A228407 -id:A228407 - cp's OEIS frontend

A377925 Records in A228407.

0, 11, 100, 101, 103, 110, 114, 120, 201, 210, 1000, 1003, 1007, 1008, 1020, 1029, 1030, 1040, 1047, 1048, 1082, 1208, 1280, 1802, 1820, 2018, 2081, 2108, 2180, 2801, 2810, 8012, 8021, 8102, 8120, 8201, 8210, 10002, 10004, 10007, 10020, 10060, 10080, 10081, 10100, 10105, 10113, 10304, 10502, 10520, 11125, 11152, 11215, 11251, 11512, 11521, 12005, 12050, 12115, 12151

Offset: 1

N. J. A. Sloane, Dec 14 2024

First 48 terms were computed by Robert G. Wilson v, Dec 31 2013: see Comments in A228407.

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms 1..514 from N. J. A. Sloane)

Cf. A228407, A229218, A377926.

from itertools import islice
from collections import Counter
def A377925_gen(): # generator of terms
    yield from (0,11)
    l, s, b, c = Counter('11'), 1, {11}, 11
    while True:
        i = s
        while True:
            if i not in b:
                li, o = Counter(str(i)), 0
                for d in (l+li).values():
                    if d % 2:
                        if o > 0:
                            break
                        o += 1
                else:
                    if i>c:
                        yield (c:=i)
                    l = li
                    b.add(i)
                    while s in b:
                        b.remove(s)
                        s += 1
                    break
            i += 1
A377925_list = list(islice(A377925_gen(),40)) # Chai Wah Wu, Dec 14 2024

A377926 Indices of records in A228407.

Original entry on oeis.org

0, 1, 4, 8, 13, 15, 22, 28, 29, 30, 31, 62, 78, 94, 108, 138, 169, 205, 238, 279, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 307, 419, 519, 611, 667, 884, 3341, 3415, 3839, 4623, 4665, 5974, 5975, 5976, 5977, 5978, 5979, 5980, 5981, 5982, 5983, 5984, 5985, 5986, 5987, 5988, 5989, 5990, 5991, 5992, 5993, 5994, 5995, 5996, 5997, 5998

Offset: 1

N. J. A. Sloane, Dec 14 2024

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms 1..514 from N. J. A. Sloane)

Cf. A228407, A229218, A377925.

from itertools import islice
from collections import Counter
def A377926_gen(): # generator of terms
    yield from (0,1)
    l, s, b, c, j = Counter('11'), 1, {11}, 11, 1
    while True:
        i = s
        while True:
            if i not in b:
                li, o = Counter(str(i)), 0
                for d in (l+li).values():
                    if d % 2:
                        if o > 0:
                            break
                        o += 1
                else:
                    j += 1
                    if i>c:
                        yield j
                        c = i
                    l = li
                    b.add(i)
                    while s in b:
                        b.remove(s)
                        s += 1
                    break
            i += 1
A377926_list = list(islice(A377926_gen(),40)) # Chai Wah Wu, Dec 14 2024

A229218 Position of n in A228407.

Original entry on oeis.org

0, 2, 6, 10, 17, 34, 45, 65, 81, 118, 3, 1, 5, 11, 18, 33, 46, 64, 80, 117, 7, 26, 9, 41, 23, 35, 44, 66, 82, 119, 14, 12, 42, 16, 54, 38, 51, 69, 85, 122, 21, 19, 24, 55, 145, 152, 197, 230, 271, 362, 179, 155, 36, 39, 153, 146, 185, 533

Offset: 0

Robert G. Wilson v, Nov 10 2013

nonn

The inverse sequence to A228407.

			a(0) = 0 since A228407(0) is 0;
a(1) = 2 since A228407(2) is 1;
a(2) = 6 since A228407(6) is 2;
a(11) = 1 since A228407(1) is 11; etc.

Cf. A228407.

(* do A228407 first, then *) lst = Array[a, 500, 0]; Flatten[ Table[ Position[lst, n], {n, 0, 57}] - 1]

A303571 a(n) = palindrome arising when A228407(n+1) is formed (if there is more than one, use the smallest).

Original entry on oeis.org

101, 111, 101, 10001, 10201, 212, 202, 10201, 12021, 232, 313, 1331, 13031, 30103, 10301, 13031, 343, 414, 1441, 14041, 40104, 14041, 14241, 2442, 12421, 12121, 12021, 102201, 102201, 102201, 1002001, 1005001, 15051, 515, 525, 2552, 15251, 15351, 3553, 13531

Offset: 0

N. J. A. Sloane, Apr 27 2018

Let A228407(n) = X, A228407(n+1) = Y. The digits of X and Y can be rearranged to form a palindrome, possibly in several ways; a(n) is the smallest such palindrome.

