A338039 Numbers m such that A338038(m) = A338038(A004086(m)) where A004086(i) is i read backwards and A338038(i) is the sum of the primes and exponents in the prime factorization of i ignoring 1-exponents; palindromes and multiples of 10 are excluded.
18, 81, 198, 576, 675, 819, 891, 918, 1131, 1304, 1311, 1818, 1998, 2262, 2622, 3393, 3933, 4031, 4154, 4514, 4636, 6364, 8181, 8749, 8991, 9478, 12441, 14269, 14344, 14421, 15167, 15602, 16237, 18018, 18449, 18977, 19998, 20651, 23843, 24882, 26677, 26892, 27225
Offset: 1
Examples
For m = 18 = 2*3^2, A338038(18) = 2 + (3+2) = 7 and for m = 81 = 3^4, A338038(81) = 7, so 18 and 81 are terms.
Links
- Michel Marcus, Table of n, a(n) for n = 1..2998
- Chris Bispels, Muhammet Boran, Steven J. Miller, Eliel Sosis, and Daniel Tsai, v-Palindromes: An Analogy to the Palindromes, arXiv:2405.05267 [math.HO], 2024.
- Muhammet Boran, Garam Choi, Steven J. Miller, Jesse Purice, and Daniel Tsai, A characterization of prime v-palindromes, arXiv:2307.00770 [math.NT], 2023.
- Daniel Tsai, A recurring pattern in natural numbers of a certain property, arXiv:2010.03151 [math.NT], 2020.
- Daniel Tsai, A recurring pattern in natural numbers of a certain property, Integers (2021) Vol. 21, Article #A32.
- Daniel Tsai, v-palindromes: an analogy to the palindromes, arXiv:2111.10211 [math.HO], 2021.
Programs
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Maple
rev:= proc(n) local L,i; L:= convert(n,base,10); add(L[-i]*10^(i-1),i=1..nops(L)) end proc: g:= proc(n) local t; add(t[1]+t[2],t=subs(1=0,ifactors(n)[2])) end proc: filter:= proc(n) local r; if n mod 10 = 0 then return false fi; r:= rev(n); r <> n and g(r)=g(n) end proc: select(filter, [$1..30000]); # Robert Israel, Oct 13 2020
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Mathematica
s[1] = 0; s[n_] := Plus @@ First /@ (f = FactorInteger[n]) + Plus @@ Select[Last /@ f, # > 1 &]; Select[Range[30000], !Divisible[#, 10] && (r = IntegerReverse[#]) != # && s[#] == s[r] &] (* Amiram Eldar, Oct 08 2020 *)
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PARI
f(n) = my(f=factor(n)); vecsum(f[,1]) + sum(k=1, #f~, if (f[k,2]!=1, f[k,2])); \\ A338038 isok(m) = my(r=fromdigits(Vecrev(digits(m)))); (m % 10) && (m != r) && (f(r) == f(m));
Comments