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.

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A061799 Smallest number with at least n divisors.

Original entry on oeis.org

1, 2, 4, 6, 12, 12, 24, 24, 36, 48, 60, 60, 120, 120, 120, 120, 180, 180, 240, 240, 360, 360, 360, 360, 720, 720, 720, 720, 720, 720, 840, 840, 1260, 1260, 1260, 1260, 1680, 1680, 1680, 1680, 2520, 2520, 2520, 2520, 2520, 2520, 2520, 2520, 5040, 5040, 5040
Offset: 1

Views

Author

Henry Bottomley, Jun 22 2001

Keywords

Comments

Smallest number which can be expressed as the least common multiple of n distinct numbers. - Amarnath Murthy, Nov 27 2002
Also smallest possible member of a set of n+1 numbers with pairwise distinct GCD's. [Following an observation by Charles R Greathouse IV] (Proof: If the smallest number min(S) of the set (with card(S)=n+1) has a distinct GCD with each of the other n numbers, then it must have at least n distinct divisors (because any GCD is a divisor). It is then easy to choose larger members of the set so that all pairs of elements have pairwise distinct GCD's, e.g., by successively multiplying by distinct and sufficiently large primes.) - M. F. Hasler, Mar 05 2013

Examples

			a(5)=12 since every number less than 12 has fewer than five divisors (1 has one; 2,3,5,7 and 11 have two each; 4 and 9 have three each; 6,8 and 10 have four each) while 12 has at least five (in fact it has six: 1,2,3,4,6 and 12).
		

Crossrefs

Programs

  • Haskell
    import Data.List (findIndex)
    import Data.Maybe (fromJust)
    a061799 n = succ $ fromJust $ findIndex (n <=) $ map a000005 [1..]
    -- Reinhard Zumkeller, Apr 01 2011
  • Mathematica
    Reap[ For[ n = 1, n <= 100, n++, s = n; While[ DivisorSigma[0, s] < n, s++]; Sow[s] ] ][[2, 1]] (* Jean-François Alcover, Feb 16 2012, after Pari *)
    With[{ds=Table[{n,DivisorSigma[0,n]},{n,6000}]},Table[SelectFirst[ds,#[[2]] >= k&],{k,60}]][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 15 2019 *)
  • PARI
    for(n=1,100,s=n; while(numdiv(s)
    				

Extensions

Replaced "factors" by "divisors" in definition and example M. F. Hasler, Oct 24 2010

A262509 Numbers n such that there is no other number k for which A155043(k) = A155043(n).

Original entry on oeis.org

0, 119143, 119147, 119163, 119225, 119227, 119921, 119923, 120081, 120095, 120097, 120101, 120281, 120293, 120349, 120399, 120707, 120747, 120891, 120895, 120903, 120917, 120919, 121443, 121551, 121823, 122079, 122261, 122263, 122273, 122277, 122813, 122961, 123205, 123213, 123223, 123237, 123257, 123765, 24660543, 24660549, 24662311, 24662329, 24663759, 24664997, 24665023, 24665351
Offset: 0

Views

Author

Antti Karttunen, Sep 25 2015

Keywords

Comments

Starting offset is zero, because a(0) = 0 is a special case in this sequence.
Numbers where A155043 takes a unique value. (Those values are given by A262508.)
Numbers n such that there does not exist any other number h from which one could reach zero in exactly the same number of steps as from n by repeatedly applying the map where k is replaced by k - A000005(k) = A049820(k). Thus in the tree where zero is the root and parent-child relation is given by A049820(child) = parent, all the numbers > n+t (where t is a small value depending on n) have n as their common ancestor. As it is guaranteed that there is at least one infinite path in such a tree, any n in this sequence can be neither a leaf nor any other vertex in a finite side-tree, as then at least one node in the infinite part would have the same distance to the root, thus it must be that n itself is in the infinite part and thus has an infinite number of descendant vertices. Also, for the same reason, the tree cannot branch to two infinite parts from any ancestor of n (which are nodes nearer to the root of the tree, zero).
From the above it follows that if this sequence is infinite, then A259934 is guaranteed to be the only infinite sequence which starts with a(0)=0 and satisfies the condition A049820(a(k)) = a(k-1) for all k>=1, where A049820(n) = n-d(n) and d(n) is the number of divisors of n (A000005). This is a sufficient condition for the uniqueness of A259934, although not necessary. See e.g. A179016 which is the unique infinite solution to a similar problem, even though A086876 contains no ones after its two initial terms.
Is it possible for any even terms to occur after zero? If not, then apart from zero, this would be a subsequence of A262517.

