A129594 -id:A129594 - cp's OEIS frontend

A242421 Fixed points of A153212: After a(1) = 1, numbers of the form p_i1^i1 * p_i2^(i2-i1) * p_i3^(i3-i2) * ... * p_ik^(ik-i_{k-1}), where p_i's are distinct primes present in the prime factorization of n, with i1 < i2 < i3 < ... < ik, and k = A001221(n) and ik = A061395(n).

1, 2, 6, 9, 30, 45, 50, 125, 210, 294, 315, 350, 441, 686, 875, 2310, 2401, 3234, 3465, 3630, 3850, 4851, 5445, 6050, 7546, 7986, 9625, 11979, 15125, 26411, 29282, 30030, 35490, 42042, 45045, 47190, 49686, 50050, 53235, 59150, 63063, 65910, 70785, 74529, 78650, 98098, 98865, 103818, 109850, 115934, 125125, 147875, 155727, 161051, 171366, 196625, 257049, 274625, 343343, 380666, 405769, 510510

Offset: 1

Antti Karttunen, May 16 2014

nonn

This sequence is closed with respect to A122111, i.e., for any n, A122111(a(n)) is either the same as a(n) or some other term a(k) of the sequence.
These numbers encode partitions in whose Young diagrams all pairs of successive horizontal and vertical segments (those pairs sharing "a common convex corner") are of equal length. Cf. the example-illustration at A153212.
Note: The seventh primorial, 510510 (= A002110(7)) occurs here as a term a(62).

			2 = p_1^1 is present, as the first prime index delta and exponent are equal.
3 = p_2^1 is not present, as 1 <> 2.
6 = p_1^1 * p_2^(2-1) is present.
9 = p_2^2 is present, as 2 = 2.
30 = p_1^1 * p_2^(2-1) * p_3^(3-2) is present, as all primorials are.
50 = p_1^1 * p_3^(3-1) is present also.

Subsequences: A002110 (primorial numbers), A062457.

Cf. A242422, A088902, A241912, A209861, A209636, A129594.

A249144 a(0) = 0, after which a(n) gives the total number of runs of the same length as the rightmost run in the binary representation of a(n-1) [i.e., A136480(a(n-1))] among the binary expansions of all previous terms, including the runs in a(n-1) itself.

Original entry on oeis.org

0, 1, 2, 4, 1, 6, 7, 1, 8, 2, 11, 3, 4, 5, 17, 19, 7, 4, 8, 5, 25, 26, 29, 31, 1, 32, 2, 35, 12, 14, 37, 41, 45, 49, 50, 52, 22, 57, 58, 61, 63, 1, 64, 2, 67, 25, 69, 73, 76, 32, 3, 33, 80, 4, 34, 87, 14, 92, 35, 36, 38, 99, 42, 105, 108, 47, 5, 114, 116, 49, 119, 23, 24, 25, 123, 54, 126, 127, 1, 128, 2

Offset: 0

Antti Karttunen, Oct 22 2014

Inspired by A248034.

			a(0) = 0 (by definition), and 0 is also '0' in binary.
For n = 1, we see that in a(0) there is one run of length 1, which is total number of runs of length 1 so far in terms a(0) .. a(n-1), thus a(1) = 1.
For n = 2, we see that the rightmost run of a(1) = 1 ('1' also in binary) has occurred two times in total (once in a(0) and a(1)), thus a(2) = 2.
For n = 3, we see that the rightmost run of a(2) = 2 ('10' in binary) is one bit long, and so far there has occurred four such runs in total (namely once in a(0) and a(1), twice in a(2)), thus a(3) = 4.
For n = 4, we see that the rightmost run of a(3) = 4 ('100' in binary) is two bits long, and it is so far the first and only two-bit run in the sequence, thus a(4) = 1.
For n = 5, we see that the rightmost run of a(4) = 1 ('1' in binary) is one bit long, and so far there has occurred 6 such one-bit runs in terms a(0) .. a(4), thus a(5) = 6.
For n = 6, we see that the rightmost run of a(5) = 6 ('110' in binary) is one bit long, and so far there has occurred 7 such one bit runs in terms a(0) .. a(5), thus a(6) = 7.

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A249143, A249146, A249148, A248034.

A227761 a(n) is the maximal difference between successive parts in the minimally runlength-encoded unordered partition of n (A227368(n)).

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Jul 26 2013

nonn

After n=3, only composites may obtain value 0. (But not all of them do; see A227762.) The first nine n for which a(n)=2 are 7, 13, 23, 33, 47, 61, 79, 97, 119, of which all are primes except 33 and 119. Conjecture: these values are given by A227786.

Are there any terms larger than 2?

