A047726 -id:A047726 - cp's OEIS frontend

A079806 Number of even numbers that can be formed by permuting the digits of n.

1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1

Offset: 0

Amarnath Murthy, Feb 06 2003

Leading zeros are allowed. - Robert Israel, Aug 27 2025

			a(246) = 6 (the numbers are 246,264,426,462,624,642). a(384) = 4 (the numbers are 384,348,438,834).

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

Cf. A047726, A079807.

f:= proc(n) local L, T, m, x, i,j;
   L:= convert(n,base,10);
   for i from 0 to 9 do T[i]:= numboccur(i,L) od;
   m:= nops(L);
   x:= 0;
   for i in [0,2,4,6,8] do
     if T[i] > 0 then x:= x + (m-1)!/(T[i]-1)!/mul(T[j]!,j={0$9} minus {i}); fi;
   od;
   x
end proc;
map(f, [$0..100]); # Robert Israel, Aug 27 2025

Table[Count[FromDigits/@Permutations[IntegerDigits[n]],?EvenQ],{n,0,120}] (* _Harvey P. Dale, Jul 25 2020 *)

More terms from Harvey P. Dale, Jul 25 2020

A086150 Number of permutations of decimal digits of n which yield nonprime numbers.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 0, 1, 1, 2, 0, 2, 0, 1, 2, 1, 0, 2, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 1, 1, 0, 1, 1, 1, 1, 2, 0, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 0, 2, 0, 1, 2, 1, 1, 2, 0, 2, 2, 2, 1, 2, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 0, 1, 1, 3, 1, 6, 3, 4, 6

Offset: 1

Labos Elemer, Aug 04 2003

From Robert Israel, Aug 13 2017: (Start)
Leading zeros are allowed.
a(n) = 0 for n in A003459. (End)

			n=117, digit-permutations={117,171,711} are all composites, so a(117)=3.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A003459, A039999, A047726.

f:= proc(L) option remember;
  nops(remove(isprime, map(t -> add(t[i]*10^(i-1),i=1..nops(t)), combinat:-permute(L))))
end proc:
seq(f(sort(convert(n,base,10))),n=1..200); # Robert Israel, Aug 13 2017

nd[x_, y_] := 10*x+y tn[x_] := Fold[nd, 0, x] Table[Count[Table[PrimeQ[tn[Part[Permutations[ IntegerDigits[w]], j]]], {j, 1, Length[Permutations[ IntegerDigits[w]]]}], False], {w, 1, 128}]
Table[Count[FromDigits/@Permutations[IntegerDigits[n]],?(!PrimeQ[#]&)],{n,110}] (* _Harvey P. Dale, Dec 24 2016 *)

a(n) + A039999(n) = A047726(n). - Robert Israel, Aug 13 2017

A276919 Number of solutions to x^3 + y^3 + z^3 + t^3 == 1 (mod n) for 1 <= x, y, z, t <= n.

Original entry on oeis.org

1, 8, 27, 64, 125, 216, 336, 512, 1296, 1000, 1331, 1728, 1794, 2688, 3375, 4096, 4913, 10368, 7410, 8000, 9072, 10648, 12167, 13824, 15625, 14352, 34992, 21504, 24389, 27000, 30225, 32768, 35937, 39304, 42000, 82944, 48396, 59280, 48438, 64000, 68921, 72576, 77529, 85184, 162000, 97336

Offset: 1

José María Grau Ribas, Sep 22 2016

It appears that a(n) = n^3 for n in A088232. See also A066498. - Michel Marcus, Oct 11 2016

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A000189, A047726, A060839, A063454, A087412, A087786, A254073, A276920.

