A355148 - OEIS

A355148

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

%I #60 Jun 10 2024 19:30:23

%S 12,14,15,16,18,21,24,25,27,28,32,35,36,42,45,48,49,54,56,63,64,72,81,

%T 3388,7744,101787,101808,111888,151848,212565,212898,232656,313464,

%U 313575,353868,383595,383838,414585,434676,454545,505808,515595,525252,555888

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

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

%C 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).

%C From _Chai Wah Wu_, Aug 23 2022: (Start)

%C Terms with 2 decompositions:

%C 12 = 3*4 = 2*6

%C 16 = 4*4 = 2*8

%C 18 = 2*9 = 3*6

%C 24 = 4*6 = 3*8

%C 36 = 4*9 = 6*6

%C 153535351846464648 = 189828981*808808808 = 172727271*888888888

%C 182919281817080718 = 303303303*603090306 = 201030102*909909909

%C 183838381816161618 = 303060303*606606606 = 202040202*909909909

%C 185676581814323418 = 306090603*606606606 = 204060402*909909909

%C 192919291807080708 = 303303303*636060636 = 212020212*909909909

%C 193838391806161608 = 303303303*639090936 = 213030312*909909909

%C 283919382716080617 = 312030213*909909909 = 303303303*936090639

%C 293656392403040304 = 461262164*636636636 = 363363363*808161808

%C 293919392706080607 = 323020323*909909909 = 303303303*969060969

%C 365838563634161436 = 603090306*606606606 = 402060204*909909909

%C 385838583614161416 = 606606606*636060636 = 424040424*909909909

%C 387676783612323216 = 606606606*639090936 = 426060624*909909909

%C 567838765432161234 = 624060426*909909909 = 606606606*936090639

%C 587838785412161214 = 646040646*909909909 = 606606606*969060969

%C Conjecture: these are an infinite number of such terms.

%C The following term has 3 decompositions:

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

%C (End)

%C 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

%H Michael S. Branicky, <a href="/A355148/b355148.txt">Table of n, a(n) for n = 1..597</a>

%H Nicolas Bělohoubek, <a href="/A355148/a355148.txt">Table of n, a(n) for n = 1..56</a>

%e 42 is the concatenation of 4 and 2, and is also 6*7 (all 1 digit).

%e 3388 is the concatenation of 33 and 88, and is also 44*77 (all 2 digits).

%e 414585 is the concatenation of 414 and 585, and is also 555*747 (all 3 digits).

%e 131080 = 232*565 is not a term since 080 begins with 0 and hence is not a three-digit palindromic number.

%e 79974224 = 8998*8888, 7999742224 = 89998*88888, 799997422224 = 899998*888888 (see comments).

%o (Python)

%o from sympy import divisors

%o from itertools import count, islice, product

%o def ispal(s): return s == s[::-1]

%o def pals(d, start0=False): # generates palindromic strings with d digits

%o digits = "0123456789"

%o if d == 1: yield from "0"*int(start0) + "123456789"; return

%o for p in product(digits, repeat=d//2):

%o if not start0 and p[0] == "0": continue

%o left = "".join(p); right = left[::-1]

%o for mid in [[""], digits][d%2]: yield left + mid + right

%o def agen(): # generator of terms

%o for d in count(1):

%o found = set()

%o for p1 in pals(d):

%o for p2 in pals(d):

%o p = int(p1)*int(p2)

%o s = str(p)

%o if len(s) != 2*d: continue

%o if ispal(s[:d]) and s[d] != "0" and ispal(s[d:]):

%o found.add(p)

%o yield from sorted(found)

%o print(list(islice(agen(), 51))) # _Michael S. Branicky_, Jun 21 2022

%K base,nonn

%O 1,1

%A _Nicolas Bělohoubek_, Jun 21 2022