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

1,53

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.

Links

T. D. Noe, Rows n = 1..15 of irregular triangle, flattened

T. D. Noe, Extremal Sums of Sequences

Example

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.

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]

Crossrefs

Cf. A083309.

Sequence in context: A253607 A212179 A257023 * A054740 A272947 A177740

Adjacent sequences: A238891 A238892 A238893 * A238895 A238896 A238897

Keyword

nonn,tabf

Author

T. D. Noe, Mar 07 2014

Status

approved