David Wasserman - cp's OEIS frontend

A325117 Irregular table read by rows: T(n,k) is the start of the first run of exactly k consecutive even integers having exactly n divisors, or 0 if no such run exists.

2, 4, 14, 0, 6, 16, 12, 18, 64, 24, 40, 182, 36, 48, 1024, 60, 198, 348, 9050, 25180, 25658650, 584558736346, 4096, 192, 144, 120, 918, 5430, 65536, 180, 17298, 262144, 240, 6640, 4413038, 576, 3072, 4194304, 360, 3400, 19548, 134044, 182644, 7126044, 359208340, 16074693138, 419893531348, 214932235538

Offset: 2

David Wasserman, Mar 27 2019

The number of terms in row n is A325116(n).
2.46*10^12 <= T(24,11) <= 299005907036986132.
T(24,14) <= 1010085195622895590495442.
T(30,3) <= 1359906389476760004389052496.
5.17*10^12 < T(36, 6) <= 13707985134823441146.
T(36, 7) <= 1678936725442128595619270138.

			T(4,3) = 6 because 6, 8, and 10 each have 4 divisors.
T(4,2) = 0. The runs 6, 8 and 8, 10 are excluded because they are part of a longer run, and there are no other consecutive even integers with 4 divisors.
T(18, 3) does not exist. This follows from the theorem: If m = 2 mod 4, and m has 18 divisors, then m-2 does not have 18 divisors.
Proof: Let d be the number of divisors function (A000005). Recall that it is multiplicative with d(p^i)=i+1. If m = 2 mod 4 and has 18 divisors, then m/2 is odd and has 9 divisors, so m=2*r^2 for some odd r. Then m-2=2(r-1)(r+1). r-1 and r+1 are even and one of them is divisible by 4, so 2^4 divides m-2. r-1 and r+1 have no prime factors in common except 2, so if they are both divisible by odd primes, call them s and t, then m-2 is divisible by 2^4*s*t and has at least 20 divisors, contrary to hypothesis. Therefore either r-1 or r+1 is a power of 2; call it 2^j. Then the exponent of 2 in m-2 is j+2, so j+3 divides 18, so j is 3 or 6. This leaves 4 possibilities for m-2: 2*6*8, 2*8*10, 2*62*64, or 2*64*66. Of these, only 2*62*64 has 18 divisors, and 2*62*64+2 does not have 18 divisors.
T(36, 11) does not exist. Proof: Suppose 11 consecutive even numbers with 36 divisors exist. Name them n_i where n_i = i (mod 32). n_16 and n_24 cannot have 36 divisors, so the 11 numbers are n_26 through n_14. Then n_8 is 8*x^2 for some odd x. Suppose 3 | x. Then 9 | n_8, so n_2 and n_14 are divisible by 3 but not 9, and by 2 but not 4. So n_2 = 6*y^2 and n_14 = 6*z^2 for some y and z, and z^2 = y^2 + 2, which is impossible. Therefore 3 doesn't divide x. Therefore x^2 = 1 (mod 3), and n_8 = 2 (mod 3). So 3 | n_6. Suppose n_6 = 0 (mod 9). Then n_26 = 6 (mod 9). So n_26 is divisible by 3 but not 9, and by 2 but not 4. So n_26 = 6*y^2. y^2 = 1 (mod 4), so n_26 = 6 (mod 8), but by definition n_26 = 2 (mod 8), a contradiction. Therefore n_6 != 0 (mod 9). Suppose n_6 = 3 (mod 9). Then n_6 is divisible by 3 but not 9, and by 2 but not 4. So n_6 = 6*y^2. y^2 = 1 (mod 3), so n_6 = 6 (mod 9), a contradiction. Therefore n_6 != 3 (mod 9), so n_6 = 6 (mod 9). Then n_26 = 3 (mod 9). So n_26 and n_6 are divisible by 3 but not 9, and by 2 but not 4. So n_26 = 6*y^2 and n_6 = 6*z^2 for some y and z, and z^2 = y^2 + 2, which is impossible.
In the table below, the following notation will be used for terms with unknown values: F: k consecutive even integers with n divisors have been found. D: Dickson's Conjecture implies the existence of k consecutive even integers with n divisors. H: Schinzel's Hypothesis H implies the existence of k consecutive even integers with n divisors. ?: It has not been proven that k consecutive even integers with n divisors do not exist. A semicolon indicates than no further terms exist.
Table begins:
   n  T(n,1), T(n,2), ...
  ==  =======================================================
   2  2;
   3  4;
   4  14, 0, 6;
   5  16;
   6  12, 18;
   7  64;
   8  24, 40, 182;
   9  36;
  10  48;
  11  1024;
  12  60, 198, 348, 9050, 25180, 25658650, 584558736346;
  13  4096;
  14  192;
  15  144;
  16  120, 918, 5430;
  17  65536;
  18  180, 17298;
  19  262144;
  20  240, 6640, 4413038;
  21  576;
  22  3072;
  23  4194304;
  24  360, 3400, 19548, 134044, 182644, 7126044, 359208340, 16074693138, 419893531348, 214932235538, F, D, D, F, D;
  25  1296;
  26  12288;
  27  900;
  28  960, 640062, 32858781246;
  29  268435456;
  30  720, 110796496, F;
  31  1073741824;
  32  840, 18088, 180726;
  33  9216;
  34  196608;
  35  5184;
  36  1260, 41650, 406780, 3237731546, 3651712573692, F, F, ?, ?, ?;

