A202463
First number of divisor symmetry n: d(n-k) = d(n+k) for 1 <= k <= n, but d(n-k-1) != d(n+k+1).
Original entry on oeis.org
4, 9, 216, 30, 20376, 432, 18000, 13338864, 15194736, 866452464, 5175273600, 35399473200
Offset: 1
8 and 10 have 4 divisors each, 7 and 11 have 2 divisors each, but 6 and 12 have different numbers of divisors; thus 9 has divisor symmetry 2. Since no smaller number has this, a(2) = 9.
-
a(n)=my(k=n);while(k++, for(i=1,n,if(numdiv(k-i)!=numdiv(k+i),next(2))); if(numdiv(k-n-1)==numdiv(k+n+1), next); return(k))
A284378
a(n) is the least integer m such that A048691 has n equal consecutive values, where A048691(k) is the number of divisors of k^2.
Original entry on oeis.org
1, 2, 14, 3728, 3854, 249293, 748909, 1151514109050
Offset: 1
A048691(2) and A048691(3) are equal (to 3).
A048691(14), A048691(15) and A048691(16) are equal (to 9).
A066480
Start of first run of exactly n consecutive integers with same number of anti-divisors.
Original entry on oeis.org
5, 1, 19, 212, 231, 353755, 7077517, 841891, 96723128, 640141432, 83101215664, 3774913237385, 29738569261171
Offset: 1
To illustrate the first 3 terms, here are the numbers of anti-divisors of the numbers 1 through 22: [0, 0, 1, 1, 2, 1, 3, 2, 2, 3, 3, 2, 4, 3, 3, 2, 5, 4, 3, 3, 3, 5] (see A066272). - _N. J. A. Sloane_, Oct 14 2019
-
a066272[n_] := Count[Range[2, n - 1], _?(Abs[Mod[n, #] - #/2] < 1 &)]
a[0] = 5;
a[n_] := For[i = 1, True, ++i, If[Equal @@ Table[a066272[j], {j, i, i + n}], Return[i]]]
a /@ Range[0, 4] (* Julien Kluge, Dec 03 2016 *)
-
nbad(n) = #select(t->n%t && tA066272
isok(k, n) = {my(nb=nbad(k)); if ((k>1) && nbad(k-1) == nb, return (0)); for (j=1, n-1, if (nbad(k+j) != nb, return(0));); nbad(k+n) != nb;}
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Oct 11 2019
A318529
a(n) begins the first run of at least n consecutive numbers with same number of exponential divisors.
Original entry on oeis.org
1, 1, 1, 242, 3624, 22020, 671346, 8870024, 49250144, 463239475, 1407472722, 82462576220, 82462576220, 5907695879319
Offset: 1
a(4) = 242 since the number of exponential divisors of 242, 243, 244, and 245 is 2, and this is the first run of 4 consecutive numbers.
- Eric Weisstein's World of Mathematics, e-Divisor
-
edivnum[1] = 1; edivnum [p_?PrimeQ] = 1; edivnum [p_?PrimeQ, e_] := DivisorSigma[ 0, e ]; edivnum [n_] := Times @@ (edivnum [#[[1]], #[[2]]] & ) /@ FactorInteger[ n ]; Seq[n_,q_] := Map[edivnum, Range[n, n + q - 1]]; findConsec[q_, nmin_, nmax_] := Module[{}, s = Seq[1, q]; n = q + 1; found = False; Do[ If[ CountDistinct[s] == 1, found = True; Break[] ]; s = Rest[AppendTo[s, edivnum[n] ]]; n++, {k, nmin, nmax}]; If[found, n - q, 0]]; seq = {1}; nmax = 100000000; Do[n1 = Last[seq]; s1 = findConsec[m, n1, nmax]; If[s1 == 0, Break[]]; AppendTo[ seq, s1 ], {m, 2, 7}]; seq (* after Jean-François Alcover in A049419 *)
A323743
Table read by rows: row n lists the numbers k for which there exist only finitely many runs of n consecutive integers whose number-of-divisors function sums to k.
Original entry on oeis.org
1, 3, 4, 5, 5, 7, 8, 9, 8, 9, 11, 12, 13, 14, 15, 10, 13, 15, 17, 18, 19, 14, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 16, 19, 20, 21, 22, 23, 25, 26, 27, 29, 30, 31, 20, 22, 24, 27, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39
Offset: 1
There is only one number with exactly 1 divisor (namely, k=1), but there are infinitely many numbers with j divisors for every j >= 2, so row 1 consists only of the single term 1.
The sequence of values tau(k) for k >= 1 is A000005, which begins 1, 2, 2, 3, 2, 4, 2, 4, 3, 4, ..., from which the sums of two consecutive terms are 1+2=3, 2+2=4, 2+3=5, 3+2=5, 2+4=6, 4+2=6, 2+4=6, 4+3=7, 3+4=7, ...; no number j < 3 appears as such a sum, every j >= 6 appears infinitely many times as such a sum, and each j in {3,4,5} appears as such a sum only finitely many times, so row 2 is {3, 4, 5}.
