A355759 Sums of the first ceiling((n+1)/2) entries on the diagonals of a square spiral with a starting value of 1 in the center, where the diagonal and the antidiagonal are used alternately.

1, 4, 6, 11, 15, 24, 32, 45, 57, 76, 94, 119, 143, 176, 208, 249, 289, 340, 390, 451, 511, 584, 656, 741, 825, 924, 1022, 1135, 1247, 1376, 1504, 1649, 1793, 1956, 2118, 2299, 2479, 2680, 2880, 3101, 3321, 3564, 3806, 4071, 4335, 4624, 4912, 5225, 5537, 5876, 6214, 6579, 6943

Offset: 1

Author

Karl-Heinz Hofmann, Aug 14 2022

Keywords

• nonn
• easy

Examples

			See the PDF in links.
		

Links

• Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000
• Karl-Heinz Hofmann, Examples of a(1..18)
• Project Euler, Problem 28. Number spiral diagonals
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,2,-2,0,2,-1).

Crossrefs

Cf. A200975, A106607, A114254.

Cf. A006527 and A208995 (bisections, see formulas).

Programs

Mathematica

CoefficientList[Series[-(x^7 - 2*x^6 + x^4 - x^3 + 2*x^2 - 2*x - 1)/((x - 1)^4*(x + 1)^2*(x^2 + 1)), {x, 0, 50}], x] (* Amiram Eldar, Aug 19 2022 *)

PARI

a(n) = (n^2 + 6*n + if(n%2,17,20))*n \ 24 + (n%4!=1); \\ Kevin Ryde, Aug 19 2022

Python

def A355759(n):  # polynomial way.
    if   n % 2 == 0: return((24 + 20*n + 6*n**2 + n**3)//24)
    elif n % 4 == 3: return((12 + 17*n + 6*n**2 + n**3)//24)
    elif n % 4 == 1: return((     17*n + 6*n**2 + n**3)//24)

Formula

G.f.: -(x^7 - 2*x^6 + x^4 - x^3 + 2*x^2 - 2*x-1)/((x - 1)^4 * (x + 1)^2 * (x^2 +1)).

a(n) = (24 + 20*n + 6*n^2 + n^3) / 24 for n even.

a(n) = (12 + 17*n + 6*n^2 + n^3) / 24 for n odd and n (mod 4) == 3.

a(n) = (17*n + 6*n^2 + n^3) / 24 for n odd and n (mod 4) == 1.

a(2*n) = A006527(n+1).

a(2*n-1) = A208995(n) - 1.

E.g.f.: ((30 + 45*x + 12*x^2 + x^3)*cosh(x) + (51 + 42*x + 12*x^2 + x^3)*sinh(x) - 6*cos(x))/24. - Stefano Spezia, Aug 19 2022