A392568 - OEIS

%I #74 Jan 31 2026 10:13:25

%S 3,4,5,6,7,8,9,148,164,167,169,178,187,196,262,334,388,393,584,668,

%T 928,937,946,955,964,973,982,1376,1655,1668,1673,2544,2654,4524,6343,

%U 10697,10764,12686,15278,15365,16524,30724,32955,41268,45634,50879,53384,54237

%N Numbers k such that there is a smaller number m > 1 such that k*m equals the concatenation of digit-wise multiplication, keeping the leading digits of k when m has fewer digits.

%C We take the digit from the greater number when no corresponding digit in the smaller number exists.

%C Greatest values for m < k are given in A392618.

%C If the multiplication per digit gives a one-digit result 0 - 9, only this single digit is used in the concatenation. So we don't allow 1124*971 = 1091404, it becomes the concatenation(1,9,14,4) = 19144 instead.

%C It is not true that m is always only one digit shorter than k, for example: 41268*998 = 41185464.

%C Numbers k and m cannot have the same number of digits unless they are both smaller than 10. - _Max Alekseyev_, Jan 30 2026

%H Max Alekseyev, <a href="/A392568/b392568.txt">Table of n, a(n) for n = 1..123</a>

%H <a href="/index/Ca#CARRYLESS">Index entries for sequences related to carryless arithmetic</a>.

%e 148 is part of the sequence because 148*92 = 13616 = concatenate(1, 4*9=36, 8*2=16).

%e This sequence was inspired by the well known internet meme corresponding to the multiplication:

%e   3  9  8  7  6  8  7  7

%e      9  5  6  4  5  4  6

%e   3|81|40|42|24|40|28|42 = 39876877*9564546

%o (PARI) /* Inefficient brute force, just for illustration. Returns a solution m > 1 if n is a term, or zero if not. */

%o is_A392568(n)={ my(dn=digits(n), L=#dn, o, dwp); forstep(m=n-1,2,-1, my(dp=digits(m*n));

%o   if (o = L - #dm=digits(m), dp[1..o] == dn[1..o] || next);

%o   #[0 | t <- dwp = [dm[i]*dn[i+o] | i<-[1..#dm]], t>9] == #dp - #dn || next;

%o   concat([if (t>9, digits(t), [t]) | t <- dwp]) == dp[o+1..-1] && return(m))}

%o select( is_A392568, [1..999]) \\ _M. F. Hasler_, Jan 25 2026

%o (Python)

%o def match(skm, sk, sm):

%o     d = len(sk) - len(sm)

%o     if skm[:d] != sk[:d]: return False

%o     return skm[d:] == "".join(str(int(sk[i+d])*int(sm[i])) for i in range(len(sm)))

%o def ok(k):

%o     sk = str(k)

%o     return any(match(str(k*m), sk, str(m)) for m in range(k-1, 1, -1))

%o print([k for k in range(2000) if ok(k)]) # _Michael S. Branicky_, Jan 30 2026

%Y Cf. A392618, A392040 (list of m*k).

%K nonn,base

%O 1,1

%A _Thomas Scheuerle_, Jan 16 2026