Gleb Ivanov - cp's OEIS frontend

A358178 a(n) is the cardinality of the set of distinct pairwise gcd's of {1! + 1, ..., n! + 1}.

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

Offset: 1

Gleb Ivanov, Nov 02 2022

nonn

			For n = 6 initial set is {1+1, 2+1, 6+1, 24+1, 120+1, 720+1} and after applying gcd for each distinct pair of elements we get {1, 7} set with cardinality of a(6) = 2.

Cf. A038507, A358127, A356371, A214799.

from math import gcd, factorial
from itertools import combinations
f, terms = [2,], []
for i in range(2,100):
    f.append(factorial(i)+1)
    terms.append(len(set([gcd(*t) for t in combinations(f, 2)])))
print(terms)

from math import gcd
from itertools import count, islice
def A358178_gen(): # generator of terms
    m, f, g = 1, [], set()
    for n in count(1):
        m *= n
        g |= set(gcd(d,m+1) for d in f)
        f.append(m+1)
        yield len(g)
A358178_list = list(islice(A358178_gen(),20)) # Chai Wah Wu, Dec 15 2022

A358154 a(n) is the smallest composite number obtained by appending one or more 1's to n.

Original entry on oeis.org

111, 21, 3111, 411, 51, 611, 711, 81, 91, 1011, 111, 121, 1311, 141, 15111, 161, 171, 18111, 1911, 201, 21111, 221, 231, 24111, 2511, 261, 27111, 2811, 291, 301, 3111, 321, 3311, 341, 351, 361, 371, 381, 391, 4011, 411, 42111, 4311, 441, 451, 4611, 471, 481, 4911, 501, 511, 5211, 531, 5411

Offset: 1

Gleb Ivanov, Nov 01 2022

a(n) is either 10*n+1, 100*n+11 or 1000*n+111, because at least one of these is divisible by 3. - Robert Israel, Nov 01 2022

Actually: exactly one of these is divisible by 3. Almost all terms are a(n) = 10n + 1: this is the case for about (k-1)/k of the terms up to 10^k (i.e., 69% ~ 2/3 up to 10^3, 76% = 3/4 up to 10^4, 80% = 4/5 up to 10^5, 83% = 5/6 up to 10^6). - M. F. Hasler, Nov 03 2022

Cf. A069568, A112386, A002808, A153275.

f:= proc(n) local x;
   x:= n;
   do
     x:= 10*x+1;
     if not isprime(x) then return x fi;
   od
end proc:
map(f, [$1..100]); # Robert Israel, Nov 01 2022

a[n_] := NestWhile[10*# + 1 &, 10*n + 1, ! CompositeQ[#] &]; Array[a, 54] (* Amiram Eldar, Nov 01 2022 *)

a(n) = my(d=digits(n), m); if (!isprime(n), d = concat(d, 1)); while(isprime(m=fromdigits(d)), d=concat(d, 1)); m; \\ Michel Marcus, Nov 01 2022

from sympy import isprime
def A358154(n):
    t = str(n)+'1'
    while isprime(int(t)):t=t+'1'
    return int(t)
print([A358154(i) for i in range(1, 100)])

A358615 Record high values in A358497.

Original entry on oeis.org

1, 12, 122, 123, 1222, 1223, 1232, 1233, 1234, 12222, 12223, 12232, 12233, 12234, 12322, 12323, 12324, 12332, 12333, 12334, 12342, 12343, 12344, 12345, 122222, 122223, 122232, 122233, 122234, 122322, 122323, 122324, 122332, 122333, 122334, 122342, 122343, 122344

Offset: 1

Gleb Ivanov, Nov 23 2022

Cf. A358497, A071159.

def A358497(n):
    d, s, k = dict(), str(n), 1
    for i in range(len(s)):
        if d.get(s[i], 0) == 0:
            d[s[i]] = str(k)
            k = (k + 1)%10
    s_t = list(s)
    for i in range(len(s)):s_t[i] = d[s[i]]
    return int(''.join(s_t))
terms = [1,]
for i in range(1, 10**6):
    if A358497(i)>terms[-1]:terms.append(A358497(i))
print(terms)

