cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-8 of 8 results.

A323508 a(n) = A323505(A156552(n)).

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 16, 24, 12, 24, 32, 48, 64, 48, 24, 120, 128, 36, 256, 96, 48, 96, 512, 240, 24, 192, 48, 192, 1024, 72, 2048, 720, 96, 384, 48, 144, 4096, 768, 192, 480, 8192, 144, 16384, 384, 96, 1536, 32768, 1440, 48, 72, 384, 768, 65536, 144, 96, 960, 768, 3072, 131072, 288, 262144, 6144, 192, 5040, 192, 288, 524288, 1536, 1536
Offset: 1

Views

Author

Antti Karttunen, Jan 17 2019

Keywords

Comments

Sequence contains only terms of A001013 and each a(n) is a multiple of A112624(n).

Crossrefs

Programs

Formula

a(n) = A323505(A156552(n)).

A323506 a(n) = A323505(n) / A246660(n).

Original entry on oeis.org

1, 2, 4, 3, 8, 12, 6, 4, 16, 24, 24, 24, 12, 18, 8, 5, 32, 48, 48, 48, 48, 72, 48, 40, 24, 36, 36, 36, 16, 24, 10, 6, 64, 96, 96, 96, 96, 144, 96, 80, 96, 144, 144, 144, 96, 144, 80, 60, 48, 72, 72, 72, 72, 108, 72, 60, 32, 48, 48, 48, 20, 30, 12, 7, 128, 192, 192, 192, 192, 288, 192, 160, 192, 288, 288, 288, 192, 288, 160, 120
Offset: 0

Views

Author

Antti Karttunen, Jan 16 2019

Keywords

Examples

			This sequence can be represented as a binary tree, as both A323505 and A246660 have similar tree structures:
                                       1
                                       |
                    ...................2....................
                   4                                        3
         8......../ \........12                  6........./ \.......4
        / \                 / \                 / \                 / \
       /   \               /   \               /   \               /   \
      /     \             /     \             /     \             /     \
    16       24         24       24         12       18          8       5
  32  48   48  48     48  72   48  40     24  36   36  36      16 24   10  6
etc.
		

Crossrefs

Programs

Formula

a(n) = A323505(n) / A246660(n).
For n > 1, a(2n) = 2*a(n).

A005940 The Doudna sequence: write n-1 in binary; power of prime(k) in a(n) is # of 1's that are followed by k-1 0's.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 9, 8, 7, 10, 15, 12, 25, 18, 27, 16, 11, 14, 21, 20, 35, 30, 45, 24, 49, 50, 75, 36, 125, 54, 81, 32, 13, 22, 33, 28, 55, 42, 63, 40, 77, 70, 105, 60, 175, 90, 135, 48, 121, 98, 147, 100, 245, 150, 225, 72, 343, 250, 375, 108, 625, 162, 243, 64, 17, 26, 39
Offset: 1

Views

Author

Keywords

Comments

A permutation of the natural numbers. - Robert G. Wilson v, Feb 22 2005
Fixed points: A029747. - Reinhard Zumkeller, Aug 23 2006
The even bisection, when halved, gives the sequence back. - Antti Karttunen, Jun 28 2014
From Antti Karttunen, Dec 21 2014: (Start)
This irregular table can be represented as a binary tree. Each child to the left is obtained by applying A003961 to the parent, and each child to the right is obtained by doubling the parent:
1
|
...................2...................
3 4
5......../ \........6 9......../ \........8
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
7 10 15 12 25 18 27 16
11 14 21 20 35 30 45 24 49 50 75 36 125 54 81 32
etc.
Sequence A163511 is obtained by scanning the same tree level by level, from right to left. Also in binary trees A253563 and A253565 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees.
A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 2 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is larger than the right child, A252750 the difference between left and right child for each node from node 2 onward.
(End)
-A008836(a(1+n)) gives the corresponding numerator for A323505(n). - Antti Karttunen, Jan 19 2019
(a(2n+1)-1)/2 [= A244154(n)-1, for n >= 0] is a permutation of the natural numbers. - George Beck and Antti Karttunen, Dec 08 2019
From Peter Munn, Oct 04 2020: (Start)
Each term has the same even part (equivalently, the same 2-adic valuation) as its index.
Using the tree depicted in Antti Karttunen's 2014 comment:
Numbers are on the right branch (4 and descendants) if and only if divisible by the square of their largest prime factor (cf. A070003).
Numbers on the left branch, together with 2, are listed in A102750.
(End)
According to Kutz (1981), he learned of this sequence from American mathematician Byron Leon McAllister (1929-2017) who attributed the invention of the sequence to a graduate student by the name of Doudna (first name Paul?) in the mid-1950's at the University of Wisconsin. - Amiram Eldar, Jun 17 2021
From David James Sycamore, Sep 23 2022: (Start)
Alternative (recursive) definition: If n is a power of 2 then a(n)=n. Otherwise, if 2^j is the greatest power of 2 not exceeding n, and if k = n - 2^j, then a(n) is the least m*a(k) that has not occurred previously, where m is an odd prime.
Example: Use recursion with n = 77 = 2^6 + 13. a(13) = 25 and since 11 is the smallest odd prime m such that m*a(13) has not already occurred (see a(27), a(29),a(45)), then a(77) = 11*25 = 275. (End)
The odd bisection, when transformed by replacing all prime(k)^e in a(2*n - 1) with prime(k-1)^e, returns a(n), and thus gives the sequence back. - David James Sycamore, Sep 28 2022