JJJ[4, n, lam] = Sum[If[Mod[a^3 + b^3 + c^3 + d^3, n] == Mod[lam, n], 1, 0], {d, 0, n - 1}, {a, 0, n - 1}, {b, 0, n - 1}, {c, 0 , n - 1}]; Table[JJJ[4, n, 1], {n, 1, 50}]

a(n) = sum(x=1, n, sum(y=1, n, sum(z=1, n, sum(t=1, n, Mod(x,n)^3 + Mod(y,n)^3 + Mod(z,n)^3 + Mod(t,n)^3 == 1)))); \\ Michel Marcus, Oct 11 2016

qperms(v) = {my(r=1,t); v = vecsort(v); for(i=1,#v-1, if(v[i]==v[i+1], t++, r*=binomial(i,t+1);t=0));r*=binomial(#v,t+1)}
a(n) = {my(t=0); forvec(v=vector(4,i,[1,n]), if(sum(i=1, 4, Mod(v[i], n)^3)==1, print1(v", "); t+=qperms(v)),1);t} \\ David A. Corneth, Oct 11 2016

def A276919(n):
    ndict = {}
    for i in range(n):
        i3 = pow(i,3,n)
        for j in range(i+1):
            j3 = pow(j,3,n)
            m = (i3+j3) % n
            if m in ndict:
                if i == j:
                    ndict[m] += 1
                else:
                    ndict[m] += 2
            else:
                if i == j:
                    ndict[m] = 1
                else:
                    ndict[m] = 2
    count = 0
    for i in ndict:
        j = (1-i) % n
        if j in ndict:
            count += ndict[i]*ndict[j]
    return count # Chai Wah Wu, Jun 06 2017

A276920 Number of solutions to x^3 + y^3 + z^3 + t^3 == 0 (mod n) for 1 <= x, y, z, t <= n.

Original entry on oeis.org

1, 8, 27, 72, 125, 216, 595, 704, 1539, 1000, 1331, 1944, 3133, 4760, 3375, 5632, 4913, 12312, 8911, 9000, 16065, 10648, 12167, 19008, 16125, 25064, 45927, 42840, 24389, 27000, 35371, 47104, 35937, 39304, 74375, 110808, 58645, 71288, 84591, 88000

Offset: 1

José María Grau Ribas, Sep 22 2016

a(n) = n^3 if n is in A074243. - Robert Israel, Oct 13 2016

			For n = 3, we see that all nondecreasing solutions {x, y, z, t} are in {{1, 1, 1, 3}, {1, 1, 2, 2}, {1, 2, 3, 3}, {2, 2, 2, 3}, {3, 3, 3, 3}}. The numbers in the sets can be ordered in 4, 6, 12, 4 and 1 ways respectively. Therefore, a(3) = 4 + 6 + 12 + 4 + 1 = 27. - _David A. Corneth_, Oct 11 2016

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms n = 1..242 from Robert Israel)

Cf. A000189, A047726, A060839, A063454, A074243, A087412, A087786, A254073, A276919.

CF:= table([[false, false, true] = 12, [true, false, false] = 12, [true, false, true] = 6, [false, false, false] = 24, [true, true, true] = 1, [false, true, true] = 4, [false, true, false] = 12, [true, true, false] = 4]):
f1:= proc(n)
  option remember;
  local count, t, x,y,z,signature;
  if isprime(n) and n mod 3 = 2 then return n^3 fi;
  count:= 0;
  for t from 1 to n do
    for x from 1 to t do
      for y from 1 to x do
        for z from 1 to y do
          if t^3 + x^3 + y^3 + z^3 mod n = 0 then
            signature:= map(evalb,[z=y,y=x,x=t]);
            count:= count + CF[signature];
          fi
  od od od od;
  count
end proc:
f:= proc(n) local t;
    mul(f1(t[1]^t[2]),t=ifactors(n)[2])
end proc:
map(f, [$1..40]); # Robert Israel, Oct 13 2016