Crossrefs

Subsequence of A259934, A261089, A262503 and A262513.
Cf. A262510 (gives the parent nodes for the terms a(1) onward), A262514, A262516, A262517.
Cf. also A086876, A179016.

Programs

  • PARI
    \\ Compute A262508 and A262509 at the same time:
    allocatemem((2^31)+(2^30));
    \\ The limits used are quite ad hoc. Beware of the horizon-effect if you change these.
    \\ As a post-check, test that A262509(n) = A259934(A262508(n)) for all the terms produced by this program.
    uplim1 = 43243200 + 672; \\ = A002182(54) + A002183(54).
    uplim2 = 36756720; \\ = A002182(53).
    uplim3 = 10810800; \\
    v155043 = vector(uplim1);
    v262503 = vector(uplim3);
    v262507 = vector(uplim3);
    v155043[1] = 1; v155043[2] = 1;
    for(i=3, uplim1, v155043[i] = 1 + v155043[i-numdiv(i)]);
    A155043 = n -> if(!n,n,v155043[n]);
    for(i=1, uplim1, v262503[v155043[i]] = i; v262507[v155043[i]]++; if(!(i%1048576),print1(i,", ")));
    A262503 = n -> if(!n,n,v262503[n]);
    A262507 = n -> if(!n,1,v262507[n]);
    k=0; for(n=0, uplim3, if((1==A262507(n)) && (A262503(n) <= uplim2), write("b262508.txt", k, " ", n); write("b262509.txt", k, " ", A262503(n)); k++));
    
  • Scheme
    (define (A262509 n) (A261089 (A262508 n)))

Formula

a(n) = A261089(A262508(n)) = A262503(A262508(n)) = A259934(A262508(n)).

A329605 Number of divisors of A108951(n), where A108951 is fully multiplicative with a(prime(i)) = prime(i)# = Product_{i=1..i} A000040(i).

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 16, 4, 9, 12, 32, 8, 64, 24, 18, 5, 128, 12, 256, 16, 36, 48, 512, 10, 27, 96, 16, 32, 1024, 24, 2048, 6, 72, 192, 54, 15, 4096, 384, 144, 20, 8192, 48, 16384, 64, 32, 768, 32768, 12, 81, 36, 288, 128, 65536, 20, 108, 40, 576, 1536, 131072, 30, 262144, 3072, 64, 7, 216, 96, 524288, 256, 1152, 72, 1048576, 18, 2097152, 6144, 48
Offset: 1

Views

Author

Antti Karttunen, Nov 18 2019

Keywords

Crossrefs

Cf. A329606 (rgs-transform), A329608, A331284 (ordinal transform).
Cf. A331285 (the position where for the first time some term has occurred n times in this sequence).

Programs

  • Mathematica
    Block[{a}, a[n_] := a[n] = Module[{f = FactorInteger[n], p, e}, If[Length[f] > 1, Times @@ a /@ Power @@@ f, {{p, e}} = f; Times @@ (Prime[Range[PrimePi[p]]]^e)]]; a[1] = 1; Array[DivisorSigma[0, a@ #] &, 75]] (* Michael De Vlieger, Jan 24 2020, after Jean-François Alcover at A108951 *)
  • PARI
    A034386(n) = prod(i=1, primepi(n), prime(i));
    A108951(n) = { my(f=factor(n)); prod(i=1, #f~, A034386(f[i, 1])^f[i, 2]) };  \\ From A108951
    A329605(n) = numdiv(A108951(n));
    
  • PARI
    A329605(n) = if(1==n,1,my(f=factor(n),e=1,m=1); forstep(i=#f~,1,-1, e += f[i,2]; m *= e^(primepi(f[i,1])-if(1==i,0,primepi(f[i-1,1])))); (m)); \\ Antti Karttunen, Jan 14 2020
    
