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%I #51 Aug 10 2019 23:54:49
%S 0,2,12,11,39,28,82,53,141,86,216,127,307,176,414,233,537,298,676,371,
%T 831,452,1002,541,1189,638,1392,743,1611,856,1846,977,2097,1106,2364,
%U 1243,2647,1388,2946,1541,3261,1702,3592,1871,3939,2048
%N a(n) is the multiplicative inverse of A001844(n+1) modulo A001844(n); where A001844 is the sequence of centered square numbers.
%C The sequence explores the relationship between the terms of A001844, the sums of consecutive squares. The sequence is an interleaving of A054552 (a number spiral arm) and (A001844-n). The gap between the lower values of A308215 and the upper values of A308217 increase by 3n; each successive gap increasing by 6.
%H Daniel Hoyt, <a href="/A308215/a308215_2.png">Graph of A308215 and A308217 in relation to A001844</a>
%F a(n) satisfies a(n)*(2*n*(n+1)+1) == 1 (mod 2*n*(n-1)+1).
%F Conjectures from _Colin Barker_, May 16 2019: (Start)
%F G.f.: x*(2 + 12*x + 5*x^2 + 3*x^3 + x^4 + x^5) / ((1 - x)^3*(1 + x)^3).
%F a(n) = (3 + (-1)^n + 2*(2+(-1)^n)*n + 2*(3+(-1)^n)*n^2) / 4 for n>0.
%F a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>6.
%F (End)
%o (Python 3)
%o import gmpy2
%o sos = [] # sum of squares
%o a=0
%o b=1
%o for i in range(50):
%o c = a**2 + b**2
%o sos.append(c)
%o a +=1
%o b +=1
%o ls = []
%o for i in range(len(sos)-1):
%o c = gmpy2.invert(sos[i+1],sos[i])
%o ls.append(int(c))
%o print(ls)
%o (PARI) f(n) = 2*n*(n+1)+1; \\ A001844
%o a(n) = lift(1/Mod(f(n+1), f(n))); \\ _Michel Marcus_, May 16 2019
%Y Cf. A001844, A033951, A054552, A308217.
%K nonn
%O 0,2
%A _Daniel Hoyt_, May 15 2019
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