Sergio Pimentel - cp's OEIS frontend

A384665 Smallest odd multiplier k such that k*n is abundant.

945, 9, 315, 3, 189, 3, 135, 3, 105, 3, 315, 1, 315, 3, 63, 3, 315, 1, 315, 1, 45, 3, 315, 1, 63, 3, 35, 3, 315, 1, 315, 3, 105, 3, 27, 1, 315, 3, 105, 1, 315, 1, 315, 3, 21, 3, 315, 1, 45, 3, 105, 3, 315, 1, 63, 1, 105, 3, 315, 1, 315, 3, 15, 3, 63, 1, 315, 3

Offset: 1

Sergio Pimentel, Jun 06 2025

nonn

a(n) <= 945 for all n. a(n) = 945 for prime numbers > 103.

The possible values appear to be: 1, 3, 5, 7, 9, 15, 21, 25, 27, 35, 45, 63, 105, 135, 189, 315, 945. - Michel Marcus, Jun 12 2025

			a(5) = 189 because 189 is the smallest odd multiplier k such that 5*k is abundant (i.e., 5*189 = 945 which is abundant).

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A005101, A005231, A023196, A254571.

f:= proc(n) local k;
  for k from 1 by 2 do if numtheory:-sigma(n*k) > 2*n*k then return k fi od
end proc:
map(f, [$1..100]); # Robert Israel, Jun 09 2025

a[n_] := Module[{k = 1}, While[DivisorSigma[-1, k*n] <= 2, k += 2]; k]; Array[a, 100] (* Amiram Eldar, Jun 06 2025 *)

a(n) = my(k=1); while (sigma(k*n,-1)<=2, k+=2); k; \\ Michel Marcus, Jun 09 2025

A382606 Natural numbers ordered by the probability (highest to lowest) to occur in the sum of repeated rolls of a fair 6-sided die.

Original entry on oeis.org

6, 5, 11, 12, 10, 16, 17, 15, 21, 22, 27, 23, 26, 28, 32, 33, 38, 37, 39, 43, 44, 49, 48, 50, 54, 55, 60, 59, 53, 65, 61, 66, 64, 70, 71, 76, 75, 77, 81, 82, 87, 72, 86, 88, 92, 93, 98, 97, 103, 99, 104, 102, 108, 109, 114, 115, 113, 110, 119, 120, 125, 124, 126

Offset: 1

Sergio Pimentel, Mar 31 2025

nonn

The asymptotic probability for large n is 2/7 since the average roll of a die is 7/2.
Only terms with probability > 2/7 occur. - Michael S. Branicky, Apr 01 2025
Of any six consecutive integers, at least one is present and gives a maximum in the sequence (i.e., all terms preceding it are smaller). - Javier Múgica, May 01 2025

			The probability of achieving a '6' in n>=6 rolls is 1/6 + 5/36 + 10/216 + 10/1296 + 5/7776 + 1/46656 which is about 36.02%.
The probability of achieving a '1' is just 1/6 (about 16.67%). 6 is the highest of all, so a(1) = 6.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Complement of A382607. Cf. A365443.

from fractions import Fraction
from math import factorial, prod
from itertools import count, islice
from sympy.utilities.iterables import partitions
def prob(n): return sum(factorial(N:=sum(p.values()))//prod(factorial(v) for v in p.values())*Fraction(1, 6**N) for p in partitions(n, k=6))
def agen(): # generator of terms
    n, vdict = 1, dict()
    for k in count(1):
        vdict[prob(k)] = k
        if k%6 == 0:
            s = [vdict[v] for v in sorted(vdict, reverse=True) if v > Fraction(2, 7)]
            yield from (s[i-1] for i in range(n, len(s)-1))
            n = len(s) - 1
print(list(islice(agen(), 20))) # Michael S. Branicky, Apr 01 2025

A382607 Natural numbers ordered by the probability (lowest to highest) to occur in the sum of repeated rolls of a fair 6-sided die.

Original entry on oeis.org

1, 2, 3, 7, 4, 8, 13, 9, 14, 19, 18, 24, 25, 20, 30, 29, 31, 35, 36, 41, 40, 34, 46, 42, 47, 45, 51, 52, 57, 56, 58, 62, 63, 68, 67, 69, 73, 74, 79, 78, 84, 80, 85, 83, 89, 90, 95, 94, 96, 100, 101, 91, 106, 105, 107, 111, 112, 117, 116, 122, 118, 123, 128, 127

Offset: 1

Sergio Pimentel, Mar 31 2025

nonn

The asymptotic probability for large n is 2/7 since the average roll of a die is 7/2.
Only terms with probability < 2/7 occur. - Michael S. Branicky, Apr 01 2025
Of any six consecutive integers, at least one is present and gives a maximum in the sequence (i.e., all terms preceding it are smaller). - Javier Múgica, May 01 2025

