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.
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.
Showing 1-3 of 3 results.
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