A373705 a(n) is the least start of a run of exactly n successive powerful numbers that are pairwise coprime, or -1 if no such run exists.
72, 1, 9, 289, 702464, 7827111875, 1321223037317, 1795433547131287
Offset: 1
Examples
a(1) = 72 because 72 is powerful, it is preceded by the powerful number 64 and gcd(64, 72) = 8 > 1, it is followed by the powerful number 81 and gcd(72, 81) = 9 > 1, and 72 is the least number with this property. a(2) = 1 because 1 and 4 are successive powerful numbers that are coprime. 8, the powerful number that follows 4, is not coprime to 4 since gcd(4, 8) = 4 > 1. a(3) = 9 because 9, 16 and 25 are 3 successive powerful numbers that are pairwise coprime: gcd(9, 16) = gcd(16, 25) = gcd(9, 25) = 1. They are not a part of a longer run since the powerful number that precedes 9 is 8 and gcd(8, 16) = 8 > 1, and the powerful number that follows 25 is 27 and gcd(9, 27) = 9 > 1. (9, 16, 25) is the run with the least start, 9, that has this property.
Crossrefs
Programs
-
Mathematica
pairCoprimeQ[s_] := Module[{ans = True}, Do[Do[If[! CoprimeQ[s[[i]], s[[j]]], ans = False; Break[]], {j, 1, i - 1}], {i, 1, Length[s]}]; ans]; pows[lim_] := Union[Flatten[Table[i^2 * j^3, {j, 1, Surd[lim, 3]}, {i, 1, Sqrt[lim/j^3]}]]]; seq[nmax_, lim_] := Module[{v = Table[0, {nmax}], s = {}, len = 0, init = 0, c = 0}, Do[len = Length[s]; AppendTo[s, k]; While[!pairCoprimeQ[s], s = Drop[s, 1]]; If[Length[s] <= len, If[len <= nmax && v[[len]] == 0, c++; v[[len]] = init]]; init = s[[1]]; If[c == nmax, Break[]], {k, pows[lim]}]; v]; seq[6, 10^10] -
PARI
iscoprime(s) = {for(i = 1, #s, for(j = 1, i-1, if(gcd(s[i], s[j]) > 1, return(0)))); 1;} pows(lim) = {my(pows = List()); for(j = 1, sqrtnint(lim, 3), for(i = 1, sqrtint(lim \ j^3), listput(pows, i^2 * j^3))); Set(pows);} lista(nmax, lim) = {my(pws = pows(lim), v = vector(nmax), s = List(), len = 0, init = 0); for(k = 1, #pws, len = #s; listput(s, pws[k]); while(!iscoprime(s), listpop(s, 1)); if(#s <= len, if(len <= nmax && v[len] == 0, v[len] = init)); init = s[1]); v;} lista(6, 10^10)
Comments