A301453 -id:A301453 - cp's OEIS frontend

A302395 a(n) is the number of ways of writing the binary expansion of n as a concatenation of distinct nonempty substrings.

1, 1, 2, 1, 3, 3, 3, 3, 6, 6, 6, 5, 5, 5, 6, 3, 9, 10, 10, 9, 9, 8, 10, 9, 9, 9, 9, 7, 9, 9, 9, 5, 14, 19, 19, 17, 17, 16, 18, 17, 19, 16, 17, 16, 17, 16, 19, 13, 15, 17, 17, 14, 15, 16, 17, 12, 18, 17, 19, 12, 15, 13, 14, 11, 25, 31, 30, 29, 27, 29, 31, 30

Offset: 0

Rémy Sigrist, Apr 07 2018

Leading zeros in the binary expansion of n are ignored.
The value a(0) = 1 corresponds to the empty concatenation.
See A301453 for similar sequences.

			For n = 7: the binary expansion of 7, "111", can be split in 3 ways into distinct nonempty substrings:
- (111),
- (11)(1),
- (1)(11).
Hence a(7) = 3.
For n = 42: the binary expansion of 42, "101010", can be split in 17 ways into distinct nonempty substrings:
- (101010),
- (10101)(0),
- (1010)(10),
- (1010)(1)(0),
- (101)(010),
- (101)(01)(0),
- (101)(0)(10),
- (10)(1010),
- (10)(101)(0),
- (10)(1)(010),
- (10)(1)(01)(0),
- (1)(01010),
- (1)(0101)(0),
- (1)(010)(10),
- (1)(01)(010),
- (1)(01)(0)(10),
- (1)(0)(1010).
Hence a(42) = 17.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Rémy Sigrist, Scatterplot of the second ordinal transform of the first 750000 terms
Index entries for sequences related to binary expansion of n

Cf. A032020, A301453.

a(n{, s=Set()}) = if (n==0, return (1), my (v=0, p=1); while (n, p=(p*2) + (n%2); n\=2; if (!setsearch(s, p), v+=a(n, setunion(s, Set(p))))); return (v))

a(2^n - 1) = A032020(n) for any n >= 0.

A302436 a(n) is the number of ways of writing the binary expansion of n as a concatenation of nonempty substrings with Hamming weight at most 1.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 2, 1, 8, 7, 6, 3, 4, 3, 2, 1, 16, 15, 14, 7, 12, 9, 6, 3, 8, 7, 6, 3, 4, 3, 2, 1, 32, 31, 30, 15, 28, 21, 14, 7, 24, 21, 18, 9, 12, 9, 6, 3, 16, 15, 14, 7, 12, 9, 6, 3, 8, 7, 6, 3, 4, 3, 2, 1, 64, 63, 62, 31, 60, 45, 30, 15, 56, 49, 42, 21

Offset: 0

Rémy Sigrist, Apr 08 2018

Leading zeros in the binary expansion of n are ignored.
The value a(0) = 1 corresponds to the empty concatenation.
See A301453 for similar sequences.

			For n = 9: the binary expansion of 9, "1001", can be split in 7 ways into nonempty substrings with Hamming weight at most 1:
- (100)(1),
- (10)(01),
- (10)(0)(1),
- (1)(001),
- (1)(00)(1),
- (1)(0)(01),
- (1)(0)(0)(1).
Hence a(9) = 7.

Rémy Sigrist, Table of n, a(n) for n = 0..10000

Cf. A301453.

a(n) = if (n==0, return (1), my (v=0, h=0); while (n, h += n%2; n\=2; if (h>1, break, v+=a(n))); return (v))

a(2^n - 1) = 1 for any n >= 0.
a(2^n) = 2^n for any n >= 0.
a(2*n) = 2*a(n) for any n > 0.

A302437 a(n) is the number of ways of writing the binary expansion of n as a concatenation of nonempty substrings with Hamming weight at most 2.

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 4, 3, 8, 8, 8, 7, 8, 7, 6, 5, 16, 16, 16, 15, 16, 15, 14, 11, 16, 15, 14, 13, 12, 11, 10, 8, 32, 32, 32, 31, 32, 31, 30, 23, 32, 31, 30, 27, 28, 25, 22, 18, 32, 31, 30, 29, 28, 27, 26, 20, 24, 23, 22, 20, 20, 18, 16, 13, 64, 64, 64, 63, 64

Offset: 0

Rémy Sigrist, Apr 08 2018

Leading zeros in the binary expansion of n are ignored.
The value a(0) = 1 corresponds to the empty concatenation.
See A301453 for similar sequences.

			For n = 14: the binary expansion of 14, "1110", can be split in 6 ways into substrings with Hamming weight at most 2:
- (11)(10),
- (11)(1)(0),
- (1)(110),
- (1)(11)(0),
- (1)(1)(10),
- (1)(1)(1)(0).
Hence a(14) = 6.

