A038369 -id:A038369 - cp's OEIS frontend

A062237 Numbers k which are (sum of digits of k) concatenated with (product of digits of k).

0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 119, 1236, 19135, 19144, 261296, 3634992, 43139968

Offset: 1

Erich Friedman, Jun 30 2001

For a d-digit number with d >= 88, the sum and product of the digits together have fewer than d digits. So every element of this sequence has 87 or fewer digits, hence it is finite. - David W. Wilson, Apr 28 2005

Fixed points of the map A380873: concatenate sum and product of digits. - M. F. Hasler, Apr 01 2025

			1236 has sum of digits 12 and product of digits 36.

Cf. A007953 (sum of digs), A007954 (product of digs), A038364, A038369, A066282, A380873, A380872 (trajectories under map).

sdpdQ[n_]:=Module[{idn=IntegerDigits[n],s,p},s=Total[idn];p=Times@@idn;n==FromDigits[Join[IntegerDigits[s],IntegerDigits[p]]]]; Select[Range[44*10^6],sdpdQ] (* Harvey P. Dale, Nov 23 2024 *)

from math import prod
from sympy.utilities.iterables import multiset_permutations as mp
from itertools import count, islice, combinations_with_replacement as mc
def c(s):
    d = list(map(int, s))
    return sorted(s) == sorted(str(sum(d)) + str(prod(d)))
def ok(s):
    d = list(map(int, s))
    return s[0] != '0' and "".join(s) == str(sum(d)) + str(prod(d))
def nd(d): yield from ("".join(m) for m in mc("0123456789", d))
def b(): yield from (s for d in count(1) for s in nd(d) if c(s))
def a(): yield from (int("".join(p)) for s in b() for p in mp(s) if ok(p))
print(list(islice(a(), 16))) # Michael S. Branicky, Jun 30 2022

More terms from Harvey P. Dale, Jul 04 2001
More terms from David W. Wilson, Apr 28 2005; he reports on May 03 2005 that there are no further terms.
Offset corrected by Altug Alkan, Apr 10 2018

A049101 Numbers m such that m divides (product of digits of m) * (sum of digits of m).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 15, 18, 24, 45, 48, 135, 144, 288, 378, 476, 756, 864, 1575, 39366, 69984, 139968

Offset: 1

Olivier Gérard

Next term if it exists is greater than 4*10^7. - Michel ten Voorde
Sequence is finite and bounded above by 10^84, since if 10^k <= n < 10^(k+1) (product of digits of n)*(sum of digits of n) <= k*9^(k+2) which is less than 10^k for k >= 84. - Henry Bottomley, May 18 2000
Numbers with a zero digit are not permitted. - Harvey P. Dale, Jul 16 2011
No further terms to 2.5*10^9. - Robert G. Wilson v, Jul 17 2011
Sequence is complete. - Giovanni Resta, Mar 20 2013
If product of digits is performed on nonzero digits only, then 1088 is also in the sequence. - Giovanni Resta, Mar 22 2013

			139968 is in the sequence since it divides (1*3*9*9*6*8) * (1+3+9+9+6+8). - _Giovanni Resta_, Mar 20 2013

Giovanni Resta, Method used to compute the full sequence

Cf. A038369, A049102, A049105, A049106.

okQ[n_]:=Module[{idn=IntegerDigits[n]},!MemberQ[idn,0] && Divisible[ (Total[idn]*Times@@idn),n]] (* Harvey P. Dale, Jul 16 2011 *)
(* full sequence *) dig[nD_] := Block[{ric, sol = {}, check}, check[mu_, minN_] := Block[{di = DigitCount@minN, k = 1, r}, While[(r = mu/k) >= minN, If[IntegerQ[r] && DigitCount[r] == di, AppendTo[sol, r]]; k++]]; ric[n_, prod_, sum_, lastd_, cnt_] := Block[{t}, If[cnt == nD, check[prod*sum, n], Do[t = nD - cnt - 1; If[n*10^(t+1) <= d*prod*9^t*(sum + d + 9*t), ric[10*n + d, d*prod, d + sum, d, cnt + 1], Break[]], {d, 9, lastd, -1}]]]; ric[0, 1, 0, 1, 0]; Print["nDig=", nD, " sol=", sol = Sort@sol]; sol]; Flatten[dig /@ Range[84]] (* Giovanni Resta, Mar 20 2013 *)

A049102 Positive numbers n such that n is a multiple of (product of digits of n) * (sum of digits of n).

