A369603 -id:A369603 - cp's OEIS frontend

A369604 T is a "boomerang sequence": adding 9 to the 1st digit of T, 10 to the 2nd digit of T, 11 to the 3rd digit of T, 12 to the 4th digit of T, 13 to the 5th digit of T, 14 to the 6th digit of T, etc., and following each result with a comma leaves T unchanged.

10, 10, 12, 12, 14, 16, 16, 18, 18, 22, 20, 26, 22, 28, 24, 32, 26, 34, 29, 30, 31, 30, 33, 38, 35, 36, 37, 44, 39, 42, 42, 42, 43, 48, 46, 48, 47, 55, 50, 48, 52, 51, 54, 52, 56, 57, 58, 64, 60, 63, 62, 66, 64, 69, 67, 68, 68, 75, 71, 70, 73, 72, 75, 74, 77, 77, 79, 84, 81, 84, 83, 88, 85, 89, 88, 89, 90, 86, 91

Offset: 1

Eric Angelini and Giorgos Kalogeropoulos, Jan 27 2024

Lexicographically earliest sequence starting with a(1) = 10.

			Adding  9 to 1 (the 1st digit of 10) gives 10
Adding 10 to 0 (the 2nd digit of 10) gives 10
Adding 11 to 1 (the 1st digit of 10) gives 12
Adding 12 to 0 (the 2nd digit of 10) gives 12
Adding 13 to 1 (the 1st digit of 12) gives 14
Adding 14 to 2 (the 2nd digit of 12) gives 16, etc.
We see that the last column above is the sequence T itself.

Éric Angelini and Giorgos Kalogeropoulos, The same sequence but differently, personal blog, Jan 24th 2024.

Cf. A103700, A369603, A369798, A369823, A369824.

a[1]=10;a[n_]:=a[n]=Flatten[IntegerDigits/@Array[a,n-1]][[n]]+8+n;Array[a,100]

from itertools import count, islice
def agen(): # generator of terms
    an, digits = 10, [0]
    for n in count(2):
        yield an
        an = n + 8 + digits.pop(0)
        digits += list(map(int, str(an)))
print(list(islice(agen(), 79))) # Michael S. Branicky, Jan 27 2024

A369798 S is a "boomerang sequence": multiply each digit d of S by the number to which d belongs: the sequence S remains identical to itself if we follow each multiplication with a comma.

Original entry on oeis.org

0, 1, 12, 24, 48, 96, 192, 384, 864, 576, 192, 1728, 384, 1152, 3072, 1536, 6912, 5184, 3456, 2880, 4032, 3456, 192, 1728, 384, 1728, 12096, 3456, 13824, 1152, 3072, 1536, 1152, 1152, 5760, 2304, 9216, 0, 21504, 6144, 1536, 7680, 4608, 9216, 41472, 62208, 6912, 13824, 25920, 5184, 41472, 20736, 10368, 13824

Offset: 1

Eric Angelini and Jean-Marc Falcoz, Feb 01 2024

S is the lexicographycally earliest nontrivial sequence of nonnegative integers with this property (if we try for a(3) the integers 1, 10 or 11, we respectively get these trivial sequences):
S = 1, 1, 1, 1, 1, 1, 1, ...
S = 1, 10, 0, 0, 0, 0, 0, ...
S = 1, 11, 1, 1, 1, 1, 1, ...

			a(1) = 0, which multiplied by 0 gives 0
a(2) = 1, which multiplied by 1 gives 1
a(3) = 12
     1st digit is 1, which multiplied by 12 gives 12
     2nd digit is 2, which multiplied by 12 gives 24
a(4) = 24
     1st digit is 2, which multiplied by 24 gives 48
     2nd digit is 4, which multiplied by 24 gives 96
a(5) = 48
     1st digit is 4, which multiplied by 48 gives 192
     2nd digit is 8, which multiplied by 48 gives 384
a(6) = 96
     1st digit is 9, which multiplied by 96 gives 864
     2nd digit is 6, which multiplied by 96 gives 576
Etc. We see that the above last column reproduces S.

Eric Angelini and Jean-Marc Falcoz, Boomerang sequences, Personal blog, Feb 1st 2024.

Cf. A369603, A369604, A369823, A369824.

Join[{0,1},Nest[Flatten[IntegerDigits@#*#]&,{12},5]] (* Giorgos Kalogeropoulos, Feb 01 2024 *)

from itertools import islice
from collections import deque
def agen(): # generator of terms
    S = deque([24])
    yield from [0, 1, 12]
    while True:
        an = S.popleft()
        yield an
        S.extend(an*d for d in map(int, str(an)))
print(list(islice(agen(), 54))) # Michael S. Branicky, Feb 01 2024

A369823 S is a "boomerang sequence": replace each digit d of S by its sixth power: the sequence S remains identical to itself if we follow each result with a comma.

Original entry on oeis.org

0, 1, 4096, 0, 531441, 46656, 0, 15625, 729, 1, 4096, 4096, 1, 4096, 46656, 46656, 15625, 46656, 0, 1, 15625, 46656, 64, 15625, 117649, 64, 531441, 1, 4096, 0, 531441, 46656, 4096, 0, 531441, 46656, 1, 4096, 0, 531441, 46656, 4096, 46656, 46656, 15625, 46656, 4096, 46656, 46656, 15625, 46656, 1

Offset: 1

Eric Angelini, Feb 02 2024

S is the lexicographycally earliest sequence of nonnegative integers with this property.

			a(1) = 0, which raised at the 6th power gives 0
a(2) = 1, which raised at the 6th power gives 1
a(3) = 4096
     1st digit is 4, which raised at the 6th power gives 4096
     2nd digit is 0, which raised at the 6th power gives 0
     3rd digit is 9, which raised at the 6th power gives 531441
     4th digit is 6, which raised at the 6th power gives 46656
Etc. We see that the above last column reproduces S.

