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Irregular triangle T(n, k), n >= 0, k = 1..2^ A092339(n), read by rows; the n-th row lists the numbers k such that n appears in the k-th row of A361644.
+20
2
0, 1, 2, 2, 3, 4, 5, 5, 5, 6, 4, 5, 6, 7, 8, 9, 10, 11, 9, 10, 10, 10, 11, 10, 11, 12, 13, 10, 13, 9, 10, 13, 14, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 17, 18, 21, 22, 18, 21, 18, 19, 20, 21, 20, 21, 21, 21, 22, 20, 21, 22, 23, 20, 21, 22, 23, 24, 25, 26, 27
COMMENTS
In other words, the n-th row contains the numbers k with the same binary length as n and for any i >= 0, if the i-th bit and the (i+1)-th bit in n are different then they are also different in k (i = 0 corresponding to the least significant bit).
EXAMPLE
Triangle T(n, k) begins (in decimal and in binary):
n n-th row bin(n) n-th row in binary
-- -------------- ------ ----------------------
0 0 0 0
1 1 1 1
2 2 10 10
3 2, 3 11 10, 11
4 4, 5 100 100, 101
5 5 101 101
6 5, 6 110 101, 110
7 4, 5, 6, 7 111 100, 101, 110, 111
8 8, 9, 10, 11 1000 1000, 1001, 1010, 1011
9 9, 10 1001 1001, 1010
10 10 1010 1010
11 10, 11 1011 1010, 1011
12 10, 11, 12, 13 1100 1010, 1011, 1100, 1101
13 10, 13 1101 1010, 1101
14 9, 10, 13, 14 1110 1001, 1010, 1101, 1110
PROG
(PARI) row(n) = { my (r = [n], m); for (e = 1, exponent(n), if (bittest(n, e-1)==bittest(n, e), m = 2^e-1; r = concat(r, [bitxor(v, m) | v <- r]); ); ); vecsort(r); }
Number of nonleading 0's in binary expansion of n.
+10
86
0, 0, 1, 0, 2, 1, 1, 0, 3, 2, 2, 1, 2, 1, 1, 0, 4, 3, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 2, 1, 1, 0, 5, 4, 4, 3, 4, 3, 3, 2, 4, 3, 3, 2, 3, 2, 2, 1, 4, 3, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 2, 1, 1, 0, 6, 5, 5, 4, 5, 4, 4, 3, 5, 4, 4, 3, 4, 3, 3, 2, 5, 4, 4, 3, 4, 3, 3, 2, 4, 3, 3, 2, 3, 2, 2, 1, 5, 4, 4, 3, 4, 3, 3, 2, 4
COMMENTS
In this version we consider the number zero to have no nonleading 0's, thus a(0) = 0. The variant A023416 has a(0) = 1. Number of steps required to reach 1, starting at n + 1, under the operation: if x is even divide by 2 else add 1. This is the x + 1 problem (as opposed to the 3x + 1 problem).
FORMULA
a(0) = 0, and for n > 0, a(n) = (a(n-1) + A007814(n) + A036987(n-1)) - 1. (End)
G.f. g(x) satisfies g(x) = (1+x)*g(x^2) + x^2/(1-x^2). - Robert Israel, Oct 26 2017
EXAMPLE
a(4) = 2 since 4 in binary is 100, which has two zeros.
a(5) = 1 since 5 in binary is 101, which has only one zero.
MAPLE
seq(numboccur(0, Bits[Split](n)), n=0..100); # Robert Israel, Oct 26 2017
MATHEMATICA
f[n_] := If[OddQ@ n, f[n -1] -1, f[n/2] +1]; f[0] = f[1] = 0; Array[f, 105, 0] (* Robert G. Wilson v, May 21 2017 *) Join[{0}, Table[Count[IntegerDigits[n, 2], 0], {n, 1, 100}]] (* Vincenzo Librandi, Oct 27 2017 *)
PROG
(PARI) a(n)=if(n, a(n\2)+1-n%2)
(Scheme) ;; with memoizing definec-macro from Antti Karttunen's IntSeq-library)
;; Alternative version based on a simple recurrence:
(Python) def a(n): return bin(n)[2:].count("0") if n>0 else 0 # Indranil Ghosh, Apr 10 2017
CROSSREFS
Cf. A000120, A007814, A023416, A029837, A036987, A080791- A080801, A048881, A054429, A092339, A102364, A120511, A233271, A233272, A233273.