Rémy Sigrist, Table of n, a(n) for n = 0..29999
Rémy Sigrist, PARI program for A303571

Cf. A228407.

PARI
```
See Links section.
```

More terms from Rémy Sigrist, Dec 27 2018

A228730 Lexicographically earliest sequence of distinct nonnegative integers such that the sum of two consecutive terms is a palindrome in base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 16, 17, 27, 28, 38, 39, 49, 50, 51, 15, 7, 26, 18, 37, 29, 48, 40, 59, 42, 13, 9, 24, 20, 35, 31, 46, 53, 58, 8, 14, 19, 25, 30, 36, 41, 47, 52, 69, 32, 12, 10, 23, 21, 34, 43, 45, 54, 57, 44, 11, 22, 33, 55, 56, 65, 66, 75, 76, 85, 86, 95, 96

Offset: 0

Paul Tek, Aug 31 2013

From M. F. Hasler, Nov 09 2013: (Start)
At each step, choose the smallest number not occurring earlier and such that a(n+1)+a(n) are palindromes, for all n.
Conjectured to be a permutation of the nonnegative integers.
See A062932 where injectivity is replaced by monotonicity; the sequences differ from a(16)=15 on.
This is an "arithmetic" analog to sequences A228407 and A228410, where instead of the sum, the union of the digits of subsequent terms is considered. (End)

			a(1) + a(2) = 3.
a(2) + a(3) = 5.
a(3) + a(4) = 7.
a(4) + a(5) = 9.
a(5) + a(6) = 11.
a(6) + a(7) = 22.
a(7) + a(8) = 33.

Paul Tek, Table of n, a(n) for n = 0..10000 (corrected by Michel Marcus, Jan 19 2019)
Paul Tek, PERL program for this sequence

Cf. A002113, A055266.

Cf. A062932 (strictly increasing variant).

{a=0;u=0; for(n=1, 99, u+=1<A002113(a+k)&&(a=k)&&next(2)))} \\ M. F. Hasler, Nov 09 2013

Perl
```
See Link section.
```

from itertools import islice
def ispal(n): s = str(n); return s == s[::-1]
def agen(): # generator of terms
    aset, an, mink = {0}, 0, 1
    yield 0
    while True:
        k = mink
        while k in aset or not ispal(an + k): k += 1
        an = k; aset.add(an); yield an
        while mink in aset: mink += 1
print(list(islice(agen(), 70))) # Michael S. Branicky, Nov 07 2022

a(0)=0 added by M. F. Hasler, Nov 15 2013

A230891 Working in base 2: a(0)=0, thereafter a(n+1) is the smallest number not already in the sequence such that the bits of a(n) and a(n+1) together can be rearranged to form a palindrome.

Original entry on oeis.org

0, 11, 1, 10, 100, 111, 1000, 101, 110, 1001, 1010, 1100, 1111, 10000, 1011, 1101, 1110, 10001, 10010, 10100, 10111, 11000, 11011, 11101, 11110, 100000, 10011, 10101, 10110, 11001, 11010, 11100, 11111, 100001, 100010, 100100, 100111, 101000, 101011, 101101, 101110, 110000, 110011, 110101, 110110, 111001

Offset: 0

N. J. A. Sloane, Nov 11 2013

A binary version of A228407.
The palindrome must be a proper binary number, i.e. must begin with 1 (if it is > 0). Also, the union of the bits of a(n) and a(n+1) cannot contain both an odd number of 0's and an odd number of 1's.
Just as for A228407, we can ask: does every number appear? The answer is yes - see the Comments in A228407.

Chai Wah Wu, Table of n, a(n) for n = 0..30000 (terms 0..1024 from Robert G. Wilson v).

Cf. A228407, A230892 (these numbers written in base 10).

a[0] = 0; a[n_] := a[n] = Block[{k = 1, idm = IntegerDigits[ a[n - 1], 2], t = a@# & /@ Range[n - 1]}, Label[ start]; While[ MemberQ[t, k], k++]; While[ Select[ Permutations[ Join[idm, IntegerDigits[k, 2]]], #[[1]] != 0 && # == Reverse@# &] == {}, k++; Goto[start]]; k]; s = Array[a, 46, 0]; FromDigits@# & /@ IntegerDigits[s, 2] (* Robert G. Wilson v, Dec 31 2013 *)

from collections import Counter
A230891_list, l, s, b = [0, 11], Counter('11'), 1, {3}
for _ in range(30001):
    i = s
    while True:
        if i not in b:
            li, o = Counter(bin(i)[2:]), 0
            for d in (l+li).values():
                if d % 2:
                    if o > 0:
                        break
                    o += 1
            else:
                A230891_list.append(int(bin(i)[2:]))
                l = li
                b.add(i)
                while s in b:
                    b.remove(s)
                    s += 1
                break
        i += 1 # Chai Wah Wu, Jun 19 2016

A230892 Terms of A230891 written in base 10: the binary expansions of a(n) and a(n+1) taken together can be rearranged to form a palindrome.

