A262038 -id:A262038 - cp's OEIS frontend

A265525 a(n) = largest base-10 palindrome m <= n such that every base-10 digit of m is <= the corresponding digit of n.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 11, 11, 11, 11, 11, 11, 11, 11, 11, 0, 11, 22, 22, 22, 22, 22, 22, 22, 22, 0, 11, 22, 33, 33, 33, 33, 33, 33, 33, 0, 11, 22, 33, 44, 44, 44, 44, 44, 44, 0, 11, 22, 33, 44, 55, 55, 55, 55, 55, 0, 11, 22, 33, 44, 55, 66, 66, 66, 66, 0, 11, 22, 33, 44, 55, 66, 77, 77, 77, 0, 11, 22, 33

Offset: 0

N. J. A. Sloane, Dec 09 2015

Reinhard Zumkeller, Table of n, a(n) for n = 0..9999

Sequences related to palindromic floor and ceiling: A175298, A206913, A206914, A261423, A262038, and the large block of consecutive sequences beginning at A265509.

Cf. A002113, A031298, A265558.

a265525 n = a265525_list !! n
a265525_list = f a031298_tabf [[]] where
   f (ds:dss) pss = y : f dss pss' where
     y = foldr (\d v -> 10 * v + d) 0 ys
     (ys:_) = dropWhile (\ps -> not $ and $ zipWith (<=) ps ds) pss'
     pss' = if ds /= reverse ds then pss else ds : pss
-- Reinhard Zumkeller, Dec 11 2015

ispal := proc(n) # test for base-b palindrome
local L, Ln, i;
global b;
L := convert(n, base, b);
Ln := nops(L);
for i to floor(1/2*Ln) do
if L[i] <> L[Ln + 1 - i] then return false end if
end do;
return true
end proc
# find max pal <= n and in base-b shadow of n, write in base 10
under10:=proc(n) global b;
local t1,t2,i,m,sw1,L2;
if n mod b = 0 then return(0); fi;
t1:=convert(n,base,b);
for m from n by -1 to 0 do
if ispal(m) then
t2:=convert(m,base,b);
L2:=nops(t2);
sw1:=1;
for i from 1 to L2 do
if t2[i] > t1[i] then sw1:=-1; break; fi;
od:
if sw1=1 then return(m); fi;
fi;
od;
end proc;
b:=10; [seq(under10(n),n=0..144)]; # Gives A265525

A265543 a(n) = smallest base-2 palindrome m >= n such that every base-2 digit of n is <= the corresponding digit of m; m is written in base 2.

Original entry on oeis.org

0, 1, 11, 11, 101, 101, 111, 111, 1001, 1001, 1111, 1111, 1111, 1111, 1111, 1111, 10001, 10001, 11011, 11011, 10101, 10101, 11111, 11111, 11011, 11011, 11011, 11011, 11111, 11111, 11111, 11111, 100001, 100001, 110011, 110011, 101101, 101101, 111111, 111111, 101101, 101101, 111111, 111111, 101101, 101101, 111111

Offset: 0

N. J. A. Sloane, Dec 09 2015

Sequences related to palindromic floor and ceiling: A175298, A206913, A206914, A261423, A262038, and the large block of consecutive sequences beginning at A265509.

See A206913 for the values of m written in base 10.

ispal:= proc(n) global b; # test if n is base-b palindrome
  local L, Ln, i;
  L:= convert(n, base, b);
  Ln:= nops(L);
for i from 1 to floor(Ln/2) do
if L[i] <> L[Ln+1-i] then return(false); fi;
od:
return(true);
end proc;
# find min pal >= n and with n in base-b shadow, write in base 10
over10:=proc(n) global b;
local t1,t2,i,m,sw1,L1;
t1:=convert(n,base,b);
L1:=nops(t1);
for m from n to 10*n do
if ispal(m) then
   t2:=convert(m,base,b);
   sw1:=1;
   for i from 1 to L1 do
      if t1[i] > t2[i] then sw1:=-1; break; fi;
                      od:
   if sw1=1 then return(m); fi;
fi;
                       od;
lprint("no solution in over10 for n = ", n);
end proc;
# find min pal >= n and with n in base-b shadow, write in base 10
overb:=proc(n) global b;
local t1,t2,i,m,mb,sw1,L1;
t1:=convert(n,base,b);
L1:=nops(t1);
for m from n to 10*n do
if ispal(m) then
   t2:=convert(m,base,b);
   sw1:=1;
   for i from 1 to L1 do
      if t1[i] > t2[i] then sw1:=-1; break; fi;
                      od:
   if sw1=1 then mb:=add(t2[i]*10^(i-1), i=1..nops(t2)); return(mb); fi;
fi;
                       od;
lprint("no solution in over10 for n = ", n);
end proc;
b:=2;
[seq(over10(n),n=0..144)]; # A175298
[seq(overb(n),n=0..144)]; # A265543

sb2p[n_]:=Module[{m=n},While[!PalindromeQ[IntegerDigits[m,2]]|| Min[ IntegerDigits[ m,2]-IntegerDigits[n,2]]<0,m++];FromDigits[ IntegerDigits[ m,2]]]; Array[sb2p,50,0] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 15 2017 *)