Antti Karttunen, Table of n, a(n) for n = 0..132

A227762 gives the positions of zeros, in other words, such n that their minimally runlength-encoded partition consists of identical parts.

Cf. also A227368 (for the concept of minimally runlength-encoded unordered partition).

(define (A227761 n) (if (< n 2) 0 (- (A043276 (A163575 (A227368 n))) 1)))
;; Alternative version which uses auxiliary functions DIFF and binexp_to_ascpart which can be found in the Program section of A129594:
(define (A227761v2 n) (if (< n 2) 0 (apply max (DIFF (binexp_to_ascpart (A227368 n))))))

a(0) = a(1) = 0, and for n>1, a(n) = A043276(A163575(A227368(n))) - 1.

A249146 a(0) = 0, after which a(n) gives the total number of runs of the same length as the maximal run in the binary representation of a(n-1) [i.e., A043276(a(n-1))] among the binary expansions of all previous terms, including the runs in a(n-1) itself.

Original entry on oeis.org

0, 1, 2, 4, 1, 6, 2, 9, 3, 4, 5, 15, 1, 16, 2, 19, 7, 1, 21, 26, 8, 2, 32, 1, 34, 3, 9, 10, 43, 11, 12, 14, 4, 15, 3, 16, 4, 17, 5, 58, 6, 18, 19, 21, 71, 8, 9, 22, 23, 10, 84, 24, 11, 26, 27, 29, 12, 31, 2, 99, 13, 34, 14, 15, 5, 108, 37, 38, 40, 16, 6, 41, 42, 130, 3, 43, 44, 46, 17, 18, 47, 7, 19, 49

Offset: 0

Antti Karttunen, Oct 22 2014

nonn

			a(0) = 0 (by definition), and 0 is also '0' in binary we consider it to contain a single run of length one.
For n = 1, we see that in a(0) there is one run of length 1, which is total number of runs of length 1 so far in terms a(0) .. a(n-1), thus a(1) = 1.
For n = 2, we see that the only and thus also the longest run of a(1) = 1 ('1' also in binary) has occurred two times in total (once in a(0) and a(1)), thus a(2) = 2.
For n = 3, we see that there are two runs in a(2) = 2 ('10' in binary), both one bit long, and so far there has occurred four such runs in total (namely once in a(0) and a(1), twice in a(2)), thus a(3) = 4.
For n = 4, we see that the longest run of a(3) = 4 ('100' in binary) is two bits long, and it is so far the first and only two-bit run in the sequence, thus a(4) = 1.
For n = 5, we see that the longest run of a(4) = 1 ('1' in binary) is one bit long, and so far there has occurred 6 such one-bit runs in terms a(0) .. a(4), thus a(5) = 6.
For n = 6, we see that the longest run of a(5) = 6 ('110' in binary) is two bits long, and so far there has occurred 2 such two bit runs (once in terms a(3) and a(5)), thus a(6) = 2.

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A043276, A249145, A249144, A248034.

A129602 In the binary expansion of n replace each run of k 0's (or 1's) with 2k-1 0's (or 1's), except in the most significant run where we double the number of 0's (or 1's).

Original entry on oeis.org

0, 3, 6, 15, 24, 13, 30, 63, 96, 49, 26, 55, 120, 61, 126, 255, 384, 193, 98, 199, 104, 53, 110, 223, 480, 241, 122, 247, 504, 253, 510, 1023, 1536, 769, 386, 775, 392, 197, 398, 799, 416, 209, 106, 215, 440, 221, 446, 895, 1920, 961, 482, 967, 488, 245, 494

Offset: 0

Antti Karttunen, May 01 2007

			a(1) = 3, as 1 is 1 in binary and doubling the number of 1's (in the only run) gives binary 11, 3 in decimal. a(9) = 49, as 9 is 1001 in binary and replacing the most significant run '1' with '11' and the center run '00' with '000' and the least significant run '1' with '1', we get 110001 in binary, 49 in decimal.

Central diagonal of array A129600, a(n) = A129600bi(n, n). Cf. A129594. For n > 0, a(n) = A004760(A129603(n)+1).

Edited definition. - N. J. A. Sloane, Dec 20 2023

A227351 Permutation of nonnegative integers: map each number by lengths of runs of zeros in its Zeckendorf expansion shifted once left to the number which has the same lengths of runs (in the same order, but alternatively of runs of 0's and 1's) in its binary representation.

