A225758 -id:A225758 - cp's OEIS frontend

A225756 Runs of consecutive numbers with the same number of divisors.

2, 3, 14, 15, 21, 22, 26, 27, 33, 34, 35, 38, 39, 44, 45, 57, 58, 75, 76, 85, 86, 87, 93, 94, 95, 98, 99, 104, 105, 116, 117, 118, 119, 122, 123, 133, 134, 135, 136, 141, 142, 143, 145, 146, 147, 148, 158, 159, 171, 172, 177, 178, 189, 190, 201, 202, 203

Offset: 1

Jean-François Alcover, May 15 2013

The first run of length 3 is {33,34,35}; the first of length 4: {116, 117, 118, 119}; of length 5: {2641, 2642, 2643, 2644, 2645}; of length 6: {1081, 1082, 1083, 1084, 1085, 1086}; of length 7: {25781, 25782, 25783, 25784, 25785, 25786, 25787}; of length 8: {77673, 77674, 77675, 77676, 77677, 77678, 77679, 77680}.

			Sequence begins:
2, 3;
14, 15;
21, 22;
26, 27;
33, 34, 35;
38, 39;
etc.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A225757 (same sum of divisors).
Cf. A225758 (same number and sum of divisors).
Cf. A140578 (first term of every run).

sel = Select[Range[300], DivisorSigma[0, #] == DivisorSigma[0, # + 1] &]; Union[sel, sel + 1]
Flatten[SequencePosition[DivisorSigma[0,Range[300]],{x_,x_}]]//Union (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 22 2021 *)

A225757 Table of consecutive numbers with the same sum of divisors.

Original entry on oeis.org

14, 15, 206, 207, 957, 958, 1334, 1335, 1364, 1365, 1634, 1635, 2685, 2686, 2974, 2975, 4364, 4365, 14841, 14842, 18873, 18874, 19358, 19359, 20145, 20146, 24957, 24958, 33998, 33999, 36566, 36567, 42818, 42819, 56564, 56565, 64665, 64666, 74918, 74919, 79826

Offset: 1

Jean-François Alcover, May 15 2013

Are 3 consecutive terms possible? There are none less than 10^12. See A002961. - T. D. Noe, May 15 2013

			Sequence begins:
14, 15;
206, 207;
957, 958;
1334, 1335;
etc.

Charles R Greathouse IV, Table of n, a(n) for n = 1..1000 (first 226 terms from T. D. Noe)

Cf. A225756 (same number of divisors), A225758 (same number and sum of divisors), A002961 (first number of each pair).

sel = Select[Range[100000], DivisorSigma[1, #] == DivisorSigma[1, # + 1] &]; Union[sel, sel + 1]
Flatten[SequencePosition[DivisorSigma[1,Range[80000]],{x_,x_}]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 13 2017 *)

v=List();t=[1,3];for(n=3,1e6,t=[t[2],sigma(n)];if(t[1]==t[2],listput(v,n-1);listput(v,n)));vecsort(Vec(v),,8) \\ Charles R Greathouse IV, May 15 2013

cp's OEIS Frontend

A225756 Runs of consecutive numbers with the same number of divisors.

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