Eric Angelini and Jean-Marc Falcoz, Boomerang sequences, Personal blog, Feb 1st 2024.

Cf. A369603, A369604, A369798, A369824, A001014.

a[1]=0;a[2]=1;a[3]=4^6;a[n_]:=a[n]=Flatten[IntegerDigits/@Array[a,n-1]][[n]]^6;Array[a,52] (* Giorgos Kalogeropoulos, Feb 04 2024 *)

A369824 S is a "boomerang sequence": replace each digit d of S by its eighth power: the sequence S remains identical to itself if we follow each result with a comma.

Original entry on oeis.org

0, 1, 256, 390625, 1679616, 6561, 43046721, 0, 1679616, 256, 390625, 1, 1679616, 5764801, 43046721, 1679616, 1, 1679616, 1679616, 390625, 1679616, 1, 65536, 6561, 0, 65536, 1679616, 5764801, 256, 1, 0, 1, 1679616, 5764801, 43046721, 1679616, 1, 1679616, 256, 390625, 1679616, 6561, 43046721

Offset: 1

Eric Angelini, Feb 02 2024

S is the lexicographycally earliest sequence of nonnegative integers with this property.

			a(1) = 0, which raised at the 8th power gives 0
a(2) = 1, which raised at the 8th power gives 1
a(3) = 256
     1st digit is 2, which raised at the 8th power gives 256
     2nd digit is 5, which raised at the 8th power gives 390625
     3rd digit is 6, which raised at the 8th power gives 1679616
Etc. We see that the above last column reproduces S.

Éric Angelini and Jean-Marc Falcoz, Boomerang sequences, Personal blog, Feb 1st 2024.

Cf. A369603, A369604, A369798, A369823, A001014, A001016.

a[1]=0;a[2]=1;a[3]=2^8;a[n_]:=a[n]=Flatten[IntegerDigits/@Array[a,n-1]][[n]]^8;Array[a,43] (* Giorgos Kalogeropoulos, Feb 04 2024 *)

A369127 S is a "boomerang sequence". Replace each digit d of S by a base-10 palindrome not yet used that contains d: the sequence S remains identical to itself if we follow each palindrome with a comma.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 101, 111, 202, 121, 131, 141, 151, 22, 303, 212, 161, 222, 171, 181, 33, 191, 313, 44, 414, 515, 55, 616, 232, 242, 323, 404, 333, 252, 717, 262, 818, 66, 919, 272, 282, 292, 1001, 77, 1111, 1221, 88, 1331, 343, 353, 1441, 99

Offset: 1

Eric Angelini, Mar 02 2024

			As a(1) to a(10) are single-digit palindromes, the replacement leaves the terms a(1) to a(10) as they were.
a(11) = 11 and we must replace the first digit d = 1 of 11 by the smallest base-10 palindrome not yet used; this is 11 (as the palindrome 1 has been used before);
a(11) = 11 and we must replace now the second digit d = 1 of 11 by the smallest base-10 palindrome not yet used; this is 101 (as 1 and 11 have been used before);
a(12) = 101 and we must replace the first digit d = 1 of 101 by the smallest base-10 palindrome not yet used; this is 111;
a(12) = 101 and we must now replace the digit d = 0 of 101 by the smallest base-10 palindrome not yet used; this is 202 (as 0 and 101 have been used before);
a(12) = 101 and we must now replace the last digit d = 1 of 101 by the smallest base-10 palindrome not yet used; this is 121; etc.

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

Cf. A002113, A369603, A369604, A369823, A369824, A369798.

from collections import deque
from itertools import count, islice
def pals(start=1): # generator of palindromes >= palindrome start
    s = str(start)
    q, r = divmod(len(s)+1, 2)
    for d in count(q):
        olst = [1, 0][int(d==q and r==1):]
        for offset in olst:
            lb = max(1, 10**(d-1)) if d>q or offset!=olst[0] else int(s[:q])
            for i in range(lb, 10**d):
                left = str(i)
                yield int(left+left[::-1][offset:])
def agen(): # generator of terms
    S = deque([101])
    head = list(range(10)) + [11]
    yield from head
    used = set(head) | {101}
    pstart = {d:0 for d in "0123456789"}
    while True:
        an = S.popleft()
        yield an
        for d in str(an):
            p = next(p for p in pals(start=pstart[d]) if p not in used and d in str(p))
            pstart[d] = p
            S.append(p)
            used.add(p)
print(list(islice(agen(), 57))) # Michael S. Branicky, Mar 03 2024

More terms from Michael S. Branicky, Mar 03 2024

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A369604 T is a "boomerang sequence": adding 9 to the 1st digit of T, 10 to the 2nd digit of T, 11 to the 3rd digit of T, 12 to the 4th digit of T, 13 to the 5th digit of T, 14 to the 6th digit of T, etc., and following each result with a comma leaves T unchanged.

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A369798 S is a "boomerang sequence": multiply each digit d of S by the number to which d belongs: the sequence S remains identical to itself if we follow each multiplication with a comma.

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A369823 S is a "boomerang sequence": replace each digit d of S by its sixth power: the sequence S remains identical to itself if we follow each result with a comma.

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A369824 S is a "boomerang sequence": replace each digit d of S by its eighth power: the sequence S remains identical to itself if we follow each result with a comma.

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Mathematica

A369127 S is a "boomerang sequence". Replace each digit d of S by a base-10 palindrome not yet used that contains d: the sequence S remains identical to itself if we follow each palindrome with a comma.

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