A275407 Let e_n(k)>=0 denote the exponent of prime(k) in the prime power representation of n. The sequence lists 1 followed by numbers n for which e_n(2*i-1)=e_n(2*i), for all i>=1.
1, 6, 35, 36, 143, 210, 216, 323, 667, 858, 1147, 1225, 1260, 1296, 1763, 1938, 2491, 3599, 4002, 4757, 5005, 5148, 5767, 6882, 7350, 7387, 7560, 7776, 9797, 10578, 11021, 11305, 11628, 12317, 14946, 16637, 19043, 20449, 21594, 22499, 23345, 24012, 25591, 28542
Offset: 1
Keywords
Examples
15 is not in the sequence, since 15 = 3*5 and the prime index of 5 is odd. 5148 is in the sequence, since 5148 = 2^2*3^2*11*13 and (1) 3 is the next prime after 2, (2) the exponents of 2 and 3 are equal, (3) the prime index of 3 is even, (4) 13 is the next prime after 11, (5) the exponents of 11 and 13 are equal, (6) the prime index of 13 is even.
Links
- Peter J. C. Moses, Table of n, a(n) for n = 1..1000
Crossrefs
Programs
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Mathematica
inA275407Q:=If[EvenQ[Length[#]],Apply[And,Join[Map[#[[1]]+1==#[[2]]&&EvenQ[#[[2]]]&,PrimePi[#[[1]]]],Map[#[[1]]==#[[2]]&,#[[2]]]]]&[Map[Partition[#,2]&,Transpose[#]]],False]&[FactorInteger[#]]&; Join[{1},Select[Range[10000],inA275407Q]] (* Peter J. C. Moses, Jul 29 2016 *)
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PARI
isok(n) = {f = factor(n); nbpok = 0; for (k=1, #f~, ip = primepi(f[k, 1]); if ((ip % 2) && (kk = vecsearch(f[,1]~, prime(ip+1))) && (f[kk, 2] == f[k,2]), nbpok++;)); nbpok == #f~/2;} \\ Michel Marcus, Jul 27 2016
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Sage
def is_A275407(n): L = list(factor(n)) if is_odd(len(L)): return False for i in range(0,len(L)//2+1,2): if L[i][1] != L[i+1][1]: return False if L[i][0] != previous_prime(L[i+1][0]): return False if is_even(len(prime_range(1, L[i+1][0]))): return False return True [n for n in (2..5000) if is_A275407(n)] # Peter Luschny, Jul 27 2016
Extensions
More terms from Peter J. C. Moses, Jul 26 2016
Comments