A327749 Natural numbers whose sum of prime factors (with repetition) is palindromic in base 10.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 24, 27, 28, 40, 45, 48, 54, 57, 62, 85, 101, 102, 106, 116, 121, 123, 131, 151, 181, 182, 191, 194, 218, 259, 260, 278, 292, 298, 305, 308, 312, 313, 351, 353, 358, 366, 370, 373, 383, 388, 403, 413, 415, 428, 440, 444, 483, 495, 498
Offset: 1
References
- Karl G. Kröber, "Palindrome, Perioden und Chaoten: 66 Streifzüge durch die palindromischen Gefilde" (1997, Deutsch-Taschenbücher; Bd. 99) ISBN 3-8171-1522-9.
- Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
-
Magma
[1] cat [k: k in [2..500]| Intseq(a) eq Reverse(Intseq(a)) where a is &+[m[1]*m[2]: m in Factorization(k)]]; // Marius A. Burtea, Sep 27 2019
-
Maple
ispali:= proc(n) option remember; local L; L:= convert(n,base,10); evalb(L = ListTools:-Reverse(L)) end proc: spf:= proc(n) add(t[1]*t[2],t=ifactors(n)[2]) end proc: select(t -> ispali(spf(t)), [$0..1000]); # Robert Israel, Nov 20 2020
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Mathematica
sopfr[1] = 0; sopfr[n_] := Plus @@ (Times @@@ FactorInteger[n]); aQ[n_] := PalindromeQ[sopfr[n]]; Select[Range[500], aQ] (* Amiram Eldar, Sep 23 2019 *)
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PARI
sopfr(n) = (n=factor(n))[, 1]~*n[, 2]; \\ A001414 isok(n) = my(d=digits(sopfr(n))); d == Vecrev(d); \\ Michel Marcus, Sep 27 2019
Comments