A294752
Squarefree products of k primes that are symmetrically distributed around their average. Case k = 5.
Original entry on oeis.org
53295, 119301, 229245, 399993, 608235, 623645, 1462731, 2324495, 3696189, 3973145, 4482879, 5356445, 5920971, 6249633, 7588977, 8270385, 10160943, 10450121, 10505373, 13185969, 13630011, 13760929, 14935029, 19095395, 20280795, 22566271, 23131549, 23408259, 24778401
Offset: 1
53295 = 3*5*11*17*19. Prime factors average is (3 + 5 + 11 + 17 + 19)/5 = 11 and 3 + 8 = 11 = 19 - 8, 5 + 6 = 11 = 17 - 6.
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with(numtheory): P:=proc(q,h) local a,b,k,n,ok;
for n from 2*3*5*7*11 to q do if not isprime(n) and issqrfree(n) then a:=ifactors(n)[2];
if nops(a)=h then b:=2*add(a[k][1],k=1..nops(a))/nops(a); ok:=1;
for k from 1 to trunc(nops(a)/2) do if a[k][1]+a[nops(a)-k+1][1]<>b then ok:=0; break; fi; od; if ok=1 then print(n); fi; fi; fi; od; end: P(10^9,5);
# Alternative:
N:= 10^8: # to get all terms <= N
M:= floor((8*N/15)^(1/3)):
P:= select(isprime, [seq(i,i=3..M,2)]): nP:= nops(P):
Res:= NULL:
for i3 from 3 to nP-2 do
p3:= P[i3];
for i1 from 1 to i3-2 do
if isprime(2*p3 - P[i1]) then
for i2 from i1+1 to i3-1 do
if isprime(2*p3 - P[i2]) then
v:=P[i1]*P[i2]*p3*(2*p3-P[i2])*(2*p3-P[i1]);
if v <= N then Res:= Res, v fi
fi
od
fi
od
od:
sort([Res]): # Robert Israel, Nov 10 2017
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isok(n, nb=5) = {if (issquarefree(n) && (omega(n)==nb), f = factor(n)[, 1]~; avg = vecsum(f)/#f; for (k=1, #f\2, if (f[k] + f[#f-k+1] != 2*avg, return(0));); return (1););} \\ Michel Marcus, Nov 10 2017
A294776
Squarefree products of k primes that are symmetrically distributed around their average. Case k = 6.
Original entry on oeis.org
1616615, 3411705, 7436429, 9408035, 10163195, 12838371, 13037385, 13844919, 14969435, 19605131, 20414121, 23783045, 24997749, 25113935, 27568145, 30478565, 31473255, 32518535, 33999455, 39946569, 43134015, 46115135, 48215255, 50907855, 56179409, 61558343
Offset: 1
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with(numtheory): P:=proc(q,h) local a,b,k,n,ok;
for n from 2*3*5*7*11*13 to q do if not isprime(n) and issqrfree(n) then a:=ifactors(n)[2];
if nops(a)=h then b:=2*add(a[k][1],k=1..nops(a))/nops(a); ok:=1;
for k from 1 to trunc(nops(a)/2) do if a[k][1]+a[nops(a)-k+1][1]<>b then ok:=0; break; fi; od; if ok=1 then print(n); fi; fi; fi; od; end: P(10^9,6);
# Alternative:
N:= 10^8: # to get all terms <= N
M:= floor(fsolve(3*5*7*(M-7)*(M-5)*(M-3) = N)):
P:= select(isprime, [seq(i,i=3..M/2,2)]): nP:= nops(P):
Res:= NULL:
for m from 10 by 2 to M do
for ix from 1 to nP-2 do
x:= P[ix];
if x >= m/2 or (x*(m-x))^3 >= N then break fi;
if not isprime(m-x) then next fi;
for iy from ix+1 to nP-1 do
y:= P[iy];
if y >= m/2 or x*(m-x)*(y*(m-y))^2 >= N then break fi;
if not isprime(m-y) then next fi;
for iz from iy+1 to nP do
z:= P[iz];
if z >= m/2 then break fi;
v:= x*(m-x)*y*(m-y)*z*(m-z);
if v > N then break fi;
if isprime(m-z) then Res:= Res, v fi;
od od od od:
sort([Res]); # Robert Israel, May 19 2019
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isok(n, nb=6) = {if (issquarefree(n) && (omega(n)==nb), f = factor(n)[, 1]~; avg = vecsum(f)/#f; for (k=1, #f\2, if (f[k] + f[#f-k+1] != 2*avg, return(0));); return (1););} \\ Michel Marcus, Nov 10 2017
A294906
a(n) is the least squarefree integer, product of n primes that are symmetrically distributed around their average.
Original entry on oeis.org
2, 6, 105, 2145, 53295, 1616615, 57998985, 3038795305, 3907126810041, 7292509103495, 66240019730740785, 82246340873964085, 1870667082297874652055, 343515424581301546805, 9160656702012692171113335, 2356596317899272514936585, 1903895854998638367242867256645
Offset: 1
The prime factors of the first terms are: [2], [2, 3], [3, 5, 7], [3, 5, 11, 13], [3, 5, 11, 17, 19], [5, 7, 11, 13, 17, 19], [3, 5, 11, 17, 23, 29, 31], ...
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isok(n, nb) = {if (issquarefree(n) && (omega(n)==nb), f = factor(n)[, 1]~; avg = vecsum(f)/#f; for (k=1, #f\2, if (f[k] + f[#f-k+1] != 2*avg, return(0));); return (1););}
a(n) = {my(k = prod(i=1, n, prime(i))); while (! isok(k, n), k++); k;}
Showing 1-3 of 3 results.