Cf. A292580 (analog for consecutive integers), A319046 (analog for consecutive odd integers), A325116.

A325116 Length of longest run of consecutive even integers having exactly n divisors.

Original entry on oeis.org

0, 1, 1, 3, 1, 2, 1, 3, 1, 1, 1, 7, 1, 1, 1, 3, 1, 2, 1, 3, 1, 1, 1

Offset: 1

David Wasserman, Mar 27 2019

The start of the first run of exactly k consecutive even integers having exactly n divisors is A325117(n,k).
If m does not divide n, then a(n) < 2^(m-1). Proof: 2^(m-1) consecutive even integers must include one congruent to 2^(m-1) mod 2^m, and the number of divisors of this one is a multiple of m.
For m > 2, if 2*(m-1) does not divide n, then a(n) < 2^(m-1). Proof: 2^(m-1) consecutive even integers must include two congruent to 2^(m-2) mod 2^(m-1). These are 2^(m-2)*x and 2^(m-2)*(x+2) for some odd x, and x and x+2 each have n/(m-1) divisors. If 2*(m-1) does not divide n, then n/(m-1) is odd. Any number with an odd number of divisors is a square, so x and x+2 are squares, but squares cannot differ by 2.
14 <= a(24) <= 15. Dickson's conjecture implies a(24)=15.

Cf. A119479 (with consecutive integers), A319045 (with consecutive odd integers).

Cf. A325117.

A187279 a(n) is the least number of terms needed to represent n as a sum of powers of the same prime.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 2, 3, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 3, 1, 2, 3, 2, 3, 2, 1, 2, 1, 2, 3, 1, 2, 2, 1, 2, 3, 3, 1, 2, 1, 2, 3, 3, 4, 4, 1, 2, 1, 2, 1, 2, 3, 2, 3, 3, 1, 2, 3, 4, 3, 2, 3, 2, 1, 2, 3, 3, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2

Offset: 1

David Wasserman, Mar 07 2011

A000961 gives all n such that a(n)=1. A024619 gives all n such that a(n) > 1.

If a(n) < m, let p be a prime such that n is a sum of < m powers of p. Then for any positive integers c_1, ... c_m that add up to n, p divides the m-nomial coefficient n!/(c_1!*c_2!*...*c_m!). If a(n) >= m then there is no prime that divides all such coefficients.

			a(15) = 3 because 15 can be expressed with powers of 3 as 3^2+3^1+3^1, or with powers of 7 as 7^1+7^1+7^0, or with powers of 13 as 13^1+13^0+13^0, but there is no such expression with fewer than three terms.

Alois P. Heinz, Table of n, a(n) for n = 1..10000
Mathematics StackExchange, The largest number y in which (x!)^(x+y)|(x^2)!

with(numtheory):
b:= proc(n, p) local c, m; m:=n; c:=0;
      while m>0 do c:= c+irem(m, p, 'm') od; c
    end:
a:= n-> min(seq(b(n, ithprime(i)), i=1..pi(n+1))):
seq(a(n), n=1..100);  # Alois P. Heinz, Nov 06 2013

Join[{1}, Table[Min[Plus @@@ IntegerDigits[n, Prime[Range[PrimePi[n]]]]], {n, 2, 110}]] (* T. D. Noe, Mar 08 2011 *)

A151897 Number of subsets of {1, 2, ..., n} such that no member is a sum of distinct other members.

