A355148 Numbers that are the concatenation of two palindromes and that have exactly two palindromic factors, all with the same number of decimal digits.

12, 14, 15, 16, 18, 21, 24, 25, 27, 28, 32, 35, 36, 42, 45, 48, 49, 54, 56, 63, 64, 72, 81, 3388, 7744, 101787, 101808, 111888, 151848, 212565, 212898, 232656, 313464, 313575, 353868, 383595, 383838, 414585, 434676, 454545, 505808, 515595, 525252, 555888

Offset: 1

Author

Nicolas Bělohoubek, Jun 21 2022

Keywords

• base
• nonn

Comments

All numbers of form (4/45)*(9*100^d - 29*10^d + 20) are terms (see example).

Also numbers of form (7/18)*(100^d - 13*10^d + 12) and are also terms (d>1) and of form (4/45)*(4*100^d - 19*10^d + 15) (d>1).

From Chai Wah Wu, Aug 23 2022: (Start)

Terms with 2 decompositions:

12 = 3*4 = 2*6

16 = 4*4 = 2*8

18 = 2*9 = 3*6

24 = 4*6 = 3*8

36 = 4*9 = 6*6

153535351846464648 = 189828981*808808808 = 172727271*888888888

182919281817080718 = 303303303*603090306 = 201030102*909909909

183838381816161618 = 303060303*606606606 = 202040202*909909909

185676581814323418 = 306090603*606606606 = 204060402*909909909

192919291807080708 = 303303303*636060636 = 212020212*909909909

193838391806161608 = 303303303*639090936 = 213030312*909909909

283919382716080617 = 312030213*909909909 = 303303303*936090639

293656392403040304 = 461262164*636636636 = 363363363*808161808

293919392706080607 = 323020323*909909909 = 303303303*969060969

365838563634161436 = 603090306*606606606 = 402060204*909909909

385838583614161416 = 606606606*636060636 = 424040424*909909909

387676783612323216 = 606606606*639090936 = 426060624*909909909

567838765432161234 = 624060426*909909909 = 606606606*936090639

587838785412161214 = 646040646*909909909 = 606606606*969060969

Conjecture: these are an infinite number of such terms.

The following term has 3 decompositions:

113131311886868688 = 279747972*404404404 = 254545452*444444444 = 252252252*448484844.

(End)

A subsequence of this sequence is {s(k)} where s(k) = (202/10989)*t(k)*u(k), t(k) = 10^(6*k + 3) - 1 and u(k) = 2099*10^(6*k + 1) + 988. s(k) can be decomposed in 2 different ways: the first is (202/333)*t(k) and (1/33)*u(k); the second is (101/111)*t(k) and (2/99)*u(k). And since {s(k)} is an infinite sequence, its existence proves Chai Wah Wu's conjecture to be true. - Nicolas Bělohoubek, May 20 2024

Examples

			42 is the concatenation of 4 and 2, and is also 6*7 (all 1 digit).
3388 is the concatenation of 33 and 88, and is also 44*77 (all 2 digits).
414585 is the concatenation of 414 and 585, and is also 555*747 (all 3 digits).
131080 = 232*565 is not a term since 080 begins with 0 and hence is not a three-digit palindromic number.
79974224 = 8998*8888, 7999742224 = 89998*88888, 799997422224 = 899998*888888 (see comments).
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..597
• Nicolas Bělohoubek, Table of n, a(n) for n = 1..56

Programs

Python

from sympy import divisors
from itertools import count, islice, product
def ispal(s): return s == s[::-1]
def pals(d, start0=False): # generates palindromic strings with d digits
    digits = "0123456789"
    if d == 1: yield from "0"*int(start0) + "123456789"; return
    for p in product(digits, repeat=d//2):
        if not start0 and p[0] == "0": continue
        left = "".join(p); right = left[::-1]
        for mid in [[""], digits][d%2]: yield left + mid + right
def agen(): # generator of terms
    for d in count(1):
        found = set()
        for p1 in pals(d):
            for p2 in pals(d):
                p = int(p1)*int(p2)
                s = str(p)
                if len(s) != 2*d: continue
                if ispal(s[:d]) and s[d] != "0" and ispal(s[d:]):
                    found.add(p)
        yield from sorted(found)
print(list(islice(agen(), 51))) # Michael S. Branicky, Jun 21 2022