Original entry on oeis.org

0, 3, 1, 2, 4, 7, 8, 5, 6, 9, 10, 12, 15, 16, 11, 13, 14, 17, 18, 20, 23, 24, 27, 29, 30, 32, 19, 21, 22, 25, 26, 28, 31, 33, 34, 36, 39, 40, 43, 45, 46, 48, 51, 53, 54, 57, 58, 60, 63, 64, 35, 37, 38, 41, 42, 44, 47, 49, 50, 52, 55, 56, 59, 61, 62, 65, 66, 68, 71

Offset: 0

N. J. A. Sloane, Nov 11 2013

See A230891 for precise definition.
Just as for A228407, we can ask: does every number appear? The answer is yes - see the Comments in A228407.
The difference d(n)=a(n)-n increases from d(3*2^(k-2)+2) = 1-2^(k-2) to d(3*2^(k-1)+1) = 1-2^(k-1), going through 0 at n=2^k+1 and n=2^k+2, cf. examples. - M. F. Hasler, Nov 12 2013
From Robert G. Wilson v, Nov 15 2013: (Start)
Beginning with k=3, each "grouping" of ever increasing terms, begins at 2^k + 3 and runs up to 2^(k+2) and includes 3*2^(k-1) terms.
Indices of powers of 2 occur at: 2, 3, 4, 6, 13, 25, 49, 97, 193, 385, 769, 1537, ..., which, except for 2, 3 & 6, is A181565: 3*2^n + 1.
When the index equals the term: 0, 4, 9, 10, 17, 18, 33, 34, 65, 66, 129, 130, 257, 258, 513, 514, 1025, 1026, 2049, 2050, ..., .
Parity of a(n) beginning at n=0: 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, ..., . (End)

			From _M. F. Hasler_, Nov 12 2013: (Start)
Let d(n)=a(n)-n, i.e., a(n)=n+d(n). Then we have, after initial values d(0..8)=(0, 2, -1, -1, 0, 2, 2, -2, -2), the
following pattern: d(9) = d(10) = 0, ..., d(13) = 3,
d(14) = -3, ..., d(17) = d(18) = 0, ..., d(25) = 7;
d(26) = -7, ..., d(33) = d(34) = 0, ..., d(49) = 15,
d(50) = -15, ..., d(65) = d(66) = 0, ..., d(97) = 31,
d(98) = -31, ..., d(129) = d(130) = 0, ..., d(193) = 63,
d(194) = -63,..., d(257) = d(258) = 0, ... (End)

Chai Wah Wu, Table of n, a(n) for n = 0..30000 (first 2051 terms from Robert G. Wilson v)
Robert G. Wilson v, Graph of the first 1025 terms

Cf. A230891, A228407.

{u=0; a=0; La=1; ha=0/*hack*/; for(n=1, 99, u+=1<=2^L,L++); bittest(ha+h=hammingweight(k),0)&&!bittest(La+L,0)&&next; !a&&k<3&&next; a=k; ha=h; La=L; break))} \\ M. F. Hasler, Nov 11 2013

from collections import Counter
A230892_list, l, s, b = [0, 3], Counter('11'), 1, {3}
for _ in range(30001):
    i = s
    while True:
        if i not in b:
            li, o = Counter(bin(i)[2:]), 0
            for d in (l+li).values():
                if d % 2:
                    if o > 0:
                        break
                    o += 1
            else:
                A230892_list.append(i)
                l = li
                b.add(i)
                while s in b:
                    b.remove(s)
                    s += 1
                break
        i += 1 # Chai Wah Wu, Jun 19 2016

A231920 Working in base 3: a(0)=0, thereafter a(n+1) is the smallest number not already in the sequence such that the bits of a(n) and a(n+1) together can be rearranged to form a palindrome.

Original entry on oeis.org

0, 11, 1, 10, 100, 12, 2, 20, 101, 22, 110, 202, 220, 1000, 102, 21, 111, 122, 212, 221, 1001, 112, 121, 200, 211, 222, 1002, 120, 201, 210, 1011, 1022, 1101, 1110, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210, 10000, 1010, 1100, 1111, 1122, 1212, 1221, 2002, 2020

Offset: 0

N. J. A. Sloane and Robert G. Wilson v, Nov 15 2013

This is a permutation of the nonnegative integers in base 3 - see the Comments in A228407 for the proof.

