Ali Sada - cp's OEIS frontend

A387242 a(n) is the least k such that A334676(k) != k and the decimal string of n appears in A334676(k).

2, 42, 26, 28, 102, 32, 85, 172, 95, 20, 22, 242, 26, 28, 302, 32, 85, 236, 95, 402, 42, 442, 115, 482, 502, 522, 162, 562, 145, 260, 62, 1615, 266, 268, 712, 272, 148, 772, 278, 802, 82, 842, 215, 882, 902, 92, 235, 962, 245, 1002, 102, 1042, 265, 1078, 1102, 112, 285, 1162

Offset: 1

Ali Sada, Aug 23 2025

			A334676(242) = 121, the first term whose decimal expansion contains the substring "12"; hence a(12) = 242.
A334676(21) = A334676(42) = 21 contains "2" but a(2) = 42 since the first does not satisfy A334676(k) != k.

Cf. A334676.

# uses A334676() and neighs() from A334676
from itertools import count, islice
def subs(s): yield from (s[i:j] for i in range(len(s)) for j in range(i+1, len(s)+1))
def agen(): # generator of terms
    adict, n = dict(), 1
    for k in count(1):
        v = A334676(k)
        if v != k:
            for t in subs(str(v)):
                if (ti:=int(t)) not in adict:
                    adict[ti] = k
                    while n in adict:
                        yield adict[n]
                        n += 1
print(list(islice(agen(), 70))) # Michael S. Branicky, Aug 23 2025

A387071 a(n) is the least k such that A387070(k) = n. If no such k exists, a(n) = -1.

Original entry on oeis.org

0, 276, 5, 6, 2, 76, 25, 26, 261, 3, 2985, 642, 3606, 462, 320, 3694, 4, 3205, 2045, 643, 493, 96, 15, 106, 318, 16, 510, 963, 1091, 503, 452, 2705, 550, 482, 1520, 1169, 19, 1462, 4512, 1428, 97, 2639, 675, 2055, 12, 216, 119, 6525, 2135, 7, 726, 246, 674, 2146, 710, 816, 74, 1026, 7401, 23, 255, 2490, 75, 2510, 8, 4782, 1125, 1923

Offset: 0

Ali Sada, Aug 15 2025

If the definition of A387070 assigned -1 in the case of all digits being removed, then a(0) would be 1245 in this sequence.
Question: What is the greatest number of distinct digits of n such that a(n) != -1?
From Andrew Howroyd, Aug 16 2025: (Start)
a(1975308269136) = 4444444 shows that it is possible for 9 distinct digits to appear in n with a(n) != -1.
On the other hand, finding a(n) for particular n is more tricky. For example: a(123) = 99756, a(1234) = 958067, a(12345) = 98980766, a(123456) = 898989898970. It is not clear if a(1234567) != -1. (End)
From David A. Corneth, Aug 17 2025: (Start)
n and a(n) can have no digits in common for n > 0.
a(n) is not divisible by 100.
Proof: If a(n) is divisible by 100 then a(n) / 10 has the same property. n and a(n) have no digits in common so if n is divisible by 10 then a(n) is not.
a(1234567) = -1.
Proof by contradiction:
If a(1234567) > -1 then a(1234567)^2 ends in 7, 8, 9 or 0 as 8, 9 and 0 are the complementary digits of 1, 2, 3, 4, 5, 6, 7 and the digit before any (0 or more) of them must be 7.
The rightmost nonzero digit of a(1234567)^2 must be 9 as 7 and 8 are not square mod 10. If the rightmost nonzero digit of a(1234567)^2 is 9 then the rightmost nonzero digit of a(1234567) is 3 or 7. It would then have a digit in common with 1234567 which is impossible. (End)
From Michael S. Branicky, Aug 18 2025: (Start)
The first term for which no such k exists is a(12463) = -1.
Proof. a(0..12462) are readily computed (see Python program and b-file).
Assume for the sake of contradiction that n = 12463 and a(n) > 0 exists.
a(n)^2 cannot end in 3, so it must end in one of n's complementary digits: 0, 5, 7, 8 or 9.
If it ends in 9, then a(n) ends in 3, which is not allowed from above since 3 is in n.
7 and 8 are not the ends of squares, so a(n)^2 must end in 5 or 0.
Continuing, a(n)^2 must end in d0 or d5, where d in {3, 5, 7, 8 or 9}.
Only 00 is possible as the two last digits of a square.
Continuing, the only last three digits possible are 500 and 000 (900 is not allowed since a(n) would end in 30, sharing a 3 with n).
Continuing, a(n)^2 can only end in 0000, and a(n) must end in 00.
Contradiction using the Comment proved above. (End)

			276 is the smallest number such that, when all of its digits are removed from the decimal representation of its square, 76176, the result is 1. Therefore, a(1) = 276.

