A009994 -id:A009994 - cp's OEIS frontend

A076166 Primes p such that sum of cubes of even digits of p equals sum of cubes of odd digits of p.

16447, 41467, 41647, 44617, 46147, 46471, 76441, 114451, 144511, 146407, 404167, 404671, 414607, 415141, 416407, 440761, 441607, 451411, 460147, 460417, 461407, 470461, 476041, 476401, 541141, 610447, 640741, 644107, 644701, 647401, 704461, 740461, 746041, 764041

Offset: 1

Zak Seidov, Nov 01 2002

Minimal number of digits in p is 5. n such that sum of even digits equals sum of odd digits in A036301.

To find terms of this sequence, one could look at zerofree positive integers having the criterion on sum of cubes of digits. Then permute the digits to see which are prime. Using those digits with 0 and permuting then only needs the check on primality. - David A. Corneth, Dec 11 2018

			16447 is OK because 1^3 + 7^3 = 6^3 + 4^3 + 4^3.
14467 has digits in nondecreasing order (is zerofree). Of the 60 permutations, 16447, 41467, 41647, 44617, 46147, 46471, 76441 are prime. - _David A. Corneth_, Dec 11 2018

Marius A. Burtea, Table of n, a(n) for n = 1..5869

Cf. A009994, A036301.

Subsequence of A076165.

oeQ[n_]:=Module[{idn = IntegerDigits[n]}, Total[Select[idn, OddQ]^3] == Total[ Select[idn, EvenQ]^3]]; Select[Range[100000], PrimeQ[#] && oeQ[#] &] (* Amiram Eldar, Dec 10 2018 after Harvey P. Dale at A076165 *)

isok(p) = isprime(p) && (d=digits(p)) && (sum(i=1, #d, d[i]^3*if(d[i]%2, 1, -1))==0); \\ Michel Marcus, Dec 13 2018

A165370 Smallest number whose sum of cubes of digits is n.

Original entry on oeis.org

0, 1, 11, 111, 1111, 11111, 111111, 1111111, 2, 12, 112, 1112, 11112, 111112, 1111112, 11111112, 22, 122, 1122, 11122, 111122, 1111122, 11111122, 111111122, 222, 1222, 11222, 3, 13, 113, 1113, 11113, 2222, 12222, 112222, 23, 123, 1123, 11123

Offset: 0

Reinhard Zumkeller, Sep 17 2009

A055012(a(n)) = n and A055012(m) <> n for m < a(n).
For all terms the digits are in nondecreasing order.
From Ondrej Kutal, Oct 06 2024: (Start)
a(n) can be formulated as a solution to an integer linear programming problem: Let c_i be the number of occurrences of digit i in the number (1 <= i <= 9). We want to minimize c_1 + c_2 + ... + c_9 (total number of digits) subject to the constraints c_1 + 8*c_2 + 27*c_3 + ... + 729*c_9 = n and c_i >= 0 for all i. To get the smallest solution among all with this number of digits, we further maximize the number of 1s (c_1), then the number of 2s (c_2), and so on up to 9s (c_9). This approach ensures we find the lexicographically smallest number with the minimum number of digits whose sum of cubes equals n.
a(n) can be computed by selecting a digit d (1 <= d <= 9) that minimizes the result of concatenating d with a(n-d^3), where n-d^3 >= 0. This can be done efficiently using dynamic programming.
As it turns out, the sequence is periodic (up to the trailing 9s) for n >= 4609, with a period of 729. Therefore, only a finite amount of values need to be computed; the rest can be derived by appending the appropriate number of 9s.
(End)

Jon E. Schoenfield, Table of n, a(n) for n = 0..10000 (first 1001 terms from Reinhard Zumkeller)

Cf. A009994, A052382, A055012, A055016.

a(n) = my(k=1); while(vecsum(apply(x->(x^3), digits(k))) != n, k++); k; \\ Michel Marcus, Sep 08 2019

