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.

A188145 Solutions of the equation n" - n' - n = 0, where n' and n" are the first and second arithmetic derivatives (see A003415).

Original entry on oeis.org

0, 20, 135, 164, 1107, 15625, 43692, 128125, 188228, 294921, 1270539, 4117715, 33765263, 34134375, 147053125, 8995560189, 19348535652, 38753462951
Offset: 1

Views

Author

Paolo P. Lava, Mar 22 2011

Keywords

Comments

Solutions of the similar equation n”-n’+n=0 are 30, 858, 1722, etc., apparently Giuga numbers whose derivative is a prime number. In fact the equation can be rewritten as n'=n+n" and if n"=1 it is the conjecture in A007850.
a(16) > 2*10^9. - Donovan Johnson, Apr 30 2011
a(19) > 10^11. - Giovanni Resta, Jun 04 2016

Examples

			n=20, n’=24, n”=44 -> 44-24-20=0;  n=135, n’=162, n”=297 -> 297-162-135=0

Crossrefs

Cf. A003415, A007850, A165558, A165561, A165562, A189639, A189710.

Programs

  • Haskell
    import Data.List (zipWith3, elemIndices)
a188145 n = a188145_list !! (n-1)
a188145_list = elemIndices 0 $ zipWith3 (\x y z -> x - y - z)
   (map a003415 a003415_list) a003415_list [0..]
-- Reinhard Zumkeller, May 10 2011
  • Maple
    readlib(ifactors):
Der:= proc(n)
local a,b,i,p,pfs;
for i from 0 to n do
  if i<=1 then a:=0;
  else pfs:=ifactors(i)[2]; a:=i*add(op(2,p)/op(1,p),p=pfs) ;
  fi;
  if a<=1 then b:=0;
  else pfs:=ifactors(a)[2]; b:=a*add(op(2,p)/op(1,p),p=pfs) ;
  fi;
  if b-a=i then lprint(i,a,b); fi;
od
end:
Der(10000000);

Extensions

a(13)-a(15) from Donovan Johnson, Apr 30 2011
Corrected a(9) and a(16)-a(18) from Giovanni Resta, Jun 04 2016

A211222 Minimum integer m for which n·m = n'·m', where n' and m' are the arithmetic derivatives of n and m. Zero if m does not exist.

Original entry on oeis.org

16, 64, 4, 1024, 20, 16384, 9, 8, 1372, 4194304, 0, 67108864, 0, 0, 2, 17179869184, 77948684, 274877906944, 6, 1000, 42417997492, 70368744177664, 1771561, 32, 300, 3125, 0, 288230376151711744, 550618520345910837374536871905139185678862401, 4611686018427387904
Offset: 2

Views

Author

Paolo P. Lava, Apr 20 2012

Keywords

Comments

The entries of the sequence are the solution of the differential equation m'=n/n'*m.
If n=n' then m'=m (A051674) and m=p^p, with p prime. Taking the minimum prime p=2, m=4.
If n'|n then m'=b*m, where b=n/n', and m=2^(2*b).
In general n and n’ can have same common factor. Let us consider the fraction k/h obtained by reduction of n/n' [GCD(k,h)=1]. If h=p, p prime, then if k>h m=4*p^(k-h) else if k
If h is composite let us make the variable substitution m=h*t. Now, m'=h'*t+h*t'=(k/h)*h*t=k*t that leads to t'=1/h*(k-h')*t. Let us reduce the fraction (k-h')/h, if necessary. If we get a prime denominator we can solve the equation by the previous formulas. Otherwise we must iterate the variable substitutions until we get:
1) a prime denominator.
2) a differential equation of the type y'=0, whose solution is y=1.
3) a differential equation of the type y'=x*y, with x<0. In this case there is no solution. This corresponds to the zeros in the sequence (12, 14, 15, 28, 35, 39, 44, 46, 50, 51, 55, 65, etc.).
In cases 1) and 2), moving backward through all the substitutions, we reach the final solution.

