A179309 -id:A179309 - cp's OEIS frontend

A178354 Numbers m such that d(1)^1 + d(2)^2 + ... + d(p)^p = d(1)^p + d(2)^(p-1) +... + d(p)^1, where d(i), i=1..p, are the digits of m.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 22, 33, 44, 55, 66, 77, 88, 99, 100, 101, 110, 111, 120, 121, 130, 131, 140, 141, 150, 151, 160, 161, 170, 171, 180, 181, 190, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 444, 454

Offset: 1

Michel Lagneau, Dec 21 2010

A179309 is included in this sequence.

All palindromes are in this sequence. - Harvey P. Dale, Mar 03 2013

			14603 is in the sequence because :
1 + 4^2 + 6^3 + 0^4 + 3^5 = 3 + 0^2 + 6^3 + 4^4 + 1^5 = 476.

Carole Dubois, Table of n, a(n) for n = 1..34794
Carole Dubois, Scatterplot of a(n) vs Sum of powers in definition.

Cf. A002113, A179309.

with(numtheory):for n from 1 to 50000 do:l:=length(n):n0:=n:s1:=0:s2:=0:for
  m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :s1:=s1+u^(l-m+1):s2:=s2+u^m:od:
  if s1=s2 then printf(`%d, `,n):else fi:od:

drQ[n_]:=Module[{id=IntegerDigits[n],len},len=Length[id];Total[ id^Range[ len]] == Total[id^Range[len,1,-1]]]; Select[Range[500],drQ] (* Harvey P. Dale, Aug 04 2018 *)

isok(m) = my(d=digits(m), p=#d); sum(k=1, p, d[k]^k) == sum(k=1, p, d[k]^(p-k+1)); \\ Michel Marcus, Mar 22 2021

def digpow(s): return sum(int(d)**i for i, d in enumerate(s, start=1))
def aupto(limit):
  alst = []
  for k in range(1, limit+1):
    s = str(k)
    if digpow(s) == digpow(s[::-1]): alst.append(k)
  return alst
print(aupto(454)) # Michael S. Branicky, Mar 23 2021

A342826 Numbers k such that d(1)^0 + d(2)^1 + ... + d(p)^(p-1) = d(1)^(p-1) + d(2)^(p-2) + ... + d(p)^0, where d(i), i=1..p, are the digits of k.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 444, 454, 464

Offset: 1

Carole Dubois, Mar 23 2021

This sequence starts off like other palindromic sequences such as A178354, A002113, A110751, and A227858 but it differs starting at the 110th term: 109th: 1001, 110th: 1011, 111th: 1101, ..., 119th: 1863, etc.
Differs from A297271 (which e.g. has 1021, 1031, 1041,.. 1091 which are absent here). - R. J. Mathar, Sep 27 2021
Contains the palindromes, and palindromes where pairs of digits have been substituted by d(i)=0, d(p-i)=1 or d(i)=1, d(p-1)=0, and "genuine" numbers like 1863, 2450, 2804, 2814, 3681, 4081, 4182, 103221, 113221, 122301, 122311, 142023,.. - R. J. Mathar, Sep 27 2021

			1863 is in this sequence because 1^0 + 8^1 + 6^2 + 3^3 = 1^3 + 8^2 + 6^1 + 3^0 = 72.

R. J. Mathar, Table of n, a(n) for n = 1..1137 (replacing older b-file which did not contain a(101))
Carole Dubois, Scatterplot of terms of A178354, A342826 vs Sum of powers in definition.

Cf. A002113 (subset), A179309, A110751, A227858.

isA342826 := proc(n)
    local dgs ;
    dgs := convert(n,base,10) ;
    if  add(op(i,dgs)^(i-1),i=1..nops(dgs)) = add(op(-i,dgs)^(i-1),i=1..nops(dgs)) then
        true;
    else
        false;
    end if;
end proc:
A342826 := proc(n)
    option remember ;
    if n =1 then
        1;
    else
        for a from procname(n-1)+ 1 do
            if isA342826(a) then
                return a;
            end if;
        end do:
    end if;
end proc: # R. J. Mathar, Sep 27 2021

Select[Range[500],Mod[#,10]!=0&&Total[IntegerDigits[#]^Range[0,IntegerLength[ #]-1]]==Total[IntegerDigits[#]^Range[IntegerLength[#]-1,0,-1]]&] (* Harvey P. Dale, Jan 18 2023 *)

def digpow(s): return sum(int(d)**i for i, d in enumerate(s))
def aupto(limit):
  alst = []
  for k in range(1, limit+1):
    s = str(k)
    if digpow(s) == digpow(s[::-1]): alst.append(k)
  return alst
print(aupto(464)) # Michael S. Branicky, Mar 23 2021

A179310 The smallest number that has more copies of some digit than all previous terms of the sequence put together.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 22, 33, 44, 55, 66, 77, 88, 99, 100, 1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999, 10000, 22222222, 33333333, 44444444, 55555555, 66666666, 77777777, 88888888, 99999999, 100000000, 1111111111

Offset: 1

Jack W Grahl, Jul 10 2010

For each natural number taken in order, we consider if we can make it using digits from as many of the previous terms as we like. If we cannot, we add it to the sequence and add its digits to the 'pool' we have for making subsequent numbers.

			This sequence is the same as A179309 up to 100. After that, we can make any three-digit number because we have had at least three of each digit so far. We can make 1000 because we have already had three 0's (in 10 and 100). So the next term is 1111 because we have only seen three 1's so far.

A subsequence of A179309.

An error in the example (pointed out by Jon E. Schoenfield) was corrected by Jack W Grahl, Jul 19 2010

More terms from Sean A. Irvine, Nov 10 2011

cp's OEIS Frontend

A178354 Numbers m such that d(1)^1 + d(2)^2 + ... + d(p)^p = d(1)^p + d(2)^(p-1) +... + d(p)^1, where d(i), i=1..p, are the digits of m.

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A342826 Numbers k such that d(1)^0 + d(2)^1 + ... + d(p)^(p-1) = d(1)^(p-1) + d(2)^(p-2) + ... + d(p)^0, where d(i), i=1..p, are the digits of k.

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Maple

Mathematica

Python

A179310 The smallest number that has more copies of some digit than all previous terms of the sequence put together.

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