A080791 -id:A080791 - cp's OEIS frontend

A253559 a(1) = 0; for n>1: a(n) = A253557(n) - 1.

0, 0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 2, 0, 1, 1, 3, 0, 2, 0, 2, 2, 1, 0, 3, 1, 1, 1, 2, 0, 2, 0, 4, 2, 1, 1, 3, 0, 1, 1, 3, 0, 3, 0, 2, 3, 1, 0, 4, 1, 2, 2, 2, 0, 2, 2, 3, 2, 1, 0, 3, 0, 1, 1, 5, 1, 3, 0, 2, 3, 2, 0, 4, 0, 1, 1, 2, 1, 2, 0, 4, 2, 1, 0, 4, 2, 1, 2, 3, 0, 4, 2, 2, 4, 1, 1, 5, 0, 2, 1, 3, 0, 3, 0, 3, 3, 1, 0, 3, 0, 3, 1, 4, 0, 3, 3, 2, 3, 1, 1, 4, 1, 1, 3, 2, 2, 2, 0, 6

Offset: 1

Antti Karttunen, Jan 12 2015

nonn

Consider the binary trees illustrated in A252753 and A252755: If we start from any n, computing successive iterations of A253554 until 1 is reached (i.e., we are traversing level by level towards the root of the tree, starting from that vertex of the tree where n is located), a(n) gives the number of even numbers > 2 encountered on the path (i.e., excluding the 2 from the count but including the starting n if it was even).

Antti Karttunen, Table of n, a(n) for n = 1..8192

Essentially, one less than A253557.
A008578 gives the positions of zeros.
Cf. A000120, A080791, A252753, A252754, A252755, A252756, A253554, A253555, A253556, A253558.
Differs from A252736 for the first time at n=21, where a(21) = 2, while A252736(21) = 1.

(define (A253559 n) (if (= 1 n) 0 (+ -1 (A253557 n))))

a(n) = A080791(A252756(n)). [Number of nonleading 0-bits in A252756(n).]
a(1) = 0; for n>1: a(n) = A253557(n) - 1.
Other identities. For all n >= 2:
a(n) = A000120(A252754(n)) - 1. [One less than the binary weight of A252754(n).]
a(n) = A253555(n) - A253558(n).

A256478 a(0) = 0; and for n >= 1, if A079559(n) = 1, then a(n) = 1 + a(A213714(n)-1), otherwise a(n) = a(A234017(n)).

Original entry on oeis.org

0, 1, 1, 2, 2, 1, 2, 3, 3, 2, 2, 3, 1, 2, 3, 4, 4, 3, 3, 3, 2, 2, 4, 2, 3, 3, 4, 1, 2, 3, 4, 5, 5, 4, 4, 4, 3, 3, 4, 3, 3, 3, 5, 2, 2, 4, 3, 4, 2, 4, 5, 3, 3, 2, 3, 4, 4, 5, 1, 2, 3, 4, 5, 6, 6, 5, 5, 5, 4, 4, 5, 4, 4, 4, 5, 3, 3, 4, 4, 4, 3, 4, 6, 3, 3, 3, 3, 5, 5, 4, 2, 2, 4, 3, 5, 3, 4, 5, 6, 2, 4, 4, 4, 5, 3, 4, 3, 3, 2, 5, 5, 3, 6, 2, 4, 4, 3, 4, 5, 5, 6, 1, 2, 3, 4, 5, 6, 7, 7

Offset: 0

Antti Karttunen, Apr 15 2015

nonn

a(n) tells how many nonzero terms of A005187 are encountered when traversing toward the root of binary tree A233276, starting from the node containing n. This count includes both n (in case it is a term of A005187) and 1 (but not 0). See also comments in A256479 and A256991.

The 1's (seem to) occur at positions given by A000325.

Antti Karttunen, Table of n, a(n) for n = 0..16384

One more than A257248.

Cf. A000120, A000325, A005187, A070939, A079559, A080791, A213714, A234017, A233275, A233276, A233277, A255559, A256479, A256991.

a(0) = 0; and for n >= 1, if A079559(n) = 1, then a(n) = 1 + a(A213714(n)-1), otherwise a(n) = a(A234017(n)).
a(n) = A000120(A233277(n)). [Binary weight of A233277(n).]
Other identities and observations. For all n >= 1:
a(n) = 1 + A257248(n) = 1 + A080791(A233275(n)).
a(n) = A070939(n) - A256479(n).
a(n) >= A255559(n).