  • PARI
    A329605(n) = if(1==n,1,my(f=factor(n),e=0,d); forstep(i=#f~,1,-1, e += f[i,2]; d = (primepi(f[i,1])-if(1==i,0,primepi(f[i-1,1]))); f[i,1] = (e+1); f[i,2] = d); factorback(f)); \\ Antti Karttunen, Jan 14 2020

Formula

a(n) = A000005(A108951(n)).
a(n) >= A329382(n) >= A329617(n) >= A329378(n).
A020639(a(n)) = A329614(n).
From Antti Karttunen, Jan 14 2020: (Start)
a(A052126(n)) = A329382(n).
a(A002110(n)) = A000142(1+n), for all n >= 0.
a(n) > A056239(n).
a(A329902(n)) = A002183(n).
A000265(a(n)) = A331286(n).
gcd(n,a(n)) = A331283(n).
If n = p(k1)^e(k1) * p(k2)^e(k2) * p(k3)^e(k3) * ... * p(kx)^e(kx), with p(n) = A000040(n) and k1 > k2 > ... > kx, then a(n) = (1+e(k1))^(k1-k2) * (1+e(k1)+e(k2))^(k2-k3) * ... * (1+e(k1)+e(k2)+...+e(kx))^kx.
A000035(a(n)) = A000035(A000005(n)) = A010052(n).
(End)

A329902 Primorial deflation of the n-th highly composite number: the unique integer k such that A108951(k) = A002182(n).

Original entry on oeis.org

1, 2, 4, 3, 6, 12, 9, 24, 10, 20, 15, 40, 30, 60, 28, 21, 56, 42, 84, 63, 168, 126, 336, 140, 66, 189, 280, 132, 99, 264, 198, 528, 220, 396, 297, 440, 792, 156, 117, 312, 234, 624, 260, 468, 351, 520, 936, 390, 1040, 1872, 780, 585, 306, 1560, 340, 612, 459, 680, 1224, 510, 1360, 2448, 1020, 765, 342, 2040, 1530, 684, 513
Offset: 1

Views

Author

Antti Karttunen, Dec 22 2019

Keywords

Crossrefs

Programs

  • Mathematica
    Map[Times @@ Prime@(TakeWhile[Reap[FixedPointList[Block[{k = 1}, While[Mod[#, Prime@ k] == 0, k++]; Sow[k - 1]; #/Product[Prime@ i, {i, k - 1}]] &, #]][[-1, 1]], # > 0 &]) &, Take[Import["https://oeis.org/b002182.txt", "Data"][[All, -1]], 69] ] (* Michael De Vlieger, Jan 13 2020, imports b-file at A002182 *)

Formula

a(n) = A329900(A002182(n)) = A319626(A002182(n)).
a(n) = A181815(A306802(n)).
A108951(a(n)) = A002182(n). [Highly composite numbers (undeflated)]
A056239(a(n)) = A112778(n). [Number of prime factors, counted with multiplicity]
A001222(a(n)) = A112779(n). [Largest exponent in the prime factorization]
A329605(a(n)) = A002183(n). [Number of divisors]
A329040(a(n)) = A324381(n).
A324888(a(n)) = A324382(n).
a(A330748(n)) = A330743(n).

Extensions

More linking formulas added by Antti Karttunen, Jan 13 2020

A108602 Number of distinct prime factors of highly composite numbers (definition 1, A002182).

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 7, 7, 8, 8, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 8, 9, 8, 9, 8, 9, 9, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9
Offset: 1

Views

Author

Jud McCranie, Jun 12 2005

Keywords

Comments

n appears A086334(n) times. - Lekraj Beedassy, Sep 02 2006

Examples

			A002182(8) = 48 = 2^4*3, which has 2 distinct prime factors, so a(8)=2.
		

Crossrefs

Formula

a(n) = A001221(A002182(n)).

Extensions

Edited by Ray Chandler, Nov 11 2005

A112778 Number of prime factors (counted with multiplicity) of highly composite numbers (definition 1, A002182).

