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Smallest number m such that GCD(a+b,a-b) = n, where a = sigma(m) and b = phi(m).
2

%I #16 Nov 14 2024 05:59:33

%S 4,1,18,21,200,14,3364,12,722,328,9801,42,25281,116,1800,15,36992,810,

%T 4414201,88,196,29161,541696,35,2928200,1413,103968,284,98942809,488,

%U 1547536,364,19602,17536,814088,370,49042009,55297,1521,440,3150464641

%N Smallest number m such that GCD(a+b,a-b) = n, where a = sigma(m) and b = phi(m).

%H Amiram Eldar, <a href="/A077102/b077102.txt">Table of n, a(n) for n = 1..82</a>

%F a(n) = Min{x; A077099(x) = n}.

%e For n = 10, a(10) = 328, sigma(328) = 630, phi(328) = 160, sigma(328) + phi(328) = 790, sigma(328) - phi(328) = 470, GCD(790,470) = 10.

%e For n = odd number, a(n) should be either a square or twice a square and so faster search for large values is possible, like e.g., for n = 97: a(97) = 435979^2 is the smallest solution.

%t f[x_] := Apply[GCD, {DivisorSigma[1, x]+EulerPhi[x], DivisorSigma[1, x]-EulerPhi[x]}]; t=Table[0, {100}]; Do[s=f[n]; If[s<101&&t[[s]]==0, t[[s]]=n], {n, 1, 10^13}]; t

%o (PARI) lista(len) = {my(v = vector(len), c = 0, k = 1, a, b, i); while(c < len, f = factor(k); a = sigma(f); b = eulerphi(f); i = gcd(a+b,a-b); if(i <= len && v[i] == 0, c++; v[i] = k); k++); v;} \\ _Amiram Eldar_, Nov 14 2024

%Y Cf. A000010, A000203, A051612, A065387, A077099, A077100, A077101.

%K nonn

%O 1,1

%A _Labos Elemer_, Nov 12 2002