A332899 a(1) = 0, and for n > 2, a(n) = a(A332893(n)) + A000035(n).

Original entry on oeis.org

0, 1, 2, 1, 3, 2, 4, 1, 2, 3, 6, 2, 5, 4, 3, 1, 7, 2, 8, 3, 4, 6, 10, 2, 3, 5, 2, 4, 9, 3, 12, 1, 6, 7, 4, 2, 11, 8, 5, 3, 13, 4, 14, 6, 3, 10, 16, 2, 4, 3, 7, 5, 15, 2, 6, 4, 8, 9, 18, 3, 17, 12, 4, 1, 5, 6, 20, 7, 10, 4, 22, 2, 19, 11, 3, 8, 6, 5, 24, 3, 2, 13, 26, 4, 7, 14, 9, 6, 21, 3, 5, 10, 12, 16, 8, 2, 23, 4, 6, 3, 25, 7, 28, 5, 4

Offset: 1

Antti Karttunen, Mar 04 2020

nonn

a(n) tells how many odd numbers are encountered when map x -> A332893(x) is used to traverse from n to 1, the root of the binary tree A332815. This count includes both the starting n itself if it is odd, but excludes 1 where the iteration ends.

a(n) also gives the index of the largest prime factor (A061395) in A332808(n), which is the inverse permutation of A108548 (see also A108546).

Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Cf. A000035, A061395, A080791, A332808, A332811, A332815, A332816, A332893, A332894, A332897, A332898.

Cf. A000079 (after its initial term, gives the positions of 1's).

A332899(n) = if(1==n,0,A332899(A332893(n)) + (n%2));

a(1) = 0, and for n > 1, a(n) = a(A332893(n)) + A000035(n).
a(n) = A000120(A332811(n)).
a(n) = A061395(A332808(n)).
a(n) = A332897(n) + A332898(n).
a(n) <= A332894(n).
For all n > 1, a(n) = 1 + A080791(A332816(n)).

A342456 A276086 applied to the primorial inflation of Doudna-tree, where A276086(n) is the prime product form of primorial base expansion of n.

Original entry on oeis.org

2, 3, 5, 9, 7, 25, 35, 15, 11, 49, 117649, 625, 717409, 1225, 55, 225, 13, 121, 1771561, 2401, 36226650889, 184877, 1127357, 875, 902613283, 514675673281, 3780549773, 1500625, 83852850675321384784127, 3025, 62004635, 21, 17, 169, 4826809, 14641, 8254129, 143, 2924207, 77, 8223741426987700773289, 59797108943, 546826709

Offset: 0

Antti Karttunen and Michael De Vlieger, Mar 14 2021

nonn

This sequence (which could be viewed as a binary tree, like the underlying A005940 and A329886) is similar to A324289, but unlike its underlying tree A283477 that generates only numbers that are products of distinct primorial numbers (i.e., terms of A129912), here the underlying tree A329886 generates all possible products of primorial numbers, i.e., terms of A025487, but in different order.

Cf. A005940, A025487, A108951, A129912, A276086, A283980, A324886, A342457 [= 2*A246277(a(n))], A342461 [= A001221(a(n))], A342462 [= A001222(a(n))], A342463 [= A342001(a(n))], A342464 [= A051903(a(n))].

Cf. A324289 (a subset of these terms, in different order).

Block[{a, f, r = MixedRadix[Reverse@ Prime@ Range@ 24]}, f[n_] :=
Times @@ MapIndexed[Prime[First[#2]]^#1 &, Reverse@ IntegerDigits[n, r]]; a[0] = 1; a[1] = 2; a[n_] := a[n] = If[EvenQ@ n, (Times @@ Map[Prime[PrimePi@ #1 + 1]^#2 & @@ # &, FactorInteger[#]] - Boole[# == 1])*2^IntegerExponent[#, 2] &[a[n/2]], 2 a[(n - 1)/2]]; Array[f@ a[#] &, 43, 0]] (* Michael De Vlieger, Mar 17 2021 *)

A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
A283980(n) = {my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); if(p==2, 6, nextprime(p+1))^e)};
A329886(n) = if(n<2,1+n,if(!(n%2),A283980(A329886(n/2)),2*A329886(n\2)));
A342456(n) = A276086(A329886(n));

a(n) = A276086(A329886(n)) = A324886(A005940(1+n)).
For all n >= 0, gcd(a(n), A329886(n)) = 1.
For all n >= 1, A055396(a(n))-1 = A061395(A329886(n)) = A290251(n) = 1+A080791(n).
For all n >= 0, a(2^n) = A000040(2+n).