# using dynamic programming
def A165370(n):
    # caching numbers,their tenth power (for fast concatenation) and cube sum
    cache = [(None, None, None)] * (n + 1)
    cache[0] = (0, 1, 0)
    cubes = [i**3 for i in range(10)]
    for i in range(1, min(n + 1, 5832)):
        for d in range(1, 10):
            if i - cubes[d] >= 0:
                sub_result, tenthpower, cubesum = cache[i - cubes[d]]
                if sub_result is not None:
                    current = d * tenthpower + sub_result
                    if cache[i][0] is None or current < cache[i][0]:
                        cache[i] = (current, 10 * tenthpower, cubesum + cubes[d])
    if n < 5832:
        return cache[n][0]
    sub_result, _, cubesum = cache[5103 + n % 729]
    nines = (n - cubesum) // 729
    return (sub_result + 1) * (10 ** nines) - 1 # Ondrej Kutal, Oct 06 2024

A274124 Numbers n such that (product of digits of n) is divisible by (sum of digits of n) and digits of n are in nondecreasing order.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 22, 36, 44, 66, 88, 123, 138, 145, 159, 167, 189, 224, 235, 246, 257, 268, 279, 333, 345, 347, 357, 369, 448, 456, 459, 466, 578, 579, 666, 678, 789, 999, 1124, 1146, 1168, 1225, 1233, 1236, 1247, 1258, 1269, 1344, 1348, 1356, 1368, 1447

Offset: 1

David A. Corneth, Jun 10 2016

Every number with a digit 0 is in A038367. Every permutation of every element of this is in A038367. These elements describe A038367 completely.

Intersection of A038367 and A009994.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A038367, A009994.

is(n) = my(v=vecsort(digits(n))); v==digits(n) && prod(i=1,#v,v[i]) % vecsum(v)==0

A327750 Numbers without zero digits such that after adding the product of its digits to it, a number with the same product of digits is obtained.

Original entry on oeis.org

28, 128, 214, 239, 266, 318, 326, 364, 494, 497, 563, 598, 613, 637, 695, 819, 1128, 1214, 1239, 1266, 1318, 1326, 1364, 1494, 1497, 1563, 1598, 1613, 1637, 1695, 1819, 2114, 2139, 2168, 2285, 2313, 2356, 2369, 2419, 2594, 2639, 2791, 3118, 3126, 3148, 3213, 3235

Offset: 1

Bernard Schott, Sep 24 2019

The idea of this sequence comes from a problem in the annual Moscow Mathematical Olympiad (MMO) in 2003: Level A, problem 2. The problem only asks to find a ten-digit number that has the property of the name.
When an integer k belongs to this sequence, the integer 111..11//k obtained by concatenation // of 111..11 and k is also a term; hence, there are primitive terms as 28, 214, 239, 266, 318, 326, ... (A340908).
A subset of it is formed by the numbers 239, 326, 364, 497, 563, 598, 613, 637, 695, 819, 1239, 1326, 1364, 1497, 1563, 1598, 1613, 1637, 1695, 1819, 2139, 2313, 2356, 2369, ... for which the number obtained after adding the product of the digits has exactly the same digits (they are obtained by permuting the digits of the initial number). So, 239 + 2*3*9 = 239 + 54 = 293, 326 + 3*2*6 = 326 + 36 = 362, 3235 + 3*2*3*5 = 3235 + 90 = 3325, 23286 + 2*3*2*8*6 = 23286 + 576 = 23862. - Marius A. Burtea, Sep 24 2019
This subset is A247888. - Bernard Schott, Jul 22 2020

			28 + 2*8 = 44 and 2*8 = 4*4 hence 28 is a term.
326 + 3*2*6 = 362 and 3*2*6 = 3*6*2 hence 326 is another term.

Michel Marcus, Table of n, a(n) for n = 1..10000
Roman Fedorov, Alexei Belov, Alexander Kovaldzhi, Ivan Yashchenko, Moscow Mathematical Olympiads, 2000-2005, Level A, Problem 2, 2003; MSRI, 2011, p. 15 and 97.
Index to sequences related to Olympiads.

Cf. A007954, A009994, A052382.

Subsequences: A247888, A340907, A340908 (primitives).