Row 3 does not contain 6 as a term because there exists no run of 3 consecutive numbers whose sum of tau values is exactly 6.
The first six rows of the table are as follows:
row 1: {1};
row 2: {3, 4, 5};
row 3: {5, 7, 8, 9};
row 4: {8, 9, 11, 12, 13, 14, 15};
row 5: {10, 13, 15, 17, 18, 19};
row 6: {14, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27}.
Cf.
A000005,
A005237,
A006558,
A048892,
A072507,
A100366,
A119479,
A141621,
A284596,
A284597,
A292580,
A319037,
A319045,
A319046.
A337158
a(n) is the start m of first run of n consecutive numbers with periodic sequence of values of number of divisors [tau(m), tau(m+1), tau(m), tau(m+1), tau(m), ...] where tau(m) != tau(m+1).
Original entry on oeis.org
1, 1, 3, 5, 182, 182, 4853, 445145, 6139163, 55804443, 55804443, 856530193, 856530193, 7709802455641
Offset: 1
[tau(5), tau(6), tau(7), tau(8)] = [2, 4, 2, 4].
[tau(182), tau(183), tau(184), tau(185), tau(186), tau(187)] = [8, 4, 8, 4, 8, 4].
A349262
a(n) is the start of the least run of exactly n consecutive numbers with the same value of A349258.
Original entry on oeis.org
1, 14, 20, 2, 91, 6850, 2302, 141, 56014, 184171, 2800171, 27805034, 35297611, 8313366182, 1791416073, 3618621410
Offset: 1
a(2) = 14 since A349258(14) = A349258(15) = 2, but A349258(13) != 2 and A349258(16) != 2.
-
f[p_, e_] := 2^DigitCount[e, 2, 1] - 1; d[1] = 0; d[n_] := Plus @@ f @@@ FactorInteger[n]; seq[len_, nmax_] := Module[{s = Table[0, {len}], dprev = 0, n = 2, c = 1, k = 1}, s[[1]] = 1; While[k < len && n < nmax, d1 = d[n]; If[d1 == dprev, c++, If[c > 0 && c <= len && s[[c]] == 0, k++; s[[c]] = n - c]; c = 1]; n++; dprev = d1]; TakeWhile[s, # > 0 &]]; seq[8, 10^4]
A349305
a(n) is the start of the least run of exactly n consecutive numbers with the same number of nonunitary divisors.
Original entry on oeis.org
4, 10, 1, 19940, 54584, 204323, 2789143044, 27092041443
Offset: 1
a(2) = 10 since A048105(10) = A048105(11) = 0, and A048105(9) != 0 and A048105(12) != 0.
-
d[n_] := DivisorSigma[0, n] - 2^PrimeNu[n]; seq[len_, nmax_] := Module[{s = Table[0, {len}], dprev = -1, n = 1, c = 0, k = 0}, While[k < len && n < nmax, d1 = d[n]; If[d1 == dprev, c++, If[c > 0 && c <= len && s[[c]] == 0, k++; s[[c]] = n - c]; c = 1]; n++; dprev = d1]; TakeWhile[s, # > 0 &]]; seq[6, 10^6]
A356766
Least number k such that k and k+2 both have exactly 2n divisors, or -1 if no such number exists.
Original entry on oeis.org
3, 6, 18, 40, 127251, 198, 26890623, 918, 17298, 6640, 25269208984375, 3400, 3900566650390623, 640062, 8418573, 18088, 1164385682220458984373, 41650, 69528379848480224609373, 128464, 34084859373, 12164094, 150509919493198394775390625, 90270, 418514293125, 64505245696
Offset: 1
For n=1, numdiv(3) = numdiv(5) = 2 = 2*1, and no number < 3 satisfies this, hence a(1) = 3.
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a={}; n=1; nmax=10; For[k=1, n<=nmax, k++, If[DivisorSigma[0, k] == DivisorSigma[0, k+2] == 2n, AppendTo[a, k]; k=1; n++]]; a (* Stefano Spezia, Aug 26 2022 *)
Flatten[Table[SequencePosition[DivisorSigma[0,Range[27*10^6]],{2n,,2n},1],{n,10}],1][[;;,1]] (* The program generates the first 10 terms of the sequence. To generate more, increase the Range constant but the program will take a long time to run. *) (* _Harvey P. Dale, Jul 01 2023 *)
-
a(n)=for(k=1,+oo,if(numdiv(k)==2*n&&numdiv(k+2)==2*n,return(k)))
A358634
a(n) is the smallest number k such that n consecutive integers starting at k have the same number of n-gonal divisors.
Original entry on oeis.org
55, 844, 16652, 844529772, 243636414, 36289272509
Offset: 3
16652 has 2 pentagonal divisors {1, 92}, 16653 has 2 pentagonal divisors {1, 5551}, 16654 has 2 pentagonal divisors {1, 22}, 16655 has 2 pentagonal divisors {1, 5}, and 16656 has 2 pentagonal divisors {1, 12}. These are the first 5 consecutive numbers with the same number of pentagonal divisors, so a(5) = 16652.
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