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

Author

Altug Alkan, Apr 22 2016

Keywords

• nonn
• easy

Comments

If there are k consecutive natural numbers and all of them are members of A000378, then the maximum value of k is 7. So if we randomly choose 2*7+1 consecutive natural numbers, at least one of them must be member of A004215. This sequence gives the average of 15 consecutive natural numbers in the case there is exactly one member from A004215 in these 15 consecutive natural numbers. In other words, this sequence gives the most isolated terms of A004215.

Numbers n which are 7 mod 16 such that n+5 and n-7 are sums of three squares, together with numbers n which are 15 mod 16 such that n+1 and n-3 are sums of three squares. - Charles R Greathouse IV, Apr 25 2016

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.
		

Crossrefs

Cf. A000378, A004215.

Programs

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 *)

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