Original entry on oeis.org

1, 12, 111, 112, 135, 144, 216, 432, 2112, 11112, 11115, 11232, 12312, 13824, 14112, 21112, 23112, 27216, 31212, 41112, 81216, 93312, 111132, 122112, 124416, 131112, 132192, 186624, 212112, 221112, 221184, 222912, 239112, 248832, 311472, 316224

Offset: 1

Olivier Gérard

Empirically, it looks as if every number of the form (10^3^n-1)/9 has this property. - David W. Wilson, Dec 12 2001
From David A. Corneth, Jan 23 2019: (Start)
Indeed, (10^3^n-1)/9 is in the sequence. It has digital sum times product of digits equal to 3^n.
Proof: (10^3^0-1)/9 = (10^1-1)/9 = 1 is in the sequence.
If (10^3^k-1)/9 is in the sequence then (10^3^(k + 1)-1)/9 = ((10^3^k-1)/9) * (10^(2*3^k) + 10^(3^k) + 1) = 3 * m * ((10^3^k-1)/9) for some m. This number is divisible by 3 * 3^k = 3^(k + 1) so (10^3^(k+1) - 1)/9 is in the sequence and so (10^3^n - 1) / 9 is in the sequence from which it follows that the sequence is infinite. (End)

			432 is a term because: 4*3*2=24, 4+3+2=9, 24*9=216 and 432/216 = 2.

David A. Corneth, Table of n, a(n) for n = 1..19637 (first 100 terms from Vincenzo Librandi, terms < 10^14)
David A. Corneth, a(n) = [product of digits of a(n)] * [sum of digits of a(n)] * [some factor]

Cf. A038369, A049101, A049105, A049106, A052382.

Select[ Range[10^6], IntegerQ[ # /(Apply[ Times, IntegerDigits[ # ]] * Apply[ Plus, IntegerDigits[ # ]] ) ] & ]

isok(n) = my(d=digits(n)); vecprod(d) && (n % (vecsum(d)*vecprod(d)) == 0); \\ Michel Marcus, Jan 23 2019

A066282 Numbers k such that k = (product of nonzero digits of k) * (sum of digits of k).

Original entry on oeis.org

0, 1, 135, 144, 1088

Offset: 1

Klaus Brockhaus, Dec 13 2001

Suppose a term k has d digits, then k > 10^(d-1), the product of nonzero digits <= 9^d, and the sum of digits <= 9*d. Since for d >= 85 we have 10^(d-1) > 9^d * (9*d), it follows that d <= 84. That is, the sequence is finite. I've further verified that there are no other terms, that is, the sequence is complete. - Max Alekseyev, Jul 29 2024

			(1+0+8+8) * (1*8*8) = 17*64 = 1088, so 1088 belongs to the sequence.

René-Louis Clerc, Nombres S+P, maxSP, minSP et |P-S|, hal-04507547 [math.nt], 2024. (In French)

Fixed points of A062331.

Cf. A023651, A038364, A038369, A062237.

function a066282(a,b: integer); var n,k,j,p,d: integer; s: string; begin for n := a to b do s := itoa(n); k := 0; p := 1; for j := 0 to length(s) - 1 do d := atoi(s[j..j]); k := k + d; if d > 0 then p := p*d; end; end; if n = p*k then write(n,","); end; end; end; a066282(0,25000).

Do[ d = Sort[ IntegerDigits[n]]; While[ First[d] == 0, d = Drop[d, 1]]; If[n == Apply[ Plus, d] Apply[ Times, d], Print[n]], {n, 0, 5*10^7} ]

a066282(a,b) = local(n,k,q,p,d); for(n=a,b,k=0; p=1; q=n; while(q>0,d=divrem(q,10); q=d[1]; k=k+d[2]; p=p*max(1,d[2])); if(n==p*k,print1(n,", ")))
a066282(0,25000)

Offset corrected by Mohammed Yaseen, Jul 21 2022

Keywords fini,full added by Max Alekseyev, Jul 29 2024

A066308 a(n) = (sum of digits of n) * (product of digits of n).