Rémy Sigrist, Table of n, a(n) for n = 0..10000

Cf. A000045, A000120, A053644, A301453.

a(n) = if (n==0, return (1), my (v=0, h=0); while (n, h += n%2; n\=2; if (h>2, break, v += a(n))); return (v))

a(2^n - 1) = A000045(n + 1) for any n >= 0.
a(2^n) = 2^n for any n >= 0.
a(2*n) = 2*a(n) for any n > 0.
If 1 <= A000120(n) <= 2 then a(n) = A053644(n).

A302439 a(n) is the number of ways of writing the binary expansion of n as a concatenation of nonempty aperiodic substrings (i.e., substrings that are not the concatenation of 2 or more equal parts).

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 3, 1, 4, 4, 6, 4, 5, 5, 4, 1, 5, 5, 10, 5, 10, 9, 11, 5, 7, 7, 10, 7, 7, 7, 5, 1, 6, 6, 14, 6, 16, 15, 16, 6, 14, 14, 19, 13, 18, 17, 16, 6, 9, 9, 17, 9, 16, 15, 17, 9, 10, 10, 14, 10, 9, 9, 6, 1, 7, 7, 18, 7, 24, 21, 21, 7, 23, 22, 32, 22

Offset: 0

Rémy Sigrist, Apr 08 2018

Leading zeros in the binary expansion of n are ignored.
The value a(0) = 1 corresponds to the empty concatenation.
See A301453 for similar sequences.
For some odd o = 2*k + 1, is there some k such that for all e > k, a(o * 2^e) = a(o + 2^(e - 1)) + c for some c? - David A. Corneth, Apr 15 2018

			For n = 20: the binary expansion of 20, "10100", can be split in 10 ways into aperiodic substrings:
- (10100),
- (101)(0)(0),
- (10)(100),
- (10)(10)(0),
- (10)(1)(0)(0),
- (1)(0100),
- (1)(010)(0),
- (1)(0)(100),
- (1)(0)(10)(0),
- (1)(0)(1)(0)(0).
Hence a(20) = 10.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n

Cf. A301453.

a(n) = if (n==0, return (1), my (v=0); for (w=1, #binary(n), my (ok=1); fordiv (w, d, if (d

a(2^n - 1) = 1 for any n >= 0.
a(2^n) = n + 1 for any n >= 0.
From David A. Corneth, Apr 15 2018: (Start)
Is a(2^n + i) >= a(2^n) for 0 <= i <= 2^n - 2?
What is the least k(n) such that
a(2^n + i) <= a(2^n + k(n)) for 1 <= i <= 2^n - 2? (End)

A330720 a(n) is the number of ways of writing the binary expansion of n as a product (or concatenation) of nonpalindromes.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 1, 1, 1, 0, 1, 2, 2, 2, 2, 0, 2, 1, 1, 2, 2, 0, 1, 1, 1, 0, 1, 3, 3, 3, 3, 3, 3, 2, 2, 3, 4, 2, 2, 2, 3, 1, 1, 3, 2, 1, 2, 2, 3, 1, 1, 2, 2, 1, 1, 1, 1, 0, 1, 4, 4, 4, 3, 5, 5, 3, 3, 4, 4, 4, 4, 5, 5, 2, 2, 5, 4, 4, 4, 0, 4

Offset: 0

Rémy Sigrist, Dec 28 2019

This sequence is a variant of A215244.

			For n = 41:
- the binary expansion of 41 is "101001",
- the possible products of nonpalindromes are "101001", "1010"."01", and "10"."10"."01",
- hence a(41) = 3.

Rémy Sigrist, Table of n, a(n) for n = 0..8192
Index entries for sequences related to binary expansion of n

Cf. A020988, A154809, A215244, A301453.

ispali:= proc(L) L = ListTools:-Reverse(L) end proc:
g:= proc(L) option remember; local m;
    add(procname(L[m+1..-1]), m= remove(t -> ispali(L[1..t]),[$1..nops(L)]))
end proc:
g([]):= 1:
seq(g(convert(n,base,2)),n=0..100); # Robert Israel, Dec 29 2019

a(n) = my (b=binary(n), v=b!=Vecrev(b)); for (s=1, #b, my (z=b[s..#b]); if (z!=Vecrev(z), v+=a(fromdigits(b[1..s-1],2)))); v

a(2^k-1) = 0 for any k >= 0.

a(A020988(k+1)) = 2^k for any k >= 0.

cp's OEIS Frontend

A302395 a(n) is the number of ways of writing the binary expansion of n as a concatenation of distinct nonempty substrings.

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A302436 a(n) is the number of ways of writing the binary expansion of n as a concatenation of nonempty substrings with Hamming weight at most 1.

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A302437 a(n) is the number of ways of writing the binary expansion of n as a concatenation of nonempty substrings with Hamming weight at most 2.

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A302439 a(n) is the number of ways of writing the binary expansion of n as a concatenation of nonempty aperiodic substrings (i.e., substrings that are not the concatenation of 2 or more equal parts).

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A330720 a(n) is the number of ways of writing the binary expansion of n as a product (or concatenation) of nonpalindromes.

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