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A386947 A variant of Recamán's sequence (A005132): a(0) = 0; for n > 0, a(n) = a(n-1) - n if nonnegative and not already in the sequence, otherwise a(n) = a(n-1) + n + 2.

0, 3, 1, 6, 2, 9, 17, 10, 20, 11, 23, 12, 26, 13, 29, 14, 32, 15, 35, 16, 38, 61, 39, 64, 40, 67, 41, 70, 42, 73, 43, 76, 44, 79, 45, 82, 46, 85, 47, 8, 50, 93, 51, 96, 52, 7, 55, 104, 56, 107, 57, 110, 58, 5, 61, 118, 62, 121, 63, 4, 66

Offset: 0

Jules Beauchamp, Aug 10 2025

When visualized in connecting semicircular lines (like the original Recamán Sequence, cf. Edmund Harriss, "The first 65 steps drawn as a spiral" in A005132), it appears to produce conical spirals.

The first repeating term is 61.

Jules Beauchamp, The first 60 steps drawn as a spiral.

Cf. A005132, A386969.

a[n_] := a[n] = If[a[n-1] >= n && FreeQ[Table[a[k], {k, 0, n-1}], a[n-1] - n], a[n-1] - n, a[n-1] + n + 2]; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Aug 22 2025 *)

A386946 a(n) is the number of imprimitive (periodic) 2n-bead balanced binary necklaces.

Original entry on oeis.org

0, 0, 1, 1, 2, 1, 5, 1, 10, 4, 27, 1, 88, 1, 247, 29, 810, 1, 2780, 1, 9260, 249, 32067, 1, 113520, 26, 400025, 2704, 1432868, 1, 5179905, 1, 18784170, 32069, 68635479, 271, 252201136, 1, 930138523, 400027, 3446168660, 1, 12817096533, 1, 47820447036, 5173304

Offset: 0

Tilman Piesk, Aug 10 2025

A003239(n) is the number of 2n-bead balanced binary necklaces. A022553(n) among them are primitive.
The remaining a(n) necklaces are periodic.
Sequences counting 2n-bead balanced binary necklaces:
                       primitive  imprimitive
                     +-----------------------+---------+
  self-complementary |  A000048     A115118  | A000013 |
   complement pairs  |  A383904     A387130  | A386388 |
                     +-----------------------+---------+
                     |  A022553      this    | A003239 |
                     +-----------------------+---------+

			  n | A003239(n) A022553(n) | a(n)
  0 |         1          1  |   0
  1 |         1          1  |   0
  2 |         2          1  |   1
  3 |         4          3  |   1
  4 |        10          8  |   2
  5 |        26         25  |   1
  6 |        80         75  |   5
  7 |       246        245  |   1
  8 |       810        800  |  10
  9 |      2704       2700  |   4
 10 |      9252       9225  |  27
 11 |     32066      32065  |   1
 12 |    112720     112632  |  88
 13 |    400024     400023  |   1
 14 |   1432860    1432613  | 247
 15 |   5170604    5170575  |  29
 16 |  18784170   18783360  | 810
There are A003239(8) = 810 balanced binary necklaces of length 16. A022553(8) = 800 of them are primitive. a(n) = 10 are not. See A387130 for a list.

Tilman Piesk, Table of n, a(n) for n = 0..1000

a(n) = A003239(n) - A022553(n).

a(n) = A115118(n) + 2 * A387130(n).

A386950 A sequence constructed by greedily sampling the pdf (H(10)-H(i))/10, where H(n) is the n-th Harmonic number and the support is [0,9], to minimize discrepancy.

Original entry on oeis.org

0, 1, 0, 2, 3, 0, 1, 4, 0, 2, 5, 1, 0, 3, 0, 1, 6, 2, 4, 0, 1, 0, 3, 2, 7, 0, 5, 1, 0, 2, 4, 1, 3, 0, 0, 1, 6, 2, 0, 3, 5, 1, 4, 8, 0, 2, 0, 1, 0, 3, 2, 1, 0, 4, 7, 0, 5, 1, 6, 2, 3, 0, 1, 0, 2, 4, 0, 1, 3, 0, 2, 5, 1, 0, 0, 3, 4, 1, 6, 2, 0, 1, 0, 7, 2, 3, 0