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 4, 5, 4, 5, 5, 6, 6, 7, 6, 6, 7, 7, 8, 8, 9, 9, 10, 9, 8, 10, 10, 9, 9, 10, 10, 11, 10, 11, 11, 11, 12, 10, 10, 11, 11, 12, 11, 12, 12, 12, 13, 12, 13, 14, 13, 13, 12, 14, 12, 13, 13, 13, 14, 13, 14, 15, 14, 14, 13, 15, 15, 14, 14, 16, 14, 15, 14, 15, 16, 15, 15, 16
Offset: 1

Views

Author

Ray Chandler, Nov 11 2005

Keywords

Comments

The values of this sequence oscillate around a slowly increasing moving average, with an amplitude roughly equal to log(a(n)): Records 1, 2, 3, ... of max(a(1..n)) - a(n) are reached at n = (9, 25, 11, 307, 1201, 7140, ...) where a(n) = (4, 8, 18, 31, 64, 169, 175, ...). - M. F. Hasler, Jan 08 2020

Examples

			A002182(8) = 48 = 2^4*3, which has 5 prime factors, counted with multiplicity, so a(8)=5.
		

Crossrefs

Programs

  • PARI
    A112778(n)=bigomega(A002182(n)) \\ or A112778(n)=v112778[n] (e.g., from b-file)
    /* To list the records of max(a(1..n)) - a(n): */
    m=r=0; for(i=1,1e4, if(mA112778(i), m=n, m-n>r, print1([i,n,r=m-n]",")))
    \\ M. F. Hasler, Jan 08 2020

Formula

a(n) = A001222(A002182(n)).

A261100 a(n) is the greatest m for which A002182(m) <= n; the least monotonic left inverse for highly composite numbers A002182.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Sep 24 2015

Keywords

Comments

Each n occurs A262501(n) times.
This is the only sequence w, which (1) satisfies w(A002182(n)) = n for all n >= 1 (thus is a left inverse of A002182), which (2) is monotonic (by necessity growing, although not strictly so), and which (3) is the lexicographically least of all sequences satisfying both (1) and (2). In other words, the largest number m for which A002182(m) <= n. - Antti Karttunen, Jun 06 2017

Crossrefs

Programs

  • Maple
    with(numtheory):
    A261100_list := proc(len) local n, k, j, b, A, tn: A := NULL; k := 0;
    for n from 1 to len do
        b := true; tn := tau(n);
        for j from 1 to n-1 while b do b := b and tau(j) < tn od:
        if b then k := k + 1 fi;
        A := A,k
    od: A end: A261100_list(120); # Peter Luschny, Jun 06 2017
  • Mathematica
    A002182 = Import["https://oeis.org/A002182/b002182.txt", "Table"];
    inter = Interpolation[Reverse /@ A002182, InterpolationOrder -> 0];
    A261100 = Rest[inter /@ Range[200]] - 1 (* Jean-François Alcover, Oct 25 2019 *)
  • PARI
    v002182 = vector(1000); v002182[1] = 1; \\ For memoization.
    A002182(n) = { my(d,k); if(v002182[n],v002182[n], k = A002182(n-1); d = numdiv(k); while(numdiv(k) <= d, k=k+1); v002182[n] = k; k); };
    A261100(n) = { my(k=1); while(A002182(k)<=n,k=k+1); (k-1); } \\ Antti Karttunen, Jun 06 2017
    
  • Scheme
    (define (A261100 n) (let loop ((k 1)) (if (> (A002182 k) n) (- k 1) (loop (+ 1 k)))))
    
  • Scheme
    ;; Requires Antti Karttunen's IntSeq-library.
    (define A261100 (LEFTINV-LEASTMONO 1 1 A002182))

Formula

a(n) = the least k for which A002182(k+1) > n.
Other identities. For all n >= 1:
a(A002182(n)) = n. [The least monotonic sequence satisfying this condition.]
A070319(n) = A002183(a(n)).

Extensions

Description clarified by Antti Karttunen, Jun 06 2017

A262511 Numbers k for which there is exactly one solution to x - d(x) = k, where d(k) is the number of divisors of k (A000005). Positions of ones in A060990.

Original entry on oeis.org

2, 3, 4, 5, 9, 10, 12, 14, 15, 16, 18, 21, 23, 26, 30, 31, 32, 41, 42, 44, 45, 47, 53, 54, 59, 60, 61, 71, 72, 73, 76, 77, 80, 82, 83, 84, 86, 89, 90, 92, 93, 94, 95, 97, 99, 101, 104, 105, 106, 110, 115, 119, 121, 122, 127, 135, 139, 146, 148, 149, 151, 154, 158, 161, 169, 171, 173, 176, 177, 183, 186, 188, 189, 190, 191, 192, 194, 195, 199, 200, 202
Offset: 1

Views

Author

Antti Karttunen, Sep 25 2015

Keywords

Crossrefs

Cf. A262512 (gives the corresponding x).
Cf. A262510 (a subsequence).
Subsequence of A236562.

