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

0, 1, 135, 144, 1088

Offset

1,3

Comments

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

Links

Table of n, a(n) for n=1..5.

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

Example

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

Mathematica

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} ]

Prog

(ARIBAS) 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).
(PARI) 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)

Crossrefs

Fixed points of A062331.

Cf. A023651, A038364, A038369, A062237.

Sequence in context: A355002 A378104 A038369 * A235809 A066176 A025363

Adjacent sequences: A066279 A066280 A066281 * A066283 A066284 A066285

Keyword

base,fini,full,nonn

Author

Klaus Brockhaus, Dec 13 2001

Extensions

Offset corrected by Mohammed Yaseen, Jul 21 2022

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

Status

approved