Original entry on oeis.org

0, 1, 3, 7, 2, 15, 6, 4, 31, 14, 12, 8, 5, 63, 30, 28, 24, 13, 16, 9, 11, 127, 62, 60, 56, 29, 48, 25, 27, 32, 17, 19, 23, 10, 255, 126, 124, 120, 61, 112, 57, 59, 96, 49, 51, 55, 26, 64, 33, 35, 39, 18, 47, 22, 20, 511, 254, 252, 248, 125, 240, 121, 123, 224

Offset: 0

Antti Karttunen, Jul 08 2013

This permutation is based on the fact that by appending one extra zero to the right of Fibonacci number representation of n (aka "Zeckendorf expansion") and then counting the lengths of blocks of consecutive (nonleading) zeros we get bijective correspondence with compositions, and thus also with the binary representation of a unique n. See the chart below:
   n   A014417(n) A014417(A022342(n+1)) Runs of    Binary number   In dec.
                  [shifted once left]   zeros      with same runs  = a(n)
  ......0              ......0    []         .....0           0
  ......1              .....10    [1]        .....1           1
  .....10              ....100    [2]        ....11           3
  ....100              ...1000    [3]        ...111           7
  ....101              ...1010    [1,1]      ....10           2
  ...1000              ..10000    [4]        ..1111          15
  ...1001              ..10010    [2,1]      ...110           6
  ...1010              ..10100    [1,2]      ...100           4
  ..10000              .100000    [5]        .11111          31
  ..10001              .100010    [3,1]      ..1110          14
  ..10010              .100100    [2,2]      ..1100          12
  ..10100              .101000    [1,3]      ..1000           8
  ..10101              .101010    [1,1,1]    ...101           5
  .100000              1000000    [6]        111111          63
Are there any other fixed points after 0, 1, 6, 803, 407483 ?

Antti Karttunen, Table of n, a(n) for n = 0..10946
OEIS Wiki, Run-length encoding
Index entries for sequences that are permutations of the natural numbers

Inverse permutation: A227352. Cf. also A003714, A014417, A006068, A048679.

Could be further composed with A075157 or A075159, also A129594.

(define (A227351 n) (A006068 (A048679 n)))

(define (A227351v2 n) (A006068 (A106151 (* 2 (A003714 n)))))

a(n) = A006068(A048679(n)) = A006068(A106151(2*A003714(n))).
This permutation effects following correspondences:
a(A000045(n)) = A000225(n-1).
a(A027941(n)) = A000975(n).
For n >=3, a(A000204(n)) = A000079(n-2).

A129603 Replace in the binary expansion of n each run of k 0's (or 1's) with 2k-1 0's (or 1's).

Original entry on oeis.org

0, 1, 2, 7, 8, 5, 14, 31, 32, 17, 10, 23, 56, 29, 62, 127, 128, 65, 34, 71, 40, 21, 46, 95, 224, 113, 58, 119, 248, 125, 254, 511, 512, 257, 130, 263, 136, 69, 142, 287, 160, 81, 42, 87, 184, 93, 190, 383, 896, 449, 226, 455, 232, 117, 238, 479, 992, 497, 250, 503

Offset: 0

Antti Karttunen, May 01 2007

			a(1) = 1, as 1 is 1 in binary and single runs stay intact (as 2*1 - 1 = 1). a(9) = 17, as 9 is 1001 in binary and keeping the most and the least significant runs as '1' and changing the center run '00' to '000', we get 10001 in binary, 17 in decimal.

Cf. A001196, A129602.

cp's OEIS Frontend

A242421 Fixed points of A153212: After a(1) = 1, numbers of the form p_i1^i1 * p_i2^(i2-i1) * p_i3^(i3-i2) * ... * p_ik^(ik-i_{k-1}), where p_i's are distinct primes present in the prime factorization of n, with i1 < i2 < i3 < ... < ik, and k = A001221(n) and ik = A061395(n).

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A249144 a(0) = 0, after which a(n) gives the total number of runs of the same length as the rightmost run in the binary representation of a(n-1) [i.e., A136480(a(n-1))] among the binary expansions of all previous terms, including the runs in a(n-1) itself.

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A227761 a(n) is the maximal difference between successive parts in the minimally runlength-encoded unordered partition of n (A227368(n)).

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A249146 a(0) = 0, after which a(n) gives the total number of runs of the same length as the maximal run in the binary representation of a(n-1) [i.e., A043276(a(n-1))] among the binary expansions of all previous terms, including the runs in a(n-1) itself.

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A129602 In the binary expansion of n replace each run of k 0's (or 1's) with 2k-1 0's (or 1's), except in the most significant run where we double the number of 0's (or 1's).

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A227351 Permutation of nonnegative integers: map each number by lengths of runs of zeros in its Zeckendorf expansion shifted once left to the number which has the same lengths of runs (in the same order, but alternatively of runs of 0's and 1's) in its binary representation.

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Scheme

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A129603 Replace in the binary expansion of n each run of k 0's (or 1's) with 2k-1 0's (or 1's).

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