A056791 Weight of binary expansion of n + length of binary expansion of n.

Original entry on oeis.org

1, 2, 3, 4, 4, 5, 5, 6, 5, 6, 6, 7, 6, 7, 7, 8, 6, 7, 7, 8, 7, 8, 8, 9, 7, 8, 8, 9, 8, 9, 9, 10, 7, 8, 8, 9, 8, 9, 9, 10, 8, 9, 9, 10, 9, 10, 10, 11, 8, 9, 9, 10, 9, 10, 10, 11, 9, 10, 10, 11, 10, 11, 11, 12, 8, 9, 9, 10, 9, 10, 10, 11, 9, 10, 10, 11, 10, 11, 11, 12, 9, 10, 10, 11, 10, 11, 11

Offset: 0

N. J. A. Sloane, Sep 01 2000

			12 = 1100 in binary, so a(12)=2+4=6.

Michael De Vlieger, Table of n, a(n) for n = 0..16384
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Index to sequences related to the complexity of n

Equals A056792 + 1.
Equals A014701 + 2.
Cf. A061313, A037861.

Table[If[n==0,1,s=IntegerDigits[n,2];Total@s+Length@s],{n,0,100}] (* Giorgos Kalogeropoulos, Sep 13 2021 *)

a(n) = if (n==0, 1, my(b=binary(n)); vecsum(b) + #b); \\ Michel Marcus, Sep 13 2021

def a(n): b = bin(n)[2:]; return b.count('1') + len(b)
print([a(n) for n in range(87)]) # Michael S. Branicky, Sep 13 2021

a(n) = a((n - n mod 2) / (2 - n mod 2)) + 1 for n>0, a(0)=1. - Reinhard Zumkeller, Jul 29 2002

a(2n) = a(n)+1, a(2n+1) = a(n)+2. G.f.: 1 + 1/(1-x) * sum(k>=0, (2t+t^2)/(1+t), t=x^2^k). For n>0, a(n) = 2*A000120(n) + A080791(n) = A000120(n) + A029837(n). - Ralf Stephan, Jun 14 2003

More terms from James Sellers, Sep 06 2000 and from David W. Wilson, Sep 07 2000

A246370 a(1)=0, a(p_n) = 1 + a(n), a(c_n) = a(n), where p_n = n-th prime = A000040(n), c_n = n-th composite number = A002808(n); Also number of nonleading 0-bits in the binary representation of A135141(n).

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Aug 27 2014

nonn

			Consider n=30. It is the 19th composite number in A002808: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, ...
Thus we consider next n=19, which is the 8th prime in A000040: 2, 3, 5, 7, 11, 13, 17, 19, ...
So we proceed with n=8, which is the 3rd composite number, and then with n=3, which is the 2nd prime, and then with n=2 which is the 1st prime, and we have finished.
All in all, it took us 5 steps (A246348(30) = 6 = 5+1) to reach 1, and on the journey, we encountered three primes, 19, 3 and 2, thus a(30) = 3.

Antti Karttunen, Table of n, a(n) for n = 1..32998

Cf. A000040, A002808, A000720, A065855, A080791, A135141, A246348, A246369, A246377.

a(1) = 1, and for n >= 1, if A010051(n) = 1 [i.e. when n is prime], a(n) = 1 + a(A000720(n)), otherwise a(n) = a(A065855(n)). [A000720(n) and A065855(n) tell the number of primes, and respectively, composites <= n].
a(n) = A080791(A135141(n)). [a(n) tells also the number of nonleading zeros in binary representation of A135141(n)].
a(n) = A000120(A246377(n))-1. [Respectively, one less than the number of 1-bits in 0/1-swapped version of that sequence].
a(n) = A246348(n) - A246369(n) - 1.

