A117720 -id:A117720 - cp's OEIS frontend

A241846 Numbers for which the cube of the sum of the digits is equal to the square of the product of their digits.

0, 1, 88, 333, 11248, 11284, 11428, 11482, 11824, 11842, 12148, 12184, 12418, 12481, 12814, 12841, 14128, 14182, 14218, 14281, 14812, 14821, 18124, 18142, 18214, 18241, 18412, 18421, 21148, 21184, 21418, 21481, 21814, 21841, 24118, 24181, 24811, 28114

Offset: 1

Michel Lagneau, Apr 30 2014

Let d_1 d_2... d_q denote the decimal expansion of a number n. The sequence lists the numbers n such that (d_1 + d_2 +...+ d_q)^3 = (d_1 * d_2 *...* d_q)^2.
The sequence is finite and contains 1419 terms because the maximum sum of the digits of a(n) is 16, the maximum product of the digits is 64 with 16^3 = 64^2 and the greatest number of the sequence is 2222221111.
The primitive values of a(n) (numbers whose decimal digits are not a permutation of another number of the sequence) are 0, 1, 88, 333, 11248, 112228, 1111444, 11112244, 111122224, 1111222222.
Nevertheless, the numbers 112228, 1111444, 11112244, 111122224, 1111222222 are not completely independent; for example, a decimal digit 4 of 1111444 becomes 22 and gives the number 11112244.

			333 is in the sequence because (3+3+3)^3 = (3*3*3)^2 = 729.
11248 is in the sequence because (1+1+2+4+8)^3 = (1*1*2*4*8)^2 = 4096.

Michel Lagneau, Table of n, a(n) for n = 1..1419

Cf. A034710, A117720.

Select[Range[30000], (Plus @@ IntegerDigits[ # ]^3) == (Times @@ IntegerDigits[ # ]^2) &]

A162840 Numbers k such that the cube of the sum of digits of k equals the product of digits of k.

Original entry on oeis.org

0, 1, 666666, 1377789, 1377798, 1377879, 1377897, 1377978, 1377987, 1378779, 1378797, 1378977, 1379778, 1379787, 1379877, 1387779, 1387797, 1387977, 1389777, 1397778, 1397787, 1397877, 1398777, 1555888, 1558588, 1558858

Offset: 0

Boris Hostnik (megpplus(AT)siol.net), Jul 14 2009

			666666 is in the sequence because (1) cubed sum of its digits is (6+6+6+6+6+6)^3 = 46656, (2) the product of its digits is 6*6*6*6*6*6=46656; 46656=46656.

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Cf. A007954, A118880, A117720, A034710. - R. J. Mathar, Jul 19 2009

A007953 := proc(n) add(d,d=convert(n,base,10)) ; end: A007954 := proc(n) mul(d,d=convert(n,base,10)) ; end: A118880 := proc(n) (A007953(n))^3; end: for n from 1 to 2000000 do if A118880(n) = A007954(n) then printf("%d,\n",n) ; fi; od: # R. J. Mathar, Jul 19 2009

Select[Range[0,156*10^4],Total[IntegerDigits[#]]^3==Times@@IntegerDigits[#]&] (* Harvey P. Dale, Jul 07 2022 *)

{n: A118880(n)=A007954(n)}. - R. J. Mathar, Jul 19 2009

A261778 Positive numbers n such that (digitsum(n))^2 equals (product of digits(n))^3.

Original entry on oeis.org

1, 11114, 11141, 11411, 14111, 41111, 111122, 111212, 111221, 112112, 112121, 112211, 121112, 121121, 121211, 122111, 211112, 211121, 211211, 212111, 221111, 1111111111111111119, 1111111111111111191, 1111111111111111911, 1111111111111119111, 1111111111111191111, 1111111111111911111

Offset: 1

K. D. Bajpai, Aug 31 2015

Sequence is infinite because it contains all the numbers made of k fours and 8^k-4k ones. - Giovanni Resta, Sep 01 2015

			11114 appears in the sequence because (1 + 1 + 1 + 1 + 4)^2 = (1*1*1*1*4)^3 = 64.
111122 appears in the sequence because (1 + 1 + 1 + 1 + 2 + 2)^2  = (1*1*1*1*2*2)^3 = 64.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Charles R Greathouse IV, GP script

Cf. A034710, A117720, A117723, A241846.

[n : n in [1..1000000] | (&+Intseq(n))^2 eq (&*Intseq(n))^3 ];

Select[Range[20000000], Plus @@ IntegerDigits[#]^2 == Times @@ IntegerDigits[#]^3 &]

for(n = 1,1000000, d = digits(n); if((sumdigits(n))^2 == prod(i = 1, #d, d[i])^3, print1(n, ", ")));

proddigits(n)=my(d=digits(n)); prod(i=1,#d,d[i])
is(n)=my(s=sumdigits(n)); if(!ispower(s,3), return(0)); s^2==proddigits(n)^3 \\ Charles R Greathouse IV, Aug 31 2015

a(22)-a(27) from Charles R Greathouse IV, Aug 31 2015

A379834 Positive integers with digits in nondescending order whose digit product is an integer power of their digit sum, given power > 1.

Original entry on oeis.org

999, 2558, 3366, 4444, 12489, 13377, 22444, 112668, 113388, 114455, 122289, 123336, 222244, 666666, 1113399, 1113468, 1114449, 1122455, 1133334, 1377789, 1555888, 2222224, 2346669, 3334689, 3344499, 4444448, 11112778, 11114466, 11122368, 11122449, 11222255, 11223333, 11467779, 12366666, 13336668

Offset: 1

Robert Kinner, Jan 03 2025

This sequence would include the terms of A274124 (digit product equal to digit sum) if the power were allowed to be 1.
This sequence is similar to A117720 (digit product is the square of the digit sum) except that powers > 2 are also included, and that A117720 does not require nondescending digits. These results are from empirical observation.
Each number of this sequence is an element of a group of permutations of its digits.
Since both addition and multiplication are commutative, it is only necessary to select one representative member of each digit permutation group. The member with digits in nondescending order is chosen.

			The product of the digits of 999 equals the sum of the digits, 27, squared.
The product of the digits of 666666 equals the cube of the sum of the digits, 36.

Robert Kinner, Pascal Program

Cf. A117720, A274124.

Pascal
```
// See Robert Kinner link.
```

cp's OEIS Frontend

A241846 Numbers for which the cube of the sum of the digits is equal to the square of the product of their digits.

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A162840 Numbers k such that the cube of the sum of digits of k equals the product of digits of k.

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Maple

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Formula

A261778 Positive numbers n such that (digitsum(n))^2 equals (product of digits(n))^3.

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Magma

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PARI

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A379834 Positive integers with digits in nondescending order whose digit product is an integer power of their digit sum, given power > 1.

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Pascal