A265558 a(n) = smallest base-10 palindrome m >= n such that every base-10 digit of n is <= the corresponding digit of m.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 11, 22, 33, 44, 55, 66, 77, 88, 99, 22, 22, 22, 33, 44, 55, 66, 77, 88, 99, 33, 33, 33, 33, 44, 55, 66, 77, 88, 99, 44, 44, 44, 44, 44, 55, 66, 77, 88, 99, 55, 55, 55, 55, 55, 55, 66, 77, 88, 99, 66, 66, 66, 66, 66, 66, 66, 77, 88, 99, 77, 77, 77, 77, 77, 77, 77, 77, 88, 99, 88, 88

Offset: 0

N. J. A. Sloane, Dec 10 2015

Rémy Sigrist, Table of n, a(n) for n = 0..10000

Sequences related to palindromic floor and ceiling: A175298, A206913, A206914, A261423, A262038, and the large block of consecutive sequences beginning at A265509.

Cf. A265525.

Maple
```
For Maple code see A265543.
```

a(n,base=10) = { my (d=digits(n,base)); for (k=1, #d\2, d[k]=d[#d+1-k]=max(d[k],d[#d+1-k])); fromdigits(d,base) } \\ Rémy Sigrist, Jun 24 2022

A333016 Smallest palindromic number >= 2^n.

Original entry on oeis.org

1, 2, 4, 8, 22, 33, 66, 131, 262, 515, 1111, 2112, 4114, 8228, 16461, 32823, 65556, 131131, 262262, 524425, 1049401, 2097902, 4194914, 8388838, 16777761, 33555533, 67111176, 134222431, 268444862, 536878635, 1073773701, 2147557412, 4294994924, 8589999858, 17179897171

Offset: 0

Eder Vanzei, Mar 05 2020

			a(10) = 1111, because 2^10 = 1024 and 1111 is the smallest palindromic number >= 1024.

Chai Wah Wu, Table of n, a(n) for n = 0..3321

Cf. A000079, A002113, A262038, A333770.

spn[n_]:=Module[{k=2^n},While[!PalindromeQ[k],k++];k]; Array[spn,40,0] (* Harvey P. Dale, Jun 12 2023 *)

a(n) = for(k=2^n, oo, if(Vecrev(v=digits(k))==v, return(k))); \\ Jinyuan Wang, Mar 05 2020

a(n) = A262038(A000079(n)). - Michel Marcus, May 04 2020

a(9) corrected by and more terms from Jinyuan Wang, Mar 05 2020

A333770 Smallest palindromic number >= 3^n.

Original entry on oeis.org

1, 3, 9, 33, 88, 252, 737, 2222, 6666, 19691, 59095, 177771, 532235, 1594951, 4783874, 14355341, 43055034, 129141921, 387424783, 1162332611, 3486886843, 10460406401, 31381118313, 94143234149, 282429924282, 847288882748

Offset: 0

Eder Vanzei, Apr 04 2020

			a(10) = 59095, because 3^10 = 59049 and 59095 is the smallest palindromic number >= 59049.

Robert Israel, Table of n, a(n) for n = 0..2093

Cf. A000244, A002113, A262038, A333016.

digrev:= proc(n) local L,i;
  L:= convert(n,base,10);
  add(L[-i]*10^(i-1),i=1..nops(L))
end proc:
f:= proc(n) local d,x,y,t;
  d:= ilog10(n)+1;
  if d::even then
    x:= floor(n/10^(d/2));
    t:= x*10^(d/2)+digrev(x);
    if t >= n then return t fi;
    (x+1)*10^(d/2)+digrev(x+1);
  else
    x:= floor(n/10^((d-1)/2));
    t:= x*10^((d-1)/2)+digrev(floor(x/10));
    if t >= n then return t fi;
    y:= x mod 10;
    if y < 9 then return t + 10^((d-1)/2) fi;
    x:= x+1;
    x*10^((d-1)/2)+digrev(floor(x/10));
  fi
end proc:
seq(f(3^i),i=0..30); # Robert Israel, May 04 2020

a(n) = for(k=3^n, oo, if(Vecrev(v=digits(k))==v, return(k)));

a(n) = A262038(A000244(n)). - Michel Marcus, May 04 2020

A357044 Lexicographic earliest sequence of distinct palindromes (A002113) such that a(n)+a(n+1) is never palindromic.