			The probability of achieving a '6' in n>=6 rolls is 1/6 + 5/36 + 10/216 + 10/1296 + 5/7776 + 1/46656 which is about 36.02%.
The probability of achieving a '1' is just 1/6 (about 16.67%). 1 is the lowest of all, so a(1)=1.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Complement of A382606. Cf. A365443.

from fractions import Fraction
from math import factorial, prod
from itertools import count, islice
from sympy.utilities.iterables import partitions
def prob(n): return sum(factorial(N:=sum(p.values()))//prod(factorial(v) for v in p.values())*Fraction(1, 6**N) for p in partitions(n, k=6))
def agen(): # generator of terms
    n, vdict = 1, dict()
    for k in count(1):
        vdict[prob(k)] = k
        if k%6 == 0:
            s = [vdict[v] for v in sorted(vdict) if v < Fraction(2, 7)]
            yield from (s[i-1] for i in range(n, len(s)-1))
            n = len(s) - 1
print(list(islice(agen(), 20))) # Michael S. Branicky, Apr 01 2025

A382237 Numbers that are not divisible by the sum of any subset of their digits.

Original entry on oeis.org

23, 29, 34, 37, 38, 43, 46, 47, 49, 53, 56, 57, 58, 59, 67, 68, 69, 73, 74, 76, 78, 79, 83, 86, 87, 89, 94, 97, 98, 203, 223, 227, 229, 233, 239, 249, 253, 257, 263, 267, 269, 277, 283, 293, 299, 307, 323, 329, 334, 337, 338, 346, 347, 349, 353, 356, 358, 359, 367, 373, 376, 377, 379, 380, 383, 386, 388, 389, 394, 397, 398, 403

Offset: 1

Sergio Pimentel, Mar 19 2025

This sequence has density zero since no numbers with the digit '1' are in it. The sequence is infinite. Example: Numbers like 23, 203, 2003, 20003, etc. are included because none of them is divisible by 2, 3, or 5.

Conjecture: after a sufficiently large n this sequence grows faster than the prime numbers.

			358 is in the sequence because it can't be divided by 3, 5, 8, (3+5)=8, (3+8)=11, (5+8)=13 or (3+5+8)=16.
289 is not in the sequence because it can be divided by (8+9)=17.

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Cf. A082943, A038772, A005349, A065877, A228017, A382239.

filter:= proc(n) local L,S;
  L:= convert(n,base,10);
  andmap(s -> s=0 or n mod s <> 0, map(convert,combinat:-choose(L),`+`))
end proc:
select(filter, [$1..1000]); # Robert Israel, Mar 19 2025

isok(k) = my(d=digits(k)); forsubset(#d, s, my(ss=sum(i=1, #s, d[s[i]])); if (ss && !(k % sum(i=1, #s, d[s[i]])), return(0))); return(1); \\ Michel Marcus, Mar 27 2025

from itertools import chain, combinations
def powerset(s): # skipping empty set
    return chain.from_iterable(combinations(s, r) for r in range(1, len(s)+1))
def ok(n): return all(n%t!=0 for s in powerset(list(map(int, str(n)))) if (t:=sum(s))>0)
print([k for k in range(1, 404) if ok(k)]) # Michael S. Branicky, Apr 01 2025

A380745 a(0) = 0; a(n) = the number of times a(n-1) has the same digits in common with a previous term, in any permutation.

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, 0, 8, 0, 9, 0, 10, 0, 11, 0, 12, 0, 13, 0, 14, 0, 15, 0, 16, 0, 17, 0, 18, 0, 19, 0, 20, 0, 21, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 11, 1, 12, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 2

Offset: 0

Sergio Pimentel, Jan 31 2025

To find a(n), let L be the multiset of the digits of a(n-1). Then a(n) is the number of terms a(i), 0 <= i <= n-2, which also have L as the multiset of its digits. - N. J. A. Sloane, Mar 27 2025

			a(43) = 1 since a(42) = 21 and previously there is only one number in the sequence that contains both a 1 and a 2.
a(104) = 3 since a(103) = 11 and previously there are 3 numbers in the sequence that contain two 1's.
a(9942) = 14 since a(9941) = 155 and previously there are 14 numbers in the sequence that contain one 1 and two 5's.

Michael S. Branicky, Table of n, a(n) for n = 0..10000

Cf. A309261, A326834, A364788, A380690.

a[0] = 0; a[n_] := a[n] = Count[Array[a, n - 1, 0], ?(Sort[IntegerDigits[a[n - 1]]] == Sort[IntegerDigits[#]] &)]; Array[a, 100, 0] (* _Amiram Eldar, Jan 31 2025 *)

from collections import Counter
from itertools import count, islice
def agen(): # generator of terms
    an, digsetcount = 0, Counter()
    while True:
        yield an
        key = "".join(sorted(str(an)))
        an = digsetcount[key]
        digsetcount[key] += 1
print(list(islice(agen(), 82))) # Michael S. Branicky, Mar 24 2025

A380690 a(0) = 0; a(n) = the number of times a(n-1) has all digits in common with a previous term.