The largest part in the unordered partition encoded in the runlengths of the binary expansion of n.
+10
8
0, 1, 1, 2, 2, 1, 2, 3, 3, 2, 1, 2, 3, 2, 3, 4, 4, 3, 2, 3, 2, 1, 2, 3, 4, 3, 2, 3, 4, 3, 4, 5, 5, 4, 3, 4, 3, 2, 3, 4, 3, 2, 1, 2, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 5, 4, 5, 6, 6, 5, 4, 5, 4, 3, 4, 5, 4, 3, 2, 3, 4, 3, 4, 5, 4, 3, 2, 3, 2, 1, 2
COMMENTS
The bijective encoding of nonordered partitions via compositions (ordered partitions) present in the binary expansion of n is explained in A227184. It appears that a(4n+2) = a(2n+1). - Ralf Stephan, Jul 20 2013
FORMULA
Defining formula:
a(0)=0; and for n>=1, a(n) = A029837(n+1) - ( A005811(n)-1). [Because the largest part in the unordered partition in this encoding scheme is computed as (c_1 + (c_2-1) + (c_3-1) + ... + (c_k-1)) where c_1 .. c_k are the parts of the k-part composition that sum together as c_1 + c_2 + ... + c_k = A029837(n+1) (the binary width of n), so we subtract from the total binary width of n the number of runs ( A005811) minus 1.] Equivalently: a(n) = A092339(n)+1 for n>0. a(n) = A005811( A129594(n)). [This just states the fact that when conjugating a partition, the largest part of an old partition will be the number of the parts in the new, conjugated partition.]
EXAMPLE
12 has binary expansion "1100", for which the lengths of runs (consecutive blocks of 0- or 1-bits) are [2,2]. Converting this to a partition in the manner explained in A227184 gives the partition {2+3}. Its largest part is 3, thus a(12)=3, which is actually the first time when this sequence differs from A043276.
MATHEMATICA
Table[Function[b, Max@ Accumulate@ Prepend[If[Length@ b > 1, Rest[b] - 1, {}], First@ b] - Boole[n == 0]]@ Map[Length, Split@ Reverse@ IntegerDigits[ n, 2]], {n, 0, 120}] // Flatten (* Michael De Vlieger, May 09 2017 *)
PROG
(Scheme)
(define (A227185v2 n) (if (zero? n) n (car (reverse (binexp_to_ascpart n))))) ;; Alternative definition, using the auxiliary functions given in A227184.
CROSSREFS
a(n) gives the rightmost nonzero term on the n-th row of A227189.
1, 2, 2, 4, 6, 2, 4, 8, 12, 6, 2, 6, 12, 4, 8, 16, 24, 12, 6, 30, 6, 2, 6, 12, 36, 12, 4, 12, 24, 8, 16, 32, 48, 24, 12, 60, 30, 6, 30, 60, 12, 6, 2, 6, 30, 6, 12, 24, 72, 36, 12, 60, 12, 4, 12, 36, 72, 24, 8, 24, 48, 16, 32, 64, 96, 48, 24, 120, 60, 12, 60, 180, 60, 30, 6, 30, 210, 30, 60, 120, 24, 12, 6, 30, 6, 2, 6, 12, 60, 30, 6, 30, 60, 12, 24, 48, 144, 72
PROG
(PARI)
A286468(n) = { my(p=((n+1)%2), i=0, m=1); while(n>0, if(((n%2)==p), m *= prime(i), p = (n%2); i = i+1); n = n\2); m };
A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); }; \\ This function from Charles R Greathouse IV, Aug 17 2011
(Scheme)
CROSSREFS
Sequences that partition N into same or coarser equivalence classes are at least these: A092339, A227185.
