1, 2, 33, 603, 602, 2522, 211673, 3405123, 3405122, 49799889, 202536181, 3195380868, 5208143601, 85843948321, 97524222465
Offset: 0
A324593
a(n) is the smallest number k such that n consecutive integers starting at k have the same number of odd divisors (A001227).
Original entry on oeis.org
1, 1, 5, 10, 10, 515, 2314, 2314, 1536863, 4053992, 4053992, 18584686, 146237365, 163039279, 4775943486, 13147233734, 86153130379
Offset: 1
515 has 4 odd divisors {1, 5, 103, 515}, 516 has 4 odd divisors {1, 3, 43, 129}, 517 has 4 odd divisors {1, 11, 47, 517}, 518 has 4 odd divisors {1, 7, 37, 259}, 519 has 4 odd divisors {1, 3, 173, 519} and 520 has 4 odd divisors {1, 5, 13, 65}. These the first 6 consecutive numbers with the same number of odd divisors, so a(6) = 515.
A324594
a(n) is the smallest number k such that n consecutive integers starting at k have the same number of nonprime divisors (A033273).
Original entry on oeis.org
1, 1, 1, 19940, 204323, 294590, 310042685, 2587701932494, 2587701932494
Offset: 1
19940 has 9 nonprime divisors {1, 4, 10, 20, 1994, 3988, 4985, 9970, 19940}, 19941 has 9 nonprime divisors {1, 51, 69, 289, 391, 867, 1173, 6647, 19941}, 19942 has 9 nonprime divisors {1, 26, 118, 169, 338, 767, 1534, 9971, 19942} and 19943 has 9 nonprime divisors {1, 49, 77, 259, 407, 539, 1813, 2849, 19943}. These the first 4 consecutive numbers with the same number of nonprime divisors, so a(4) = 19940.
A323253
a(n) is the smallest number k such that factorizations of n consecutive integers starting at k have the same excess of number of primes counted with multiplicity over number of primes counted without multiplicity (A046660).
Original entry on oeis.org
1, 1, 1, 844, 74849, 671346, 8870025
Offset: 1
671346 = 2 * 3^2 * 13 * 19 * 151,
671347 = 17^2 * 23 * 101,
671348 = 2^2 * 47 * 3571,
671349 = 3 * 7^2 * 4567,
671350 = 2 * 5^2 * 29 * 463,
671351 = 53^2 * 239.
These the first 6 consecutive numbers with the same number of proper prime power divisors, so a(6) = 671346.
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Do[find = 0; k = 0; While[find == 0, k++; If[Length[Union[Table[PrimeOmega[j] - PrimeNu[j], {j, k, k + n - 1}]]] == 1, find = 1; Print[k]]], {n, 1, 5}] (* Vaclav Kotesovec, Sep 01 2019 *)
(* faster program *) fak = Table[f = FactorInteger[j]; Total[Transpose[f][[2]]] - Length[f], {j, 1, 10000000}]; m = Max[fak]; Table[Min[Table[SequencePosition[fak, ConstantArray[j, n]], {j, 0, m}]], {n, 1, 7}] (* Vaclav Kotesovec, Sep 01 2019 *)
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excess(n) = bigomega(n) - omega(n);
score(n) = my(t=excess(n)); for(k=1, oo, if(excess(n+k) != t, return(k)));
upto(nn) = my(n=1); for(k=1, nn, while(score(k) >= n, print1(k, ", "); n++)); \\ Daniel Suteu, Sep 01 2019
A338628
a(n) is the smallest number k such that n consecutive integers starting at k have the same number of square divisors (A046951).
Original entry on oeis.org
1, 1, 1, 844, 3624, 22020, 671346, 8870024, 264459172, 463239475, 1407472722, 108494875170, 12385053656370, 145065154350545
Offset: 1
844 has 2 square divisors {1, 4}, 845 has 2 square divisors {1, 169}, 846 has 2 square divisors {1, 9} and 847 has 2 square divisors {1, 121}. These are the first 4 consecutive numbers with the same number of square divisors, so a(4) = 844.
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Do[find = 0; k = 0; While[find == 0, k++; If[Length[Union[Table[Length[Select[Divisors[j], IntegerQ[Sqrt[#]] &]], {j, k, k + n - 1}]]] == 1, find = 1; Print[k]]], {n, 1, 7}]
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isok(n, k) = #Set(apply(x->sumdiv(x, d, issquare(d)), vector(n, i, k+i-1))) == 1;
a(n) = my(k=1); while(! isok(n, k), k++); k; \\ Michel Marcus, Nov 05 2020
A358044
a(n) is the smallest number k such that n consecutive integers starting at k have the same number of triangular divisors (A007862).
Original entry on oeis.org
1, 1, 55, 5402, 2515069
Offset: 1
55 has 2 triangular divisors {1, 55}, 56 has 2 triangular divisors {1, 28} and 57 has 2 triangular divisors {1, 3}. These are the first 3 consecutive numbers with the same number of triangular divisors, so a(3) = 55.
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