JJJ[4, n, lam] = Sum[If[Mod[a^3 + b^3 + c^3 + d^3, n] == Mod[lam, n], 1, 0], {d, 0, n - 1}, {a, 0, n - 1}, {b, 0, n - 1}, {c, 0 , n - 1}]; Table[JJJ[4, n, 0], {n, 1, 50}]

a(n) = sum(x=1, n, sum(y=1, n, sum(z=1, n, sum(t=1, n, Mod(x,n)^3 + Mod(y,n)^3 + Mod(z,n)^3 + Mod(t,n)^3 == 0)))); \\ Michel Marcus, Oct 11 2016

qperms(v) = {my(r=1,t); v = vecsort(v); for(i=1,#v-1, if(v[i]==v[i+1], t++, r*=binomial(i, t+1);t=0));r*=binomial(#v,t+1)}
a(n) = {my(t=0); forvec(v=vector(4,i,[1,n]), if(sum(i=1,4,Mod(v[i],n)^3)==0, t+=qperms(v)),1);t} \\ David A. Corneth, Oct 11 2016

def A276920(n):
    ndict = {}
    for i in range(n):
        i3 = pow(i,3,n)
        for j in range(i+1):
            j3 = pow(j,3,n)
            m = (i3+j3) % n
            if m in ndict:
                if i == j:
                    ndict[m] += 1
                else:
                    ndict[m] += 2
            else:
                if i == j:
                    ndict[m] = 1
                else:
                    ndict[m] = 2
    count = 0
    for i in ndict:
        j = (-i) % n
        if j in ndict:
            count += ndict[i]*ndict[j]
    return count # Chai Wah Wu, Jun 06 2017

A319274 Osiris or Digit re-assembly numbers: numbers that are equal to the sum of permutations of subsamples of their own digits.

Original entry on oeis.org

132, 264, 396, 8991, 10545, 35964, 255530, 1559844, 9299907, 47755078, 89599104, 167264994, 283797162, 473995260, 3929996070, 6379993620, 10009998999, 11111111110, 22222222220, 33333333330, 44444444440, 55555555550, 66666666660, 77777777770, 88888888880, 99999999990

Offset: 1

Pieter Post, Sep 16 2018

This sequence differs from A241754 because this sequence uses permutations only once.
Permutations are of the same length k, leading zeros are allowed.
The k's in the sequence are: 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 7, 10, 10, 10, 10, 10, 10, 10, 10, 10, 6, 7, 7, 8, 7, 9, 9.

			10545 = 014 + 015 + 041 + 045 + 051 + 054 + 055 + 104 + 105 + 140 + 145 + 150 + 154 + 155 + 401 + 405 + 410 + 415 + 450 + 451 + 455 + 501 + 504 + 505 + 510 + 514 + 515 + 540 + 541 + 545 + 550 + 551 + 554.

Giovanni Resta, Table of n, a(n) for n = 1..33 (terms < 10^16)

Cf. A047726, A179239, A241754, A241899.

import itertools
def getData(a, b):
    dig = (itertools.permutations(str(a), b))
    for d in dig:
        yield d
for w in range(2, 6):
    kk=int(w*'1')
    for i in range (kk, 10**(w+3), kk):
        m=[]
        get = getData(i, w)
        while True:
            try:
                n = next(get)
                ee=int("".join((n)))
                if ee not in m:
                    m.append(ee)
            except StopIteration:
                if sum (m)==i and len(m)>1:
                    m.sort()
                    print (sum(m), len(m), m, i)
                break

a(12)-a(26) from Giovanni Resta, Sep 16 2018

A276354 Palindromes n > 0 such that the sum of all distinct permutations of the digits of n is also a palindrome.

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, 191, 202, 212, 222, 232, 242, 252, 272, 303, 313, 323, 333, 353, 404, 414, 434, 444, 515, 555, 666, 777, 787, 868, 888, 949, 999, 1001, 1111, 1221, 2002, 2112, 2222, 2992

Offset: 1

Altug Alkan, Sep 05 2016

Values of A002113(n) such that A045876(A002113(n)) is in A002113 (n > 0).
If n has a zero digit then the permutations starting with 0 are included in the sum (i.e., 0010 is taken to be 10, 01 is taken to be 1, etc.).
A010785(n) is an obvious term of this sequence for all n > 0.

			101 is a term because 11 + 101 + 110 = 222 is also a palindrome.
232 is a term because 223 + 232 + 322 = 777 is also a palindrome.
2002 is a term because 22 + 202 + 220 + 2002 + 2020 + 2200 = 6666 is also a palindrome.

