A346705 -id:A346705 - cp's OEIS frontend

A329888 a(n) = A329900(A329602(n)); Heinz number of the even bisection (even-indexed parts) of the integer partition with Heinz number n.

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

Offset: 1

Antti Karttunen, Dec 22 2019

nonn

From Gus Wiseman, Aug 05 2021 and Antti Karttunen, Oct 13 2021: (Start)
Also the product of primes at even positions in the weakly decreasing list (with multiplicity) of prime factors of n. For example, the prime factors of 108 are (3,3,3,2,2), with even bisection (3,2), with product 6, so a(108) = 6.
Proof: A108951(n) gives a number with the same largest prime factor (A006530) and its exponent (A071178) as in n, and with each smaller prime p = 2, 3, 5, 7, ... < A006530(n) having as its exponent the partial sum of the exponents of all prime factors >= p present in n (with primes not present in n having the exponent 0). Then applying A000188 replaces each such "partial sum exponent" k with floor(k/2). Finally, A319626 replaces those halved exponents with their first differences (here the exponent of the largest prime present stays intact, because the next larger prime's exponent is 0 in n). It should be easy to see that if prime q is not present in n (i.e., does not divide it), then neither it is present in a(n). Moreover, if the partial sum exponent of q is odd and only one larger than the partial sum exponent of the next larger prime factor of n, then q will not be present in a(n), while in all other cases q is present in a(n). See also the last example.
(End)

			From _Gus Wiseman_, Aug 15 2021: (Start)
The list of all numbers with image 12 and their corresponding prime factors begins:
  144: (3,3,2,2,2,2)
  216: (3,3,3,2,2,2)
  240: (5,3,2,2,2,2)
  288: (3,3,2,2,2,2,2)
  336: (7,3,2,2,2,2)
  360: (5,3,3,2,2,2)
(End)
The positions from the left are indexed as 1, 2, 3, ..., etc, so e.g., for 240 we pick the second, the fourth and the sixth prime factor, 3, 2 and 2, to obtain a(240) = 3*2*2 = 12. For 288, we similarly pick the second (3), the fourth (2) and the sixth (2) to obtain a(288) = 3*2*2 = 12. - _Antti Karttunen_, Oct 13 2021
Consider n = 11945934 = 2*3*3*3*7*11*13*13*17. Its primorial inflation is A108951(11945934) = 96478365991115908800000 = 2^9 * 3^8 * 5^5 * 7^5 * 11^4 * 13^3 * 17^1. Applying A000188 to this halves each exponent (floored down if the exponent is odd), leaving the factors 2^4 * 3^4 * 5^2 * 7^2 * 11^2 * 13^1 = 2497294800. Then applying A319626 to this number retains the largest prime factor (and its exponent), and subtracts from the exponent of each of the rest of primes the exponent of the next larger prime, so from 2^4 * 3^4 * 5^2 * 7^2 * 11^2 * 13^1 we get 2^(4-4) * 3^(4-2) * 5^(2-2) * 7^(2-2) * 11^(2-1) * 13^1 = 3^2 * 11^1 * 13^1 = 1287 = a(11945934), which is obtained also by selecting every second prime from the list [17, 13, 13, 11, 7, 3, 3, 3, 2] and taking their product. - _Antti Karttunen_, Oct 15 2021

Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
Index entries for sequences related to Heinz numbers
Index entries for sequences related to primorial numbers

Cf. A000188, A108951, A319626, A329602, A329900.
A left inverse of A000290.
Positions of 1's are A008578.
Positions of primes are A168645.
The sum of prime indices of a(n) is A346700(n).
The odd version is A346701.
The odd non-reverse version is A346703.
The non-reverse version is A346704.
The version for standard compositions is A346705, odd A346702.
A001221 counts distinct prime factors.
A001222 counts all prime factors.
A001414 adds up prime factors, row sums of A027746.
A027187 counts partitions of even length, ranked by A028260.
A056239 adds up prime indices, row sums of A112798.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A346633 adds up the even bisection of standard compositions.
A346698 adds up the even bisection of prime indices.
Cf. A000097, A035363, A102750, A236913, A253553, A344606, A344617, A344653, A345957, A345958, A345959.

Table[Times@@Last/@Partition[Reverse[Flatten[Apply[ConstantArray,FactorInteger[n],{1}]]],2],{n,100}] (* Gus Wiseman, Oct 13 2021 *)

PARI
```
A329888(n) = A329900(A329602(n));
```

A329888(n) = if(1==n,n,my(f=factor(n),m=1,p=0); forstep(k=#f~,1,-1,while(f[k,2], m *= f[k,1]^(p%2); f[k,2]--; p++)); (m)); \\ (After Wiseman's new interpretation) - Antti Karttunen, Sep 21 2021

a(n) = A329900(A329602(n)) = A329900(A000188(A108951(n))).
A108951(a(n)) = A329602(n).
a(n^2) = n for all n >= 1.
a(n) * A346701(n) = n. - Gus Wiseman, Aug 07 2021
A056239(a(n)) = A346700(n). - Gus Wiseman, Aug 07 2021
Antti Karttunen, Sep 21 2021
From Antti Karttunen, Oct 13 2021: (Start)
a(n) = A319626(A329602(n)) = A319626(A000188(A108951(n))).
For all x in A102750, a(x) = a(A253553(x)). (End)

Name amended with Gus Wiseman's new interpretation - Antti Karttunen, Oct 13 2021

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

A329888 a(n) = A329900(A329602(n)); Heinz number of the even bisection (even-indexed parts) of the integer partition with Heinz number n.

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