Examples

			n=18; n'=11. m'=18/11*m. Here n>n', n'=p and therefore m=4*11^7.
n=24; n'=44. m'=24/44*m. Simplifying the fraction we have m'=6/11*m. Here k<k' e k'=p and therefore m=116.
n=44; n'=48. m'=44/48*m. Simplifying the fraction we have m'=11/12*m. Let m=12t, m’=16t+12t'=11t. We have t'=-5/12*t that has no solutions. Therefore a(44)=0.
n=62; n'=33. m'=62/33*m. Let m=33*t, m'=14*t+33*t'=62*t. We have t’=48/33*t=16/11*t. therefore t=4*11^5 and m=33*4*11^5.

Links

Crossrefs

Cf. A003415, A051674, A165558.

A294153 Numbers k = a * b, such that k' = a' * b' where k', a' and b' are the arithmetic derivatives of k, a and b.

Original entry on oeis.org

0, 1, 256, 512, 1152, 1728, 1920, 3072, 3456, 7776, 11664, 12800, 12960, 20736, 23328, 52488, 72000, 78732, 81920, 86400, 87480, 100352, 110208, 124800, 139968, 153216, 157464, 200000, 219520, 263424, 294912, 321024, 336000, 354294, 400000, 486000, 486720, 531441
Offset: 1

Views

Author

Paolo P. Lava, Oct 24 2017

Keywords

Comments

A046311 is a subset of this sequence.
Some numbers admit more than one couple of divisors a, b: 3456 = 8 * 432 = 54 * 64 and 3456' = 15552 = 8' * 432' = 12 * 1296 = 54' * 64' = 81 * 192.
Apart from the first term, squares of A165558 are part of the sequence. In A165558 n' = 2 * n and therefore (n^2)' = 2 * n * n' = 2 * n * 2 * n = (2 * n)^2. Thus n^2 = n * n and (n^2)' = n' * n'.

Examples

			a(0) = 0 because 0 = 0 * b and 0' = 0' * b' = 0;
a(1) = 1 because 1 = 1 * 1 and 1' = 1' * 1' = 0;
a(2) = 256 because 256 = 16 * 16 and 256' = 16' * 16' = 32 * 32 = 1024;
a(3) = 512 because 512 = 8 * 64 and 512' = 8' * 64' = 12 * 192 = 2304.

Links

Crossrefs

Cf. A003415, A046311, A165558.

Programs

  • Maple
    with(numtheory): P:=proc(q) local a,b,c,j,k,n,p;
for n from 1 to q do j:=sort([op(divisors(n))]);
for k from 2 to trunc((nops(j)+1)/2) do
a:=j[k]*add(op(2,p)/op(1,p), p=ifactors(j[k])[2]);
b:=(n/j[k])*add(op(2,p)/op(1,p), p=ifactors(n/j[k])[2]);
c:=n*add(op(2,p)/op(1,p), p=ifactors(n)[2]);
if c=a*b then print(n); break; fi; od; od; end: P(10^6);

A282771 Integers that are one third of their arithmetic derivatives.

Original entry on oeis.org

0, 64, 432, 2916, 19683, 50000, 337500, 2278125, 13176688, 39062500, 88942644, 263671875, 600362847, 10294287500, 30517578125, 69486440625, 2712892291396, 4564986729776, 8042412109375, 18312022966923, 30813660425988, 207992207875419, 2119447102653125, 3566395882637500
Offset: 1

Views

Author

Paolo P. Lava, Mar 02 2017

Keywords

Comments

All integers of the form p^p*q^q*r^r, with p, q and r primes, are in the sequence.

Examples

			337500 = 2^2*3^3*5^5,  337500’ = 1012500 and 337500 = 3*1012500.

Crossrefs

Cf. A003415, A072873, A165558.

Programs

  • Maple
    with(numtheory): with(combinat): P:=proc(q) local a,k,n,x,y;
x:=[]; y:=[]; for n from 1 to q do for k from 1 to 3 do x:=[op(x),ithprime(n)^ithprime(n)]; od; od;
a:=choose(x,3); for k from 1 to nops(a) do y:=[op(y),convert(a[k],`*`)]; od; y:=sort([op(y)]); print(0);
for k from 1 to 100 do print(y[k]); od; end: P(10);

Formula

Solution of the equation n’ = 3*n.
Showing 1-4 of 4 results.

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