Offset: 1

Jwalin Bhatt, Aug 10 2025

The pdf of the sequence is the probability to see i as the remainder when you divide 1,2,3... with a random number from [1,10].
The respective probabilities are:
  p(0) = 7381/25200 = 0.292896...
  p(1) = 4861/25200 = 0.192896...
  p(2) = 3601/25200 = 0.142896...
  p(3) = 2761/25200 = 0.109563...
  p(4) = 2131/25200 = 0.084563...
  p(5) = 1627/25200 = 0.064563...
  p(6) = 1207/25200 = 0.047896...
  p(7) = 121/3600   = 0.033611...
  p(8) = 19/900     = 0.021111...
  p(9) = 1/100      = 0.01
The sequence repeats after 25200 terms. If the random number is picked from [1,n] the period of the sequence is given by n*lcm{1,2,...,n} (A081528).
The arithmetic mean of this sequence is 2.25.
For a sequence with length n, the n-th root of the product of nonzero terms is (2**(5333/12600))*(3**(559/3150))*(5**(1627/25200))*(7**(121/3600)) which is 1.9302557640832...

			Let p(k) denote the probability of k and c(k) denote the number of occurrences of k among the first n-1 terms.
We take the ratio of the actual occurrences c(k)+1 to the probability and pick the one with the lowest value.
Since p(k) is monotonic decreasing, we only need to compute c(k) once we see c(k-1).
| n | (c(0)+1)/p(0) | (c(1)+1)/p(1) | (c(2)+1)/p(2) | choice |
|---|---------------|---------------|---------------|--------|
| 1 |     3.414     |       -       |       -       |   0    |
| 2 |     6.828     |     5.181     |       -       |   1    |
| 3 |     6.828     |    10.362     |     6.998     |   0    |
| 4 |    10.242     |    10.362     |     6.998     |   2    |

Jwalin Bhatt, Table of n, a(n) for n = 1..25200

Cf. A081528.

samplePDF[numCoeffs_] := Module[{coeffs, counts},
  coeffs = {}; counts = ConstantArray[0, 10];
  Do[
    minTime = Infinity;
    Do[
      time = 10*(counts[[i]] + 1)/(HarmonicNumber[10] - HarmonicNumber[i - 1]);
      If[time < minTime,minIndex = i;minTime = time],{i, 1, 10}];
    counts[[minIndex]] += 1;
    coeffs = Append[coeffs, minIndex - 1],
    {numCoeffs}
   ];
  coeffs
 ]
A386950 = samplePDF[120]

from sympy import harmonic
def sample_pdf(num_coeffs):
  coeffs, counts = [], [0]*10
  for _ in range(num_coeffs):
    min_time = float('inf')
    for i, count in enumerate(counts):
      time = 10*(count+1) / (harmonic(10)-harmonic(i))
      if time < min_time:
        min_index, min_time = i, time
    counts[min_index] += 1
    coeffs.append(min_index)
  return coeffs
A386950 = sample_pdf(120)

A386953 Total number of entries in rows 0,1,...,n of Pascal's triangle not divisible by 9.

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 28, 36, 45, 49, 57, 69, 78, 90, 105, 119, 135, 153, 160, 174, 195, 209, 228, 252, 273, 297, 324, 328, 336, 348, 360, 378, 402, 422, 450, 486, 495, 513, 540, 560, 588, 624, 655, 693, 738, 752, 780, 822, 850, 888, 936, 978, 1026, 1080, 1087

Offset: 0

Chai Wah Wu, Aug 10 2025

nonn

James G. Huard, Blair K. Spearman, and Kenneth S. Williams, Pascal's triangle (mod 9), Acta Arithmetica, 78 (1997), 331-349.

Cf. A001316, A006046-A006048, A194458, A194459, A382720-A382731.

Partial sums of A386952.

import re
from gmpy2 import digits
def A386953(n):
    c = 0
    for m in range(n+1):
        s = digits(m,3)
        n1 = s.count('1')
        n2 = s.count('2')
        n01 = s.count('10')
        n02 = s.count('20')
        n11 = len(re.findall('(?=11)',s))
        n12 = s.count('21')
        c += ((3*((1+n01<<2)+n11)+((n02<<2)+n12<<2))*3**n2<>2
    return c

A386952 Number of entries in the n-th row of Pascal's triangle not divisible by 9.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 4, 8, 12, 9, 12, 15, 14, 16, 18, 7, 14, 21, 14, 19, 24, 21, 24, 27, 4, 8, 12, 12, 18, 24, 20, 28, 36, 9, 18, 27, 20, 28, 36, 31, 38, 45, 14, 28, 42, 28, 38, 48, 42, 48, 54, 7, 14, 21, 20, 31, 42, 33, 48, 63, 14, 28, 42, 31, 44, 57, 48

Offset: 0

Chai Wah Wu, Aug 10 2025

nonn

James G. Huard, Blair K. Spearman, and Kenneth S. Williams, Pascal's triangle (mod 9), Acta Arithmetica, 78 (1997), 331-349.