Programs

  • PARI
    allocatemem(123456789);
    uplim = 14414400 + 504; \\ = A002182(49) + A002183(49).
    v060990 = vector(uplim);
    for(n=3, uplim, v060990[n-numdiv(n)]++);
    A060990 = n -> if(!n,2,v060990[n]);
    uplim2 = 14414400;
    n=0; k=1; while(n <= uplim2, if(1==A060990(n), write("b262511_big.txt", k, " ", n); k++); n++;);
    
  • Scheme
    ;; With Antti Karttunen's IntSeq-library.
    (define A262511 (ZERO-POS 1 1 (COMPOSE -1+ A060990)))

Formula

Other identities. For all n >= 1:
a(n) = A049820(A262512(n)).

A046952 Sets record for f(n) = |{(a,b):a*b=n and a|b}|. Also squares of highly composite numbers A002182.

Original entry on oeis.org

1, 4, 16, 36, 144, 576, 1296, 2304, 3600, 14400, 32400, 57600, 129600, 518400, 705600, 1587600, 2822400, 6350400, 25401600, 57153600, 101606400, 228614400, 406425600, 635040000, 768398400, 2057529600, 2540160000, 3073593600
Offset: 1

Views

Author

Simon Colton (simonco(AT)cs.york.ac.uk)

Keywords

Comments

Invented by the HR automatic theory formation program.
Also, integers whose number of square divisors sets a new record. - Bernard Schott, Jan 14 2022
As a(n) is the square of n-th highly composite number (A002182), the record number of square divisors of a(n) is A046951(a(n)) = tau(A002182(n)) = A002183(n) where tau is the divisor counting function (A000005). - Bernard Schott, Jan 15 2022
Integers m for which number of solutions (A353282) to the Diophantine equation S(x,y) = (x+y) + (x-y) + (x*y) + (x/y) = m sets a new record; these records are respectively 0, 1, 2, 3, 5, 7, ... Example: the 5 solutions for S(x,y) = 144 are (36,1), (32,2), (27,3), (20,5), (11,11). - Bernard Schott, Apr 19 2022

Examples

			f(1)=1, (first with 1), f(4)=2 (first with 2), f(16)=3 (first with 3).
		

Crossrefs

Cf. A350756 (similar, with triangular divisors).

Formula

a(n) = A002182(n)^2. - Bernard Schott, Jan 14 2022

A112779 Largest exponent in the prime factorization of highly composite numbers (definition 1, A002182).

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 2, 4, 2, 3, 2, 4, 3, 4, 3, 2, 4, 3, 4, 3, 5, 4, 6, 4, 3, 4, 5, 4, 3, 5, 4, 6, 4, 5, 4, 5, 6, 4, 3, 5, 4, 6, 4, 5, 4, 5, 6, 4, 6, 7, 5, 4, 4, 6, 4, 5, 4, 5, 6, 4, 6, 7, 5, 4, 4, 6, 5, 5, 4, 6, 5, 6, 4, 6, 7, 5, 4, 6, 5, 7, 6, 5, 6, 4, 4, 6, 7, 5, 5, 4, 6, 6, 5, 7, 6, 5, 6, 4, 7, 6, 7, 5, 7, 6, 8
Offset: 1

Views

Author

Ray Chandler, Nov 11 2005

Keywords

Comments

Each highly composite number can be written as the product of primorials (A002110); a(n) is also the number of primorials used in the product.
a(i) is the exponent of 2 in the prime factorization of A002182(i), cf. formula. - David A. Corneth, Aug 16 2015; edited by M. F. Hasler, Jan 03 2020

Examples

			A002182(8) = 48 = 2^4*3, which has largest exponent 4, so a(8)=4.
		

Crossrefs

Programs

Formula

a(n) = A007814(A002182(n)). - David A. Corneth, Aug 16 2015
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