A358497 Replace each new digit in n with index 1, 2, ..., 9, 0 in order in which that digit appears in n, from left to right.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 12, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 11, 12

Offset: 0

Gleb Ivanov, Nov 19 2022

a(n) gives the canonical form which enumerates digits in order of their first occurrence in n, from left to right. a(n) uniquely defines the pattern of identical digits in n. - Dmytro Inosov, Jul 16 2024

			n = 10 has 2 different digits; replace first encountered digit 1 -> 1, replace second digit 0 -> 2, therefore a(10) = 12.
n = 141 has 3 digits, but only 2 different ones; replace first encountered digit 1 -> 1, replace second encountered digit 4 -> 2, therefore a(141) = 121.

David A. Corneth, Table of n, a(n) for n = 0..9999
Dmytro S. Inosov and Emil Vlasák, Cryptarithmically unique terms in integer sequences, arXiv:2410.21427 [math.NT], 2024. See pp. 3, 18.

Cf. A071159, A227362, A358615 (record high values).

A358497[k_] := With[{pI = Values@PositionIndex@IntegerDigits@k}, MapIndexed[#1 -> Mod[#2[[1]], 10] &, pI, {2}] // Flatten // SparseArray // FromDigits]; (* Dmytro Inosov, Jul 15 2024 *)

a(n) = {if(n == 0, return(1)); my(d = digits(n), m = Map(), t = 0); for(i = 1, #d, if(mapisdefined(m, d[i]), d[i] = mapget(m, d[i]) , t++; if(t == 10, t = 0); mapput(m, d[i], t); d[i] = t ) ); fromdigits(d) } \\ David A. Corneth, Nov 23 2022

def A358497(n):
    d,s,k = dict(),str(n),1
    for i in range(len(s)):
        if d.get(s[i],0) == 0:
            d[s[i]] = str(k)
            k = (k + 1)%10
    s_t = list(s)
    for i in range(len(s)):s_t[i] = d[s[i]]
    return int(''.join(s_t))
print([A358497(i) for i in range(100)])

def A358497(n):
    s, c, i  = str(n), {}, 49
    for d in map(ord,s):
        if d not in c:
            c[d] = i
            i += 1
    return int(s.translate(c)) # Chai Wah Wu, Jul 09 2024

a(a(n)) = a(n). - Dmytro Inosov, Jul 16 2024

A358127 a(n) is the cardinality of the set of pairwise gcd's of {prime(1)+1, ..., prime(n)+1}.

Original entry on oeis.org

1, 3, 4, 5, 5, 5, 5, 7, 8, 8, 8, 9, 9, 11, 12, 14, 14, 14, 14, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20, 21, 22, 22, 23, 23, 23, 23, 23, 24, 24, 26, 27, 29, 29, 30, 32, 32, 33, 35, 36, 36, 37, 37, 37, 37, 38, 38, 39, 39, 39, 39, 40, 40, 42, 42, 43, 43, 43, 44, 45, 45, 48, 48, 48, 48, 50, 50, 50, 50

Offset: 2

Gleb Ivanov, Oct 30 2022

nonn

			For n = 3 initial set is {2+1, 3+1, 5+1} and after applying gcd for each distinct pair of elements we get {1, 2, 3} set with cardinality of a(3) = 3.

Cf. A008864, A356371, A214799.

from sympy import nextprime
from math import gcd
from itertools import combinations
pr, terms = [2,3], []
for i in range(100):
    terms.append(len(set([gcd(t[0]+1, t[1]+1) for t in combinations(pr,2)])))
    pr.append(nextprime(pr[-1]))
print(terms)

from math import gcd
from itertools import count, islice
from sympy import prime
def A358127_gen(): # generator of terms
    a, b = [3], set()
    for n in count(2):
        q = prime(n)+1
        b |= set(gcd(p,q) for p in a)
        yield len(b)
        a.append(q)
A358127_list = list(islice(A358127_gen(),100)) # Chai Wah Wu, Nov 02 2022