Michael S. Branicky, Table of n, a(n) for n = 0..12462 (terms 0..10000 from Andrew Howroyd)

Cf. A387070.

f[k_] := FromDigits[Select[IntegerDigits[k^2], FreeQ[IntegerDigits[k], #] &]]; seq[max_] := Module[{v = Array[f, max, 0], s = {}, k = 0, i}, While[NumberQ[(i = FirstPosition[v, k][[1]])], AppendTo[s, i - 1]; k++]; s]; seq[10000] (* Amiram Eldar, Aug 16 2025 *)

\\ here b(n) is A387070(n).
b(n)={my(S=Set(digits(n))); fromdigits(select(x->!setsearch(S,x), digits(n^2)))}
a(n)={for(k=0, oo, if(b(k)==n, return(k)))} \\ Andrew Howroyd, Aug 15 2025

# uses code in A387070
from itertools import count, product
def A387071(n):
    if n == 0: return 0
    s = sorted(set("0123456789") - set(str(n)))
    if s == [] or s == ["0"]: return -1
    return next(k for d in count(1) for t in product(s, repeat=d) if A387070(k:=int("".join(t)))==n)
print([A387071(n) for n in range(68)]) # Michael S. Branicky, Aug 16 2025

A387070 Remove every digit that appears in n from the decimal representation of n^2. If no digits remain, set a(n) = 0.

Original entry on oeis.org

0, 0, 4, 9, 16, 2, 3, 49, 64, 81, 0, 2, 44, 69, 96, 22, 25, 289, 324, 36, 4, 44, 484, 59, 576, 6, 7, 9, 74, 841, 9, 96, 104, 1089, 1156, 122, 129, 169, 1444, 1521, 16, 68, 176, 189, 1936, 202, 211, 2209, 230, 201, 2, 260, 704, 2809, 2916, 302, 313, 3249, 3364, 3481, 3, 372, 3844, 99, 9, 422, 435

Offset: 0

Ali Sada, Aug 15 2025

			a(25) = 6 since 25^2 = 625 and once we remove the 2 and 5, we are left with 6.
a(26) = 7 since 26^2 = 676 and once we remove the 6, we are left with 7.

Cf. A029793, A258682.

a[n_] := FromDigits[Select[IntegerDigits[n^2], FreeQ[IntegerDigits[n], #] &]]; Array[a, 100, 0] (* Amiram Eldar, Aug 16 2025 *)

a(n)={my(S=Set(digits(n))); fromdigits(select(x->!setsearch(S,x), digits(n^2)))} \\ Andrew Howroyd, Aug 15 2025

def A387070(k):
    s = set(str(k))
    t = "".join(d for d in str(k**2) if d not in s)
    return int(t) if t != "" else 0
print([A387070(n) for n in range(67)]) # Michael S. Branicky, Aug 16 2025

a(n) = A258682(n) for n <= 19.
From David A. Corneth, Aug 16 2025: (Start)
a(A029793(k)) = 0.
a(A029783(k)) = A029783(k)^2. (End)

A386395 Smallest final number reachable from n by successively replacing a substring of digits with its square root.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 8, 3, 10, 11, 12, 13, 12, 15, 2, 17, 18, 13, 20, 21, 22, 23, 22, 5, 26, 27, 28, 23, 30, 31, 32, 33, 32, 35, 6, 37, 38, 33, 20, 21, 22, 23, 22, 5, 26, 27, 28, 7, 50, 51, 52, 53, 52, 55, 56, 57, 58, 53, 60, 61, 62, 63, 8, 65, 66, 67, 68, 63, 70, 71, 72, 73, 72, 75, 76, 77, 78, 73, 80, 3, 82, 83, 82

Offset: 1

Ali Sada, Jul 20 2025

The substring replaced must be a square and cannot begin with a 0 digit.

There may be multiple different replacements possible and the aim is whatever steps reach the smallest final value.