Original entry on oeis.org

1, 2, 4, 7, 13, 22, 37, 60, 100, 155, 249, 381, 591, 889, 1365, 2009, 3047, 4453, 6602, 9567, 14151, 20228, 29654, 42302, 61369, 87108, 126066, 177580, 256039, 360304, 515740, 724069, 1036860, 1448746, 2069526, 2893311, 4117725, 5749540, 8186555

Offset: 0

David Wasserman, Apr 16 2008

nonn

This sequence and A085489 first differ at n = 7. a(7) = 60, A085489(7) = 61. A085489(7) includes {1, 2, 4, 7}, which is excluded from a(7) because 1+2+4 = 7.

If this sequence counts sum-free sets, A326080 counts sum-closed sets, which are different from sum-full sets (A093971). - Gus Wiseman, Jun 07 2019

			a(4) = 13, including all subsets of {1, 2, 3, 4} except {1, 2, 3} (excluded
because 1+2 = 3), {1, 3, 4} (excluded because 1+3 = 4), and {1, 2, 3, 4} (excluded for both reasons.)
From _Gus Wiseman_, Jun 07 2019: (Start)
The a(0) = 1 through a(4) = 13 subsets:
  {}  {}   {}     {}     {}
      {1}  {1}    {1}    {1}
           {2}    {2}    {2}
           {1,2}  {3}    {3}
                  {1,2}  {4}
                  {1,3}  {1,2}
                  {2,3}  {1,3}
                         {1,4}
                         {2,3}
                         {2,4}
                         {3,4}
                         {1,2,4}
                         {2,3,4}
(End)

Fausto A. C. Cariboni, Table of n, a(n) for n = 0..85

Cf. A007865, A050291, A085489, A103580, A308546, A326080, A326083, A326117.

Table[Length[Select[Subsets[Range[n]],Intersection[#,Plus@@@Subsets[#,{2,Length[#]}]]=={}&]],{n,0,10}] (* Gus Wiseman, Jun 07 2019 *)

a(0) = 1 prepended by Gus Wiseman, Jun 07 2019

A152957 Numbers n such that LS(n) divides n, where LS is the "Look and Say" function (A045918).

Original entry on oeis.org

22, 777, 4444, 200000, 333000, 333333, 555555, 660000, 666666, 700000, 4444400, 7700000, 22333000, 44445555, 55556666, 77700000, 88888888, 200000000, 777770000, 999900000, 999999000, 2222220000, 22222200000, 22333338888, 28800000000, 35555555505, 111111000000, 111777000000

Offset: 1

David Wasserman, Dec 15 2008

			777 is a member because it is three 7's, so LS(777) = 37 which divides 777.

Cf. A045918, A079342.

Definition corrected and more terms from Sean A. Irvine, Mar 17 2011

A140665 Self-erasure surviving numbers, second version. See A108652 for the definition.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 23, 24, 25, 26, 27, 28, 29, 30, 32, 36, 37, 38, 39, 40, 42, 45, 46, 47, 48, 49, 50, 51, 54, 58, 59, 60, 62, 64, 67, 68, 69, 70, 73, 76, 80, 83, 84, 86, 89, 90, 91, 92, 93, 94, 95, 98, 128, 144, 146, 148, 161, 168, 170, 173, 177, 180, 185

Offset: 1

David Wasserman, May 20 2008

The largest member that I've found is 402119190.

Cf. A108652.

A135058 Least m such that both m and m+n have exactly n distinct prime divisors, ignoring multiplicity.

Original entry on oeis.org

1, 2, 10, 102, 1326, 96135, 607614, 159282123, 9617162170, 1110180535035, 28334309296920, 16513791577659519, 271518698440871310

Offset: 0

David Wasserman, Feb 11 2008

Note that here the m and m+n may be divisible by squares (compare A097978).
a(13) <= 592357638037885411965.
If we change "exactly n" to "at least n", the sequence is still the same at least through a(12).

			a(2) = 10 because 10=2*5 and 12=3*2^2 have two distinct prime factors.
a(3) = 102 because 102=2*3*17 and 105=3*5*7 each have three distinct prime factors.
a(5) = 96135 because 96135 = 3*5*13*17*29 and 96140 = 2^2*5*11*19*23 each have 5 distinct prime factors.

Cf. A097978, A098515.

a(n) = min{m: A001221(m) = A001221(m+n) = n}. - R. J. Mathar, Mar 01 2017

A137905 Numbers that appear as binomial coefficients exactly twice.

