A020485 -id:A020485 - cp's OEIS frontend

A050782 Smallest positive multiplier m such that m*n is palindromic (or zero if no such m exists).

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 21, 38, 18, 35, 17, 16, 14, 9, 0, 12, 1, 7, 29, 21, 19, 37, 9, 8, 0, 14, 66, 1, 8, 15, 7, 3, 13, 15, 0, 16, 6, 23, 1, 13, 9, 3, 44, 7, 0, 19, 13, 4, 518, 1, 11, 3, 4, 13, 0, 442, 7, 4, 33, 9, 1, 11, 4, 6, 0, 845, 88, 4, 3, 7, 287, 1, 11, 6, 0, 12345679, 8

Offset: 0

Patrick De Geest, Oct 15 1999

Multiples of 81 require the largest multipliers.
From Jon E. Schoenfield, Jan 15 2015: (Start)
In general, a(n) is large when n is a multiple of 81. E.g., for n in [1..10000], of the 9000 terms where a(n)>0, 111 are at indices n that are multiples of 81; of the remaining 8889 terms,
     755 are in [1..9],
    1760 are in [10..99],
    3439 are in [100..999],
    2180 are in [1000..9999],
     708 are in [10000..99999],
      36 are in [100000..999999],
       6 are in [1000000..9999999],
       2 are in [10000000..99999999],
       2 are in [100000000..999999999],
and 1 (the largest) is a(8891) = 8546948927,
but the smallest of the 111 terms whose indices are multiples of 81 is a(2997)=333667. (End)
a(n) = 0 iff 10 | n. a(n) = 1 iff n is a palindrome. If k | a(n) then a(k*n) = a(n)/k. - Robert Israel, Jan 15 2015

			E.g., a(81) -> 81 * 12345679 = 999999999 and a palindrome.

Giovanni Resta, Table of n, a(n) for n = 0..10000 (first 8181 terms from Chai Wah Wu)
Patrick De Geest, World!Of Numbers

Cf. A002113, A020485, A050810.

digrev:= proc(n) local L,d,i;
  L:= convert(n,base,10);
  d:= nops(L);
  add(L[i]*10^(d-i),i=1..d);
end proc:
f:= proc(n)
local d,d2,x,t,y;
if n mod 10 = 0 then return 0 fi;
if n < 10 then return 1 fi;
for d from 2 do
  if d::even then
    d2:= d/2;
    for x from 10^(d2-1) to 10^d2-1 do
       t:= x*10^d2 + digrev(x);
       if t mod n = 0 then return(t/n) fi;
    od
  else
    d2:= (d-1)/2;
    for x from 10^(d2-1) to 10^d2-1 do
      for y from 0 to 9 do
        t:= x*10^(d2+1)+y*10^d2+digrev(x);
        if t mod n = 0 then return(t/n) fi;
      od
    od
  fi
od;
end proc:
seq(f(n),n=0 .. 100); # Robert Israel, Jan 15 2015

t={0}; Do[i=1; If[IntegerQ[n/10],y=0,While[Reverse[x=IntegerDigits[i*n]]!=x,i++]; y=i]; AppendTo[t,y],{n,80}]; t (* Jayanta Basu, Jun 01 2013 *)

from _future_ import division
def palgen(l,b=10): # generator of palindromes in base b of length <= 2*l
    if l > 0:
        yield 0
        for x in range(1,l+1):
            n = b**(x-1)
            n2 = n*b
            for y in range(n,n2):
                k, m = y//b, 0
                while k >= b:
                    k, r = divmod(k,b)
                    m = b*m + r
                yield y*n + b*m + k
            for y in range(n,n2):
                k, m = y, 0
                while k >= b:
                    k, r = divmod(k,b)
                    m = b*m + r
                yield y*n2 + b*m + k
def A050782(n, l=10):
    if n % 10:
        x = palgen(l)
        next(x)  # replace with x.next() in Python 2.x
        for i in x:
            q, r = divmod(i, n)
            if not r:
                return q
        else:
            return 'search limit reached.'
    else:
        return 0 # Chai Wah Wu, Dec 30 2014

A109924 Least palindromic multiple of concatenation 123...n.

Original entry on oeis.org

1, 252, 8118, 28382, 536797635, 6180330816, 85770307758, 2889123219882, 535841353148535, 135444949494445310, 1522312136776312132251, 2111913320628668260233191112, 6690072525779588859775252700966, 202511080654222947749222456080115202, 538412926804799527505725997408629214835

Offset: 1

Amarnath Murthy, Jul 16 2005

When n is a multiple of 10, any multiple of 123...n has trailing zeros, therefore it cannot be palindromic. The terms listed as a(10k) are therefore the least palindromic multiples with "invisible leading zeros allowed", or equivalently, trailing zeros ignored.

