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%I #20 Jan 25 2024 07:33:49
%S 1,1,2,4,5,10,11,23,25,26,54,57,59,122,133,142,147,304,330,351,362,
%T 747,806,880,931,957,1968,2105,2275,2391,2450,5022,5336,5733,6155,
%U 6444,6591,13486,14267,15252,16295,17008,17370,35487,37402,39835,42452,45220
%N Square spiral of sums of selected preceding terms, starting at 1.
%C Enter 1 into center position of the spiral. Repeat: Go to next position of the spiral and enter into that position the sum of the numbers in those already filled positions that are horizontally, vertically or diagonally adjacent to it.
%C Clockwise and counterclockwise construction of the spiral result in the same sequence.
%H Klaus Brockhaus, <a href="/A141481/b141481.txt">Table of n, a(n) for n=1..961</a>
%H Eric Wastl, <a href="https://adventofcode.com/2017/day/3">Day 3 Spiral Memory Part Two</a>, Advent of Code 2017.
%e Clockwise constructed spiral begins
%e .
%e 362--747--806--880--931
%e |
%e 351 11---23---25---26
%e | | |
%e 330 10 1----1 54
%e | | | |
%e 304 5----4----2 57
%e | |
%e 147--142--133--122---59
%o (PARI) {m=5; h=2*m-1; A=matrix(h, h); print1(A[m, m]=1, ","); T=[[1, 0], [1, -1], [0, -1], [ -1, -1], [ -1, 0], [ -1, 1], [0, 1], [1, 1]]; for(n=1, (h-2)^2-1, g=sqrtint(n); r=(g+g%2)\2; q=4*r^2; d=n-q; if(n<=q-2*r, j=d+3*r; k=r, if(n<=q, j=r; k=-d-r, if(n<=q+2*r, j=r-d; k=-r, j=-r; k=d-3*r))); j=j+m; k=k+m; s=0; for(c=1, 8, v=[j, k]; v+=T[c]; s=s+A[v[1], v[2]]); A[j, k]=s; print1(s, ","))} \\ _Klaus Brockhaus_, Aug 27 2008
%Y Cf. A063826, A094767, A094768, A094769, A126937.
%K nonn
%O 1,3
%A Niclas Rantala (nrantala(AT)hotmail.com), Aug 09 2008
%E Edited and extended beyond a(9) by _Klaus Brockhaus_, Aug 27 2008
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