			425 -> 25 -> 5.
1004 -> 102.
6251 -> 251 -> 51.
9616 -> 3616 -> 196 -> 136 -> 16 -> 4 -> 2.
9996 -> 9936 -> 996 -> 936 -> 96 -> 36 -> 6.

Michael S. Branicky, Table of n, a(n) for n = 1..10000
Michael S. Branicky, Scatter plot of n, a(n) for n = 1..10^6.

Cf. A000290, A048385.

from math import isqrt
def children(n):
    s = str(n)
    return set(int(s[:i]+str(r)+s[j:]) for i in range(len(s)) for j in range(i+1, len(s)+1) if (w:=s[i:j])[0]!='0' and (r:=isqrt(t:=int(w)))**2 == t)
def a(n):
    reach, expand = {n}, [n]
    while expand:
        q = expand.pop()
        for c in children(q):
            if c not in reach:
                reach.add(c)
                expand.append(c)
    return min(reach)
print([a(n) for n in range(1, 85)]) # Michael S. Branicky, Jul 20 2025

A385342 Square array A(n,k) read by descending antidiagonals, where each row n starts with an ordered list of positive integers then we swap A(n,k) with the number in the position A(n, A(n,k*n)).

Original entry on oeis.org

1, 2, 2, 3, 1, 3, 4, 6, 6, 4, 5, 8, 1, 8, 5, 6, 10, 12, 12, 10, 6, 7, 12, 15, 1, 15, 3, 7, 8, 14, 18, 20, 20, 2, 14, 8, 9, 16, 21, 24, 1, 24, 21, 4, 9, 10, 18, 24, 28, 30, 30, 28, 24, 18, 10, 11, 20, 27, 32, 35, 1, 35, 2, 27, 5, 11, 12, 22, 30, 36, 40, 42, 42, 40, 36, 30, 22, 12

Offset: 1

Ali Sada, Jun 26 2025

The lower triangular array is the multiplication table.
The second row is A073675.
The first superdiagonal and the first subdiagonal are both A002378.

			The square array begins:
  1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ...
  2, 1, 6, 8, 10, 3, 14, 4, 18, 5, 22, 24, 26, 7, 30, 32, 34, ...
  3, 6, 1, 12, 15, 2, 21, 24, 27, 30, 33, 4, 39, 42, 5, 48, 51, ...
  4, 8, 12, 1, 20, 24, 28, 2, 36, 40, 44, 3, 52, 56, 60, 64, 68, ...
  5, 10, 15, 20, 1, 30, 35, 40, 45, 2, 55, 60, 65, 70, 3, 80, 85, ...
  6, 12, 18, 24, 30, 1, 42, 48, 54, 60, 66, 2, 78, 84, 90, 96, 102, ...
  7, 14, 21, 28, 35, 42, 1, 56, 63, 70, 77, 84, 91, 2, 105, 112, 119, ...
  8, 16, 24, 32, 40, 48, 56, 1, 72, 80, 88, 96, 104, 112, 120, 2, 136, ...
  9, 18, 27, 36, 45, 54, 63, 72, 1, 90, 99, 108, 117, 126, 135, 144, 153, ...
  10, 20, 30, 40, 50, 60, 70, 80, 90, 1, 110, 120, 130, 140, 150, 160, 170, ...

Cf. A002378, A073675.

A384908 Start with a list L of positive integers. At each step n, let the center be the smallest number that has not been used as a center before with index m > 1. For all i < m, swap L(i) with L(i+m). a(n) = L(1).

Original entry on oeis.org

3, 4, 2, 5, 7, 1, 3, 9, 11, 4, 12, 6, 16, 15, 17, 5, 12, 19, 21, 8, 20, 23, 25, 10, 24, 27, 23, 21, 11, 31, 33, 34, 32, 35, 37, 31, 16, 33, 15, 41, 43, 17, 44, 18, 12, 47, 49, 50, 48, 51, 53, 47, 49, 55, 57, 50, 22, 59, 61, 26, 60, 29, 64, 66, 61, 67, 69, 70, 68, 65, 72, 28, 64, 75, 54, 27, 78, 80, 56

Offset: 1

Ali Sada, Jun 12 2025

nonn

			Start with 1,2,3,4,5,6,.. The first number with index > 1 that has not been used as a center is 2, so it becomes the center and the list becomes 3,2,1,4,5,6.. and a(1) = 3. Now, the center is 1 and the list becomes 4,5,1,3,2,6,7.. and a(n) = 4. Now, the center is 3 (since 1 has already been used as a center) and the list becomes 2,6,7,3,1,5,4,8,9.. and a(n) = 2, and so on.