Original entry on oeis.org

1, 9, 3, 7, 5, 8, 2, 11, 4, 6, 22, 88, 44, 66, 77, 33, 99, 55, 101, 909, 111, 191, 121, 181, 131, 171, 141, 161, 151, 252, 262, 242, 272, 232, 282, 222, 292, 212, 393, 313, 494, 323, 383, 333, 373, 343, 363, 353, 454, 464, 444, 474, 434, 484, 424

Offset: 1

Eric Angelini and M. F. Hasler, Sep 14 2022

Obviously the sequence cannot contain 0.

It is easy to prove that the sequence is a permutation of the nonzero palindromes (in the sense that it contains each of them exactly once).

Eric Angelini, Sums with palindromes, personal blog "Cinquante signes" on blogspot.com, and post to the math-fun list, Sep 12 2022
Eric Angelini, Sums with palindromes, personal blog "Cinquante signes" on blogspot.com, and post to the math-fun list, Sep 12 2022 [Cached copy]

Cf. A002113 (palindromes), A029742 (non-palindromes), A262038 (next palindrome), A357045 (non-palindromes with palindromic sum of neighbors).

A357044_first(n, U=[0], a=9)={vector(n,k, k=U[1]; while(is_A002113(a+k=A262038(k+1)) || setsearch(U, k), ); U=setunion(U,[a=k]); while(#U>1 && U[2]==A262038(U[1]+1), U=U[^1]); a)}

from itertools import count, islice
def ispal(n): s = str(n); return s == s[::-1]
def nextpal(p): # next largest palindrome after palindrome p
    d = str(p)
    if set(d) == {'9'}: return int('1' + '0'*(len(d)-1) + '1')
    h = str(int(d[:(len(d)+1)//2]) + 1)
    return int(h + h[:-1][::-1]) if len(d)&1 else int(h + h[::-1])
def agen():
    aset, pal, minpal = {1}, 1, 2
    while True:
        an = pal; yield an; aset.add(an); pal = minpal
        while pal in aset or ispal(an+pal): pal = nextpal(pal)
        while minpal in aset: minpal = nextpal(minpal)
print(list(islice(agen(), 55))) # Michael S. Branicky, Sep 14 2022

A265511 a(n) = largest base-3 palindrome m <= n such that every base-3 digit of m is <= the corresponding digit of n; m is written in base 10.

Original entry on oeis.org

0, 1, 2, 0, 4, 4, 0, 4, 8, 0, 10, 10, 0, 13, 13, 0, 16, 16, 0, 10, 20, 0, 13, 23, 0, 16, 26, 0, 28, 28, 0, 28, 28, 0, 28, 28, 0, 28, 28, 0, 40, 40, 0, 40, 40, 0, 28, 28, 0, 40, 40, 0, 52, 52, 0, 28, 56, 0, 28, 56, 0, 28, 56, 0, 28, 56, 0, 40, 68, 0, 40, 68, 0, 28, 56, 0, 40, 68, 0, 52, 80, 0, 82, 82, 0, 82, 82, 0, 82

Offset: 0

N. J. A. Sloane, Dec 09 2015

Robert Israel, Table of n, a(n) for n = 0..10000

Sequences related to palindromic floor and ceiling: A175298, A206913, A206914, A261423, A262038, and the large block of consecutive sequences beginning at A265509.

F:= proc(n) local L;
  L:= convert(n,base,3);
  if L[1] = 0 then return 0 fi;
  add(min(L[i],L[-i])*3^(i-1),i=1..nops(L))
end proc:
map(F, [$0..100]); # Robert Israel, Jan 13 2020

A265523 a(n) = largest base-9 palindrome m <= n such that every base-9 digit of m is <= the corresponding digit of n; m is written in base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 10, 10, 10, 10, 10, 10, 10, 10, 0, 10, 20, 20, 20, 20, 20, 20, 20, 0, 10, 20, 30, 30, 30, 30, 30, 30, 0, 10, 20, 30, 40, 40, 40, 40, 40, 0, 10, 20, 30, 40, 50, 50, 50, 50, 0, 10, 20, 30, 40, 50, 60, 60, 60, 0, 10, 20, 30, 40, 50, 60, 70, 70, 0, 10, 20, 30, 40, 50, 60, 70, 80, 0, 82, 82

Offset: 0

N. J. A. Sloane, Dec 09 2015

Robert Israel, Table of n, a(n) for n = 0..10000

Sequences related to palindromic floor and ceiling: A175298, A206913, A206914, A261423, A262038, and the large block of consecutive sequences beginning at A265509.