Examples

			From _N. J. A. Sloane_, Aug 22 2022: (Start)
Let c_i = number of 1's in binary expansion of n-1 that have i 0's to their right, and let p(j) = j-th prime.  Then a(n) = Product_i p(i+1)^c_i.
If n=9, n-1 is 1000, c_3 = 1, a(9) = p(4)^1 = 7.
If n=10, n-1 = 1001, c_0 = 1, c_2 = 1, a(10) = p(1)*p(3) = 2*5 = 10.
If n=11, n-1 = 1010, c_1 = 1, c_2 = 1, a(11) = p(2)*p(3) = 15. (End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A103969. Inverse is A005941 (A156552).
Cf. A125106. [From Franklin T. Adams-Watters, Mar 06 2010]
Cf. A252737 (gives row sums), A252738 (row products), A332979 (largest on row).
Related permutations of positive integers: A163511 (via A054429), A243353 (via A006068), A244154, A253563 (via A122111), A253565, A332977, A334866 (via A225546).
A000120, A003602, A003961, A006519, A053645, A070939, A246278, A250246, A252753, A253552 are used in a formula defining this sequence.
Formulas for f(a(n)) are given for f = A000265, A003963, A007949, A055396, A056239.
Numbers that occur at notable sets of positions in the binary tree representation of the sequence: A000040, A000079, A002110, A070003, A070826, A102750.
Cf. A106737, A290077, A323915, A324052, A324054, A324055, A324056, A324057, A324058, A324114, A324335, A324340, A324348, A324349 for various number-theoretical sequences applied to (i.e., permuted by) this sequence.
k-adic valuation: A007814 (k=2), A337821 (k=3).
Positions of multiples of 3: A091067.
Primorial deflation: A337376 / A337377.
Sum of prime indices of a(n) is A161511, reverse version A359043.
A048793 lists binary indices, ranked by A019565.
A066099 lists standard comps, partial sums A358134 (ranked by A358170).

Programs

  • Haskell
    a005940 n = f (n - 1) 1 1 where
       f 0 y _          = y
       f x y i | m == 0 = f x' y (i + 1)
               | m == 1 = f x' (y * a000040 i) i
               where (x',m) = divMod x 2
    -- Reinhard Zumkeller, Oct 03 2012
    (Scheme, with memoization-macro definec from Antti Karttunen's IntSeq-library)
    (define (A005940 n) (A005940off0 (- n 1))) ;; The off=1 version, utilizing any one of three different offset-0 implementations:
    (definec (A005940off0 n) (cond ((< n 2) (+ 1 n)) (else (* (A000040 (- (A070939 n) (- (A000120 n) 1))) (A005940off0 (A053645 n))))))
    (definec (A005940off0 n) (cond ((<= n 2) (+ 1 n)) ((even? n) (A003961 (A005940off0 (/ n 2)))) (else (* 2 (A005940off0 (/ (- n 1) 2))))))
    (define (A005940off0 n) (let loop ((n n) (i 1) (x 1)) (cond ((zero? n) x) ((even? n) (loop (/ n 2) (+ i 1) x)) (else (loop (/ (- n 1) 2) i (* x (A000040 i)))))))
    ;; Antti Karttunen, Jun 26 2014
    
  • Maple
    f := proc(n,i,x) option remember ; if n = 0 then x; elif type(n,'even') then procname(n/2,i+1,x) ; else procname((n-1)/2,i,x*ithprime(i)) ; end if; end proc:
    A005940 := proc(n) f(n-1,1,1) ; end proc: # R. J. Mathar, Mar 06 2010
  • Mathematica
    f[n_] := Block[{p = Partition[ Split[ Join[ IntegerDigits[n - 1, 2], {2}]], 2]}, Times @@ Flatten[ Table[q = Take[p, -i]; Prime[ Count[ Flatten[q], 0] + 1]^q[[1, 1]], {i, Length[p]}] ]]; Table[ f[n], {n, 67}] (* Robert G. Wilson v, Feb 22 2005 *)
    Table[Times@@Prime/@(Join@@Position[Reverse[IntegerDigits[n,2]],1]-Range[DigitCount[n,2,1]]+1),{n,0,100}] (* Gus Wiseman, Dec 28 2022 *)
  • PARI
    A005940(n) = { my(p=2, t=1); n--; until(!n\=2, n%2 && (t*=p) || p=nextprime(p+1)); t } \\ M. F. Hasler, Mar 07 2010; update Aug 29 2014
    
