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 1-4 of 4 results.

A073904 Smallest multiple k*n of n having n divisors.

Original entry on oeis.org

1, 2, 9, 8, 625, 12, 117649, 24, 36, 80, 25937424601, 60, 23298085122481, 448, 2025, 384, 48661191875666868481, 180, 104127350297911241532841, 240, 35721, 11264, 907846434775996175406740561329, 360, 10000, 53248, 26244, 1344
Offset: 1

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Author

Amarnath Murthy, Aug 18 2002

Keywords

Comments

Smallest refactorable number, m, such that m=k*n has n divisors. - Robert G. Wilson v, Oct 31 2005

Examples

			Smallest multiple a(n)=k*n; a(1)=1*1, a(2)=1*2, a(3)=3*3, a(4)=2*4, a(5)=125*5, a(6)=2*6, ... having d(k*n)=n divisors; d(1)=1, d(2)=2, d(3^2)=3, d(2^3)=4, d(5^4)=5, d(2^2*3)=3*2=6, ...
		

Crossrefs

Cf. A033950 (refactorable numbers, also known as tau numbers).
Cf. A110821 (SuperRefactorable numbers).

Programs

  • Mathematica
    f[n_] := Block[{k = 1}, If[ PrimeQ[n], n^(n - 1), While[d = DivisorSigma[0, k*n]; d != n, k++ ]; k*n]]; Table[ f[n], {n, 28}] (* Robert G. Wilson v *)

Formula

If p is a prime then a(p) = p^(p-1). If n = p^2 then a(n) = 2^(p-1)*p^(p-1).
a(p^r) = (2*3*5*...*p_r)^(p-1) for r < p <= p_r. a(p^r) = (2*3*...*p_(r-1))^(p-1)*p^(p-1) for p > p_r. Else a(p^r) = ...? for r >= p. Problem a(2^r) = ...? Cf. A005179(p^n)=(2*3*...*p_n)^(p-1) for p_n < 2^p. - Thomas Ordowski, Aug 20 2005
a(p^r) = (2*3...*p_(r-1)*p)^(p-1) for p > p_r; else a(p^r) = (2*3...*p...*p_m)^(p-1)*p^(p^k-p) for p <= p_r and p_m < 2^p, where m=r-k+1 for smallest k such that p^k > r, so k=floor(log(r)/log(p))+1 and p > log(p_m)/log(2). Examples: If k=1 then a(p^r) = (2*3*...*p_r)^(p-1) for r < p <= p_r. If p=2 then a(2^r) = (2*3*...*p_m)*2^(2^k-2) for r < 5. For instance, let r=4 so k=3, m=2 and a(2^4)=384. - Thomas Ordowski, Aug 22 2005
If p is a prime and n=p^r then a(p^r) = (s_1*s_2*...*s_r)^(p-1) where (s_r) is a permutation of the (ascending sequence) numbers of the form q^(p^j) for every prime q and j>=0; permutation such that s_(p^j)=p^(p^j) and shifted remainder. For example, if p=3 then (s_r): 3, 2, 3^3, 5, 7, 2^3, 11, 13, 3^9, 17, 19, ... so a(3^r) = (3*2*27*5*...*s_r)^2. - Thomas Ordowski, Aug 29 2005
If n=2^r then a(2^r) is the product of the first r members of the A109429 sequence. - Thomas Ordowski, Aug 29 2005
a(n) = n * A076931(n). - Thomas Ordowski, Oct 07 2005
a(4) = 8; a(2*prime(n)) = A299795(n), for n>1. - Bernard Schott, Nov 06 2022

Extensions

a(12) corrected by Thomas Ordowski, Aug 18 2005
Further corrections from Thomas Ordowski, Oct 07 2005
a(21), a(27) & a(28) from Robert G. Wilson v, Oct 31 2005

A117897 Number of labeled trees on prime numbers of nodes through n-th prime.

Original entry on oeis.org

1, 4, 129, 16936, 2357964627, 1794518358664, 2862424846028174457, 5483249282630830360396, 39471589603944768518079950019, 3053134546009996125349281528007992109928
Offset: 1

Views

Author

Jonathan Vos Post, May 03 2006

Keywords

Comments

A000178 = Sum_{k=1..n} k^(k-1). A001923 = Sum_{k=1..n} k^k. A061789 = Sum_{k=1..n} prime(k)^prime(k), prime(k) = k-th prime.
First differences a(n+1) - a(n) for n=1,...,9 are A076931(j) at j=3, 5, 7, 11, 13, 17, 19, 23 and 29. - R. J. Mathar, May 01 2007

Examples

			a(1) = number of labeled trees on prime(1) numbers of nodes = number of labeled trees on 2 nodes = A000272(2) = 2^0 = 1.
a(2) = number of labeled trees on prime(1) or prime(2) numbers of nodes = number of labeled trees on 2 or 3 nodes = A000272(2)+A000272(3) = 2^0 + 3^1 = 4.
a(3) = number of labeled trees on prime(1) or prime(2) or prime(3) numbers of nodes = number of labeled trees on 2 or 3 or 5 nodes = A000272(2)+A000272(3)+A000272(5) = 2^0 + 3^1 + 5^3 = 129.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Prime[k]^(Prime[k] -2), {k,n}], {n,20}] (* G. C. Greubel, Sep 27 2021 *)
  • Sage
    [sum( nth_prime(k)^(nth_prime(k) -2) for k in (1..n)) for n in (1..20)] # G. C. Greubel, Sep 27 2021

Formula

a(n) = Sum_{k=1..n} prime(k)^(prime(k)-2).
a(n) = Sum_{k=1..n} A000272(A000040(k)).

A076932 Largest k such that n*k has n divisors, or 0 if there are no possibilities for k or infinitely many.

Original entry on oeis.org

1, 1, 3, 2, 125, 3, 16807, 0, 0, 125, 2357947691, 0, 1792160394037, 16807, 375, 0, 2862423051509815793, 0, 5480386857784802185939, 0, 50421, 2357947691, 39471584120695485887249589623, 0, 0, 1792160394037, 0, 0
Offset: 1

Views

Author

Amarnath Murthy, Oct 18 2002

Keywords

Examples

			a(6) = 3 (and not 2) as 18 has six divisors though 12 also has 6 divisors. a(8) = 0 as for every prime p > 2 8p has 8 divisors.
		

Crossrefs

Extensions

More terms from Sascha Kurz, Jan 21 2003

A275463 Least k such that n divides d(k*n) (d = A000005).

Original entry on oeis.org

1, 1, 3, 2, 16, 2, 64, 3, 4, 8, 1024, 5, 4096, 32, 48, 24, 65536, 10, 262144, 12, 192, 512, 4194304, 15, 400, 2048, 900, 48, 268435456, 24, 1073741824, 60, 3072, 32768, 5184, 35, 68719476736, 131072, 12288, 42, 1099511627776, 96, 4398046511104
Offset: 1

Views

Author

Altug Alkan, Jul 28 2016

Keywords

Examples

			a(5) = 16 because 5 divides A000005(16*5) = 10.
a(40) = 42 because 40 divides A000005(42*40) = 40.
		

Crossrefs

Programs

  • PARI
    a(n) = if(isprime(n) && n>3, 2^(n-1), {my(k=1); while(numdiv(k*n) % n != 0, k++); k; })
Showing 1-4 of 4 results.