F:= proc(n) local L;
  L:= convert(n,base,9);
if L[1] = 0 then return 0 fi;
  add(min(L[i],L[-i])*9^(i-1),i=1..nops(L))
end proc:
map(F, [$0..100]); # Robert Israel, Jan 13 2020

A265526 Largest base-2 palindrome m <= n, written in base 2.

Original entry on oeis.org

0, 1, 1, 11, 11, 101, 101, 111, 111, 1001, 1001, 1001, 1001, 1001, 1001, 1111, 1111, 10001, 10001, 10001, 10001, 10101, 10101, 10101, 10101, 10101, 10101, 11011, 11011, 11011, 11011, 11111, 11111, 100001, 100001, 100001, 100001, 100001, 100001, 100001, 100001, 100001, 100001, 100001, 100001, 101101, 101101, 101101

Offset: 0

N. J. A. Sloane, Dec 09 2015

Sequences related to palindromic floor and ceiling: A175298, A206913, A206914, A261423, A262038, and the large block of consecutive sequences beginning at A265509.

ispal:= proc(n) global b; # test for base-b palindrome
  local L, Ln, i;
  L:= convert(n, base, b);
  Ln:= nops(L);
for i from 1 to floor(Ln/2) do
if L[i] <> L[Ln+1-i] then return(false); fi;
od:
return(true);
end proc;
# find max pal <= n, write in base 10
less10:=proc(n) global b;
local t1,t2,i,m,sw1,L2;
t1:=convert(n,base,b);
for m from n by -1 to 0 do
if ispal(m) then return(m); fi;
                        od;
end proc;
# find max pal <= n, write in base b
lessb:=proc(n) global b;
local t1,t2,i,m,mb,sw1,L2;
t1:=convert(n,base,b);
for m from n by -1 to 0 do
if ispal(m) then
   t2:=convert(m,base,b);
   L2:=nops(t2);
   mb:=add(t2[i]*10^(i-1), i=1..L2); return(mb); fi;
                        od;
end proc;
b:=2;
[seq(less10(n),n=0..100)]; # A206913
[seq(lessb(n),n=0..100)]; # A265526
[seq(less10(2*n),n=0..100)]; # A265527
[seq(lessb(2*n),n=0..100)]; # A265528
b:=10;
[seq(less10(n),n=0..100)]; # A261423

A265527 Largest base-2 palindrome m <= 2n, written in base 10.

Original entry on oeis.org

0, 1, 3, 5, 7, 9, 9, 9, 15, 17, 17, 21, 21, 21, 27, 27, 31, 33, 33, 33, 33, 33, 33, 45, 45, 45, 51, 51, 51, 51, 51, 51, 63, 65, 65, 65, 65, 73, 73, 73, 73, 73, 73, 85, 85, 85, 85, 93, 93, 93, 99, 99, 99, 99, 107, 107, 107, 107, 107, 107, 119, 119, 119, 119, 127, 129, 129, 129, 129, 129, 129, 129, 129, 129, 129, 129

Offset: 0

N. J. A. Sloane, Dec 09 2015

Sequences related to palindromic floor and ceiling: A175298, A206913, A206914, A261423, A262038, and the large block of consecutive sequences beginning at A265509.

cp's OEIS Frontend

A265525 a(n) = largest base-10 palindrome m <= n such that every base-10 digit of m is <= the corresponding digit of n.

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Maple

A265543 a(n) = smallest base-2 palindrome m >= n such that every base-2 digit of n is <= the corresponding digit of m; m is written in base 2.

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Mathematica

A265558 a(n) = smallest base-10 palindrome m >= n such that every base-10 digit of n is <= the corresponding digit of m.

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Maple

PARI

A333016 Smallest palindromic number >= 2^n.

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A333770 Smallest palindromic number >= 3^n.

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A357044 Lexicographic earliest sequence of distinct palindromes (A002113) such that a(n)+a(n+1) is never palindromic.

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Python

A265511 a(n) = largest base-3 palindrome m <= n such that every base-3 digit of m is <= the corresponding digit of n; m is written in base 10.

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A265523 a(n) = largest base-9 palindrome m <= n such that every base-9 digit of m is <= the corresponding digit of n; m is written in base 10.

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Maple

A265526 Largest base-2 palindrome m <= n, written in base 2.

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