Irregular triangle T(n, k), n >= 0, k = 1..max(1, 2^( A005811(n)-1)), read by rows; the n-th row lists the integers with the same binary length as n and whose partial sums of run lengths are included in those of n.
+10
5
0, 1, 2, 3, 3, 4, 7, 4, 5, 6, 7, 6, 7, 7, 8, 15, 8, 9, 14, 15, 8, 9, 10, 11, 12, 13, 14, 15, 8, 11, 12, 15, 12, 15, 12, 13, 14, 15, 14, 15, 15, 16, 31, 16, 17, 30, 31, 16, 17, 18, 19, 28, 29, 30, 31, 16, 19, 28, 31, 16, 19, 20, 23, 24, 27, 28, 31
COMMENTS
In other words, the n-th row contains the numbers k with the same binary length as n and for any i >= 0, if the i-th bit and the (i+1)-th bit in k are different then they are also different in n (i = 0 corresponding to the least significant bit).
EXAMPLE
Triangle begins (in decimal and in binary):
n n-th row bin(n) n-th row in binary
-- ------------ ------ ------------------
0 0 0 0
1 1 1 1
2 2, 3 10 10, 11
3 3 11 11
4 4, 7 100 100, 111
5 4, 5, 6, 7 101 100, 101, 110, 111
6 6, 7 110 110, 111
7 7 111 111
8 8, 15 1000 1000, 1111
9 8, 9, 14, 15 1001 1000, 1001, 1110, 1111
.
For n = 9:
- the binary expansion of 9 is "1001",
- the corresponding run lengths are 1, 2, 1,
- so the 9th row contains the values with the following run lengths:
1, 2, 1 -> 9 ("1001" in binary)
1, 2+1 -> 8 ("1000" in binary)
1+2, 1 -> 14 ("1110" in binary)
1+2+1 -> 15 ("1111" in binary)
PROG
(PARI) row(n) = { my (r = []); while (n, my (v = valuation(n+n%2, 2)); n \= 2^v; r = concat(v, r)); my (s = [if (#r, 2^r[1]-1, 0)]); for (k = 2, #r, s = concat(s * 2^r[k], [(h+1)*2^r[k]-1|h<-s]); ); vecsort(s); }
a(n) = a(floor(n/2)) + (-1)^(n*(n+1)/2) with a(1)=0.
+10
1
0, -1, 1, 0, -2, 0, 2, 1, -1, -3, -1, 1, -1, 1, 3, 2, 0, -2, 0, -2, -4, -2, 0, 2, 0, -2, 0, 2, 0, 2, 4, 3, 1, -1, 1, -1, -3, -1, 1, -1, -3, -5, -3, -1, -3, -1, 1, 3, 1, -1, 1, -1, -3, -1, 1, 3, 1, -1, 1, 3, 1, 3, 5, 4, 2, 0, 2, 0, -2, 0, 2, 0, -2, -4, -2, 0, -2, 0, 2, 0
COMMENTS
For 2^(2*k-1) - 1 < n < 2^(2*k), k>0, there's no n such that a(n)=0.
For 2^(2*k) - 1 < n < 2^(2*k+1), k>0, there are A000984(k+1) n's such that a(n)=0.
FORMULA
a(1) = 0, a(n) = a(floor(n/2)) + (-1)^(n*(n+1)/2).
EXAMPLE
a(9) = a(4) + (-1)^45 = -1, a(10) = a(5) + (-1)^55 = -3.
For 7 < n < 16, there's no n such that a(n)=0; for 15 < n < 32, there are 6 n's such that a(n)=0.
MAPLE
a:=proc(n) `if`(n=1, 0, a(floor(n/2))+(-1)^(n*(n+1)/2)) end: seq(a(n), n=1..100); # Muniru A Asiru, Oct 07 2018
MATHEMATICA
a[1] = 0; a[n_] := a[n] = a[Floor[n/2]] + (-1)^(n*(n + 1)/2); Table[a@n, {n, 1, 50}]
PROG
(PARI) a(n) = if (n==1, 0, a(n\2) + (-1)^(n*(n+1)/2)); \\ Michel Marcus, Oct 05 2018
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