Original entry on oeis.org
1372, 465125, 4879688, 6272006419, 3533294646441, 405211279147678088
Offset: 1
1372 = 2^2 * 7^3 is a term since it is a term of A349062 (the gap to the next powerful number, 1444, is 72, which is a record) and it is not a square.
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seq[max_] := Module[{pows = Union[Flatten[Table[i^2*j^3, {j, 1, Surd[max, 3]}, {i, 1, Sqrt[max/j^3]}]]], gapmax = 0, gap, s = {}}, Do[gap = pows[[k+1]] - pows[[k]]; If[gap > gapmax, gapmax = gap; If[!IntegerQ[Sqrt[pows[[k]]]], AppendTo[s, pows[[k]]]]], {k, 1, Length[pows] - 1}]; s]; seq[10^10]
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lista(mx) = {my(s = List(), gap, gapmax = 0); for(j = 1, sqrtnint(mx, 3), for(i = 1, sqrtint(mx\j^3), listput(s, i^2 * j^3))); s = Set(s); for(k = 1, #s - 1, gap = s[k+1] - s[k]; if(gap > gapmax, gapmax = gap; if(!issquare(s[k]), print1(s[k], ", "))));}
A363191
a(n) is the least start of a run of exactly n consecutive powerful numbers (A001694) that are even, or -1 if no such run exists.
Original entry on oeis.org
16, 4, 196, 968, 8712, 437400, 85730400, 5030690600, 264615012500, 5239012864, 550886816376, 2494017320776852
Offset: 1
a(1) = 16, since 16 = 2^4 is an even powerful number, preceded by an odd powerful number, 9 = 3^2, and followed by an odd powerful number, 25 = 5^2.
a(2) = 4, since 4 = 2^2 and 8 = 2^3 are two consecutive even powerful numbers, preceded by an odd powerful number, 1, and followed by an odd powerful number, 9 = 3^2.
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seq[lim_] := Module[{pow = Union[Flatten[Table[i^2*j^3, {j, 1, lim^(1/3)}, {i, 1, Sqrt[lim/j^3]}]]], s = {}, rem, ind}, rem = Mod[pow, 2]; Do[ind = SequencePosition[rem, Join[{1}, Table[0, {k}], {1}], 1]; If[ind == {}, Break[]]; AppendTo[s, pow[[ind[[1, 1]] + 1]]], {k, 1, Infinity}]; s]; seq[10^10]
A363192
a(n) is the least start of a run of exactly n consecutive powerful numbers (A001694) that are odd, or -1 if no such run exists.
Original entry on oeis.org
1, 25, 2187, 703125, 93096125, 10229709861, 197584409639, 32044275110699, 164029657560618375
Offset: 1
a(1) = 1, since 1 is an odd powerful number, followed by an even powerful number, 4 = 2^2.
a(2) = 25, since 25 = 5^2 and 27 = 3^3 are two consecutive odd powerful numbers, preceded by an even powerful number, 16 = 2^4, and followed by an even powerful number, 32 = 2^5.
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seq[lim_] := Module[{pow = Union[Flatten[Table[i^2*j^3, {j, 1, lim^(1/3)}, {i, 1, Sqrt[lim/j^3]}]]], s = {}, rem, ind}, rem = Join[{0}, Mod[pow, 2]]; Do[ind = SequencePosition[rem, Join[{0}, Table[1, {k}], {0}], 1]; If[ind == {}, Break[]]; AppendTo[s, pow[[ind[[1, 1]]]]], {k, 1, Infinity}]; s]; seq[1.1*10^10]
A363014
Cubefull numbers (A036966) with a record gap to the next cubefull number.
Original entry on oeis.org
1, 8, 16, 32, 81, 128, 343, 512, 729, 864, 1024, 1331, 3456, 4096, 6912, 8192, 12167, 25000, 32768, 35937, 43904, 46656, 55296, 70304, 93312, 110592, 117649, 140608, 186624, 287496, 331776, 357911, 373248, 592704, 707281, 889056, 1000000, 1124864, 1157625, 1296000
Offset: 1
The sequence of cubefull numbers begins with 1, 8, 16, 27, 32, 64, 81 and 125. The differences between these terms are 7, 8, 11, 5, 32, 17 and 44. The record values, 7, 8, 11, 32 and 44 occur after the cubefull numbers 1, 8, 16, 32 and 81, the first 5 terms of this sequence.
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cubQ[n_] := Min[FactorInteger[n][[;; , 2]]] > 2; seq[kmax_] := Module[{s = {}, k1 = 1, gapmax = 0, gap}, Do[If[cubQ[k], gap = k - k1; If[gap > gapmax, gapmax = gap; AppendTo[s, k1]]; k1 = k], {k, 2, kmax}]; s]; seq[10^6]
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iscubefull(n) = n==1 || vecmin(factor(n)[, 2]) > 2;
lista(kmax) = {my(gapmax = 0, gap, k1 = 1); for(k = 2, kmax, if(iscubefull(k), gap = k - k1; if(gap > gapmax, gapmax = gap; print1(k1, ", ")); k1 = k));}
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.
Original entry on oeis.org
72, 1, 9, 289, 702464, 7827111875, 1321223037317, 1795433547131287
Offset: 1
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.
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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]
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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)
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