A357195 a(n) is the smallest palindrome of the form k(2n+k-1)/2 where k is a positive integer.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 33, 11, 969, 222, 99, 66, 33, 242, 282, 424, 161, 66, 22, 212, 252, 646, 171, 55, 252, 414, 555, 525, 99, 33, 474, 1001, 111, 5005, 77, 484, 1111, 1881, 414, 808, 44, 606, 141, 404, 303, 99, 101, 555, 444, 333, 222, 55, 171, 484

Offset: 1

Gleb Ivanov, Sep 17 2022

Cf. A020485, A002113, A262038.

ispal(p) = my(d=digits(p)); d==Vecrev(d);
a(n) = my(k=1); while(!ispal(x=k*(2*n+k-1)/2), k++); x; \\ Michel Marcus, Sep 17 2022

pal10 = lambda n: str(n) == str(n)[::-1]
def seq(n):
    k = 1
    while not pal10(k*(2*n+k-1)//2):k+=1
    return k*(2*n+k-1)//2
print([seq(n) for n in range(1, 100)])

from itertools import count
def A357195(n): return next(filter(lambda n:(s := str(n))[:(t:=len(s)+1>>1)]==s[:-t-1:-1],(k*((n<<1)+k-1)>>1 for k in count(1)))) # Chai Wah Wu, Oct 29 2022

A357126 a(n) is the smallest positive integer k such that k > n and A071364(k) = A071364(n).

Original entry on oeis.org

3, 5, 9, 7, 10, 11, 27, 25, 14, 13, 20, 17, 15, 21, 81, 19, 50, 23, 28, 22, 26, 29, 40, 49, 33, 125, 44, 31, 42, 37, 243, 34, 35, 38, 100, 41, 39, 46, 56, 43, 66, 47, 45, 52, 51, 53, 80, 121, 75, 55, 63, 59, 250, 57, 88, 58, 62, 61, 84, 67, 65, 68, 729, 69, 70, 71, 76, 74, 78, 73, 200, 79, 77, 98

Offset: 2

Gleb Ivanov, Oct 26 2022

nonn

			a(12) = 20 as 12 has (2, 1) sequence of exponents in canonical prime factorization via 12 = 2^2 * 3^1 and the smallest positive integer > 12 with the same sequence of exponents in canonical prime factorization being (2, 1) is 20 as 20 = 2^2 * 5^1. - _David A. Corneth_, Oct 26 2022

Cf. A000961, A003961, A071364, A065642, A081382, A081761.

f4(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = prime(i)); factorback(f); \\ A071364
a(n) = my(k=n+1, f=f4(n)); while (f4(k) != f, k++); k; \\ Michel Marcus, Oct 26 2022

first(n) = { my(res = vector(n + 1), todo = n, m = Map(), u = precprime(n)); for(e = 2, logint(n, 2), u = max(u, nextprime(sqrtnint(n, e) + 2)^e) ); forfactored(i = 2, u, cs = i[2][,2]; if(mapisdefined(m, cs), ci = mapget(m, cs); if(ci <= n + 1, res[ci] = i[1]; mapput(m, cs, i[1]); todo--; if(todo <= 0, res = res[^1]; return(res) ) ) , if(i[1] <= n + 1, mapput(m, cs, i[1]) ) ) ) } \\ David A. Corneth, Oct 26 2022

from sympy import factorint
to_s_exp = lambda n: tuple(i[1] for i in sorted(factorint(n).items()))
terms = []
for i in range(2, 100):
    k = i+1;t = to_s_exp(i)
    while t != to_s_exp(k):k+=1
    terms.append(k)
print(terms)

a(A000961(k)) = a(A003961(A000961(k))) for k > 1. - David A. Corneth, Oct 26 2022

a(n) >= A081761(n). - Rémy Sigrist, Feb 16 2023

A357314 a(1) = 1; a(n) is the second smallest number k such that k > a(n-1) and concatenation of a(1), ..., a(n-1), k is a palindrome.

Original entry on oeis.org

1, 21, 1121, 1211121, 2111211211121, 112112111212111211211121, 12111212111211211121112112111212111211211121, 211121121112111211211121211121121112112111212111211211121112112111212111211211121

Offset: 1

Gleb Ivanov, Sep 23 2022

Conjecture: Length A055642(a(n)) = A000073(n+2), and A305393 is a sequence of digits in the concatenation of all terms in this sequence.

