A391114 - OEIS

%I #17 Jan 05 2026 16:55:59

%S 4,1,10,10,1,26,2,28,1,66,16,1,76,1,162,3,46,1,196,1,386,22,2,126,1,

%T 484,1,898,4,64,1,326,1,1156,1,2050,28,2,176,1,806,1,2692,1,4610,5,82,

%U 1,456,1,1926,1,6148,1,10242,34,3,226,1,1128,1,4486,1,13828,1,22530

%N Array read by antidiagonals upwards: A(n, k) = if(n mod 2 = 0, floor((n+2^k-2) / 2^k), n * (k*2^k+1) + 1).

%C Does for every column k iterating n -> A(n, k) always reach 1 (one cycle 1 -> k*2^k+2 -> k+1 -> ... -> 1)? For k=1 we have the Collatz or 3x+1 problem (see A006370).

%C Answer: no, because for the column k = 10 the sequence starts with 10242, 11, 112652, 111, 1136752, 1111, 11377752, 11112, 11, ... showing a cycle and doesn't reach 1. - _Werner Schulte_, Jan 05 2026

%F A(2*n+1, k) = A(2*n-1, k) + 2 * A002064(k).

%F A(n, k) = A(n-2, k) + A(n-2^k, k) - A(n-2-2^k, k).

%F G.f. of column k: ((k*2^k+2) * x + x^2 + (2*k*2^k+2) * (x^3 - x^(2^k+1)) / (1 - x^2) + k*2^k * x^(2^k+1)) / ((1 - x^2) * (1 - x^(2^k))).

%e The array A(n, k) for k > 0 and n > 0 starts:

%e n \ k :  1   2   3   4    5    6     7     8     9     10     11     12      13

%e ===============================================================================

%e    1  :  4  10  26  66  162  386   898  2050  4610  10242  22530  49154  106498

%e    2  :  1   1   1   1    1    1     1     1     1      1      1      1       1

%e    3  : 10  28  76 196  484 1156  2692  6148 13828  30724  67588 147460  319492

%e    4  :  2   1   1   1    1    1     1     1     1      1      1      1       1

%e    5  : 16  46 126 326  806 1926  4486 10246 23046  51206 112646 245766  532486

%e    6  :  3   2   1   1    1    1     1     1     1      1      1      1       1

%e    7  : 22  64 176 456 1128 2696  6280 14344 32246  71688 157704 344072  745480

%e    8  :  4   2   1   1    1    1     1     1     1      1      1      1       1

%e    9  : 28  82 226 586 1450 3466  8074 18442 41482  92170 202762 442378  958474

%e   10  :  5   3   2   1    1    1     1     1     1      1      1      1       1

%e   11  : 34 100 276 716 1772 4236  9868 22540 50700 112652 247820 540684 1171468

%e   12  :  6   3   2   1    1    1     1     1     1      1      1      1       1

%e   13  : 40 118 326 846 2094 5006 11662 26638 59918 133134 292878 638990 1384462

%e   etc.

%t A[n_, k_] := If[EvenQ[n], Floor[(n + 2^k - 2)/2^k], n*(k*2^k + 1) + 1]; Table[A[n - k + 1, k], {n, 1, 11}, {k, 1, n}] // Flatten (* _Amiram Eldar_, Nov 29 2025 *)

%o (PARI) A(n, k) = if(n%2==0, floor((n+2^k-2)/2^k), n*(k*2^k+1)+1)

%Y Cf. A000012 (row 2), A002064, A006370 (column 1), A391054 (column 2), A391075 (column 3).

%K nonn,easy,tabl

%O 1,1

%A _Werner Schulte_, Nov 29 2025