Subsequence of A020485.

			123*j is not palindromic for j < 66 and 123*66 = 8118, hence a(3) = 8118.

P. De Geest, Smallest multipliers to make a number palindromic.

Cf. A020485, A050782, A061816, A109929.

f[n_] := Block[{k = 1, p = FromDigits[ Flatten[ IntegerDigits /@ Range[n]]]}, While[ If[ Mod[p, 10] == 0, p/=10]; While[k*p != FromDigits[ Reverse[ IntegerDigits[k*p]]], k++ ]]; k*p]; Table[ f[n], {n, 11}] (* Robert G. Wilson v, Jul 19 2005 *)

intreverse(n) = local(d, rev); rev=0; while(n>0, d=divrem(n, 10); n=d[1]; rev=10*rev+d[2]);
{s="";for(n=1,10,s=concat(s,n);k=eval(s);if(n%10==0,m=0, j=1;while((m=k*j)!=intreverse(m),j++));print1(m,","))}

A109924(n)={ n=eval(concat(vector(n,i,Str(i))));forstep(i=n/10^valuation(n,10),9e99,n/10^valuation(n,10), (m=Vec(Str(i)))==vecextract(m,"-1..1")&return(i*10^valuation(n,10)))} \\ M. F. Hasler, Jun 19 2011

Edited and extended (a(5) to a(10)) by Klaus Brockhaus, Jul 19 2005
a(10)-a(11) from Robert G. Wilson v, Jul 19 2005
Definition of a(10k) clarified by M. F. Hasler, Jun 19 2011.
a(12)-a(14) from Giovanni Resta, Sep 22 2019
a(15) from Giovanni Resta, Sep 24 2019

A112725 Smallest positive palindromic multiple of 3^n.

Original entry on oeis.org

1, 3, 9, 999, 999999999, 29799999792, 39789998793, 39989598993, 68899199886, 68899199886, 68899199886, 68899199886, 68899199886, 2699657569962, 146189959981641, 191388777883191, 191388777883191, 18641845754814681

Offset: 0

Farideh Firoozbakht, Nov 12 2005

a(0)=1; a(1)=3 and it is easily shown that for n>1, 10^3^(n-2)-1 is a palindromic multiple of 3^n(see comments line of A062567). So for each n, a(n) exists and for n>1, a(n)<=10^3^(n-2)-1. This sequence is a subsequence of A020485(a(n)=A020485(3^n)) and for all n, A062567(3^n)<=a(n) because for all n, A062567(n)<= A020485(n). Jud McCranie conjectures that for n>1 A062567(3^n) =10^3^(n-2)-1, if his conjecture were true then from the above facts we conclude that for n>1 a(n)=10^3^(n-2)-1, but we see that for 4

			a(18)=18771463736417781 because 18771463736417781=3^18*48452429 is the smallest positive palindromic multiple of 3^18.

Cf. A020485, A062567, A112726.

b[n_]:=(For[m=1, !FromDigits[Reverse[IntegerDigits[m*n]]]==m*n, m++ ]; m*n);Do[Print[b[3^n]], {n, 0, 18}]

A307278 Record values in A050782.

Original entry on oeis.org

0, 1, 21, 38, 66, 518, 845, 12345679, 172839506, 445372913, 1173450678278939, 111222333444555666777889

Offset: 1

N. J. A. Sloane, Apr 05 2019

Cf. A020485, A050782, A307279.

a(11)-a(12) from Giovanni Resta, Oct 15 2022

A307279 Indices of record values in A050782.

Original entry on oeis.org

0, 1, 12, 13, 32, 54, 71, 81, 162, 8081, 8181, 8991

Offset: 1

N. J. A. Sloane, Apr 05 2019

Cf. A020485, A050782, A307278.

a(11)-a(12) from Giovanni Resta, Oct 15 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

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cp's OEIS Frontend

A050782 Smallest positive multiplier m such that m*n is palindromic (or zero if no such m exists).

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A109924 Least palindromic multiple of concatenation 123...n.

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Mathematica

PARI

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A112725 Smallest positive palindromic multiple of 3^n.

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Mathematica

A307278 Record values in A050782.

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A307279 Indices of record values in A050782.

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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

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