  • PARI
    a(n)=my(p=2, t=1); for(i=0,exponent(n), if(bittest(n,i), t*=p, p=nextprime(p+1))); t \\ Charles R Greathouse IV, Nov 11 2021
    
  • Python
    from sympy import prime
    import math
    def A(n): return n - 2**int(math.floor(math.log(n, 2)))
    def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
    print([b(n - 1) for n in range(1, 101)]) # Indranil Ghosh, Apr 10 2017
    
  • Python
    from math import prod
    from itertools import accumulate
    from collections import Counter
    from sympy import prime
    def A005940(n): return prod(prime(len(a)+1)**b for a, b in Counter(accumulate(bin(n-1)[2:].split('1')[:0:-1])).items()) # Chai Wah Wu, Mar 10 2023

Formula

From Reinhard Zumkeller, Aug 23 2006, R. J. Mathar, Mar 06 2010: (Start)
a(n) = f(n-1, 1, 1)
where f(n, i, x) = x if n = 0,
= f(n/2, i+1, x) if n > 0 is even
= f((n-1)/2, i, x*prime(i)) otherwise. (End)
From Antti Karttunen, Jun 26 2014: (Start)
Define a starting-offset 0 version of this sequence as:
b(0)=1, b(1)=2, [base cases]
and then compute the rest either with recurrence:
b(n) = A000040(1+(A070939(n)-A000120(n))) * b(A053645(n)).
or
b(2n) = A003961(b(n)), b(2n+1) = 2 * b(n). [Compare this to the similar recurrence given for A163511.]
Then define a(n) = b(n-1), where a(n) gives this sequence A005940 with the starting offset 1.
Can be also defined as a composition of related permutations:
a(n+1) = A243353(A006068(n)).
a(n+1) = A163511(A054429(n)). [Compare the scatter plots of this sequence and A163511 to each other.]
This permutation also maps between the partitions as enumerated in the lists A125106 and A112798, providing identities between:
A161511(n) = A056239(a(n+1)). [The corresponding sums ...]
A243499(n) = A003963(a(n+1)). [... and the products of parts of those partitions.]
(End)
From Antti Karttunen, Dec 21 2014 - Jan 04 2015: (Start)
A002110(n) = a(1+A002450(n)). [Primorials occur at (4^n - 1)/3 in the offset-0 version of the sequence.]
a(n) = A250246(A252753(n-1)).
a(n) = A122111(A253563(n-1)).
For n >= 1, A055396(a(n+1)) = A001511(n).
For n >= 2, a(n) = A246278(1+A253552(n)).
(End)
From Peter Munn, Oct 04 2020: (Start)
A000265(a(n)) = a(A000265(n)) = A003961(a(A003602(n))).
A006519(a(n)) = a(A006519(n)) = A006519(n).
a(n) = A003961(a(A003602(n))) * A006519(n).
A007814(a(n)) = A007814(n).
A007949(a(n)) = A337821(n) = A007814(A003602(n)).
a(n) = A225546(A334866(n-1)).
(End)
a(2n) = 2*a(n), or generally a(2^k*n) = 2^k*a(n). - Amiram Eldar, Oct 03 2022
If n-1 = Sum_{i} 2^(q_i-1), then a(n) = Product_{i} prime(q_i-i+1). These are the Heinz numbers of the rows of A125106. If the offset is changed to 0, the inverse is A156552. - Gus Wiseman, Dec 28 2022

Extensions

More terms from Robert G. Wilson v, Feb 22 2005
Sign in a formula switched and Maple program added by R. J. Mathar, Mar 06 2010
Binary tree illustration and keyword tabf added by Antti Karttunen, Dec 21 2014

A156552 Unary-encoded compressed factorization of natural numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 7, 6, 9, 16, 11, 32, 17, 10, 15, 64, 13, 128, 19, 18, 33, 256, 23, 12, 65, 14, 35, 512, 21, 1024, 31, 34, 129, 20, 27, 2048, 257, 66, 39, 4096, 37, 8192, 67, 22, 513, 16384, 47, 24, 25, 130, 131, 32768, 29, 36, 71, 258, 1025, 65536, 43, 131072, 2049, 38, 63, 68, 69, 262144
Offset: 1

Views

Author

Leonid Broukhis, Feb 09 2009

Keywords

Comments

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.
From Antti Karttunen, Jun 27 2014: (Start)
The odd bisection (containing even terms) halved gives A244153.
The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.
(End)
Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

Examples

			For 84 = 2*2*3*7 -> 1*1 + 1*2 + 2*4 + 8*8 =  75.
For 105 = 3*5*7 -> 2*1 + 4*2 + 8*4 = 42.
For 137 = p_33 -> 2^32 = 4294967296.
For 420 = 2*2*3*5*7 -> 1*1 + 1*2 + 2*4 + 4*8 + 8*16 = 171.
For 147 = 3*7*7 = p_2 * p_4 * p_4 -> 2*1 + 8*2 + 8*4 = 50.
		