A253558 a(n) = A253556(n) + 1.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 4, 1, 2, 3, 5, 2, 6, 4, 3, 1, 7, 2, 8, 3, 2, 5, 9, 2, 3, 6, 4, 4, 10, 3, 11, 1, 3, 7, 4, 2, 12, 8, 5, 3, 13, 2, 14, 5, 2, 9, 15, 2, 4, 3, 3, 6, 16, 4, 3, 4, 4, 10, 17, 3, 18, 11, 6, 1, 5, 3, 19, 7, 3, 4, 20, 2, 21, 12, 7, 8, 5, 5, 22, 3, 5, 13, 23, 2, 4, 14, 4, 5, 24, 2, 4, 9, 2, 15, 6, 2, 25, 4, 8, 3, 26, 3

Offset: 1

Antti Karttunen, Jan 12 2015

nonn

Consider the binary trees illustrated in A252753 and A252755: If we start from any n, computing successive iterations of A253554 until 1 is reached (i.e., we are traversing level by level towards the root of the tree, starting from that vertex of the tree where n is located at), a(n) gives the number of odd numbers encountered on the path (i.e., including both the final 1 and the starting n if it was odd).

Antti Karttunen, Table of n, a(n) for n = 1..8192

One more than A253556.
Powers of two, A000079, gives the positions of ones.
Cf. A000040, A253555, A253557, A253559.
After n=1, differs from A061395 for the first time at n=21, where a(21) = 2, while A061395(21) = 4.

Scheme
```
(define (A253558 n) (+ 1 (A253556 n)))
```

a(n) = A253556(n) + 1.
a(n) = A080791(A252754(n)) + 1. [One more than the number of nonleading 0-bits in A252754(n).]
Other identities.
For all n >= 1:
a(A000040(n)) = n.
For all n >= 2:
a(n) = A000120(A252756(n)). [Binary weight of A252756(n).]
a(n) = A253555(n) - A253559(n).

A254045 a(1) = 0, for n > 1: a(n) = a(A253889(n)) + floor((n modulo 3)/2).

Original entry on oeis.org

0, 1, 0, 1, 2, 0, 1, 1, 1, 3, 2, 2, 2, 3, 1, 1, 1, 0, 0, 2, 3, 3, 2, 2, 2, 2, 1, 2, 4, 2, 1, 3, 4, 1, 3, 4, 3, 3, 3, 4, 4, 2, 2, 2, 3, 1, 2, 2, 3, 2, 4, 3, 1, 2, 2, 1, 2, 2, 3, 5, 3, 4, 1, 3, 4, 0, 3, 3, 5, 5, 3, 3, 4, 3, 4, 4, 3, 2, 3, 2, 1, 3, 3, 4, 2, 5, 3, 2, 3, 3, 3, 2, 2, 2, 4, 3, 1, 5, 5, 4, 2, 2, 1, 4, 1, 3, 5, 1, 5, 4, 3, 3, 4, 1, 3, 4, 3, 6, 5, 3, 1, 5, 3, 2, 3, 3, 5, 3

Offset: 1

Antti Karttunen, Jan 23 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 1..8192

Cf. A007051, A064216, A080791, A253786, A253889, A253894, A254044.

a(1) = 0, for n > 1: a(n) = a(A253889(n)) + floor((n modulo 3)/2).
a(1) = 0, thereafter, if n = 3k+2, then a(3k+2) = 1 + a(k+1), otherwise a(n) = a(A253889(n)).
a(n) = A080791(A064216(n)). [Number of nonleading zeros in binary representation of terms of A064216.]
a(n) = A253894(n) - A254044(n).
Other identities and observations:
a(A007051(n)) = n for all n >= 0.
a(n) >= A253786(n) for all n >= 1.

A256479 a(1) = 0, and for n > 1, if A079559(n) = 0, then a(n) = 1 + a(A234017(n)), otherwise a(n) = a(A213714(n)-1).

Original entry on oeis.org

0, 1, 0, 1, 2, 1, 0, 1, 2, 2, 1, 3, 2, 1, 0, 1, 2, 2, 2, 3, 3, 1, 3, 2, 2, 1, 4, 3, 2, 1, 0, 1, 2, 2, 2, 3, 3, 2, 3, 3, 3, 1, 4, 4, 2, 3, 2, 4, 2, 1, 3, 3, 4, 3, 2, 2, 1, 5, 4, 3, 2, 1, 0, 1, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 4, 4, 3, 3, 3, 4, 3, 1, 4, 4, 4, 4, 2, 2, 3, 5, 5, 3, 4, 2, 4, 3, 2, 1, 5, 3, 3, 3, 2, 4, 3, 4, 4, 5, 2, 2, 4, 1, 5, 3, 3, 4, 3, 2, 2, 1, 6, 5, 4, 3, 2, 1, 0, 1

Offset: 1

Antti Karttunen, Apr 15 2015

nonn

a(n) tells how many terms of A055938 are encountered when traversing toward the root of binary tree A233276, starting from the node containing n. This count includes also n in case it itself is a term of A055938. See also comments in A256478 and A256991.