from itertools import count, islice
def agen(): # generator of terms
    L, nextL, mink, cset = [1, 2], 3, 1, set() # cset is set of centers used
    while True:
        c, m = next((k, i) for k in count(mink) if k not in cset and (i:=L.index(k))>0)
        if len(L) < 2*m+1:
            L += list(range(nextL, 2*m+2))
            nextL = 2*m+2
        for i in range(m):
            L[i], L[i+m+1] = L[i+m+1], L[i]
        while mink in cset:
            mink += 1
        yield L[0]
        cset.add(c)
print(list(islice(agen(), 80))) # Michael S. Branicky, Jun 12 2025

A384034 Irregular triangle read by rows. Start with T(1,1) = 1. For each subsequent row, traverse the array so far. For each value m, insert m new values from the next unused integers immediately to the right of m. The process is repeated row by row, where each number in the array dictates how many new values are added after it.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 4, 5, 1, 6, 3, 7, 8, 9, 2, 10, 11, 4, 12, 13, 14, 15, 5, 16, 17, 18, 19, 20, 1, 21, 6, 22, 23, 24, 25, 26, 27, 3, 28, 29, 30, 7, 31, 32, 33, 34, 35, 36, 37, 8, 38, 39, 40, 41, 42, 43, 44, 45, 9, 46, 47, 48, 49, 50, 51, 52, 53, 54, 2, 55, 56

Offset: 1

Ali Sada, May 21 2025

			Triangle begins:
  1;
  1, 2;
  1, 3, 2, 4, 5;
  1, 6, 3, 7, 8, 9, 2, 10, 11, 4, 12, 13, 14, 15, 5, 16, 17, 18, 19, 20;
  ...
Row 1 = [1].
For m = 1, insert 1 new integer (next unused is 2) to the right of 1.
Row 2 = [1, 2].
For m = 1, insert 1 new integer (next unused is 3) to the right of 1.
For m = 2, insert 2 new integers (next unused are 4, 5) to the right of 2.
Row 3 = [1, 3, 2, 4, 5].
For m = 1, insert 1 new integer (6).
For m = 3, insert 3 new integers (7, 8, 9).
For m = 2, insert 2 new integers (10, 11).
For m = 4, insert 4 new integers (12, 13, 14, 15).
For m = 5, insert 5 new integers (16, 17, 18, 19, 20).
Row 4 = [1, 6, 3, 7, 8, 9, 2, 10, 11, 4, 12, 13, 14, 15, 5, 16, 17, 18, 19, 20].
And so on.

A383789 a(1) = 1; for n > 1, a(n) is the smallest positive integer not already in the sequence such that it shares at least one digit with a(n-1), and it has a different number of digits from a(n-1).

Original entry on oeis.org

1, 10, 100, 11, 101, 12, 2, 20, 102, 13, 3, 23, 103, 14, 4, 24, 104, 15, 5, 25, 105, 16, 6, 26, 106, 17, 7, 27, 107, 18, 8, 28, 108, 19, 9, 29, 109, 21, 110, 30, 113, 31, 111, 41, 112, 22, 120, 32, 121, 42, 114, 34, 123, 33, 130, 35, 115, 45, 124, 40, 134, 36, 116, 46, 126, 51, 117, 37, 127, 47, 137

Offset: 1

Ali Sada, May 09 2025

			a(6) = 12, and 2 is the smallest number not already in the sequence that shares a digit with 12 and has a different number of digits. So, a(7) = 2.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A184992, A303848.

function MySequence(N) a := [1]; while #a lt N do prev := a[#a]; d := Intseq(prev); m := #d; k := 2; repeat is_used := k in a; inter := &and[not x in d : x in Intseq(k)]; len_eq := #Intseq(k) eq m; k +:= 1; until not is_used and not inter and not len_eq; Append(~a, k - 1); end while; return a; end function; MySequence(100); // Vincenzo Librandi, May 17 2025

a[1] = 1; a[n_] := a[n] = Module[{k = 2, s = Array[a, n-1], d = IntegerDigits[a[n-1]], m = IntegerLength[a[n-1]]}, While[! FreeQ[s, k] || Intersection[IntegerDigits[k], d] == {} || IntegerLength[k] == m, k++]; k]; Array[a, 100] (* Amiram Eldar, May 10 2025 *)

isok(k, nbd, s, va) = if (#select(x->(x==k), va), return(0)); my(d=digits(k)); if (#setintersect(Set(d), s) && (#d != nbd), return(1));
find(va, n) = my(k=2, d=digits(va[n-1]), nbd=#d, s=Set(d)); while (!isok(k, nbd, s, va), k++); k;
lista(nn) = my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = find(va, n);); va; \\ Michel Marcus, May 13 2025

A383787 Largest number obtainable by either keeping each decimal digit d in n or replacing it with 9-d.

