A227761 -id:A227761 - cp's OEIS frontend

A227762 Numbers in whose minimally runlength-encoded unordered partition all parts are equal; positions of zeros in A227761.

0, 1, 2, 3, 4, 6, 8, 9, 10, 12, 15, 16, 20, 24, 25, 28, 30, 35, 36, 42, 48, 49, 54, 56, 63, 64, 72, 80, 81, 88, 90, 99, 100, 110, 120, 121, 130, 132

Offset: 1

Antti Karttunen, Jul 26 2013

After 3 no more primes. First missing composites are: 14, 18, 21, 22, 26, 27, 32, 33, 34, 38, 39, 40, ...

			The first row in A227739 (please see its Example section) that sums to 6 occurs as its row 8 (= A227368(6)). The corresponding partition {3+3} contains only equal parts, thus 6 is a member of this sequence. The first row in A227739 that sums to 5 occurs as its row 9 (= A227368(5)). The corresponding partition {1+2+2} contains more than just one kind of summands, thus 5 do not occur in this sequence.

Cf. A227761, A227739, A227368.

A227368 a(n) = Index k where A227183(k) for the first time gets value n; the runlength binary code for minimally runlength-encoded unordered partition of size n.

Original entry on oeis.org

0, 1, 2, 5, 4, 9, 8, 17, 16, 23, 32, 39, 40, 71, 72, 87, 80, 151, 144, 167, 160, 295, 288, 327, 320, 351, 576, 607, 640, 671, 672, 1183, 1184, 1311, 1312, 1375, 1344, 2399, 2368, 2655, 2624, 2719, 2688, 4767, 4736, 5279, 5248, 5407, 5376, 5503, 9472, 9599, 10496

Offset: 0

Antti Karttunen, Jul 08 2013

nonn

The word "minimally" in the description means that the integer in whose binary representation some unordered partition of n is encoded should be as small as possible. This sequence gives such a minimal integer for each n, which encodes an unordered partition whose sum is n. The details of the encoding system are explained in A227183.
Also, a(n) gives the index of the first row of A227189/A227739 which sums to n.
Project: Find an algorithm which computes a(n) with a more sophisticated method than just by a blind search. This is a kind of an optimization problem for representing n as a special "bit-packed" sum: the smallest summand of size x costs x bits, and its any subsequent usage in the sum costs just one bit each time. Using any additional summand y > x costs (y-x)+1 bits when used first time, and then again additional usages cost only one bit each. Goal: minimize the number of bits needed. If multiple candidates with the same number of bits are found, then the one which results the smallest integer (when interpreted as a binary number) wins.
For any composite n = t*u, the upper bound for the size of a(n) is t+u-1 bits.
A000267(n) seems to give the binary width of a(n+1). Compare to the conjecture given at A227370.

			   n   a(n)   binary   corresponding partition    sum = n
                       (cf. A227183 for details)
   0     0         0   (0)                          0
   1     1         1   (1)                          1
   2     2        10   (1 + 1)                      2
   3     5       101   (1 + 1 + 1)                  3
   4     4       100   (2 + 2)                      4
   5     9      1001   (1 + 2 + 2)                  5
   6     8      1000   (3 + 3)                      6
   7    17     10001   (1 + 3 + 3)                  7
   8    16     10000   (4 + 4)                      8
   9    23     10111   (3 + 3 + 3)                  9
  10    32    100000   (5 + 5)                     10
  11    39    100111   (3 + 4 + 4)                 11
  12    40    101000   (3 + 3 + 3 + 3)             12
  13    71   1000111   (3 + 5 + 5)                 13
  14    72   1001000   (3 + 3 + 4 + 4)             14
  15    87   1010111   (3 + 3 + 3 + 3 + 3)         15
  16    80   1010000   (4 + 4 + 4 + 4)             16
  17   151  10010111   (3 + 3 + 3 + 4 + 4)         17
  18   144  10010000   (4 + 4 + 5 + 5)             18
  19   167  10100111   (3 + 4 + 4 + 4 + 4)         19
  20   160  10100000   (5 + 5 + 5 + 5)             20
a(5) = 9, because 5 occurs for the first time in A227183 as A227183(9).
Note that for 20, there is for example also a code 175, "10101111" in binary, which results a partition (4 + 4 + 4 + 4 + 4) (= 20), but as 160 < 175, and there are no other partitions of 20 which would result even smaller code number, 160 is the winner (the minimal code), and thus a(20)=160.
A227761 gives the maximum difference between successive parts that occurs in these partitions.

Antti Karttunen, Table of n, a(n) for n = 0..132
Index entries for sequences related to partitions

Same sequence sorted into ascending order: A227369.

Cf. also A227183, A227761, A227762.

def A227368(n):
  '''Index k where A227183(k) for the first time gets value n. A naive implementation.'''
  k = 0
  while(A227183(k) != n): k += 1
  return(k)

a(n) = A227369(A227370(n)) [See comments and conjecture at A227370]

A227786 Take squares larger than 1, subtract 3 from even squares and 2 from odd squares; a(n) = a(n-1) + A168276(n+1) (with a(1) = 1).

Original entry on oeis.org

1, 7, 13, 23, 33, 47, 61, 79, 97, 119, 141, 167, 193, 223, 253, 287, 321, 359, 397, 439, 481, 527, 573, 623, 673, 727, 781, 839, 897, 959, 1021, 1087, 1153, 1223, 1293, 1367, 1441, 1519, 1597, 1679, 1761, 1847, 1933, 2023, 2113, 2207, 2301, 2399, 2497, 2599, 2701

Offset: 1

Antti Karttunen, Jul 31 2013

nonn

Conjecture: from n>=2 onward, a(n) gives the positions of 2's in A227761.

a(29) = 897 = 3*13*23 is the first term which is neither prime nor semiprime, that is, has more than two prime divisors.

Antti Karttunen, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Bisections: A082109, A073577. Cf. also A227761.

a(n) = A000290(n+1) - 2 - (n mod 2).
a(1)=1, and for n>1, a(n) = a(n-1)+A168276(n+1).
a(n) = (1/2) * (2*n^2 + 4*n -3 + (-1)^n) = 2*A116940(n-1) + 1. a(n-1) = 2*ceiling(n^2/2) - 3 = 2*A000985(n) - 3. G.f.: x*(-x^3 - x^2 + 5*x + 1)/((1-x)^3 * (1+x)). - Ralf Stephan, Aug 10 2013

cp's OEIS Frontend

A227762 Numbers in whose minimally runlength-encoded unordered partition all parts are equal; positions of zeros in A227761.

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A227368 a(n) = Index k where A227183(k) for the first time gets value n; the runlength binary code for minimally runlength-encoded unordered partition of size n.

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A227786 Take squares larger than 1, subtract 3 from even squares and 2 from odd squares; a(n) = a(n-1) + A168276(n+1) (with a(1) = 1).

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