Crossrefs

One less than A005941.
Inverse permutation: A005940 with starting offset 0 instead of 1.
Cf. also A297106, A297112 (Möbius transform), A297113, A153013, A290308, A300827, A323243, A323244, A323247, A324201, A324812 (n for which a(n) is a square), A324813, A324822, A324823, A324398, A324713, A324815, A324819, A324865, A324866, A324867.

Programs

  • Mathematica
    Table[Floor@ Total@ Flatten@ MapIndexed[#1 2^(#2 - 1) &, Flatten[ Table[2^(PrimePi@ #1 - 1), {#2}] & @@@ FactorInteger@ n]], {n, 67}] (* Michael De Vlieger, Sep 08 2016 *)
  • PARI
    a(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ David A. Corneth, Mar 08 2019
    
  • PARI
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
    A156552(n) = if(1==n, 0, if(!(n%2), 1+(2*A156552(n/2)), 2*A156552(A064989(n)))); \\ (based on the given recurrence) - Antti Karttunen, Mar 08 2019
    
  • Perl
    # Program corrected per instructions from Leonid Broukhis. - Antti Karttunen, Jun 26 2014
    # However, it gives correct answers only up to n=136, before corruption by a wrap-around effect.
    # Note that the correct answer for n=137 is A156552(137) = 4294967296.
    $max = $ARGV[0];
    $pow = 0;
    foreach $i (2..$max) {
    @a = split(/ /, `factor $i`);
    shift @a;
    $shift = 0;
    $cur = 0;
    while ($n = int shift @a) {
    $prime{$n} = 1 << $pow++ if !defined($prime{$n});
    $cur |= $prime{$n} << $shift++;
    }
    print "$cur, ";
    }
    print "\n";
    (Scheme, with memoization-macro definec from Antti Karttunen's IntSeq-library, two different implementations)
    (definec (A156552 n) (cond ((= n 1) 0) (else (+ (A000079 (+ -2 (A001222 n) (A061395 n))) (A156552 (A052126 n))))))
    (definec (A156552 n) (cond ((= 1 n) (- n 1)) ((even? n) (+ 1 (* 2 (A156552 (/ n 2))))) (else (* 2 (A156552 (A064989 n))))))
    ;; Antti Karttunen, Jun 26 2014
    
  • Python
    from sympy import primepi, factorint
    def A156552(n): return sum((1<Chai Wah Wu, Mar 10 2023