Cf. A230891, A231922, A231924, A231926, A231928, A231930, A231932, A228407, A231921.

a[0] = 0; a[n_] := a[n] = Block[{k = 1, idm = IntegerDigits[ a[n - 1], 3], t = a@# & /@ Range[n - 1]}, Label[ start]; While[ MemberQ[t, k], k++];  While[ Select[ Permutations[ Join[ idm, IntegerDigits[k, 3]]], #[[1]] != 0 && # == Reverse@# &] == {}, k++; Goto[ start]]; k]; s = Array[a, 60, 0]; FromDigits@# & /@ IntegerDigits[s, 3]

A231933 Terms of A231932 written in base 10: the binary expansions of a(n) and a(n+1) taken together can be rearranged to form a palindrome.

Original entry on oeis.org

0, 10, 1, 9, 81, 11, 2, 18, 82, 20, 3, 12, 28, 84, 27, 90, 30, 4, 13, 37, 85, 36, 94, 22, 38, 92, 19, 83, 99, 163, 171, 729, 86, 14, 5, 23, 47, 95, 32, 48, 93, 21, 29, 100, 24, 6, 15, 55, 87, 54, 96, 33, 57, 109, 31, 39, 112, 120, 256, 280, 336, 352, 732, 91, 16, 7, 25

Offset: 0

N. J. A. Sloane and Robert G. Wilson v, Nov 15 2013

See A231931 for precise definition.

This is a permutation of the nonnegative integers - see the Comments in A228407 for the proof.

Cf. A230892, A231921, A231923, A231925, A231927, A231929, A231931, A228407, A231932.

a[0] = 0; a[n_] := a[n] = Block[{k = 1, idm = IntegerDigits[ a[n - 1], 9], t = a@# & /@ Range[n - 1]}, Label[ start]; While[ MemberQ[t, k], k++];  While[ Select[ Permutations[ Join[ idm, IntegerDigits[k, 9]]], #[[1]] != 0 && # == Reverse@# &] == {}, k++; Goto[ start]]; k]; s = Array[a, 60, 0]

A231880 The digits of a(n) and a(n+1) together can be reordered to form a square; lexicographically earliest sequence of distinct nonnegative integers with this property.

Original entry on oeis.org

0, 10, 24, 3, 6, 1, 8, 14, 4, 9, 16, 12, 15, 21, 25, 2, 5, 22, 45, 18, 28, 81, 34, 20, 29, 7, 48, 13, 27, 19, 26, 11, 52, 30, 42, 39, 61, 33, 46, 17, 23, 40, 32, 49, 60, 57, 64, 35, 92, 43, 36, 31, 63, 54, 67, 41, 38, 44, 62, 47, 56, 70, 65, 74, 124, 69, 37

Offset: 0

N. J. A. Sloane, Nov 17 2013, based on a posting to the Sequence Fans Mailing List by Andrew Weimholt, Nov 12 2013

A231880 and A231881 eventually merge: A231880(2540) = 2536; A231881(2539) = 2541; A231880(2541,2542,..) = A231881(2540,2541,..) = 2544,2551; ... Hans Havermann, Nov 17 2013

Hans Havermann, Table of n, a(n) for n = 0..10000

A variant of A228407. Cf. A231881.

a[0] = 0; a[n_] := a[n] = Block[{k = 1, idm = IntegerDigits@ a[n - 1], t = a@# & /@ Range[n - 1]}, Label[ start]; While[ MemberQ[t, k], k++]; While[ Select[ Permutations[ Join[idm, IntegerDigits[ k]]], #[[1]] != 0 && IntegerQ[ Sqrt[ FromDigits[ #]]] &] == {}, k++; Goto[ start]]; k]; Array[a, 100, 0] (* Robert G. Wilson v, Nov 17 2013 *)

Corrected and extended by Hans Havermann, Nov 17 2013

More terms added (from b-file) by Jon E. Schoenfield, Dec 22 2013

cp's OEIS Frontend

A377925 Records in A228407.

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A377926 Indices of records in A228407.

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A229218 Position of n in A228407.

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A303571 a(n) = palindrome arising when A228407(n+1) is formed (if there is more than one, use the smallest).

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A228730 Lexicographically earliest sequence of distinct nonnegative integers such that the sum of two consecutive terms is a palindrome in base 10.

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PARI

Perl

Python

Extensions

A230891 Working in base 2: a(0)=0, thereafter a(n+1) is the smallest number not already in the sequence such that the bits of a(n) and a(n+1) together can be rearranged to form a palindrome.

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Mathematica

Python

A230892 Terms of A230891 written in base 10: the binary expansions of a(n) and a(n+1) taken together can be rearranged to form a palindrome.

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PARI

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A231920 Working in base 3: a(0)=0, thereafter a(n+1) is the smallest number not already in the sequence such that the bits of a(n) and a(n+1) together can be rearranged to form a palindrome.

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Mathematica

A231933 Terms of A231932 written in base 10: the binary expansions of a(n) and a(n+1) taken together can be rearranged to form a palindrome.