A238894 Irregular triangle of the number of times that sums +- 3 +- 5 +- 7 +- 11 +-...+- prime(2n+1) equal an even number in the range -d to d, where d = 3 + 5 + 7 + 11 +...+ prime(2n+1).

1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0

Offset: 1

Author

T. D. Noe, Mar 07 2014

Keywords

• nonn
• tabf

Comments

Because the value at odd numbers is zero, we count only the values at even numbers. This sequence, a generalization of A083309, is more interesting plotted.

The rows of the irregular triangle begin at positions 1, 10, 37, 94, 193, 352, 589, 916, 1355, 1922, 2633, 3506, 4565, 5828, and 7307. having lengths 9, 27, 57, 99, 159, 237, 327, 439, 567, 711, 873, 1059, 1263, 1479, and 1719.

Examples

			The first row of the irregular triangle is {1, 0, 0, 1, 0, 1, 0, 0, 1} because the sums +- 3 +- 5 form the numbers -8, -2, 2, and 8. The odd numbers are suppressed.
		

Links

• T. D. Noe, Rows n = 1..15 of irregular triangle, flattened
• T. D. Noe, Extremal Sums of Sequences

Crossrefs

Cf. A083309.

Programs

Mathematica

nMax = 10; d = {1, 0, 0, 1}; t = {}; Do[p = Prime[n + 1]; d = PadLeft[d, Length[d] + p] + PadRight[d, Length[d] + p]; If[0 == Mod[n, 2], AppendTo[t, d]], {n, 2, nMax}]; Flatten[t]