Formula

From Antti Karttunen, Jun 26 2014: (Start)
a(1) = 0, a(n) = A000079(A001222(n)+A061395(n)-2) + a(A052126(n)).
a(1) = 0, a(2n) = 1+2*a(n), a(2n+1) = 2*a(A064989(2n+1)). [Compare to the entanglement recurrence A243071].
For n >= 0, a(2n+1) = 2*A244153(n+1). [Follows from the latter clause of the above formula.]
a(n) = A005941(n) - 1.
As a composition of related permutations:
a(n) = A003188(A243354(n)).
a(n) = A054429(A243071(n)).
For all n >= 1, A005940(1+a(n)) = n and for all n >= 0, a(A005940(n+1)) = n. [The offset-0 version of A005940 works as an inverse for this permutation.]
This permutations also maps between the partition-lists A112798 and A125106:
A056239(n) = A161511(a(n)). [The sums of parts of each partition (the total sizes).]
A003963(n) = A243499(a(n)). [And also the products of those parts.]
(End)
From Antti Karttunen, Oct 09 2016: (Start)
A161511(a(n)) = A056239(n).
A029837(1+a(n)) = A252464(n). [Binary width of terms.]
A080791(a(n)) = A252735(n). [Number of nonleading 0-bits.]
A000120(a(n)) = A001222(n). [Binary weight.]
For all n >= 2, A001511(a(n)) = A055396(n).
For all n >= 2, A000120(a(n))-1 = A252736(n). [Binary weight minus one.]
A252750(a(n)) = A252748(n).
a(A250246(n)) = A252754(n).
a(A005117(n)) = A277010(n). [Maps squarefree numbers to a permutation of A003714, fibbinary numbers.]
A085357(a(n)) = A008966(n). [Ditto for their characteristic functions.]
For all n >= 0:
a(A276076(n)) = A277012(n).
a(A276086(n)) = A277022(n).
a(A260443(n)) = A277020(n).
(End)
From Antti Karttunen, Dec 30 2017: (Start)
For n > 1, a(n) = Sum_{d|n, d>1} 2^A033265(a(d)). [See comments.]
More linking formulas:
A106737(a(n)) = A000005(n).
A290077(a(n)) = A000010(n).
A069010(a(n)) = A001221(n).
A136277(a(n)) = A181591(n).
A132971(a(n)) = A008683(n).
A106400(a(n)) = A008836(n).
A268411(a(n)) = A092248(n).
A037011(a(n)) = A010052(n) [conjectured, depends on the exact definition of A037011].
A278161(a(n)) = A046951(n).
A001316(a(n)) = A061142(n).
A277561(a(n)) = A034444(n).
A286575(a(n)) = A037445(n).
A246029(a(n)) = A181819(n).
A278159(a(n)) = A124859(n).
A246660(a(n)) = A112624(n).
A246596(a(n)) = A069739(n).
A295896(a(n)) = A053866(n).
A295875(a(n)) = A295297(n).
A284569(a(n)) = A072411(n).
A286574(a(n)) = A064547(n).
A048735(a(n)) = A292380(n).
A292272(a(n)) = A292382(n).
A244154(a(n)) = A048673(n), a(A064216(n)) = A244153(n).
A279344(a(n)) = A279339(n), a(A279338(n)) = A279343(n).
a(A277324(n)) = A277189(n).
A037800(a(n)) = A297155(n).
For n > 1, A033265(a(n)) = 1+A297113(n).
(End)
From Antti Karttunen, Mar 08 2019: (Start)
a(n) = A048675(n) + A323905(n).
a(A324201(n)) = A000396(n), provided there are no odd perfect numbers.
The following sequences are derived from or related to the base-2 expansion of a(n):
A000265(a(n)) = A322993(n).
A002487(a(n)) = A323902(n).
A005187(a(n)) = A323247(n).
A324288(a(n)) = A324116(n).
A323505(a(n)) = A323508(n).
A079559(a(n)) = A323512(n).
A085405(a(n)) = A323239(n).
The following sequences are obtained by applying to a(n) a function that depends on the prime factorization of its argument, which goes "against the grain" because a(n) is the binary code of the factorization of n, which in these cases is then factored again:
A000203(a(n)) = A323243(n).
A033879(a(n)) = A323244(n) = 2*a(n) - A323243(n),
A294898(a(n)) = A323248(n).
A000005(a(n)) = A324105(n).
A000010(a(n)) = A324104(n).
A083254(a(n)) = A324103(n).
A001227(a(n)) = A324117(n).
A000593(a(n)) = A324118(n).
A001221(a(n)) = A324119(n).
A009194(a(n)) = A324396(n).
A318458(a(n)) = A324398(n).
A192895(a(n)) = A324100(n).
A106315(a(n)) = A324051(n).
A010052(a(n)) = A324822(n).
A053866(a(n)) = A324823(n).
A001065(a(n)) = A324865(n) = A323243(n) - a(n),
A318456(a(n)) = A324866(n) = A324865(n) OR a(n),
A318457(a(n)) = A324867(n) = A324865(n) XOR a(n),
A318458(a(n)) = A324398(n) = A324865(n) AND a(n),
A318466(a(n)) = A324819(n) = A323243(n) OR 2*a(n),
A318467(a(n)) = A324713(n) = A323243(n) XOR 2*a(n),
A318468(a(n)) = A324815(n) = A323243(n) AND 2*a(n).
(End)

Extensions

More terms from Antti Karttunen, Jun 28 2014

A283477 If 2n = 2^e1 + 2^e2 + ... + 2^ek [e1 .. ek distinct], then a(n) = A002110(e1) * A002110(e2) * ... * A002110(ek).

Original entry on oeis.org

1, 2, 6, 12, 30, 60, 180, 360, 210, 420, 1260, 2520, 6300, 12600, 37800, 75600, 2310, 4620, 13860, 27720, 69300, 138600, 415800, 831600, 485100, 970200, 2910600, 5821200, 14553000, 29106000, 87318000, 174636000, 30030, 60060, 180180, 360360, 900900, 1801800, 5405400, 10810800, 6306300, 12612600, 37837800, 75675600
Offset: 0

Views

Author

Antti Karttunen, Mar 16 2017

Keywords

Comments

a(n) = Product of distinct primorials larger than one, obtained as Product_{i} A002110(1+i), where i ranges over the zero-based positions of the 1-bits present in the binary representation of n.
This sequence can be represented as a binary tree. Each child to the left is obtained as A283980(k), and each child to the right is obtained as 2*A283980(k), when their parent contains k:
1
|
...................2....................
6 12
30......../ \........60 180......../ \......360
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
210 420 1260 2520 6300 12600 37800 75600
etc.