Indranil Ghosh, Table of n, a(n) for n = 1..385

Cf. A002113, A007953, A010785, A045876, A047726, A055642.

A047726(n) = n=eval(Vec(Str(n))); (#n)!/prod(i=0, 9, sum(j=1, #n, n[j]==i)!);
A055642(n) = #Str(n);
A007953(n) = sumdigits(n);
A045876(n) = ((10^A055642(n)-1)/9)*(A047726(n)*A007953(n)/A055642(n));
isA002113(n) = n=digits(n); for(i=1, #n\2, if(n[i]!=n[#n+1-i], return(0))); 1;
lista(nn) = for(n=1, nn, if(isA002113(n) && isA002113(A045876(n)), print1(n", ")));

A276413 Non-repdigit numbers k that divide A045876(k).

Original entry on oeis.org

370, 407, 481, 518, 592, 629, 2727, 13008, 14634, 16260, 19512, 22764, 29268, 39024, 87804, 101010, 102564, 103896, 104895, 105820, 108262, 109890, 113960, 115830, 116883, 124740, 125356, 125874, 126984, 128205, 129870, 132275, 134680, 135135, 136752

Offset: 1

Altug Alkan, Sep 05 2016

A161020 is a subsequence.

			2727 is a term because 2277 + 2727 + 2772 + 7227 + 7272 + 7722 = 29997 is divisible by 2727.

Robert Israel, Table of n, a(n) for n = 1..373

Cf. A161020, A045876.

filter:= proc(x) local L, D, n, M, s, j;
  L:= convert(x, base, 10);
  D:= [seq(numboccur(j, L), j=0..9)];
  if numboccur(0,D) = 9 then return false fi;
  n:= nops(L);
  M:= n!/mul(d!, d=D);
    s:= add(j*D[j+1], j=0..9);
  evalb(((10^n-1)*M/9/n*s) mod x = 0)
end proc:
select(filter, [$1..2*10^5]); # Robert Israel, Sep 12 2016

A047726(n) = n=eval(Vec(Str(n))); (#n)!/prod(i=0, 9, sum(j=1, #n, n[j]==i)!);
A055642(n) = #Str(n);
A007953(n) = sumdigits(n);
A045876(n) = ((10^A055642(n)-1)/9)*(A047726(n)*A007953(n)/A055642(n));
isA010785(n) = {1==#Set(digits(n))}
lista(nn) = for(n=1, nn, if(A045876(n) % n == 0 && !isA010785(n), print1(n", ")));

A276502 Least k > 0 such that A045876(n) divides A045876(n*10^k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 6

Offset: 1

Altug Alkan, Sep 10 2016

Corresponding values of A045876(n*10^a(n))/A045876(n) are 11, 11, 11, 11, 11, 11, 11, 11, 11, 101, 303, 303, 303, 303, 303, 303, 303, 303, 303, 101, 303, 303, 303, 303, 303, 303, 303, 303, 303, 101, ...
From Charlie Neder, Jul 16 2018: (Start)
From the formula for A045876(n) we make the following modifications:
- A (the mean of the digits) becomes S/D (sum of digits / # of digits)
- N (# of arrangements of digits) becomes R*Z (# of arrangements of nonzero digits * # of ways to insert the proper number of zeros)
Appending zeros to n does not change S or R, so if (S*R*Z*I/D)(n) divides (S*R*Z*I/D)(n*10^k), then (Z*I/D)(n) divides (Z*I/D)(n*10^k). However, Z, I, and D are completely determined by the number of digits of n and the number of those digits which are zero, so a(n) = a(A136400(n)). (End)

			a(10) = 2 because A045876(10) = 1+10 = 11 does not divide A045876(100) = 1+10+100 = 111 and 11 divides A045876(1000) = 1+10+100+1000 = 1111.

Cf. A045876.