Original entry on oeis.org

8, 7, 6, 5, 5, 6, 7, 8, 9, 89, 88, 87, 86, 85, 85, 86, 87, 88, 89, 79, 78, 77, 76, 75, 75, 76, 77, 78, 79, 69, 68, 67, 66, 65, 65, 66, 67, 68, 69, 59, 58, 57, 56, 55, 55, 56, 57, 58, 59, 59, 58, 57, 56, 55, 55, 56, 57, 58, 59, 69, 68, 67, 66, 65, 65, 66, 67, 68, 69, 79, 78, 77, 76, 75, 75, 76, 77, 78

Offset: 1

Ali Sada, May 09 2025

			To find a(129) we replace 1 with 8 and 2 with 7. So, a(129) = 879.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Log log scatterplot of a(n), n = 1..10^6.

Cf. A061601, A383788.

a[n_] := FromDigits[IntegerDigits[n] /. d_?(# < 5 &) -> 9 - d]; Array[a, 100] (* Amiram Eldar, May 10 2025 *)

a(n) = fromdigits(apply(x->(if (x<5, 9-x, x)), digits(n))); \\ Michel Marcus, May 12 2025

def a(n): return int("".join(d if d>"4" else str(9-int(d)) for d in str(n)))
print([a(n) for n in range(1, 79)]) # Michael S. Branicky, May 10 2025

A383788 Smallest number obtainable by either keeping each decimal digit d in n or replacing it with 9-d.

Original entry on oeis.org

1, 2, 3, 4, 4, 3, 2, 1, 0, 10, 11, 12, 13, 14, 14, 13, 12, 11, 10, 20, 21, 22, 23, 24, 24, 23, 22, 21, 20, 30, 31, 32, 33, 34, 34, 33, 32, 31, 30, 40, 41, 42, 43, 44, 44, 43, 42, 41, 40, 40, 41, 42, 43, 44, 44, 43, 42, 41, 40, 30, 31, 32, 33, 34, 34, 33, 32, 31, 30, 20, 21, 22, 23, 24, 24, 23, 22

Offset: 1

Ali Sada, May 09 2025

			To find a(346), we replace 6 with 3. So, a(346) = 343.

Cf. A061601, A383787.

a[n_] := FromDigits[IntegerDigits[n] /. d_?(# > 4 &) -> 9 - d]; Array[a, 100] (* Amiram Eldar, May 10 2025 *)

a(n) = fromdigits(apply(x->(if (x>4, 9-x, x)), digits(n))); \\ Michel Marcus, May 12 2025

def a(n): return int("".join(d if d<"5" else str(9-int(d)) for d in str(n)))
print([a(n) for n in range(1, 78)]) # Michael S. Branicky, May 10 2025

cp's OEIS Frontend

User: Ali Sada

A387242 a(n) is the least k such that A334676(k) != k and the decimal string of n appears in A334676(k).

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A387071 a(n) is the least k such that A387070(k) = n. If no such k exists, a(n) = -1.

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A387070 Remove every digit that appears in n from the decimal representation of n^2. If no digits remain, set a(n) = 0.

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A386395 Smallest final number reachable from n by successively replacing a substring of digits with its square root.

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A385342 Square array A(n,k) read by descending antidiagonals, where each row n starts with an ordered list of positive integers then we swap A(n,k) with the number in the position A(n, A(n,k*n)).

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A384908 Start with a list L of positive integers. At each step n, let the center be the smallest number that has not been used as a center before with index m > 1. For all i < m, swap L(i) with L(i+m). a(n) = L(1).

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A383789 a(1) = 1; for n > 1, a(n) is the smallest positive integer not already in the sequence such that it shares at least one digit with a(n-1), and it has a different number of digits from a(n-1).

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A383787 Largest number obtainable by either keeping each decimal digit d in n or replacing it with 9-d.

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