Crossrefs

Programs

  • Mathematica
    Table[Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e == 1 :> {Times @@ Prime@ Range@ PrimePi@ p, e}] &[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2]], {n, 0, 43}] (* Michael De Vlieger, Mar 18 2017 *)
  • PARI
    A283477(n) = prod(i=0,exponent(n),if(bittest(n,i),vecprod(primes(1+i)),1)) \\ Edited by M. F. Hasler, Nov 11 2019
    
  • Python
    from sympy import prime, primerange, factorint
    from operator import mul
    from functools import reduce
    def P(n): return reduce(mul, [i for i in primerange(2, n + 1)])
    def a108951(n):
        f = factorint(n)
        return 1 if n==1 else reduce(mul, [P(i)**f[i] for i in f])
    def a019565(n): return reduce(mul, (prime(i+1) for i, v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1 # after Chai Wah Wu
    def a(n): return a108951(a019565(n))
    print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 22 2017
    
  • Python
    from sympy import primorial
    from math import prod
    def A283477(n): return prod(primorial(i) for i, b in enumerate(bin(n)[:1:-1],1) if b =='1') # Chai Wah Wu, Dec 08 2022
  • Scheme
    (define (A283477 n) (A108951 (A019565 n)))
    ;; Recursive "binary tree" implementation, using memoization-macro definec:
    (definec (A283477 n) (cond ((zero? n) 1) ((even? n) (A283980 (A283477 (/ n 2)))) (else (* 2 (A283980 (A283477 (/ (- n 1) 2)))))))
    

Formula

a(0) = 1; a(2n) = A283980(a(n)), a(2n+1) = 2*A283980(a(n)).
Other identities. For all n >= 0 (or for n >= 1):
a(2n+1) = 2*a(2n).
a(n) = A108951(A019565(n)).
A097248(a(n)) = A283475(n).
A007814(a(n)) = A051903(a(n)) = A000120(n).
A001221(a(n)) = A070939(n).
A001222(a(n)) = A029931(n).
A048675(a(n)) = A005187(n).
A248663(a(n)) = A006068(n).
A090880(a(n)) = A283483(n).
A276075(a(n)) = A283984(n).
A276085(a(n)) = A283985(n).
A046660(a(n)) = A124757(n).
A056169(a(n)) = A065120(n). [seems to be]
A005361(a(n)) = A284001(n).
A072411(a(n)) = A284002(n).
A007913(a(n)) = A284003(n).
A000005(a(n)) = A284005(n).
A324286(a(n)) = A324287(n).
A276086(a(n)) = A324289(n).
A267263(a(n)) = A324341(n).
A276150(a(n)) = A324342(n). [subsequences in the latter are converging towards this sequence]
G.f.: Product_{k>=0} (1 + prime(k + 1)# * x^(2^k)), where prime()# = A002110. - Ilya Gutkovskiy, Aug 19 2019

Extensions

More formulas and the binary tree illustration added by Antti Karttunen, Mar 19 2017
Four more linking formulas added by Antti Karttunen, Feb 25 2019

A246660 Run Length Transform of factorials.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 24, 1, 1, 1, 2, 1, 1, 2, 6, 2, 2, 2, 4, 6, 6, 24, 120, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 24, 2, 2, 2, 4, 2, 2, 4, 12, 6, 6, 6, 12, 24, 24, 120, 720, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 24, 1, 1, 1
Offset: 0

Views

Author

Peter Luschny, Sep 07 2014

Keywords

Comments

For the definition of the Run Length Transform see A246595.
Only Jordan-Polya numbers (A001013) are terms of this sequence.

Crossrefs

Cf. A003714 (gives the positions of ones).
Run Length Transforms of other sequences: A001316, A071053, A227349, A246588, A246595, A246596, A246661, A246674.

Programs

  • Mathematica
    Table[Times @@ (Length[#]!&) /@ Select[Split[IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 83}] (* Jean-François Alcover, Jul 11 2017 *)
  • PARI
    A246660(n) = { my(i=0, p=1); while(n>0, if(n%2, i++; p = p * i, i = 0); n = n\2); p; };
    for(n=0, 8192, write("b246660.txt", n, " ", A246660(n)));
    \\ Antti Karttunen, Sep 08 2014
    
  • Python
    from operator import mul
    from functools import reduce
    from re import split
    from math import factorial
    def A246660(n):
        return reduce(mul,(factorial(len(d)) for d in split('0+',bin(n)[2:]) if d)) if n > 0 else 1 # Chai Wah Wu, Sep 09 2014
  • Sage
    def RLT(f, size):
        L = lambda n: [a for a in Integer(n).binary().split('0') if a != '']
        return [mul([f(len(d)) for d in L(n)]) for n in range(size)]
    A246660_list = lambda len: RLT(factorial, len)
    A246660_list(88)
    