Original entry on oeis.org

3, 4, 5, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 22, 23, 24, 25, 26, 27, 29, 30, 31, 32, 33, 34, 37, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 51, 52, 53, 54, 57, 58, 59, 60, 61, 62, 63, 64, 65, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 79, 80, 81, 82, 83, 85, 86, 87, 88

Offset: 1

David Wasserman, Feb 21 2008

Complement of A006987; a(n) = A058084(a(n)). - Reinhard Zumkeller, Mar 20 2009

			7 is a member because 7 = binomial(7, 1) = binomial(7, 6) and no other binomial coefficient equals 7. [clarified by _Jonathan Sondow_, Jan 12 2018]

Cf. A006987, A058084, A062527, A098564, A098566.

isok(n) = (sum(i=0, n, sum(j=0, i, binomial(i,j)==n)) == 2) \\ Michel Marcus, Jun 16 2013

a(n) = A185024(n+1). - Elijah Beregovsky, May 14 2019

A131103 Rectangular array read by antidiagonals: a(n, k) is the number of ways to put k labeled objects into n labeled boxes so that there are no boxes with exactly one object (n, k >= 1).

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 0, 3, 2, 1, 0, 4, 3, 8, 1, 0, 5, 4, 21, 22, 1, 0, 6, 5, 40, 63, 52, 1, 0, 7, 6, 65, 124, 243, 114, 1, 0, 8, 7, 96, 205, 664, 969, 240, 1, 0, 9, 8, 133, 306, 1405, 3196, 3657, 494, 1, 0, 10, 9, 176, 427, 2556, 7425, 15712, 12987, 1004, 1, 0, 11, 10, 225, 568, 4207

Offset: 1

David Wasserman, Jun 14 2007, Jun 15 2007

Problem suggested by Brandon Zeidler. Columns four and five are A000567 and A051874. Second row is A130102.

			Array begins:
0 1 1 1 1 1 1
0 2 2 8 22 52 114
0 3 3 21 63 243 969

Cf. A131104, A131105, A131106, A131107.

a(n, k) = sum_{j=1..min(floor(k/2), n)} A008299(k, j)*n!/(n-j)!.

A131104 Rectangular array read by antidiagonals: a(n, k) is the number of ways to put k labeled objects into n labeled boxes so that there is one box with exactly one object (n, k >= 1).

Original entry on oeis.org

1, 2, 0, 3, 0, 0, 4, 0, 6, 0, 5, 0, 18, 8, 0, 6, 0, 36, 24, 10, 0, 7, 0, 60, 48, 120, 12, 0, 8, 0, 90, 80, 420, 396, 14, 0, 9, 0, 126, 120, 1000, 1512, 1092, 16, 0, 10, 0, 168, 168, 1950, 3720, 6804, 2736, 18, 0, 11, 0, 216, 224, 3360, 7380, 23240, 31008, 6480, 20, 0, 12, 0

Offset: 1

David Wasserman, Jun 14 2007, Jun 15 2007

Problem suggested by Brandon Zeidler. Columns 3 through 5 are A028896, A033996, 10*A007586.

			Array begins:
1 0 0 0 0 0 0
2 0 6 8 10 12 14
3 0 18 24 120 396 1092

Cf. A131103, A131105, A131106, A131107.

a(n, 1) = n. For k > 1, a(n, k) = sum_{j=1..min(floor((k-1)/2), n-1)} A008299(k-1, j)*n!*k*/(n-j-1)!.

cp's OEIS Frontend

User: David Wasserman

A325117 Irregular table read by rows: T(n,k) is the start of the first run of exactly k consecutive even integers having exactly n divisors, or 0 if no such run exists.

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A325116 Length of longest run of consecutive even integers having exactly n divisors.

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A187279 a(n) is the least number of terms needed to represent n as a sum of powers of the same prime.

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Maple

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A151897 Number of subsets of {1, 2, ..., n} such that no member is a sum of distinct other members.

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Mathematica

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A152957 Numbers n such that LS(n) divides n, where LS is the "Look and Say" function (A045918).

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A140665 Self-erasure surviving numbers, second version. See A108652 for the definition.

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A135058 Least m such that both m and m+n have exactly n distinct prime divisors, ignoring multiplicity.

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Formula

A137905 Numbers that appear as binomial coefficients exactly twice.

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PARI

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A131103 Rectangular array read by antidiagonals: a(n, k) is the number of ways to put k labeled objects into n labeled boxes so that there are no boxes with exactly one object (n, k >= 1).

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A131104 Rectangular array read by antidiagonals: a(n, k) is the number of ways to put k labeled objects into n labeled boxes so that there is one box with exactly one object (n, k >= 1).

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