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 0, 6, 16, 30, 48, 70, 96, 126, 160, 198, 0, 12, 30, 54, 84, 120, 162, 210, 264, 324, 0, 20, 48, 84, 128, 180, 240, 308, 384, 468, 0, 30, 70, 120, 180, 250, 330, 420, 520, 630, 0, 42, 96, 162, 240, 330

Offset: 1

Labos Elemer, Dec 13 2001

a(n) can be greater than, less than, or equal to n; see Example section.

			For n = 12, a(12) = (1 + 2)*(1*2) = 3*2 = 6 < n;
for n = 19, a(19) = (1 + 9)*(1*9) = 90 > n;
for n = 135, a(135) =(1 + 3 + 5)*(1*3*5) = 135 = n.

Harry J. Smith, Table of n, a(n) for n = 1..1000

Cf. A007953, A007954.
Cf. A038369 (fixed points), A034710, A061672.
Cf. A049101, A049102, A049103, A049104, A049105, A049106.

asum[x_] := Apply[Plus, IntegerDigits[x]] apro[x_] := Apply[Times, IntegerDigits[x]] a[n]=asum[n]*apro[n]
sdpd[n_]:=Module[{idn=IntegerDigits[n]},Total[idn]Times@@idn]; Array[ sdpd,70] (* Harvey P. Dale, Dec 31 2011 *)

a(n) = my(d = digits(n)); vecsum(d) * vecprod(d); \\ Michel Marcus, Feb 24 2017

Edited by Jon E. Schoenfield, Jul 09 2018

A023651 Numbers k such that (product of digits of k) * (sum of digits of k) = 2k.

Original entry on oeis.org

0, 2, 15, 24, 1575, 39366

Offset: 1

Jason Earls, Dec 11 2001

Except for k = 0, this sequence is a subsequence of A049101. - Jason Yuen, Feb 26 2024

Cf. A007953, A007954, A049101.

Cf. A038364, A038369, A062237, A066282.

Do[ If[ 2n == Apply[ Times, IntegerDigits[n]] Apply[ Plus, IntegerDigits[n]], Print[n]], {n, 0, 10^7} ]

isok(n) = if(n, factorback(digits(n)), 0) * sumdigits(n) == 2*n \\ Mohammed Yaseen, Jul 22 2022

from math import prod
def s(n): return sum(map(int, str(n)))
def p(n): return prod(map(int, str(n)))
for n in range(0, 10**6):
  if p(n)*s(n)==2*n:
    print(n) # Mohammed Yaseen, Jul 22 2022

Offset corrected by Arkadiusz Wesolowski, Oct 17 2012

A051004 Numbers divisible both by their individual digits and by the sum of their digits.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 24, 36, 48, 111, 112, 126, 132, 135, 144, 162, 216, 222, 224, 264, 288, 312, 315, 324, 333, 336, 396, 432, 444, 448, 555, 612, 624, 648, 666, 735, 777, 864, 888, 936, 999, 1116, 1122, 1128, 1164, 1212, 1224, 1236, 1296, 1332

Offset: 1

Eric W. Weisstein

No zero digits permitted. [Harvey P. Dale, Dec 18 2011]

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Digit

Intersection of A005349 and A034838.

Cf. A038369, A087142.

a051004 n = a051004_list !! (n-1)
a051004_list =  [x | x <- a005349_list,
                     x == head (dropWhile (< x) a034838_list)]
-- Reinhard Zumkeller, Mar 03 2012

ddQ[n_]:=Module[{idn=IntegerDigits[n]},!MemberQ[idn,0] && Divisible[ n,Total[idn]]&& And @@ Divisible[n,idn]]; Select[Range[1400],ddQ] (* Harvey P. Dale, Dec 18 2011 *)

Offset corrected by Reinhard Zumkeller, Mar 03 2012

A066310 Numbers k such that k < (product of digits of k) * (sum of digits of k).

