7, 9, 7, 7, 7, 9, 7, 15,   3, 5, 3, 5, 5, 5, 5, 3, 5, 3, 1, 1, 1,
4  | 5, 7, 9, 7, 7, 7, 9, 7, 23,   3, 5, 3, 5, 5, 5, 5, 3, 5, 3, 1, 1, 1, 1,
5  | 7, 9, 7, 7, 7, 9, 7, 39,   3, 5, 3, 5, 5, 5, 5, 3, 5, 3, 1, 1, 1, 1, 1,
6  | 9, 7, 7, 7, 9, 7, 71,   3, 5, 3, 5, 5, 5, 5, 3, 5, 3, 1, 1, 1, 1, 1, 1,
7  | 7, 7, 7, 9, 7, 135,  3, 5, 3, 5, 5, 5, 5, 3, 5, 3, 1, 1, 1, 1, 1, 1, 1,
8  | 7, 7, 9, 7, 391,  3, 5, 3, 5, 5, 5, 5, 3, 5, 3, 1, 1, 1, 1, 1, 1, 1, 1,
9  | 7, 9, 7, 647,  3, 5, 3, 5, 5, 5, 5, 3, 5, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1,
10 | 9, 7, 1671, 3, 5, 3, 5, 5, 5, 5, 3, 5, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
11 | 7, 2695, 3, 5, 3, 5, 5, 5, 5, 3, 5, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,

Cf. A265745, A371094, A372282, A372288, A372555, A372560.

\\ Needs also program from A372560.
up_to = 91;
A001045(n) = (2^n - (-1)^n) / 3;
A130249(n) = (#binary(3*n+1)-1);
A265745(n) = { my(s=0); while(n,s++; n -= A001045(A130249(n))); (s); };
A372561sq(n,k) = A265745(A372560sq(n,k));
A372561list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A372561sq((a-(col-1)),col))); (v); };
v372561 = A372561list(up_to);
A372561(n) = v372561[n];

cp's OEIS Frontend

A372282 Array read by upward antidiagonals: A(n, k) = A371094(A(n-1, k)) for n > 1, k >= 1; A(1, k) = 2*k-1.

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A265745 a(n) is the number of Jacobsthal numbers (A001045) needed to sum to n using the greedy algorithm.

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A372283 Array read by upward antidiagonals: A(n, k) = R(A(n-1, k)) for n > 1, k >= 1; A(1, k) = 2*k-1, where Reduced Collatz function R(n) gives the odd part of 3n+1.

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A372287 Array read by upward antidiagonals: A(n, k) = A371092(A372283(n, k)), n,k >= 1.

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A372360 Array read by upward antidiagonals: A(n, k) = A000120(A372361(n, k)), n,k >= 1; Binary weights of terms of arrays A372359 and A372361.

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A372561 Array read by upward antidiagonals: A(n, k) = A265745(A372560(n, k)) for n > 1, k >= 1.

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