  • Scheme
    ;; A stand-alone loop version, like the Pari-program above:
    (define (A246660 n) (let loop ((n n) (i 0) (p 1)) (cond ((zero? n) p) ((odd? n) (loop (/ (- n 1) 2) (+ i 1) (* p (+ 1 i)))) (else (loop (/ n 2) 0 p)))))
    ;; One based on given recurrence, utilizing memoizing definec-macro from my IntSeq-library:
    (definec (A246660 n) (cond ((zero? n) 1) ((even? n) (A246660 (/ n 2))) (else (* (A007814 (+ n 1)) (A246660 (/ (- n 1) 2))))))
    ;; Yet another implementation, using fold:
    (define (A246660 n) (fold-left (lambda (a r) (* a (A000142 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
    (definec (A000142 n) (if (zero? n) 1 (* n (A000142 (- n 1)))))
    ;; Other functions are as in A227349 - Antti Karttunen, Sep 08 2014
    

Formula

a(2^n-1) = n!.
a(0) = 1, a(2n) = a(n), a(2n+1) = a(n) * A007814(2n+2). - Antti Karttunen, Sep 08 2014
a(n) = A112624(A005940(1+n)). - Antti Karttunen, May 29 2017
a(n) = A323505(n) / A323506(n). - Antti Karttunen, Jan 17 2019

A322827 A permutation of A025487: Sequence of least representatives of distinct prime signatures obtained from the run lengths present in the binary expansion of n.

Original entry on oeis.org

1, 2, 6, 4, 36, 30, 12, 8, 216, 180, 210, 900, 72, 60, 24, 16, 1296, 1080, 1260, 5400, 44100, 2310, 6300, 27000, 432, 360, 420, 1800, 144, 120, 48, 32, 7776, 6480, 7560, 32400, 264600, 13860, 37800, 162000, 9261000, 485100, 30030, 5336100, 1323000, 69300, 189000, 810000, 2592, 2160, 2520, 10800, 88200, 4620, 12600
Offset: 0

Views

Author

Antti Karttunen, Jan 16 2019

Keywords

Comments

A101296(a(n)) gives a permutation of natural numbers.

Examples

			The sequence can be represented as a binary tree:
                                      1
                                      |
                   ...................2...................
                  6                                       4
       36......../ \........30                 12......../ \........8
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
   216      180         210    900         72       60         24       16
etc.
Both children are multiples of their common parent, see A323503, A323504 and A323507.
The value of a(n) is computed from the binary expansion of n as follows: Starting from the least significant end of the binary expansion of n (A007088), we record the successive run lengths, subtract one from all lengths except the first one, and use the reversed partial sums of these adjusted values as the exponents of successive primes.
For 11, which is "1011" in base 2, we have run lengths [2, 1, 1] when scanned from the right, and when one is subtracted from all except the first, we have [2, 0, 0], partial sums of which is [2, 2, 2], which stays same when reversed, thus a(11) = 2^2 * 3^2 * 5^2 = 900.
For 13, which is "1101" in base 2, we have run lengths [1, 1, 2] when scanned from the right, and when one is subtracted from all except the first, we have [1, 0, 1], partial sums of which is [1, 1, 2], reversed [2, 1, 1], thus a(13) = 2^2 * 3^1 * 5^1 = 60.
Sequence A227183 is based on the same algorithm.
		

Crossrefs

Cf. A000079 (right edge), A000400 (left edge, apart from 2), A005811, A046523, A101296, A227183, A322585, A322825, A323503, A323504, A323507.
Other rearrangements of A025487 include A036035, A059901, A063008, A077569, A085988, A086141, A087443, A108951, A181821, A181822.
Cf. A005940, A283477, A323505 for other similar trees.

Programs

  • Mathematica
    {1}~Join~Array[Times @@ MapIndexed[Prime[First@ #2]^#1 &, Reverse@ Accumulate@ MapIndexed[Length[#1] - Boole[First@ #2 > 1] &, Split@ Reverse@ IntegerDigits[#, 2]]] &, 54] (* Michael De Vlieger, Feb 05 2020 *)
  • PARI
    A322827(n) = if(!n,1,my(bits = Vecrev(binary(n)), rl=1, o = List([])); for(i=2,#bits,if(bits[i]==bits[i-1], rl++, listput(o,rl))); listput(o,rl); my(es=Vecrev(Vec(o)), m=1); for(i=1,#es,m *= prime(i)^es[i]); (m));

Formula

a(n) = A046523(a(n)) = A046523(A322825(n)).
A001221(a(n)) = A005811(n).
A001222(a(n)) = A227183(n).
A322585(a(n)) = 1.

A106831 Define a triangle in which the entries are of the form +-1/(b!c!d!e!...), where the order of the factorials is important; read the triangle by rows and record and expand the denominators.

