1,3

Enter 1 into center position of the spiral. Repeat: Add to the number in the present position the numbers in all those already filled positions that are horizontally or vertically adjacent to it, go to next position of the spiral and enter the sum into it.

a(1) = 1, a(n) = a(n-1) + Sum_{i < n-1 and a(i) is adjacent to a(n-1)} a(i).

Here only four positions are considered adjacent, eight however in A094767.

Clockwise and counterclockwise construction of the spiral result in the same sequence.

Klaus Brockhaus, Table of n, a(n) for n=1..729

Clockwise constructed spiral begins

13770..22272..36058..58355..94455

.8502.....16.....25.....42.....68

.5252......9......1......1....110

.3241......6......3......2....179

.2005...1236....763....470....291

where

a(2) = a(1) = 1,

a(3) = a(2)+a(1) = 2,

a(4) = a(3)+a(2) = 3,

a(5) = a(4)+a(3)+a(1) = 6,

a(6) = a(5)+a(4) = 9,

a(7) = a(6)+a(5)+a(1) = 16.

(PARI) {m=5; h=2*m-1; A=matrix(h, h); print1(A[m, m]=1, ", "); pj=m; pk=m; T=[[1, 0], [0, -1], [ -1, 0], [0, 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=A[pj, pk]; for(c=1, 4, v=[pj, pk]; v+=T[c]; s=s+A[v[1], v[2]]); A[j, k]=s; print1(s, ", "); pj=j; pk=k)} [From Klaus Brockhaus, Aug 27 2008]

Cf. A063826, A094767, A094769, A126937, A141481.

Sequence in context: A007865 A052812 A062114 this_sequence A093830 A118033 A048810

Adjacent sequences: A094765 A094766 A094767 this_sequence A094769 A094770 A094771

nonn

Yasutoshi Kohmoto (zbi74583(AT)boat.zero.ad.jp), Jun 10 2004

Edited and extended beyond a(14) by Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Aug 27 2008