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 8, 9, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 33, 34, 35, 36, 37, 38, 39, 42, 43, 44, 45, 46, 47, 48, 49, 52, 53, 54, 55, 56, 57, 58, 59, 62, 63, 64, 65, 66, 67, 68, 69, 72, 73, 74, 75, 76, 77, 78, 79, 82, 83, 84, 85, 86, 87, 88, 89, 92, 93, 94, 95

Offset: 1

Labos Elemer and Klaus Brockhaus, Dec 13 2001

			14 < (1*4)*(1+4) = 20, so 14 is a term of this sequence.
For n=199, (1+9+9)*1*9*9 = 1539 > 199, so 199 is here.

Harry J. Smith, Table of n, a(n) for n=1..1000

Cf. A038369, A049101-A049106, A034710, A061672, A066306-A066309.

function a066311(a,b: integer); var n,k,j,p,d: integer; s: string; begin for n := a to b do s := itoa(n); k := 0; p := 1; for j := 0 to length(s) - 1 do d := atoi(s[j..j]); k := k + d; p := p*d; end; if n < p*k then write(n,","); end; end; end; a066311(0,120);

asum[x_] := Apply[Plus, IntegerDigits[x]] apro[x_] := Apply[Times, IntegerDigits[x]] sz[x_] := asu[x]*apro[x] Do[s=sz[n]; If[Greater[s, n], Print[n]], {n, 1, 200}]

isok(m) = my(d=digits(m)); m < vecprod(d)*vecsum(d); \\ Michel Marcus, Mar 23 2020

A066024 Numbers k such that the product of the digits of k minus the sum of the digits of k is prime.

Original entry on oeis.org

24, 25, 27, 29, 33, 34, 35, 37, 38, 42, 43, 45, 47, 49, 52, 53, 54, 56, 57, 59, 65, 67, 72, 73, 74, 75, 76, 78, 79, 83, 87, 92, 94, 95, 97, 125, 126, 128, 133, 144, 146, 148, 152, 162, 164, 166, 182, 184, 188, 215, 216, 218, 222, 223, 225, 227, 229, 232, 245, 247

Offset: 1

Enoch Haga, Dec 11 2001

			a(1)=24 because 2 + 4 = 6, 2*4 = 8, and 8 - 6 = 2, which is prime. [corrected by _Harry J. Smith_, Nov 07 2009]

Harry J. Smith, Table of n, a(n) for n = 1..1000

Cf. A038369, A066027.

isok(k) = {my(d=digits(k)); isprime(vecprod(d)-vecsum(d))} \\ Harry J. Smith, Nov 07 2009

A066027 Numbers k such that the sum of digits of k minus the product of digits of k is prime.

Original entry on oeis.org

20, 30, 50, 70, 101, 102, 104, 106, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 131, 140, 141, 151, 160, 161, 171, 181, 191, 200, 201, 203, 205, 209, 210, 211, 230, 250, 290, 300, 302, 304, 308, 311, 320, 340, 380, 401, 403, 407, 409, 410

Offset: 1

Enoch Haga, Dec 11 2001

			a(11)=112 because 1 + 1 + 2 = 4, 2*1*1 = 2, and 4 - 2 = 2, which is prime. [corrected by _Harry J. Smith_, Nov 07 2009]

Harry J. Smith, Table of n, a(n) for n = 1..1000

Cf. A066024, A038369.

isok(k) = {my(d=digits(k)); isprime(vecsum(d) - vecprod(d))} \\ Harry J. Smith, Nov 07 2009

cp's OEIS Frontend

A062237 Numbers k which are (sum of digits of k) concatenated with (product of digits of k).

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A049101 Numbers m such that m divides (product of digits of m) * (sum of digits of m).

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A049102 Positive numbers n such that n is a multiple of (product of digits of n) * (sum of digits of n).

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A066282 Numbers k such that k = (product of nonzero digits of k) * (sum of digits of k).

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A066308 a(n) = (sum of digits of n) * (product of digits of n).

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A023651 Numbers k such that (product of digits of k) * (sum of digits of k) = 2k.

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A051004 Numbers divisible both by their individual digits and by the sum of their digits.

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