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, 0, 8, 0, 9, 0, 10, 0, 11, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 11, 12, 0, 12, 1, 13, 0, 13, 1, 14, 0, 14, 1, 15, 0, 15, 1, 16, 0, 16, 1, 17, 0, 17, 1, 18, 0, 18, 1, 19, 0, 19, 1, 20, 0, 20, 1, 21

Offset: 0

Sergio Pimentel, Jan 30 2025

Every number should appear a limited number of times in the sequence as opposed as in A326834.

			a(24) = 2 since a(23) = 1 and previously there are two numbers that only have the digit '1': a(22) = 11 and a(2) = 1.
a(4062) = 113 since a(4061) = 112 and previously there are 113 occurrences of numbers that only have the digits '1' and '2' such as 12,21,112,121,122.

Cf. A309261, A326834, A364788.

a[0] = 0; a[n_] := a[n] = Count[Array[a, n - 1, 0], ?(Union[IntegerDigits[a[n - 1]]] == Union[IntegerDigits[#]] &)]; Array[a, 100, 0] (* _Amiram Eldar, Jan 30 2025 *)

from collections import Counter
from itertools import count, islice
def agen(): # generator of terms
    an, digsetcount = 0, Counter()
    while True:
        yield an
        key = "".join(sorted(set(str(an))))
        an = digsetcount[key]
        digsetcount[key] += 1
print(list(islice(agen(), 80))) # Michael S. Branicky, Jan 30 2025

A378024 Smallest k >= 10 whose decimal representation read in base (n+10) is a multiple of k.

Original entry on oeis.org

1913268348854, 11788, 1557, 126357, 145, 1038, 1757, 116, 153, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 10, 21, 11, 23, 12, 25, 13, 27, 14, 29, 10, 31, 16, 11, 17, 35, 12, 37, 19, 13, 10, 41, 14, 43, 11, 15, 23, 47, 12, 49, 10, 17, 13, 53, 18, 11, 14, 19, 29, 59, 10, 61

Offset: 1

Sergio Pimentel, Nov 14 2024

This is a variation of A139285 that has terms beyond n=10.

			a(5)=145 because 145 read in base 15 is 1*15^2+4*15+5=290 = 2*145 and no other number below 145 has this property except for the trivial values < 10.

Cf. A139285.

a(n) = my(k=10); while(fromdigits(digits(k), n+10) % k, k++); k; \\ Michel Marcus, Nov 16 2024

More terms from Michel Marcus, Nov 16 2024

A371561 Numbers with multiplicative digital root of 5 that are free of 1s and have their digits in ascending order.

Original entry on oeis.org

5, 35, 57, 355, 359, 557, 579, 3335, 3357, 5579, 5777, 33557, 35559, 333555, 357799, 557779, 3335779, 3355777, 33333577

Offset: 1

Sergio Pimentel, Mar 27 2024

Conjectured to be complete.

If it exists, a(20) > 10^500. - Michael S. Branicky, Apr 18 2024

Cf. A034052, A263473, A263479, A031347.

A031347 = Table[NestWhile[Times @@ IntegerDigits[#] &, n, # > 9 &], {n, 1, 100000}]; Select[Range[100000], A031347[[#]] == 5 && DigitCount[#, 10, 1] == 0 && Sort[IntegerDigits[#]] == IntegerDigits[#] &] (* Vaclav Kotesovec, Apr 17 2024 *)

from math import prod
from itertools import count, islice, combinations_with_replacement as mc
def A031347(n):
    while n > 9: n = prod(map(int, str(n)))
    return n
def bgen(): yield from (m for d in count(1) for m in mc((3,5,7,9), d))
def agen(): yield from (int("".join(map(str, t))) for t in bgen() if A031347(prod(t)) == 5)
print(list(islice(agen(), 19))) # Michael S. Branicky, Apr 17 2024, edited Apr 18 2024 after Chai Wah Wu

from math import prod
from itertools import count, islice
def A371561_gen(): # generator of terms
    for l in count(1):
        for a in range(l,-1,-1):
            a3 = 3**a
            for b in range(l-a,-1,-1):
                b3 = a3*5**b
                for c in range(l-a-b,-1,-1):
                    d = l-a-b-c
                    d3 = b3*7**c*9**d
                    while d3 > 9:
                        d3 = prod(int(x) for x in str(d3))
                    if d3==5:
                        yield (10**(a+b+c+d)-1)//3+(10**d*(10**c*(10**b+1)+1)-3)*2//9
A371561_list = list(islice(A371561_gen(),19)) # Chai Wah Wu, Apr 17 2024

A370748 a(0)=1; a(n) = sum of all previous terms, eliminating repeated digits (starting from the left).