Cf. A001316, A006047, A194459, A382720-A382725.

import re
from gmpy2 import digits
def A386952(n):
    s = digits(n,3)
    n1 = s.count('1')
    n2 = s.count('2')
    n01 = s.count('10')
    n02 = s.count('20')
    n11 = len(re.findall('(?=11)',s))
    n12 = s.count('21')
    return ((3*((1+n01<<2)+n11)+((n02<<2)+n12<<2))*3**n2<>2

A385956 Intersection of A025487 and A002378.

Original entry on oeis.org

2, 6, 12, 30, 72, 210, 240, 420, 1260, 6480, 50400, 147840, 510510, 4324320

Offset: 1

Ken Clements, Aug 10 2025

These numbers are the products of two consecutive integers that are also Hardy-Ramanujan integers; that is, of the form 2^k1*3^k2*...*p_n^k_n, where k1 >= k2 >= ... >= k_n. This sequence is finite with last term a(14) = 2079*2080 = 4324320.

			a(1) = 2 = 1*2 = 2^1.
a(2) = 6 = 2*3 = 2^1 * 3^1.
a(3) = 12 = 3*4 = 2^2 * 3^1.
a(4) = 30 = 5*6 = 2^1 * 3^1 * 5^1.
a(5) = 72 = 8*9 = 2^3 * 3^2.
a(6) = 210 = 14*15 = 2^1 * 3^1 * 5^1 * 7^1.

Cf. A055932, A025487, A002378, A385189, A386951.

Select[FactorialPower[Range[0, 3000], 2], (Max@Differences[(f = FactorInteger[#])[[;; , 2]]] < 1 && f[[-1, 1]] == Prime[Length[f]]) &] (* Amiram Eldar, Aug 10 2025 *)

from sympy import prime, factorint
def is_Hardy_Ramanujan(n):
    factors = factorint(n)
    p_idx = len(factors)
    if list(factors.keys())[-1] != prime(p_idx):
        return False
    expos = list(factors.values())
    e = expos[0]
    for i in range(1, p_idx):
        if expos[i] > e:
            return False
        e = expos[i]
    return True
print([ n*(n+1) for n in range(1, 10_000) if is_Hardy_Ramanujan(n*(n+1))])

A386951 Intersection of A025487 and A007531.

Original entry on oeis.org

6, 24, 60, 120, 210, 720, 3360, 9240, 166320, 970200, 43243200

Offset: 1

Ken Clements, Aug 10 2025

These numbers are the products of three consecutive integers that are also Hardy-Ramanujan integers; that is, of the form 2^k1*3^k2*...*p_n^k_n, where k1 >= k2 >= ... >= k_n. This sequence is finite with last term a(11) = 350*351*352 = 43243200.

			a(1) = 6 = 1*2*3 = 2^1 * 3^1.
a(2) = 24 = 2*3*4 = 2^3 * 3^1.
a(3) = 60 = 3*4*5 = 2^2 * 3^1 * 5^1.
a(4) = 120 = 4*5*6 = 2^3 * 3^1 * 5^1.
a(5) = 210 = 5*6*7 = 2^1 * 3^1 * 5^1 * 7^1.
a(6) = 720 = 8*9*10 = 2^4 * 3^2 * 5^1.

Cf. A055932, A025487, A007531, A385189, A385956.

Select[FactorialPower[Range[0, 1000], 3], (Max@ Differences[(f = FactorInteger[#])[[;; , 2]]] < 1 && f[[-1, 1]] == Prime[Length[f]]) &] (* Amiram Eldar, Aug 10 2025 *)

from sympy import prime, factorint
def is_Hardy_Ramanujan(n):
    factors = factorint(n)
    p_idx = len(factors)
    if list(factors.keys())[-1] != prime(p_idx):
        return False
    expos = list(factors.values())
    e = expos[0]
    for i in range(1, p_idx):
        if expos[i] > e:
            return False
        e = expos[i]
    return True
print([ n*(n+1)*(n+2) for n in range(1, 1000) if is_Hardy_Ramanujan(n*(n+1)*(n+2))])

A386949 Irregular triangle whose n-th row lists the nonzero terms of the n-th column of A386755.

