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A225620 Indices of partitions in the table of compositions of A228351. 67
1, 2, 3, 4, 6, 7, 8, 10, 12, 14, 15, 16, 20, 24, 26, 28, 30, 31, 32, 36, 40, 42, 48, 52, 56, 58, 60, 62, 63, 64, 72, 80, 84, 96, 100, 104, 106, 112, 116, 120, 122, 124, 126, 127, 128, 136, 144, 160, 164, 168, 170, 192, 200, 208, 212, 224, 228, 232, 234, 240, 244, 248, 250, 252, 254, 255 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Also triangle read by rows in which T(n,k) is the decimal representation of a binary number whose mirror represents the k-th partition of n according with the list of juxtaposed reverse-lexicographically ordered partitions of the positive integers (A026792).
In order to construct this sequence as a triangle we use the following rules:
- In the list of A026792 we replace each part of size j of the k-th partition of n by concatenation of j - 1 zeros and only one 1.
- Then replace this new set of parts by the concatenation of its parts.
- Then replace this string by its mirror version which is a binary number.
T(n,k) is the decimal value of this binary number, which represents the k-th partition of n (see example).
The partitions of n are represented by a subsequence with A000041(n) integers starting with 2^(n-1) and ending with 2^n - 1, n >= 1. The odd numbers of the sequence are in A000225.
First differs from A065609 at a(23).
This sequence is a sorted version of b(n) where b(2^k) = 2^k for k >= 0, b(n) = A080100(n)*(2*b(A053645(n)) + 1) otherwise. - Mikhail Kurkov, Oct 21 2023 [verification needed]
LINKS
FORMULA
a(A000070(m) - k) = 2^m - A228354(k) for m > 0, 0 < k <= A000041(m). - Mikhail Kurkov, Oct 20 2023 [verification needed]
EXAMPLE
T(6,8) = 58 because 58 in base 2 is 111010 whose mirror is 010111 which is the concatenation of 01, 01, 1, 1, whose number of digits are 2, 2, 1, 1, which are also the 8th partition of 6.
Illustration of initial terms:
The sequence represents a table of partitions (see below):
--------------------------------------------------------
. Binary Partitions
n k T(n,k) number Mirror Diagram (A026792)
. 1 2 3 4 5 6
--------------------------------------------------------
. _
1 1 1 1 1 | 1,
. _ _
1 1 2 10 01 _ | 2,
2 2 3 11 11 | | 1,1,
. _ _ _
3 1 4 100 001 _ _ | 3,
3 2 6 110 011 _ | | 2,1,
3 3 7 111 111 | | | 1,1,1,
. _ _ _ _
4 1 8 1000 0001 _ _ | 4,
4 2 10 1010 0101 _ _|_ | 2,2,
4 3 12 1100 0011 _ _ | | 3,1,
4 4 14 1110 0111 _ | | | 2,1,1,
4 5 15 1111 1111 | | | | 1,1,1,1,
. _ _ _ _ _
5 1 16 10000 00001 _ _ _ | 5,
5 2 20 10100 00101 _ _ _|_ | 3,2,
5 3 24 11000 00011 _ _ | | 4,1,
5 4 26 11010 01011 _ _|_ | | 2,2,1,
5 5 28 11100 00111 _ _ | | | 3,1,1,
5 6 30 11110 01111 _ | | | | 2,1,1,1,
5 7 31 11111 11111 | | | | | 1,1,1,1,1,
. _ _ _ _ _ _
6 1 32 100000 000001 _ _ _ | 6
6 2 36 100100 001001 _ _ _|_ | 3,3,
6 3 40 101000 000101 _ _ | | 4,2,
6 4 42 101010 010101 _ _|_ _|_ | 2,2,2,
6 5 48 110000 000011 _ _ _ | | 5,1,
6 6 52 110100 001011 _ _ _|_ | | 3,2,1,
6 7 56 111000 000111 _ _ | | | 4,1,1,
6 8 58 111010 010111 _ _|_ | | | 2,2,1,1,
6 9 60 111100 001111 _ _ | | | | 3,1,1,1,
6 10 62 111110 011111 _ | | | | | 2,1,1,1,1,
6 11 63 111111 111111 | | | | | | 1,1,1,1,1,1,
.
Triangle begins:
1;
2, 3;
4, 6, 7;
8, 10, 12, 14, 15;
16, 20, 24, 26, 28, 30, 31;
32, 36, 40, 42, 48, 52, 56, 58, 60, 62, 63;
...
From Gus Wiseman, Apr 01 2020: (Start)
Using the encoding of A066099, this sequence ranks all finite nonempty multisets, as follows.
1: {1}
2: {2}
3: {1,1}
4: {3}
6: {1,2}
7: {1,1,1}
8: {4}
10: {2,2}
12: {1,3}
14: {1,1,2}
15: {1,1,1,1}
16: {5}
20: {2,3}
24: {1,4}
26: {1,2,2}
28: {1,1,3}
30: {1,1,1,2}
31: {1,1,1,1,1}
(End)
MATHEMATICA
stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n, 2]], 1], 0]]//Reverse;
Select[Range[0, 100], LessEqual@@stc[#]&] (* Gus Wiseman, Apr 01 2020 *)
PROG
(PARI) b1(n) = my(L = logint(n, 2), A = n - 2^L); if(A > 0, L - logint(A, 2))
upto(n) = my(v1, v2); v1 = vector(n, i, if(i < 3, i)); v2 = [1, 2]; my(i = 2, j = 1, k = 0); for(s = 3, n, if(i == j, i++; v1[s] = 2^(i - 1); j = (i + 1) \ 2; k = 0; v2 = concat(v2, s), my(A = v2[j] + k); v1[s] = v1[A] + 2^(i - 1); if((i - j - 1) >= b1(v1[A + 1]), j++; k = 0, k++))); v1 \\ Mikhail Kurkov, Oct 15 2023 [verification needed]
CROSSREFS
Column 1 is A000079. Row n has length A000041(n). Right border gives A000225.
The case covering an initial interval is A333379 or A333380.
All of the following pertain to compositions in the order of A066099.
- The weakly increasing version is this sequence.
- The weakly decreasing version is A114994.
- The strictly increasing version is A333255.
- The strictly decreasing version is A333256.
- The unequal version is A233564.
- The equal version is A272919.
- The case covering an initial interval is A333217.
- Initial intervals are ranked by A164894.
- Reversed initial intervals are ranked by A246534.
Sequence in context: A328595 A065609 A334274 * A335041 A335042 A257282
KEYWORD
nonn,tabf,changed
AUTHOR
Omar E. Pol, Aug 03 2013
STATUS
approved

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Last modified April 23 02:23 EDT 2024. Contains 371906 sequences. (Running on oeis4.)