[k:k in [1..3500]| not 0 in Intseq(k) and &*Intseq(k) eq &*(Intseq(k+&*Intseq(k)))]; // Marius A. Burtea, Sep 24 2019

pd[n_] := Times @@ IntegerDigits[n]; aQ[n_] := (p = pd[n]) > 0 && pd[n + p] == p; Select[Range[5000], aQ] (* Amiram Eldar, Sep 24 2019 *)

isok(n) = my(d = digits(n), p); vecmin(d) && (p=vecprod(d)) && (vecprod(digits(n+p)) == p); \\ Michel Marcus, Sep 24 2019

def test(n):
    m, p = n, 1
    while m > 0:
        m, p = m//10, p*(m%10)
    if p == 0:
        return 0
    m, q = n+p, 1
    while m > 0:
        m, q = m//10, q*(m%10)
    return p == q
n, a = 0, 0
while n < 100:
    a = a+1
    if test(a):
        n = n+1
    print(n,a) # A.H.M. Smeets, Sep 25 2019

More terms from Amiram Eldar, Sep 24 2019

A342264 Lexicographically earliest sequence of distinct nonnegative terms such that both a(n) and a(n) + a(n+1) have digits in nondecreasing order.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 11, 12, 14, 15, 18, 16, 17, 19, 25, 22, 23, 24, 33, 26, 29, 27, 28, 38, 39, 49, 66, 45, 34, 35, 44, 55, 56, 57, 58, 59, 67, 46, 68, 47, 69, 48, 77, 36, 78, 37, 79, 88, 89, 99, 123, 111, 112, 113, 114, 115, 118, 116, 117, 119, 125, 122, 124, 133, 126, 129, 127, 128, 138

Offset: 1

Eric Angelini and Carole Dubois, Mar 07 2021

10 is obviously the first integer not present in the sequence as 1 > 0.

			a(10) = 9 and a(11) = 13 sum up to 22: the three numbers have digits in nondecreasing order;
a(11) = 13 and a(12) = 11 sum up to 24 (same property);
a(12) = 11 and a(13) = 12 sum up to 23 (same property);
etc.

Robert Israel, Table of n, a(n) for n = 1..4000

Cf. A009994 (numbers with digits in nondecreasing order), A342265 and A342266 (variations on the same idea).

ND[1]:= [$1..9]:
for d from 2 to 5 do
   ND[d]:= map(proc(t) local j; seq(10*t + j,j=(t mod 10) .. 9) end proc, ND[d-1])
od:
S:= [seq(op(ND[i]),i=1..5)]): nS:= nops(S):
isnd:= proc(x) member(x,ND[ilog10(x)+1]) end proc:
R:= 0: t:= 0:
for count from 2 to 100 do
  found:= false;
  for i from 1 to nS do
    if isnd(t + S[i]) then
      R:= R, S[i];
      t:= S[i];
      S:= subsop(i=NULL, S);
      nS:= nS-1;
      found:= true;
      break
    fi;
  od;
  if not found then break fi;
od:
R; # Robert Israel, Jul 14 2025

def nondec(n): s = str(n); return s == "".join(sorted(s))
def aupton(terms):
  alst = [0]
  for n in range(2, terms+1):
    an = 1
    while True:
      while an in alst: an += 1
      if nondec(an) and nondec(alst[-1]+an): alst.append(an); break
      else: an += 1
  return alst
print(aupton(74)) # Michael S. Branicky, Mar 07 2021

A342265 Lexicographically earliest sequence of distinct nonnegative terms such that both a(n) and the cumulative sum a(1)+a(2)+...+a(n) have digits in nondecreasing order.

Original entry on oeis.org

0, 1, 2, 3, 5, 4, 7, 6, 8, 9, 11, 12, 44, 13, 14, 16, 22, 45, 15, 18, 23, 55, 24, 88, 33, 77, 34, 78, 111, 333, 17, 19, 79, 29, 89, 25, 99, 199, 112, 444, 26, 28, 56, 35, 188, 113, 119, 556, 114, 999, 122, 1199, 888, 123, 4444, 36, 66, 124, 118, 222, 445, 115, 129, 67, 133, 667, 134, 68, 889, 223

Offset: 1

Eric Angelini and Carole Dubois, Mar 07 2021

10 is obviously the first integer not present in the sequence as 1 > 0.

