A342707 T(n, k) is the result of replacing 2's by k's in the hereditary base-2 expansion of n; square array T(n, k) read by antidiagonals upwards, n, k >= 0.
0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 2, 2, 1, 0, 2, 1, 3, 3, 1, 0, 1, 2, 4, 4, 4, 1, 0, 2, 2, 5, 27, 5, 5, 1, 0, 0, 3, 6, 28, 256, 6, 6, 1, 0, 1, 1, 7, 30, 257, 3125, 7, 7, 1, 0, 0, 2, 8, 31, 260, 3126, 46656, 8, 8, 1, 0, 1, 2, 9, 81, 261, 3130, 46657, 823543, 9, 9, 1, 0
Offset: 0
Examples
Array T(n, k) begins:
n\k| 0 1 2 3 4 5 6 7 8 9
---+-------------------------------------------------------------------
0| 0 0 0 0 0 0 0 0 0 0
1| 1 1 1 1 1 1 1 1 1 1
2| 0 1 2 3 4 5 6 7 8 9
3| 1 2 3 4 5 6 7 8 9 10
4| 1 1 4 27 256 3125 46656 823543 16777216 387420489
5| 2 2 5 28 257 3126 46657 823544 16777217 387420490
6| 1 2 6 30 260 3130 46662 823550 16777224 387420498
7| 2 3 7 31 261 3131 46663 823551 16777225 387420499
8| 0 1 8 81 1024 15625 279936 5764801 134217728 3486784401
9| 1 2 9 82 1025 15626 279937 5764802 134217729 3486784402
10| 0 2 10 84 1028 15630 279942 5764808 134217736 3486784410
Links
- Wikipedia, Hereditary base-n notation
Crossrefs
Programs
-
PARI
T(n,k) = { my (v=0, e); while (n, n-=2^e=valuation(n,2); v+=k^T(e,k)); v }
Formula
T(n, n) = A343255(n).
T(n, 0) = A345021(n).
T(n, 1) = A000120(n).
T(n, 2) = n.
T(n, 3) = A222112(n-1).
T(0, k) = 0.
T(1, k) = 1.
T(2, k) = k.
T(3, k) = k + 1.
T(4, k) = k^k = A000312(k).
T(5, k) = k^k + 1 = A014566(k).
T(6, k) = k^k + k = A066068(k).
T(7, k) = k^k + k + 1 = A066279(k).
T(16, k) = k^k^k = A002488(k).
T(m + n, k) = T(m, k) + T(n, k) when m AND n = 0 (where AND denotes the bitwise AND operator).