Original entry on oeis.org

2, 6, 4, 24, 12, 12, 8, 120, 48, 36, 24, 48, 24, 24, 16, 720, 240, 144, 96, 144, 72, 72, 48, 240, 96, 72, 48, 96, 48, 48, 32, 5040, 1440, 720, 480, 576, 288, 288, 192, 720, 288, 216, 144, 288, 144, 144, 96, 1440, 480, 288, 192, 288, 144, 144, 96, 480, 192, 144, 96, 192
Offset: 0

Views

Author

N. J. A. Sloane, May 22 2005

Keywords

Comments

Row n has 2^n terms. Row 0 is +1/2!. An entry +-1/b!c!d!... has two children, a left child -+1/(a+1)!b!c!... and a right child +-1/2!b!c!d!...
Let S_n = sum of entries in row n of the triangle. Then for n > 0, n!S_{n-1} is the Bernoulli number B_n.

Examples

			Woon's "Bernoulli Tree" begins like this (see also the given Wikipedia-link). This sequence gives the values of the denominators:
                                     +1
                                    ────
                                     2!
                 -1                 /  \                  +1
                ──── ............../    \.............. ─────
                 3!                                      2!2!
        +1        .         -1                 -1         .         +1
       ────      / \       ────               ────       / \      ──────
        4! ...../   \..... 2!3!               3!2! ...../   \.... 2!2!2!
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
    -1      +1         +1       -1         +1      -1          -1       +1
   ────    ────       ────     ──────     ────   ──────      ──────  ────────
    5!     2!4!       3!3!     2!2!3!     4!2!   2!3!2!      3!2!2!  2!2!2!2!
etc.
		

Crossrefs

Cf. A242179 (numerators), A050925, A050932, A000142.
Cf. A323505 (mirror image), and also A005940, A283477, A322827 for other similar trees.

Programs

  • Haskell
    a106831 n k = a106831_tabf !! n !! n
    a106831_row n = a106831_tabf !! n
    a106831_tabf = map (map (\(, , left, right) -> left * right)) $
       iterate (concatMap (\(x, f, left, right) -> let f' = f * x in
       [(x + 1, f', f', right), (3, 2, 2, left * right)])) [(3, 2, 2, 1)]
    -- Reinhard Zumkeller, May 05 2014
    
  • Maple
    Contribution from Peter Luschny, Jun 12 2009: (Start)
    The routine computes the triangle row by row and gives the numbers with their sign.
    Thus A(1)=[2]; A(2)=[ -6,4]; A(3)=[24,-12,-12,8]; etc.
    A := proc(n) local k, i, j, m, W, T; k := 2;
    W := array(0..2^n); W[1] := [1,`if`(n=0,1,2)];
    for i from 1 to n-1 do for m from k by 2 to 2*k-1 do
    T := W[iquo(m,2)]; W[m] := [ -T[1],T[2]+1,seq(T[j],j=3..nops(T))];
    W[m+1] := [T[1],2,seq(T[j],j=2..nops(T))]; od; k := 2*k; od;
    seq(W[i][1]*mul(W[i][j]!,j=2..nops(W[i])),i=iquo(k,2)..k-1) end:
    seq(print(A(i)),i=1..5); (End)
  • Mathematica
    a [n_] := Module[{k, i, j, m, w, t}, k = 2; w = Array[0&, 2^n]; w[[1]] := {1, If[n == 0, 1, 2]}; For[i = 1, i <= n-1, i++, For[m = k, m <= 2*k-1 , m = m+2, t = w[[Quotient[m, 2]]]; w[[m]] = {-t[[1]], t[[2]]+1, Sequence @@ Table[t[[j]], {j, 3, Length[t]}]}; w[[m+1]] = {t[[1]], 2, Sequence @@ Table[t[[j]], {j, 2, Length[t]}]}]; k = 2*k]; Table[w[[i, 1]]*Product[w[[i, j]]!, {j, 2, Length[w[[i]]]}], {i, Quotient[k, 2], k-1}]]; Table[a[i] , {i, 1, 6}] // Flatten // Abs (* Jean-François Alcover, Dec 20 2013, translated from Maple *)
  • PARI
    A106831off1(n) = if(!n,1, my(rl=1, m=1); while(n,if(!(n%2), rl++, m *= ((1+rl)!); rl=1); n >>= 1); (m));
    A106831(n) = A106831off1(1+n); \\ Antti Karttunen, Jan 16 2019
    
  • PARI
    A001511(n) = (1+valuation(n,2));
    A106831r1(n) = if(!n,1,if(n%2, 2*A106831r1((n-1)/2), (1+A001511(n))*A106831r1(n/2))); \\ Implements the given recurrence.
    A106831(n) = A106831r1(1+n); \\ Antti Karttunen, Jan 16 2019

Formula

From Antti Karttunen, Jan 16 2019: (Start)
If sequence is shifted one term to the right, then the following recurrence works:
a(0) = 1; and for n > 0, a(2n) = (1+A001511(2n))*a(n), a(2n+1) = 2*a(n).
(End)

Extensions

More terms from Franklin T. Adams-Watters, Apr 28 2006
Example section reillustrated by Antti Karttunen, Jan 16 2019
Showing 1-8 of 8 results.