The last term is a(173) = 122222, at which point the cumulative sum is 12467999, and the sequence cannot be extended. - Michael S. Branicky, Feb 05 2024

			Terms a(1) = 0 to a(5) = 5 sum up to 11: those six numbers have digits in nondecreasing order;
terms a(1) = 0 to a(6) = 4 sum up to 15: those seven numbers have digits in nondecreasing order;
terms a(1) = 0 to a(7) = 7 sum up to 22: those eight numbers have digits in nondecreasing order; etc.

Michael S. Branicky, Table of n, a(n) for n = 1..173

Cf. A009994 (numbers with digits in nondecreasing order), A342264 and A342266 (variations on the same idea).

def nondec(n): s = str(n); return s == "".join(sorted(s))
def aupton(terms):
  alst = [0]
  for n in range(2, terms+1):
    an, cumsum = 1, sum(alst)
    while True:
      while an in alst: an += 1
      if nondec(an) and nondec(cumsum + an): alst.append(an); break
      else: an += 1
  return alst
print(aupton(100)) # Michael S. Branicky, Mar 07 2021

A342266 Lexicographically earliest sequence of distinct nonnegative terms such that both a(n) and a(n) * a(n+1) have digits in nondecreasing order.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 8, 7, 5, 9, 13, 12, 14, 16, 18, 26, 44, 27, 17, 15, 23, 29, 46, 28, 48, 47, 24, 19, 117, 38, 36, 33, 34, 37, 67, 35, 127, 114, 39, 57, 78, 146, 236, 58, 77, 1444, 177, 157, 2477, 144, 247, 45, 25, 49, 227, 147, 257, 1777, 12568, 116, 68, 66, 118, 113, 59, 226, 148, 166, 134, 167, 334

Offset: 1

Eric Angelini and Carole Dubois, Mar 07 2021

10 is obviously the first integer not present in the sequence as 1 > 0; 11 will never show either because the result of a(n) * 11 is already in the sequence or because the said result has digits in contradiction with the definition.
It would be good to have a proof that there is an infinite sequence with the desired property. It could happen then any choice for any number of initial terms will eventually fail. - David A. Corneth and N. J. A. Sloane, Mar 07 2021
The authors agree, but are unable to give the desired proof. So it is indeed possible that this sequence is wrong from the first term on.

			a(5) = 4 and a(6) = 6 have product 24: the three numbers have digits in nondecreasing order;
a(6) = 6 and a(7) = 8 have product 48: the three numbers have digits in nondecreasing order;
a(7) = 8 and a(7) = 7 have product 56: the three numbers have digits in nondecreasing order; etc.

Carole Dubois, Table of n, a(n) for n = 1..138 (terms calculated with a(n+1) < a(n)*1000, (backtracking when not found inside this limit)).

Cf. A009994 (numbers with digits in nondecreasing order), A342264 and A342265 (variations on the same idea).

A045910 Numbers with digits nondecreasing and their reciprocals sum to 1/(positive integer).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 22, 36, 44, 66, 88, 236, 244, 333, 488, 666, 999, 2488, 2666, 3366, 3446, 4444, 6999, 8888, 26999, 28888, 33999, 34688, 36666, 44488, 44666, 55555, 366999, 368888, 446999, 448888, 466688, 666666, 3999999, 4688999, 4888888, 6666999, 6668888, 7777777, 66999999, 68888999, 88888888, 999999999

Offset: 1

Len Smiley

Intersection of A037264 and A009994, A214949(a(n)) = 1. - Reinhard Zumkeller, Aug 02 2012

			1/3 + 1/4 + 1/6 + 1/8 + 1/8 = 1/1, so 34688 is in the sequence.

Condensation (by ordering digits) and completion of A037264.

a045910 n = a045910_list !! (n-1)
a045910_list =  [x | x <- takeWhile (<= 999999999) $ a009994_list,
                     a214949 x == 1]
-- Reinhard Zumkeller, Aug 02 2012

A046811 Numbers (with nonzero digits only) where A046810 increases.

