cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

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%I A355148 #60 Jun 10 2024 19:30:23
%S A355148 12,14,15,16,18,21,24,25,27,28,32,35,36,42,45,48,49,54,56,63,64,72,81,
%T A355148 3388,7744,101787,101808,111888,151848,212565,212898,232656,313464,
%U A355148 313575,353868,383595,383838,414585,434676,454545,505808,515595,525252,555888
%N A355148 Numbers that are the concatenation of two palindromes and that have exactly two palindromic factors, all with the same number of decimal digits.
%C A355148 All numbers of form (4/45)*(9*100^d - 29*10^d + 20) are terms (see example).
%C A355148 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 A355148 From _Chai Wah Wu_, Aug 23 2022: (Start)
%C A355148 Terms with 2 decompositions:
%C A355148   12 = 3*4 = 2*6
%C A355148   16 = 4*4 = 2*8
%C A355148   18 = 2*9 = 3*6
%C A355148   24 = 4*6 = 3*8
%C A355148   36 = 4*9 = 6*6
%C A355148   153535351846464648 = 189828981*808808808 = 172727271*888888888
%C A355148   182919281817080718 = 303303303*603090306 = 201030102*909909909
%C A355148   183838381816161618 = 303060303*606606606 = 202040202*909909909
%C A355148   185676581814323418 = 306090603*606606606 = 204060402*909909909
%C A355148   192919291807080708 = 303303303*636060636 = 212020212*909909909
%C A355148   193838391806161608 = 303303303*639090936 = 213030312*909909909
%C A355148   283919382716080617 = 312030213*909909909 = 303303303*936090639
%C A355148   293656392403040304 = 461262164*636636636 = 363363363*808161808
%C A355148   293919392706080607 = 323020323*909909909 = 303303303*969060969
%C A355148   365838563634161436 = 603090306*606606606 = 402060204*909909909
%C A355148   385838583614161416 = 606606606*636060636 = 424040424*909909909
%C A355148   387676783612323216 = 606606606*639090936 = 426060624*909909909
%C A355148   567838765432161234 = 624060426*909909909 = 606606606*936090639
%C A355148   587838785412161214 = 646040646*909909909 = 606606606*969060969
%C A355148 Conjecture: these are an infinite number of such terms.
%C A355148 The following term has 3 decompositions:
%C A355148   113131311886868688 = 279747972*404404404 = 254545452*444444444 = 252252252*448484844.
%C A355148 (End)
%C A355148 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 A355148 Michael S. Branicky, <a href="/A355148/b355148.txt">Table of n, a(n) for n = 1..597</a>
%H A355148 Nicolas Bělohoubek, <a href="/A355148/a355148.txt">Table of n, a(n) for n = 1..56</a>
%e A355148 42 is the concatenation of 4 and 2, and is also 6*7 (all 1 digit).
%e A355148 3388 is the concatenation of 33 and 88, and is also 44*77 (all 2 digits).
%e A355148 414585 is the concatenation of 414 and 585, and is also 555*747 (all 3 digits).
%e A355148 131080 = 232*565 is not a term since 080 begins with 0 and hence is not a three-digit palindromic number.
%e A355148 79974224 = 8998*8888, 7999742224 = 89998*88888, 799997422224 = 899998*888888 (see comments).
%o A355148 (Python)
%o A355148 from sympy import divisors
%o A355148 from itertools import count, islice, product
%o A355148 def ispal(s): return s == s[::-1]
%o A355148 def pals(d, start0=False): # generates palindromic strings with d digits
%o A355148     digits = "0123456789"
%o A355148     if d == 1: yield from "0"*int(start0) + "123456789"; return
%o A355148     for p in product(digits, repeat=d//2):
%o A355148         if not start0 and p[0] == "0": continue
%o A355148         left = "".join(p); right = left[::-1]
%o A355148         for mid in [[""], digits][d%2]: yield left + mid + right
%o A355148 def agen(): # generator of terms
%o A355148     for d in count(1):
%o A355148         found = set()
%o A355148         for p1 in pals(d):
%o A355148             for p2 in pals(d):
%o A355148                 p = int(p1)*int(p2)
%o A355148                 s = str(p)
%o A355148                 if len(s) != 2*d: continue
%o A355148                 if ispal(s[:d]) and s[d] != "0" and ispal(s[d:]):
%o A355148                     found.add(p)
%o A355148         yield from sorted(found)
%o A355148 print(list(islice(agen(), 51))) # _Michael S. Branicky_, Jun 21 2022
%K A355148 base,nonn
%O A355148 1,1
%A A355148 _Nicolas Bělohoubek_, Jun 21 2022