Original entry on oeis.org

1, 1, 2, 2, 1, 3, 3, 3, 3, 1, 2, 2, 4, 4, 1, 5, 5, 5, 5, 5, 5, 1, 2, 2, 3, 3, 6, 6, 6, 6, 1, 7, 7, 7, 7, 7, 7, 7, 7, 1, 2, 2, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 1, 3, 3, 3, 3, 9, 9, 9, 9, 9, 9, 9, 9, 1, 2, 2, 5, 5, 5, 5, 10, 10, 10, 10, 10, 10, 1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11

Offset: 1

Michel Marcus, Aug 10 2025

			Triangle begins:
  1;
  1, 2, 2;
  1, 3, 3, 3, 3;
  1, 2, 2, 4, 4;
  1, 5, 5, 5, 5, 5, 5;
  1, 2, 2, 3, 3, 6, 6, 6, 6;
  1, 7, 7, 7, 7, 7, 7, 7, 7;
  1, 2, 2, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8;
  1, 3, 3, 3, 3, 9, 9, 9, 9, 9, 9, 9, 9;
  1, 2, 2, 5, 5, 5, 5, 10, 10, 10, 10, 10, 10;
  ...

Cf. A386755 (original triangle), A386520 (row sums).

Cf. A027750.

orow(n) = my(v=vector(n), m=n); for(k=1, n, my(keepm = m); while(m%k, m--); if (m == 0, keepm=m, v[m] = k; m--); ); v; \\ A386755
nrow(n) = my(ok=1, k=1, last=-1, list=List(), r); while(ok, r=row(k); if ((#r >= n) && r[n], listput(list, r[n])); k++; if (#r>=n, if ((last==n) && (r[n]==0), ok = 0, last = r[n]))); Vec(list);

A386636 Triangle read by rows where T(n,k) is the number of inseparable type set partitions of {1..n} into k blocks.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 4, 0, 0, 0, 1, 5, 0, 0, 0, 0, 1, 21, 15, 0, 0, 0, 0, 1, 28, 21, 0, 0, 0, 0, 0, 1, 92, 196, 56, 0, 0, 0, 0, 0, 1, 129, 288, 84, 0, 0, 0, 0, 0, 0, 1, 385, 1875, 1380, 210, 0, 0, 0, 0, 0, 0, 1, 561, 2860, 2145, 330, 0, 0, 0, 0, 0, 0

Offset: 0

Gus Wiseman, Aug 10 2025

A set partition is of inseparable type iff the underlying set has no permutation whose adjacent elements always belong to different blocks. Note that this only depends on the sizes of the blocks.
A set partition is also of inseparable type iff its greatest block size is at least 2 more than the sum of all its other block sizes.
This is different from inseparable partitions (A325535) and partitions of inseparable type (A386638 or A025065).

			Row n = 6 counts the following set partitions:
  .  {123456}  {1}{23456}  {1}{2}{3456}  .  .  .
               {12}{3456}  {1}{2345}{6}
               {13}{2456}  {1}{2346}{5}
               {14}{2356}  {1}{2356}{4}
               {15}{2346}  {1}{2456}{3}
               {16}{2345}  {1234}{5}{6}
               {1234}{56}  {1235}{4}{6}
               {1235}{46}  {1236}{4}{5}
               {1236}{45}  {1245}{3}{6}
               {1245}{36}  {1246}{3}{5}
               {1246}{35}  {1256}{3}{4}
               {1256}{34}  {1345}{2}{6}
               {1345}{26}  {1346}{2}{5}
               {1346}{25}  {1356}{2}{4}
               {1356}{24}  {1456}{2}{3}
               {1456}{23}
               {12345}{6}
               {12346}{5}
               {12356}{4}
               {12456}{3}
               {13456}{2}
Triangle begins:
    0
    0    0
    0    1    0
    0    1    0    0
    0    1    4    0    0
    0    1    5    0    0    0
    0    1   21   15    0    0    0
    0    1   28   21    0    0    0    0
    0    1   92  196   56    0    0    0    0
    0    1  129  288   84    0    0    0    0    0
    0    1  385 1875 1380  210    0    0    0    0    0

For separable partitions we have A386583, sums A325534, ranks A335433.
For inseparable partitions we have A386584, sums A325535, ranks A335448.
For separable type partitions we have A386585, sums A336106, ranks A335127.
For inseparable type partitions we have A386586, sums A386638 or A025065, ranks A335126.
Row sums are A386634.
The complement is counted by A386635, row sums A386633.
A000110 counts set partitions, row sums of A048993.
A000670 counts ordered set partitions.
A003242 and A335452 count anti-runs, ranks A333489, patterns A005649.
A279790 counts disjoint families on strongly normal multisets.
A335434 counts separable factorizations, inseparable A333487.
A336103 counts normal separable multisets, inseparable A336102.
A386587 counts disjoint families of strict partitions of each prime exponent.
Cf. A008284, A072233, A097805, A106351, A116861, A124762, A129506, A190945, A239455, A335125, A374252, A386575.

sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
stnseps[stn_]:=Select[Permutations[Union@@stn],And@@Table[Position[stn,#[[i]]][[1,1]]!=Position[stn,#[[i+1]]][[1,1]],{i,Length[#]-1}]&]
Table[Length[Select[sps[Range[n]],Length[#]==k&&stnseps[#]=={}&]],{n,0,5},{k,0,n}]

A386635 Triangle read by rows where T(n,k) is the number of separable type set partitions of {1..n} into k blocks.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, 3, 1, 0, 0, 3, 6, 1, 0, 0, 10, 25, 10, 1, 0, 0, 10, 75, 65, 15, 1, 0, 0, 35, 280, 350, 140, 21, 1, 0, 0, 35, 770, 1645, 1050, 266, 28, 1, 0, 0, 126, 2737, 7686, 6951, 2646, 462, 36, 1, 0, 0, 126, 7455, 32725, 42315, 22827, 5880, 750, 45, 1

Offset: 0

Gus Wiseman, Aug 10 2025

A set partition is of separable type iff the underlying set has a permutation whose adjacent elements always belong to different blocks. Note that this only depends on the sizes of the blocks.
A set partition is also of separable type iff its greatest block size is at most one more than the sum of all its other blocks sizes.
This is different from separable partitions (A325534) and partitions of separable type (A336106).

			Row n = 4 counts the following set partitions:
  .  .  {{1,2},{3,4}}  {{1},{2},{3,4}}  {{1},{2},{3},{4}}
        {{1,3},{2,4}}  {{1},{2,3},{4}}
        {{1,4},{2,3}}  {{1},{2,4},{3}}
                       {{1,2},{3},{4}}
                       {{1,3},{2},{4}}
                       {{1,4},{2},{3}}
Triangle begins:
    1
    0    1
    0    0    1
    0    0    3    1
    0    0    3    6    1
    0    0   10   25   10    1
    0    0   10   75   65   15    1
    0    0   35  280  350  140   21    1

Column k = 2 appears to be A128015.
For separable partitions we have A386583, sums A325534, ranks A335433.
For inseparable partitions we have A386584, sums A325535, ranks A335448.
For separable type partitions we have A386585, sums A336106, ranks A335127.
For inseparable type partitions we have A386586, sums A386638 or A025065, ranks A335126.
Row sums are A386633.
The complement is counted by A386636, row sums A386634.
A000110 counts set partitions, row sums of A048993.
A000670 counts ordered set partitions.
A003242 and A335452 count anti-runs, ranks A333489, patterns A005649.
A239455 counts Look-and-Say partitions, ranks A351294, conjugate A381432.
A335434 counts separable factorizations, inseparable A333487.
A336103 counts normal separable multisets, inseparable A336102.
A351293 counts non-Look-and-Say partitions, ranks A351295, conjugate A381433.
A386587 counts disjoint families of strict partitions of each prime exponent.
Cf. A072233, A097805, A106351, A116861, A190945, A279790, A333382, A335125, A374252, A386575, A386578.

sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
stnseps[stn_]:=Select[Permutations[Union@@stn],And@@Table[Position[stn,#[[i]]][[1,1]]!=Position[stn,#[[i+1]]][[1,1]],{i,Length[#]-1}]&];
Table[Length[Select[sps[Range[n]],Length[#]==k&&stnseps[#]!={}&]],{n,0,5},{k,0,n}]

cp's OEIS Frontend

User: ,10,

A386947 A variant of Recamán's sequence (A005132): a(0) = 0; for n > 0, a(n) = a(n-1) - n if nonnegative and not already in the sequence, otherwise a(n) = a(n-1) + n + 2.

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A386946 a(n) is the number of imprimitive (periodic) 2n-bead balanced binary necklaces.

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A386950 A sequence constructed by greedily sampling the pdf (H(10)-H(i))/10, where H(n) is the n-th Harmonic number and the support is [0,9], to minimize discrepancy.

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A386953 Total number of entries in rows 0,1,...,n of Pascal's triangle not divisible by 9.

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A386952 Number of entries in the n-th row of Pascal's triangle not divisible by 9.

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A385956 Intersection of A025487 and A002378.

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A386951 Intersection of A025487 and A007531.

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A386949 Irregular triangle whose n-th row lists the nonzero terms of the n-th column of A386755.

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PARI

A386636 Triangle read by rows where T(n,k) is the number of inseparable type set partitions of {1..n} into k blocks.

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