Original entry on oeis.org

1, 2, 13, 113, 149, 1123, 1237, 11234, 11239, 12347, 12359, 12367, 12379, 13459, 13789, 111389, 112279, 112337, 112346, 112348, 112349, 112379, 112679, 113789, 123457, 123467, 123469, 123479, 1112347, 1112357, 1112359, 1113479, 1122349

Offset: 1

David W. Wilson

			A046810(*) reaches 3 for the first time at 113.

Robert Israel, Table of n, a(n) for n = 1..86

Cf. A009994, A046810, A052382.

S[1]:= [seq([i],i=1..9)]:
for d from 2 to 7 do S[d]:= map(t -> seq([i,op(t)],i=t[1]..9),S[d-1]) od:
f:= proc(L) local t,i;
  add(`if`(isprime(add(t[i]*10^(i-1),i=1..nops(t))),1,0),t=combinat:-permute(L))
end proc:
R:= NULL: m:= -1:
for d from 1 to 7 do
for L in S[d] do
    v:= f(L);
    if v > m then m:= v;  x:= add(L[i]*10^(i-1),i=1..nops(L)); R:= R,x; fi
od od:
R; # Robert Israel, Mar 31 2025

Offset corrected by Robert Israel, Mar 31 2025

A055481 Numbers k for which there exists some m such that k = Sum_{i=1..1+floor(log_10(k))} binomial(m, d_i), where d_i is the i-th digit of k.

Original entry on oeis.org

1, 10, 18, 21, 72, 100, 101, 111, 134, 231, 246, 505, 682, 1000, 1010, 1100, 1122, 2210, 3103, 4006, 6008, 10000, 10001, 10012, 11101, 15453, 20101, 29358, 34698, 56576, 84304, 100000, 100010, 100011, 100100, 100101, 100110, 100303, 101000, 101001, 101010

Offset: 1

Erich Friedman, Jun 27 2000

Contains numbers of the form 10^k, k >= 0 so the sequence is infinite. - David A. Corneth, Oct 30 2018

			3103 = C(22, 3) + C(22, 1) + C(22, 0) + C(22, 3).
C(k, 1) + C(k, 1) + C(k, 1) + C(k, 0) + C(k, 0) + C(k, 0) = 3k + 3 so all 6-digit numbers with 3 ones and 3 zeros are in the sequence. - _David A. Corneth_, Oct 30 2018

Giovanni Resta, Table of n, a(n) for n = 1..500

Cf. A007088, A009994, A055482.

ok[n_] := Block[{d = IntegerDigits@n, k=1, v, x}, If[ Max@d <= 3, False =!= Reduce[ Total@ Binomial[x, d] == n && x>0, x, Integers], While[(v = Total@ Binomial[k, d]) < n, k++]; v == n]]; Select[ Range[10^5], ok] (* Giovanni Resta, Oct 30 2018 *)

is(n) = my(d = digits(n)); for(i = 1, n, s = sum(j = 1, #d, binomial(i, d[j])); if(s >= n, return(s == n))) \\ David A. Corneth, Oct 30 2018

a(31)-a(41) from Giovanni Resta, Oct 30 2018

cp's OEIS Frontend

A076166 Primes p such that sum of cubes of even digits of p equals sum of cubes of odd digits of p.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A165370 Smallest number whose sum of cubes of digits is n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Python

A274124 Numbers n such that (product of digits of n) is divisible by (sum of digits of n) and digits of n are in nondecreasing order.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

A327750 Numbers without zero digits such that after adding the product of its digits to it, a number with the same product of digits is obtained.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Python

Extensions

A342264 Lexicographically earliest sequence of distinct nonnegative terms such that both a(n) and a(n) + a(n+1) have digits in nondecreasing order.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Python

A342265 Lexicographically earliest sequence of distinct nonnegative terms such that both a(n) and the cumulative sum a(1)+a(2)+...+a(n) have digits in nondecreasing order.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

A342266 Lexicographically earliest sequence of distinct nonnegative terms such that both a(n) and a(n) * a(n+1) have digits in nondecreasing order.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A045910 Numbers with digits nondecreasing and their reciprocals sum to 1/(positive integer).

Views

Author

Keywords

Comments

Examples