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 653, 6189, 7238, 7961, 875, 8452, 9604, 10658, 176, 17342, 13468, 14852, 16304, 179308, 35861, 3947, 39842, 43826, 48209, 5301, 53602, 589204, 17840, 196248, 139246, 153742, 16854

Offset: 0

Sergio Pimentel, Mar 07 2024

Duplicate digits are eliminated as in A137564.
No term can be greater than 9876543210.
a(260390084) = 9876543210. - Michael S. Branicky, Apr 09 2024

			a(17) = 653 since the sum of a(0) through a(16) is 65536, and eliminating repeated digits gives a(17) = A137564(65536) = 653.

Alois P. Heinz, Table of n, a(n) for n = 0..20000

Cf. A010784, A137564, A370254.

Cf. A371863 (records), A371864 (indices of records).

seq={1};Do[s=Total[seq];AppendTo[seq,FromDigits[DeleteDuplicates[IntegerDigits[s]]]],43];seq (* James C. McMahon, Apr 09 2024 *)

f(n) = my(d=digits(n)); fromdigits(vecextract(d, vecsort(vecsort(d, , 9)))) \\ A137564
lista(nn) = my(va=vector(nn)); va[1] = 1; for (n=2, nn, va[n] = f(sum(j=1, n-1, va[j]));); va; \\ Michel Marcus, Mar 08 2024

from itertools import islice
def A137564(n):
    seen, out, s = set(), "", str(n)
    for d in s:
        if d not in seen: out += d; seen.add(d)
    return int(out)
def A370748gen(): # generator of terms
    an, s = 1, 0
    while True:
        yield an
        s += an
        an = A137564(s)
print(list(islice(A370748gen(), 45))) # Michael S. Branicky, Apr 09 2024

A370254 a(0) = 1, a(n) = result of eliminating the digit 7 from the sum of all previous terms for n>=1.

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 3268, 36036, 202, 224, 2498, 4996, 9992, 89984, 19968, 199936, 39982, 439854, 8908, 888616, 1232, 18464, 196928, 1993856, 39812, 402524, 4430048, 8860096, 120192, 1840384, 1968068

Offset: 0

Sergio Pimentel, Feb 13 2024

			a(16) = 3268 since the sum of a(0)..a(15) = 32768. Eliminating the "7" we get 3268.

Alois P. Heinz, Table of n, a(n) for n = 0..16384

Cf. A004182, A011782, A069638, A108566.

a:= proc(n) option remember; `if`(n=0, 1, (l-> add(l[i]*10^(i-1),
      i=1..nops(l)))(subs(7=NULL, convert(s(n-1), base, 10))))
    end:
s:= proc(n) option remember; `if`(n<0, 0, s(n-1)+a(n)) end:
seq(a(n), n=0..40);  # Alois P. Heinz, Feb 15 2024

Module[{nmax=50, s=1, a}, NestList[(s+=(a=FromDigits[DeleteCases[IntegerDigits[s], 7]]); a) &, s, nmax]] (* Paolo Xausa, Feb 19 2024 *)

lista(nn) = my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = sum(k=1, n-1, va[k]); my(d=digits(va[n])); if (vecsearch(Set(d), 7), my(list = List()); for (i=1, #d, if (d[i] !=7, listput(list, d[i]));); va[n] = fromdigits(Vec(list)););); va; \\ Michel Marcus, Feb 13 2024

a(n) = A004182(Sum_{i=0..n-1} a(i)) for n >= 1, a(0) = 1.

a(n) = A011782(n) for n <= 15.

cp's OEIS Frontend

User: Sergio Pimentel

A384665 Smallest odd multiplier k such that k*n is abundant.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

A382606 Natural numbers ordered by the probability (highest to lowest) to occur in the sum of repeated rolls of a fair 6-sided die.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

A382607 Natural numbers ordered by the probability (lowest to highest) to occur in the sum of repeated rolls of a fair 6-sided die.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

A382237 Numbers that are not divisible by the sum of any subset of their digits.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Python

A380745 a(0) = 0; a(n) = the number of times a(n-1) has the same digits in common with a previous term, in any permutation.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python

A380690 a(0) = 0; a(n) = the number of times a(n-1) has all digits in common with a previous term.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

A378024 Smallest k >= 10 whose decimal representation read in base (n+10) is a multiple of k.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Extensions

A371561 Numbers with multiplicative digital root of 5 that are free of 1s and have their digits in ascending order.

Views

Author