Antti Karttunen, Table of n, a(n) for n = 1..16384

One less than A257249.
Cf. A000120, A055938, A070939, A079559, A080791, A213714, A234017, A233275, A233276, A233277, A256478, A256991.
Cf. also A000225 (gives the positions zeros).

a(1) = 0, and for n > 1, if A079559(n) = 0, then a(n) = 1 + a(A234017(n)), otherwise a(n) = a(A213714(n)-1).
a(n) = A080791(A233277(n)). [Number of nonleading zeros in the binary representation of A233277(n).]
Other identities. For all n >= 1:
a(n) = A257249(n) - 1 = A000120(A233275(n)) - 1.
a(n) = A070939(n) - A256478(n).
a(A000225(n)) = 0.

A346702 The a(n)-th composition in standard order is the odd bisection of the n-th composition in standard order.

Original entry on oeis.org

0, 1, 2, 1, 4, 2, 1, 3, 8, 4, 2, 5, 1, 3, 6, 3, 16, 8, 4, 9, 2, 5, 10, 5, 1, 3, 6, 3, 12, 6, 3, 7, 32, 16, 8, 17, 4, 9, 18, 9, 2, 5, 10, 5, 20, 10, 5, 11, 1, 3, 6, 3, 12, 6, 3, 7, 24, 12, 6, 13, 3, 7, 14, 7, 64, 32, 16, 33, 8, 17, 34, 17, 4, 9, 18, 9, 36, 18

Offset: 0

Gus Wiseman, Aug 12 2021

nonn

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again.

a(n) is the row number in A066099 of the odd bisection of the n-th row of A066099.

			Composition number 741 in standard order is (2,1,1,3,2,1), with odd bisection (2,1,2), which is composition number 22 in standard order, hence a(741) = 22.

Length of the a(n)-th standard composition is A000120(n)/2 rounded up.
Positions of 1's are A003945.
Positions of 2's (and zero) are A083575.
Sum of the a(n)-th standard composition is A209281(n+1).
Positions of first appearances are A290259.
The version for prime indices is A346703.
The version for even bisection is A346705, with sums A346633.
A000120 and A080791 count binary digits 1 and 0, with difference A145037.
A011782 counts compositions.
A029837 gives length of binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A345197 counts compositions by sum, length, and alternating sum.
Cf. A000009, A000302, A000346, A025047, A088218, A124754, A294175, A346697 (reverse: A346699).

Table[Total[2^Accumulate[Reverse[First/@Partition[Append[ Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse,0],2]]]]/2,{n,0,100}]

A029837(a(n)) = A209281(n).

cp's OEIS Frontend

A253559 a(1) = 0; for n>1: a(n) = A253557(n) - 1.

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A256478 a(0) = 0; and for n >= 1, if A079559(n) = 1, then a(n) = 1 + a(A213714(n)-1), otherwise a(n) = a(A234017(n)).

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A332899 a(1) = 0, and for n > 2, a(n) = a(A332893(n)) + A000035(n).

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A342456 A276086 applied to the primorial inflation of Doudna-tree, where A276086(n) is the prime product form of primorial base expansion of n.

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A056791 Weight of binary expansion of n + length of binary expansion of n.

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A246370 a(1)=0, a(p_n) = 1 + a(n), a(c_n) = a(n), where p_n = n-th prime = A000040(n), c_n = n-th composite number = A002808(n); Also number of nonleading 0-bits in the binary representation of A135141(n).

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A253558 a(n) = A253556(n) + 1.

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A254045 a(1) = 0, for n > 1: a(n) = a(A253889(n)) + floor((n modulo 3)/2).

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A256479 a(1) = 0, and for n > 1, if A079559(n) = 0, then a(n) = 1 + a(A234017(n)), otherwise a(n) = a(A213714(n)-1).

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