			For n = 3 concatenation of the previous terms is 121. Numbers that would make it a palindrome if concatenated to it are 121, 1121, ... and the second smallest of them is a(3) = 1121.

Cf. A000073, A002275, A055642, A083122, A305393.

pal = lambda s: s == s[::-1]
up_to = 10
terms = [1, ]
for i in range(up_to-1):
    c, r = ''.join(map(str, terms)), 0
    for j in range(len(str(terms[-1])), len(c)+1):
        found, p  = False, int(c[:j][::-1])
        if p > terms[-1] and pal(c + c[:j][::-1]):
            r+=1
            if r == 2:
                terms.append(p);found = True;break
    if found: continue
    j = 0
    while 1:
        j+=1
        r+=1
        if r == 2:
            terms.append(int(str(j) + c[::-1]))
            break
print(terms)

A356371 a(n) is the smallest positive integer k, such that set of pairwise gcd of k, k+1, ..., k+n has a cardinality of n.

Original entry on oeis.org

1, 2, 3, 8, 15, 24, 35, 48, 63, 270, 440, 528, 780, 1078, 2925, 1440, 8160, 2142, 5472, 34560, 23919, 235598, 64239, 42480, 158400, 1255800, 1614600, 1247400, 16442971, 8233650, 41021370, 21561120, 127327167, 439824000, 439824000, 24504444, 1329112224, 1653775162

Offset: 1

Gleb Ivanov, Oct 17 2022

nonn

n | a(n). - David A. Corneth, Oct 17 2022

Giovanni Resta, Table of n, a(n) for n = 1..60 (first 42 terms from Chai Wah Wu)

Cf. A214799.

a[n_] := Module[{k = 1}, While[Length[Union[GCD @@@ Subsets[k + Range[0, n], {2}]]] != n, k++]; k]; Array[a, 20] (* Amiram Eldar, Oct 17 2022 *)

from math import gcd
from itertools import count
def A356371(n):
    for k in count(n,n):
        if len(set(gcd(i,j) for i in range(k,n+k+1) for j in range(i+1,n+k+1))) == n:
            return k # Chai Wah Wu, Oct 18 2022

a(31)-a(38) from Giovanni Resta, Oct 17 2022

A355966 Number of 3-dimensional polyominoes (or polycubes) with n cells that have cavities (inclusions of empty space).

Original entry on oeis.org

20, 404, 6164, 75917, 835491

Offset: 11

Gleb Ivanov, Jul 21 2022

Even if two polycubes are mirror images of each other, they are considered different for this sequence.
Polycubes with less than 11 cells can't have cavities.
Largest enclosed volume >= A355880(n-5) for polycubes with n cells.

Gleb Ivanov, Python program for n = 11..14
Wikipedia, Polycube

Cf. A001419, A000162, A038119, A355880.

Cf. A357083 (without distinguished reflections).

cp's OEIS Frontend

User: Gleb Ivanov

A358178 a(n) is the cardinality of the set of distinct pairwise gcd's of {1! + 1, ..., n! + 1}.

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A358154 a(n) is the smallest composite number obtained by appending one or more 1's to n.

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Maple

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A358615 Record high values in A358497.

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A358497 Replace each new digit in n with index 1, 2, ..., 9, 0 in order in which that digit appears in n, from left to right.

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Mathematica

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A358127 a(n) is the cardinality of the set of pairwise gcd's of {prime(1)+1, ..., prime(n)+1}.

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Python

Python

A357195 a(n) is the smallest palindrome of the form k*(2*n+k-1)/2 where k is a positive integer.

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PARI

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Python

A357126 a(n) is the smallest positive integer k such that k > n and A071364(k) = A071364(n).

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PARI

PARI

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A357314 a(1) = 1; a(n) is the second smallest number k such that k > a(n-1) and concatenation of a(1), ..., a(n-1), k is a palindrome.

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A357195 a(n) is the smallest palindrome of the form k(2n+k-1)/2 where k is a positive integer.