A045876[n_] := Total[FromDigits /@ Permutations[IntegerDigits[n]]]; a[n_] := For[k = 1, True, k++, If[Divisible[A045876[n*10^k], A045876[n]], Return[k] ] ]; Array[a, 101] (* Jean-François Alcover, Jul 26 2017 *)

A047726(n) = n=eval(Vec(Str(n))); (#n)!/prod(i=0, 9, sum(j=1, #n, n[j]==i)!);
A055642(n) = #Str(n);
A007953(n) = sumdigits(n);
A045876(n) = ((10^A055642(n)-1)/9)*(A047726(n)*A007953(n)/A055642(n));
a(n) = {my(k = 1); while (A045876(n*(10^k)) % A045876(n), k++); k; }

A276510 Numbers k such that the sum of all the different permutations of the digits of k (A045876(k)) is a pandigital number (a term of A171102).

Original entry on oeis.org

10234567, 10234576, 10234579, 10234597, 10234657, 10234675, 10234678, 10234687, 10234756, 10234759, 10234765, 10234768, 10234786, 10234795, 10234867, 10234876, 10234957, 10234975, 10235467, 10235476, 10235479, 10235497, 10235647, 10235674, 10235746, 10235749

Offset: 1

Altug Alkan, Sep 06 2016

			10234759 is a term because A045876(10234759) = 1567999984320, which contains every digit from 0 to 9.

Cf. A045876, A171102.

A047726(n) = n=eval(Vec(Str(n))); (#n)!/prod(i=0, 9, sum(j=1, #n, n[j]==i)!);
A055642(n) = #Str(n);
A007953(n) = sumdigits(n);
A045876(n) = ((10^A055642(n)-1)/9)*(A047726(n)*A007953(n)/A055642(n));
isA171102(n) = 9<#vecsort(Vecsmall(Str(n)), , 8);
is(n) = isA171102(A045876(n));

A276597 Least k such that n divides A045876(k).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 10, 39, 49, 59, 69, 79, 89, 102, 199, 109, 106, 13, 599, 103, 799, 139, 108, 149, 2999, 104, 4999, 169, 12, 179, 8999, 105, 100, 289, 139, 389, 10000, 106, 79999, 13, 159, 689, 299999, 107, 100006, 1789, 179, 2789, 899999, 108, 14, 4789, 199, 5789, 5999999, 109

Offset: 1

Altug Alkan, Sep 08 2016

Corresponding values of A045876(a(n))/n are 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 11, 11, 11, 11, 11, 11, 37, 111, 111, 74, 2, 111, 37, 111, 111, 74, 111, 1111, 37, 1111, 111, 1, 111, 1111, 37, 3, 111, 74, 111, 271, 37, 11111, 1, 74, 111, 111111, 37, 79365, 3333, ...

			a(10) = 19 because 19+91 = 110 is divisible by 10.
a(18) = 102 because 12+21+102+120+201+210 is divisible by 18.

Giovanni Resta, Table of n, a(n) for n = 1..150

Cf. A045876.

A047726(n) = n=eval(Vec(Str(n))); (#n)!/prod(i=0, 9, sum(j=1, #n, n[j]==i)!);
A055642(n) = #Str(n);
A007953(n) = sumdigits(n);
A045876(n) = ((10^A055642(n)-1)/9)*(A047726(n)*A007953(n)/A055642(n));
a(n) = {my(k = 1); while (A045876(k) % n, k++); k; }

cp's OEIS Frontend

A079806 Number of even numbers that can be formed by permuting the digits of n.

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A086150 Number of permutations of decimal digits of n which yield nonprime numbers.

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A276919 Number of solutions to x^3 + y^3 + z^3 + t^3 == 1 (mod n) for 1 <= x, y, z, t <= n.

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A276920 Number of solutions to x^3 + y^3 + z^3 + t^3 == 0 (mod n) for 1 <= x, y, z, t <= n.

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A319274 Osiris or Digit re-assembly numbers: numbers that are equal to the sum of permutations of subsamples of their own digits.

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A276354 Palindromes n > 0 such that the sum of all distinct permutations of the digits of n is also a palindrome.

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A276413 Non-repdigit numbers k that divide A045876(k).

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