A272189 Values of A004215(n) such that A004215(n+1) = A004215(n) + 8 = A004215(n-1) + 16.
15, 39, 47, 71, 79, 103, 135, 143, 167, 175, 199, 207, 231, 263, 271, 295, 303, 327, 335, 359, 391, 399, 423, 431, 463, 487, 519, 527, 551, 559, 583, 591, 615, 647, 655, 679, 687, 711, 719, 743, 775, 783, 807, 815, 839, 847, 871, 903, 911, 935, 943, 975, 999, 1031, 1039, 1063, 1071, 1095
Offset: 1
Examples
15 is a term because 8, 9, 10, 11, 12, 13, 14 and 16, 17, 18, 19, 20, 21, 22 are consecutive members of A000378.
Programs
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Mathematica
Take[#, {2}] & /@ Select[#, Union@ Differences@ # == {8} &] &@ Partition[#, 3, 1] &@ Select[Range@ 1200, Mod[#/4^IntegerExponent[#, 4], 8] == 7 &] // Flatten (* Michael De Vlieger, Apr 25 2016, after Ant King at A004215 *)
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PARI
isA004215(n)=(n>>(2*valuation(n, 4)))%8==7 is(n)=my(m=n%16); n>9 && if(m==7, !isA004215(n+5) && !isA004215(n-7), m==15 && !isA004215(n+1) && !isA004215(n-3)) \\ Charles R Greathouse IV, Apr 25 2016
Comments