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A355148
Numbers that are the concatenation of two palindromes and that have exactly two palindromic factors, all with the same number of decimal digits.
2
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
1,1
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
LINKS
Michael S. Branicky, Table of n, a(n) for n = 1..597
Nicolas Bělohoubek, Table of n, a(n) for n = 1..56
EXAMPLE
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).
PROG
(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
CROSSREFS
Sequence in context: A275113 A043651 A043701 * A344980 A344883 A290001
KEYWORD
base,nonn
AUTHOR
Nicolas Bělohoubek, Jun 21 2022
STATUS
approved